Download - Lab manual for various experiments in chemical engineering

Chemical Engineering Department, Birla Institute of Technology & Science Pilani, Pilani Campus

1

A LABORATORY MANUAL

FOR

Chemical Engineering

Laboratory –I

Edited By

Dr Suresh Gupta & Dr Hare Krishna Mohanta

Department of Chemical Engineering,

Birla Institute of Technology & Science (BITS), Pilani

September 2014


2

CONTENTS

CYCLE - I

S. No. Experiment Page No.

1. a. Flow through Fluidized Bed (Gas and Water)

b. Flow through Packed Bed (Gas and Water)

3

2 . a. Losses due to pipe fittings

b. Losses due to friction in pipes

c. Drag Coefficient determination

10

3. a. Bernoulli’s Theorem verification

b. Discharge through venturi, orifice and rotameter

c. Flow through tubular pipe

22

4. a. Pitot tube experiment (Air and Water)

b. Reynolds Apparatus

34

5. a. Centrifugal pump characteristics

b. Reciprocating pump characteristics

42

6. a. Heat Pipe demonstrator

b. Thermal Conductivity of solids

c. Thermal conductivity of liquids

52

7. a. Drop wise and film wise condensation

b. Unsteady state heat transfer unit

61

8. a. Heat Transfer in agitated vessel

b. Fluidized bed heat transfer unit

76

9. a. Parallel flow & Counter flow heat exchanger

b. Shell and Tube heat exchanger

83

10. a. Plate type Heat Exchanger

b. Finned tube heat exchanger

92

CYCLE - II

S. No. Experiment Page No.

11. Stefan-Boltzmann Apparatus

151

12. Cross-circulation drying apparatus 110

13. a. Vapour in air diffusion

b. Open pan evaporator

114

14. Simple/Differential distillation setup 120

15. Batch crystallizer 125

16. a. Steam distillation setup

b. Vapor liquid equilibrium setup

128

17. Two phase flow 136

18. Mass transfer with chemical reaction 141

19. Adsorption in packed bed 144

20. Humidification in wetted wall column 148


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EXPERIMENT NO. 1(a)

FLOW THROUGH FLUIDIZED BED (AIR & WATER)

1. Aim

Study of hydrodynamic and bed characteristics for flow through fluidized beds.

2. Objective

To determine the minimum fluidization velocity experimentally as well as

theoretically.

To study the bed expansion characteristics of the fluidized bed (plot log NRe vs.

porosity and pressure drop vs. NRe).

3. Apparatus

Stop watch, graduated cylinders, beakers

4. Theory When a liquid or a gas is passed at very low velocity up through a bed of solid particles, the

particles do not move, and the pressure drop is given by the Ergun equation. If the fluid

velocity is steadily increased, the pressure drop and the drag on individual particles increase,

and eventually the particles start to move and become suspended in the fluid. The terms

‘fluidization’ and ‘fluidized bed’ are used to describe the condition of fully suspended

particles, since the suspension behaves like a dense fluid.

Fluidized beds are used extensively in the chemical process industries, particularly for the

cracking of high-molecular-weight petroleum fractions. Such beds inherently possess

excellent heat transfer and mixing characteristics. These are devices in which a large surface

area of contact between a liquid and a gas, or a solid and a gas or liquid is obtained for

achieving rapid mass and heat transfer and for chemical reactions.

The fluidized bed is one of the best known contacting methods used in the processing

industry, for instance in oil refinery plants. Among its chief advantages are that the particles

are well mixed leading to low temperature gradients, they are suitable for both small and

large scale operations and they allow continuous processing. There are many well established

operations that utilize this technology, including cracking and reforming of hydrocarbons,

coal carbonization and gasification, ore roasting, Fisher-Tropsch synthesis, coking,

aluminium production, melamine production, and coating preparations. Nowadays, you will

find fluidized beds used in catalyst regeneration, solid-gas reactors, combustion of coal,

roasting of ores, drying, and gas adsorption operations. The application of fluidization is also

well recognized in nuclear engineering as a unit operation for example, in uranium extraction,

nuclear fuel fabrication, reprocessing of fuel and waste disposal.

When a fluid is admitted at the bottom of a packed bed of solids at a low flow rate, it passes

upward through bed without causing any particle motion. If the particles are quite small, flow

in the channels between the particles will be laminar and the pressure drop across the bed will

be proportional to the superficial velocity (Vo) and for turbulent situations, pressure drop

across the bed increase nonlinearly with the increase in the superficial velocity. As the

velocity is gradually increased, the pressure drop increases, but particles do not move and the

bed height remains the same. At a certain velocity, the pressure drop across the bed

counterbalances the force of gravity on the particles or the weight of the bed, and any other

further increase in velocity causes the particles to move and the true fluidization begins. For a


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high enough fluid velocity, the friction force is large enough to lift the particles. This

represents the onset of fluidization once the bed is fluidized pressure drop across the bed

remains constant, but the bed height continues to increase with increasing flow.

Figure 1 Fluidization regimes

In order to determine the pressure drop through a fixed bed for any flow condition, the Ergun

equation (1952) can be used:

dp is the size of particles (µm)

L is the height of the bed (m)

µ is the viscosity of sir (N/m2.s)

U is the average superficial velocity (m/s)

ϵ is the bed voidage or porosity

ρ is the density of air/water (kg/m3)

∆P is the pressure drop across the bed (N/m2)

The average Reynolds number based on superficial velocity which is given by,

If the Reynolds number is less than 10 then it is laminar flow and is greater than 2000 it is

turbulent flow. The rest of the values lie in the transition regime. If the flow rate of air/water

Q is measured in litres, A is the bed cross-sectional area and U is the superficial velocity in

m/s, then


5

Theoretically at incipient fluidization (the stage in the fluidized bed where the force on the

solid is enough to balance the weight of the solid material),

∆P is in mm of manometer

The pressure drop at fluidization can also be predicted by the equation,

ρp is the particle density (kg/m3)

ρ is the fluid density (kg/m3)

g is the gravitational force (m/s2)

Minimum fluidization velocity

Umf, the minimum fluidizing velocity, is frequently used in fluid-bed calculations and in

quantifying one of the particle properties. This parameter is best measured in small-scale

equipment at ambient conditions. The correlation given below can then be used to back

calculate dp. This gives a particle size that takes into account effects of size distribution and

sphericity. The correlation can then be used to estimate Umf at process conditions. If Umf

cannot be determined experimentally, use the expression below directly.

Assumption: - Consider the uniform particle size.

Remf = (1135.7 +0.0408Ar) 0.5

- 33.7 (Wen and Yu correlation for particles dp> 100 µm)

/Re mffpmf Ud

23 / gdAr fsfp (Ar is the Archimedes Number)

For particles of dp < 100 µm, Baeyens and Geldart (1977) can be used,

5. Experimental procedure

1. The height of the static bed Z1 i.e. when there is no flow of water/air (porosity 1) was

noted.

2. The flow of air/water in the column is started and the flow rates from the rotameter

were noted.

3. The corresponding bed heights and pressure drop values were noted.

4. The flow rates were increased steadily and similar data were collected at different

intervals.

5. 6 to 8 readings of flow rates were varied and reading were taken.

6. Steady state flow rate of water was ensured at each point.


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6. Observations and calculations

Particle size = 8 mm

Porosity = 0.6

Inside diameter of the column (D) = 0.055m

Cross-sectional area of fluidized column

= 0.002375 m²

Density of water ρ (at T0C) = 1000 kg/m

3

Viscosity of water µ (at T0C) = 0.798*10

-3 kg/m-s

Density of air = 1.1687 kg/m3

Viscosity of air = 1.8633x10-5

Pa.s

Velocity V (m/s) = volumetric flow rate of water / cross section area of column

Initial height of static bed in the column (Z1)

Porosity of the fluidized Bed: If Z1 and 1 are the height and porosity of the static bed and Z2

and 2 are the height and porosity of fluidized bed, the

)1(

)1(

2

112

ZZ (Total volume of column equal to total volume of solid)

Z2 = 2 Z2 = Z1 (1- 1)

Z2 - Z1 (1- 1) = 2 Z2

2 = Z2 - Z1 (1- 1) / Z2

Note: Keep the observation tables for air and water separately. The sample calculations

for the same should be shown separately.

Obs No. Height of

Bed (Z2)

Volumetric

Flow rate of

water/air in

m³/s

Velocity(v)

m/s

Porosity

()

Pressure

drop (P) NRe

7. Results & Discussions

References:

1. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical

Engineering, 6th

edition, McGraw Hill, New Delhi, India, 2000.

2. Kunii, Levenspeil O, Fluidized Engineering, Robert E Krieger Publishing Company,

Huntington, New York, 1977

http://www.libreriauniversitaria.it/goto/author_Warren+L.+McCabe/shelf_BUS/Warren_L__McCabe.html

http://www.libreriauniversitaria.it/goto/author_Julian+Smith/shelf_BUS/Julian_Smith.html


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EXPERIMENT NO. 1(b)

FLOW THROUGH PACKED BED (AIR & WATER)

1. Aim

a) To plot Modified Reynolds’s No. (NRe,m) vs. Modified friction factor (f) on a log-log

graph

b) To verify Ergun’s equation

2. Apparatus

Stop watch, Graduated cylinder, packing as Hollow rings of Glass, Manometer, beakers etc.

3. Theory

A packed bed is a hollow tube, pipe, or other vessel that is filled with a packing material. The

packing can be randomly filled with small objects like Raschig rings, Pall rings or else it can

be a specifically designed structured packing. Packed column is a pressure vessel that has a

packed section. In general, packed towers are used for bringing two phases in contact with

one another and there will be strong interaction between the fluids in this case between solid

and fluid. As the fluid passes through the bed, it does so through the voids presents in the

bed. The voids form continuous channels throughout the bed. The flow may be laminar

through some channels and turbulent in other channels.

Just as with straight pipes, Ergun relates the flows and pressure drops to a Reynolds Number

and friction factor respectively. The Reynolds number for packed beds, Rep , depends upon

the controlled variable i.e. Superficial velocity Vo and the system parameters ρ, ε, μ, and Dp

and is defined as (Bird et. al., 1996),

Dp is the equivalent spherical diameter of the particle, V0 is the superficial velocity defined as

the volumetric flow rate divided by the cross-sectional area of the column, ρ is the fluid

density, ε is the dimensionless void fraction defined as the volume of void space over the

total volume of packing, and μ is the fluid viscosity.

The friction factor, fp, depends upon V0 and the pressure drop, ΔP, and system parameters,

and is defined as (Bird et al., 1996)

P = Pressure drop in kg/cm2

= porosity

L = effectively length of the bed, cm

V = Superficial-fluid velocity, based on empty cross-section of pipe, m/sec

= density of the fluid

= viscosity of fluid

https://en.wikipedia.org/wiki/Tubing_(material)

https://en.wikipedia.org/wiki/Raschig_ring

https://en.wikipedia.org/wiki/Structured_packing

https://en.wikipedia.org/wiki/Pressure_vessel


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4. Working principle

Figure 1 Schematic of Packed Bed Column

The packed column shown above is filled with packings, in our case the packing is Raschig

glass rings. Now, in this the water is passed through the bottom of column and the flow

happens through the packing thus creating friction , this friction leads to pressure drop in the

column which is measured by the manometer attached to the two ends of packed column.

Also, the flow here is regulated by the rotameter attached to the column. An increase in flow

increases the friction and thus leads to an increase in pressure drop.

5. Procedure

a. Water/air was allowed to flow from bottom to top in a packed bed.

b. Flow of water/air is regulated to vary the flow rate.

c. The corresponding pressure drop values were noted by means of the manometer for 6

to 8 different rates of flow of fluid.

d. The flow rate values were noted and it was ensured that the steady state flow rates

have been achieved.

6. Observations

Inside diameter of the tube = 50 mm

Volume of each particle = 1.57x10-6

m3

Surface area of the particle = 0.000628 m2

Porosity of the bed = 0.3

Effective length of packing = 750 mm

Temperature of fluid = ambient

Manometer fluid, Hg = 13,534 kg/m³

Density of water ρ (at T0C) = 1000 kg/m

3

Viscosity of water µ (at T0C) = 0.798*10

-3 kg/m-s

Density of air = 1.1687 kg/m3

Viscosity of air = 1.8633x10-5

Pa.s Manometer Reading h (metres)


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S.

No.

Volumetric

flow rate of

water. Q

m3/sec

Manometer

reading.

Hgofmh )(

Superficial Velocity of

water = Vo (Q /Cross

sectional area of the

column m/s)

Pressure drop/length

2/)(

mNL

mgh

L

P

Note: Calculate friction factor and modified Reynolds No. and plot on a log-log graph

separately for air and water. The observation columns and sample calculations for both

of them should be shown separately.

Calculations

Certain assumptions are carried out before calculating the Friction factor and Reynolds

number. First, we assume that there is no channeling in the packed bed. Channeling occurs

when the fluid flowing through the packed bed finds a “preferred path” through the bed. We

also assume that the diameter of the packing is much smaller than the diameter of the column

as well. The maximum recommended particle diameter is one-fifth of the column diameter.

We assume that velocity, particle diameter and void fraction behaves as a bulk behavior and

hence we can use an average values.

Initially, we calculate diameter of the particle,

DP =

Calculate friction factor as,

and modified Reynolds number as,

7. Results and Discussion

Modified Reynold’s number, friction factor are calculated accordingly to the changes in

variables affecting and Ergun’s equation is verified.

References: a) Bird, R. Byron, Transport Phenomena. Madison, Wisconsin: John Wiley & Sons,

1996.

b) Geankoplis, Christie J., Transport Process and Unit Operations. 4th ed., New Jersey:

Prentice Hall, 2003.

c) Ergun, Sabri, “Fluid Flow through Packed Columns.” Chemical Engineering

Progress, Vol. 48, No 2. , 89-94 , 1952.


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EXPERIMENT NO. 2(a)

LOSSES DUE TO PIPE FITTINGS, SUDDEN ENLARGEMENT &

CONTRACTION

1 Aim

To determine the loss co-efficient for following sections

(a) Sudden expansion

(b) Sudden contraction

(c) Tee – junction

(d) Pipe bend

(e) Elbow

2. Apparatus

FHP (Fractional Horse Power) pump driven pipe system consisting of elbow bend, Tee-

junction, sudden enlargement and contraction, manometer with tapping into pipe for head

readings, sump and discharge tanks, stopwatch.

3. Theory

In Hydraulic Engineering practice, it is frequently necessary to estimate the head loss

incurred by a fluid as it flows along a pipeline. For example, it may be desired to predict the

rate of flow along a proposed pipe connecting two reservoirs at different levels. Or it may be

necessary to calculate what additional head would be required to double the rate of flow

along an existing pipeline.

Loss of head is incurred by fluid mixing, which occurs at fittings such as bends or valves, and

by frictional resistance at the pipe wall. Where there are numerous fittings and the pipes are

short, the major part of the head loss will be due to the local mixing near the fittings. For a

long pipeline, on the other hand, skin friction at the pipe wall will predominate.

Loss of head due to change in cross-section, bends, elbows, valves and fittings of all types

fall into the category of minor losses in pipe lines. In long pipe lines the friction losses are

much larger than these minor losses and hence the latter are often neglected. But, in shorter

pipelines their consideration is necessary for the correct estimate of losses.

When there is any type of bend in pipe, the velocity of flow changes, due to which the

separation of the flow from the boundary and also formation of eddies, takes place. Thus the

energy is lost. The losses of head due to bend in pipe:

g

VKh LL

2

2

The minor losses in contraction can be expressed as:

g

VKh LL

2

2

1

The minor losses in enlargement can be expressed as:

g

VVKh LL

2

)( 2

21

Where, hL = minor loss or head loss

KL = Loss coefficient

V = velocity of fluid.


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V1 = velocity of fluid in pipe of small Diameter.

V2 = velocity of fluid in pipe of large Diameter.

4.Experimental set-up

The apparatus consist of a ½” smooth and sharp bends and elbow, a sudden expansion from

½” to 1”, sudden contraction from 1” to ½” and ½” ball valve, gate valve, tee fitting and pipe

union joint. Pressure tapings are provided at inlet and outlet of these fittings at suitable

distance. A differential manometer fitted in the line gives pressure loss due to fittings. Supply

to the pipeline is made through centrifugal pump, which deliver water from sump tank. The

flow of water in pipeline is regulated by means of Control valve & By-Pass valve. Discharge

is measured with the help of measuring tank and stop watch.

4.1 Utilities Required:

a) Power supply: Single Phase, 220 Volts, 50 Hz, 5 Amp with Earth.

b) De-mineralized Water.

c) Drain.

d) Space required: 1.6m x 0.6m

Schematic Diagram for Losses due Pipe Fittings,

Sudden Enlargement & Contraction Apparatus

5 Experimental procedure:

a) Close the drain valves provided.

b) Fill Sump tank ¾ with clean water and ensure that no foreign particles are there.

c) Close all flow control valves given on the water line, pressure taps of manometer

connected to different pipe -fittings and open by-pass valve.

d) Now switch on the main power supply (220 Volts AC, 50 Hz) and start the pump.

e) Operate the flow control valve to regulate the flow of water in the desired test section.

f) Open the pressure taps of manometer of related test section, very slowly to avoid the

blow of water on manometer fluid.


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g) Now open the air release valve provided on the manometer, slowly to release the air

in manometer. When there is no air in the manometer, close the air release valves.

h) Adjust water flow rate in desired section with the help of control valve and record the

manometer reading.

i) Measure the flow of water, discharged through desired test section, using stop watch

and measuring tank.

j) Repeat same procedure for different flow rates of water, operating control valve and

by-pass valve.

k) When experiment is over for one desired test section, open the By-Pass Valve fully.

Then close the flow control valve of running test section and open the Control valve

of secondly desired test section.

l) Repeat same procedure for the 9 selected test sections of the two bends, elbow,

contraction, expansion, tee and union joint and both gate and ball valves.

m) When experiment is over, close all manometers pressure taps first, switch off pump,

and switch off power supply to panel.

6 Observation & Calculations

6.1 Specification:

Sudden Enlargement : From 16.5mm to 27mm

Sudden Contraction : From 27mm to 16.5mm.

Bend : 1/2"

Elbow : 1/2"

Ball valve : 1/2"

Gate valve : 1/2"

Tee joint : 1/2"

Union joint : 1/2"

Water Circulation : FHP Pump.

Flow Measurement : Using Measuring Tank with Piezometer, Capacity 40 Ltrs.

Sump Tank : Capacity 80 Ltrs.

Stop Watch : Electronic.

Control Panel Comprises of: Standard make On/Off Switch, Mains Indicator, etc.

6.2 Formulae:

(a) Discharge:

100*

*

t

RAQ ---------------- (1)

(b) Velocity:

1

1a

QV (Velocity in ½” pipe) ---------------- (2)

2

2a

QV (Velocity in 1” pipe) ---------------- (2a)

(c) Loss of Head (for Contraction):

g

VKh LL

2

2

1 ---------------- (3)

(d) Loss of Head (for Expansion):

CKg

VVKh LLL

2

)( 2

21 ---------------- (4)


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(e) Experimental loss of head

100

6.13*hhL --------------- (5)

(f) Loss Co-efficient:

2

1

2

V

ghK LL ---------------- (6)

6.3 Data:

A = Area of measuring tank = 0.1m2

s = Specific gravity of Hg = 13.6

g = Acceleration due to gravity = 9.81 m/sec2

d1 = Dia. of smaller pipe = 0.0165 m

d2 = Dia. of larger pipe = 0.027m.

a1 = Cross-sectional area of Small Dia. pipe = 2.14 x 10-4

m2

a2 = Cross-sectional area of Large Dia. pipe = 5.73 x 10-4

m2

6.4 Observation Table:

S.

No.

Type of Fitting Pressure

Difference, h

(cm)

Rise of water level in

measuring tank R (cm)

Time taken for R

(sec.)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

6.5 Calculation Table:

S.

No

Discharge,

Q

Velocity,

V1

Velocity,

V2

Loss of Kinetic

Energy, C

Loss of

head

hL

Loss

coefficient K L

6.6 Nomenclature: Q = discharge, m

3/s

V1 = velocity of fluid in pipe of Small Diameter (m).

V2 = velocity of fluid in pipe of Large Diameter (m).

a1 = Cross-sectional area of Small Dia. pipe

a2 = Cross-sectional area of Large Dia. pipe

KL = loss coefficient.


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h = pressure difference, m

hL = head loss, m

R = Rise of water level in measuring tank (m).

t = Time taken for R (sec.).

7. Result &discussion

8. Conclusion

9. Precaution & maintenance instructions:

a) Do not run the pump at low voltage i.e. less than 180 Volts.

b) Never fully close, the delivery line and by-pass line valves simultaneously.

c) Always keep apparatus free from dust. To prevent clogging of moving parts, run

pump at least once in a fortnight.

d) Frequently Grease/Oil the rotating parts, once in three months. Always use clean

water.

e) It apparatus will not in use for more than one month, drain the apparatus completely,

and fill pump with cutting oil.

f)

9.1 Troubleshooting:

a) If pump gets jam, open the back cover of pump and rotate the shaft manually.

b) If pump gets heat up, switch off the main power for 15 minutes, avoid closing the

flow control valve, and by pass valve simultaneously during operation.

References:

a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”,

4th

ed., McGraw Hill, NY, 1985.

b) Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering

Vol. - 1”, 5th

ed., Asian Books ltd., ND, 1996.

c) Brown, G.G., “Unit Operations”, 1st ed., CBS, ND, 1995.

d) Foust, A.S., et. al., “Principles of Unit Operations”, 2nd

ed., John Wiley, NY, 1980.


15

EXPERIMENT NO. 2(b)

LOSSES DUE TO PIPE FRICTION

1. Aim To determine the losses due to friction in pipes and friction factor for Darcy - Weisbach

equation

2. Apparatus

FHP (Fractional Horse Power) pump driven pipe system consisting of Galvanized Iron (GI)

pipe of ½”, 1” and Stainless Steel (SS) pipe of ½” diameter, manometer with tapping into

pipe for head readings, sump and discharge tanks, stopwatch.

3. Theory

When a fluid is flowing through a pipe, it experiences some resistance due to which some of

the energy is lost. This loss of energy in the pipelines comes under major energy losses and

minor energy losses. In long pipelines, the friction losses are much larger than the minor

losses and hence, the latter are often neglected. The losses due to friction in the pipelines are

known as major energy losses. The friction in the pipeline is due to a viscous drag between

the stream bands of fluid. The stream bands adjacent to the solid surface are always at rest

relative to the wetted surface. The viscous drag is due to the molecular attractions between

the molecular of the fluid.

It is found that the total friction resistance to fluid flow depends on the following:

1. The area of the wetted surface

2. The density of the fluid

3. The surface roughness

4. It is independent of the fluid pressure

5. It increase with the square of the velocity

It depends on a frictional co-efficient. The loss of head in pipe due to friction is calculated

from Darcy-Weisbach equation which has been given by:

Where:

h f = loss of head due to friction

f = Co-efficient of friction

L = distance between pressure point

V = mean velocity of fluid

d = diameter of pipe

g = Acceleration due to Gravity

4. Experimental setup The apparatus consist of three pipes of varying diameters and materials for which common

inlet connection is provided with control valve to regulate the flow, near the down stream end

of the pipe. Pressure tapings are taken at suitable distance apart between which a manometer

is provided to study the pressure loss due to the friction. Discharge is measured with the help

of measuring tank and stopwatch.

d 2g

LV f 4h

2

f


16

Schematic Diagram for Friction in Pipe Lines Apparatus

4.1 Utilities Required

a) Power supply: Single Phase, 220 Volts, 50 Hz, 5 Amp with Earth.

b) De-mineralized Water Supply.

c) Drain.

d) Space required : 1.6m x 0.6m

5. Experimental procedure a) Close the drain valves provided.

b) Fill Sump tank ¾ with clean water and ensure that no foreign particles are there.

c) Close all flow control valves given on the water line, pressure taps of manometer

connected to different pipe -fittings and open by-pass valve.

d) Now switch on the main power supply (220 Volts AC, 50 Hz) and start the pump.

e) Operate the flow control valve to regulate the flow of water in the desired test section.

f) Open the pressure taps of manometer of related test section, very slowly to avoid the

blow of water on manometer fluid.

g) Now open the valve provided on the manometer, slowly to release the air in

manometer. When there is no air in the manometer, close the valves.

h) Adjust water flow rate in desired section with the help of control valve and record the

manometer reading.

i) Measure the flow of water, discharged through desired test section, using stop watch

and measuring tank.

j) Repeat same procedure for different flow rates of water, operating control valve and

by-pass valve.

k) When experiment is over for one desired test section, open the By-Pass Valve fully.



l) Repeat same procedure for different flow rates of water, operating Control Valve and

By-Pass valve.


17

m) When experiment is over, close all manometers pressure taps first, switch off pump,

and switch off power supply to panel.

6. Observation & calculation

6.1 Specification

Pipes (3 Nos.) : Material GI of ½” & 1” diameter.

SS of ½” diameter

Pipe Test Section : Length 1.1 m for each Pipe


Flow Measurement : Using Measuring Tank with Piezometer, Capacity,

40 Ltrs.



Control Panel Comprises of:

Standard make On/Off Switch, Mains Indicator, etc.

Tanks will be made of Stainless Steel.

The whole set-up is well designed and arranged in a good quality painted structure.

6.2 Formulae

1. Head losses,

1

100 W

mf

hh

, m of water.

2. Co-efficient of Friction:

2

f

V

gd

L 4

2hf

3. Discharge (Q):

smt

AxRQ /

100*

3

4. Velocity of Fluid:

sma

QV /

6.3 Data:

Area of measuring tank, A = 0.1m2

Sp. gravity of Hg = 13.6

Acceleration due to gravity, g = 9.81m/sec2

Inside Diameter of Pipe, d

For GI pipe (1”) = 0.027 m

For GI pipe (1/2”) = 0.016.5 m

For SS pipe (1/2”) = 0.016 m

Cross-section area of pipe, a

For GI pipe (1”) = 5.70 x 10-4

m2

For GI pipe (1/2”) = 2.14 x 10-4

m2

For SS pipe (1/2”) = 2.01 x 10-4

m2

Distance between pressure points, L = 1.1 m each for all pipes

6.4 Observation Table: (for For GI pipe 1”, GI pipe 1/2”, and SS pipe 1/2”)


18

Extent of

Valve open

Pressure difference

h (cm)

Rise of water level in

measuring Tank R (cm)

Time taken, t

(sec.)

½ Open

Full Open


Extent of Valve

open

Pressure

head, hf

(m)

Discharge,

Q

(m3/s)

Velocity of

Fluid, V

(m/s)

Fanning Friction

factor, f

½ Open

Full Open

6.6 Nomenclature:

A = Area of measuring tank

a = Cross-section area of pipe

d = Inside Diameter of Pipe

g = Acceleration due to gravity

h = manometer reading, cm

hf = Pressure head of water (m)

R = Rise of water level in measuring Tank (m)

t = Time taken for R (sec)

L = distance between two points

7. Result & discussion

8. Precautions

1. Do not run the pump at low voltage i.e. less than 180 Volts.

2. Never fully close the Delivery line and By-Pass line Valves simultaneously.

3. Always keep apparatus free from dust.

4. To prevent clogging of moving parts, Run Pump at least once in a fortnight.

5. Frequently Grease/Oil the rotating parts, once in three months.

6. Always use clean water.

7. If apparatus will not in use for more than one month, drain the apparatus completely,


8.1 Troubleshooting:

1. If pump gets jam, open the back cover of pump and rotate the shaft manually.

2. If pump gets heat up, switch off the main power for 15 minutes and avoid closing the

flow control valve and by pass valve simultaneously during operation.

References e) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”,

4th

ed, McGraw Hill, NY, 1985.

f) Foust, A.S., et. al., “Principles of Unit Operations”, 2nd

ed., John Wiley, NY, 1980.

g) Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol.

- 1”, 5th



19

EXPERIMENT NO. 2(c)

DRAG COEFFICIENT

1. Aim & Apparatus

To verify Stoke’s law and to study the variation of the drag coefficient with Reynolds number

(Re) for sphere.

Experimental setup of 3 cylinders filled with three different liquids – demineralised water,

glycerine and hydraulic oil, spherical balls, measuring scales and stopwatch.

2. Theory

When a body moves through any fluid, it experiences a resistance, which acts in a direction

opposite to that of the motion of the body. This resistance is called the drag force (FD) and it

depends on the size of the body, velocity with which it moves and the viscosity of the fluid.

According to stoke, the drag force acting on a sphere moving through a fluid due to its weight

is given by the following expression:

FD = 3πµU₀Dp (i)

Where Dp is the diameter of sphere, µ is the viscosity of fluid, and U₀ is called the terminal

fall velocity. Terminal velocity is defined as the velocity attained by a body in falling through

a fluid at rest, when the drag force on the body is equal to the submerged weight of the body.

For equilibrium condition in the case of a freely falling sphere through a liquid, drag force

plus the buoyant force must be equal to the weight of the sphere and is given by the

expression

(ii)

Where Fg is the drag force, g is acceleration due to gravity (9.8 m/sec2), ρp is density of

spherical particle, and ρf is the density of fluid stream.

The drag coefficient predicted by Stoke’s law is given by the expression

, where

(iii)

where ν is the kinematic viscosity of fluid stream (m2/sec). Thus, coefficient of drag CD varies

with Reynolds number.

Experiments have shown that Eq. (ii) holds well for Re < 0.2, and the sphere is falling in an

infinite fluid. If the fluid is not infinite in extent but is confined within a container (finite

dimensions), then the resistance to motion is increased, and in such a case the modified value

of drag coefficient, as given by the following expression, should be used:

CD =

, (1+2.1

) (vi)

Where D is the smallest lateral dimensions of the container and D is the diameter of sphere.

Also the observed fall velocity U is corrected in Eq. (iii) by using the following expression in

order to get the fall velocity corresponding to infinite fluid medium:

Corrected velocity, U₀ = U (1+2.4

)

Where, D is the diameter of the container.

3. Experimental set-up

The set-up consists of three transparent vertical cylinders of diameter 100 mm and height 1

m. A hopper with a valve is provided at the bottom of the cylinder to collect the spheres. The

cylinder is supported by four vertical posts and fixed to a MS table. A vertical scale is fixed


20

on the surface of a cylinder. The cylinders are filled with fluids of varying viscosities such as

glycerine, hydraulic oil and water.


1. Measure the diameters of spheres and note down their materials.

2. Determine the mass of the spheres on electronic balance.

3. Mark two lines on the cylinder for measurement of the vertical distance (L) for the

determination of terminal velocity. The upper line should be at a depth of 100 mm or

more from the free surface so that the terminal velocity is achieved.

4. Hold the sphere with a finger and a thumb and bring it up to the fluid level. Leave the

sphere gently (wet the sphere before dropping it).

5. Note down the time taken by the sphere in falling through distance L.

6. Repeat steps (4) and (5) for other diameters of spheres (Set 1).

7. Repeat steps (4) and (6) for spheres of other material (Set 2).

Fig 1: Stoke’s Apparatus


5.1 Specification:

Liquid in the cylinder =

Mass density of liquid, ρf = 1261kg/m³ (glycerol),

1080kg/m³ (hydraulic oil),

1000kg/m³ (water)

Mass density of sphere = 2070 kg/m3

Kinematic viscosity of liquid at T 20°C, v = 1.006x10-6

m²/s (water),

1.180x10-6

m²/s (glycerol)

1.25*10-4

m2/s (hydraulic oil)

Diameter of cylinder, D = 87mm

Liquid 2 Liquid 3 Liquid 1


21

Distance of fall, L =

Diameter of sphere = 0.819cm

Time of falling for sphere (t) = ____ sec

Terminal Velocity (U) (Length/time for fall) = ______ (m/sec)

5.2 Observation Table:

S.No Fluid type Length of liquid column (m) Sphere falling time (s)

1. Water

2. Glycerol

3. Hydraulic Oil


Fluid

type

Terminal

velocity

Corrected

terminal

velocity

Reynolds

number

Discharge

coefficient

Corrected

Discharge

coefficient

Fg FD

6. Result & discussions

Plot a log-log graph of CD (corrected discharge coefficient) vs. NRe (Reynolds number)

7. Conclusion

8. Precautions

The sphere needs to be dropped from the centre of the cylinder.

Extreme care should be taken while taking out the sphere from the cylinder to avoid

loss of excess fluid.

Wet the sphere with the specific fluid before dropping it in the cylinder.

References:

h) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”,

7th


i) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”, 26th

reprint,

McGraw Hill, NY, 2011.


22

EXPERIMENT NO. 3(a)

BERNOULLI’S THEOREM

1. Aim & Objective

To experimentally verify Bernoulli equation for fluid flow in horizontal pipe.

2. Apparatus

FHP pump run setup consisting of 4 rectangular tanks, horizontal pipe section, 8 piezometric

tubes with measuring scales, stopwatch.

3. Theory

Energy of a Liquid in Motion

The energy, in general, may be defined as the capacity to do work. Though the energy exists

in many forms, yet the following are important from the subject point of view:

a) Potential Energy

b) Kinetic Energy, and

c) Pressure Energy.

Potential Energy of a Liquid in Motion

It is the energy possessed by a liquid particle, by virtue of its position. If a liquid particle is Z

meters above the horizontal datum (arbitrarily chosen), the potential energy of the particle

will be Z meter-kilogram per kg of liquid. Potential head of the liquid, at that point, will be Z

meters of the liquid.

Kinetic Energy of a Liquid Particle in Motion:

It is the energy, possessed by a liquid particle, by virtue of its motion or velocity. If a liquid

particle is flowing with a mean velocity of v meter per second, then the kinetic energy of the

particle will be v2/ 2g m-kg per kg of liquid. Velocity head of the liquid, at that velocity, will

be v2/ 2g meter of the liquid.

Pressure Energy of a Liquid Particle in Motion:

It is the energy, possessed by a liquid particle, by virtue of its existing pressure. If a liquid

particle is under a pressure of p kg / m2, then the pressure energy of the particle will be p/w

m-kg per kg of liquid, where w is the specific weight of the liquid

Total Energy of a Liquid Particle in Motion:

The total energy of a liquid particle, in motion, is the sum of its potential energy, kinetic

energy and pressure energy. Mathematically,

Total Energy,

w

p

g

vZE

2

2

m-kg/ kg of liquid

BERNOULLI’S EQUATION:

It states, “For a perfect incompressible liquid, flowing in a continuous stream, the total

energy of a particle remains the same; while the particle moves from one point to another.”

This statement is based on the assumption that there are no losses due to friction in pipe.

Mathematically,

w

p

g

vZE

2

2

= Constant --------- (1)

Bernoulli Equation:


23

---------- (2)

This is the basic from of Bernoulli equation for steady incompressible inviscid flows. It may

be written for any two points 1 and 2 on the same streamline as

--------- (3)

The constant of Bernoulli equation, can be named as total head (ho) has different values on

different streamlines.

-------- (4)

The total head may be regarded as the sum of the piezometric head h* = p/(ρg) + z and the

kinetic head v2/2g.

Bernoulli equation is arrived from the following assumptions:

a) Steady flow - common assumption applicable to many flows.

b) Incompressible flow - acceptable if the flow Mach number is less than 0.3.

c) Frictionless flow - very restrictive; solid walls introduce friction effects.

d) Valid for flow along a single streamline; i.e., different streamlines may have different

ho.

e) No shaft work - no pump or turbines on the streamline.

f) No transfer of heat - either added or removed.

Range of validity of the Bernoulli Equation:

Bernoulli equation is valid along any streamline in any steady, inviscid, incompressible flow.

There are no restrictions on the shape of the streamline or on the geometry of the overall

flow. The equation is valid for flow in one, two or three dimensions.

Modifications on Bernoulli equation:

Bernoulli equation can be corrected and used in the following form for real cases.

Where 'q' is the work done by pump and 'w' is the work done by the fluid and h is the head

loss by friction.

4. Experimental Set-up

The apparatus is made from transparent acrylic and has both the convergent and divergent

sections. Water is supplied from the constant head tank attached to the test section. Constant

level is maintained in the supply tank. Piezometric tubes are attached at different distance on

the test section. Water discharges to the discharge tank attached at the far end of the test

section and from there it goes to the measuring tank through valve. The entire setup is

mounted on a stand.


24

Figure 1: Schematic diagram of experimental setup


a) Fill the sump tank with water.

b) Before starting the pump, make sure that the bypass valve is fully open and the

discharge valve is closed.

c) Switch ON the Pump.

d) Flow control valve to the supply tank is opened and bypass valve is closed slowly

and simultaneously. Keep drain valve of the discharge tank fully open.

e) After we get steady height of liquid in the supply tank adjust the drain valve on

the discharge tank so as to get steady level there also.

f) Collect the predetermined quantity of water in the measuring tank and measure

the time required for the same.

g) Also record the height of liquid in each of the piezometric tube.

6. Observations

Sr. No.

Height. diff. in

measuring tank

(m)

H1

H2

ΔH

Volume (m3) V

Time (sec) T

Vol. Flow rate

(m3/sec)

Q

pie

zom

e

tric

tub

e

posi

tion

1

H1=P1/w(m)

v1 = Q/A1

v12/ 2g (m)


25

Total Head

2

H2=P2/w(m)

v2 = Q/A2

v22/ 2g (m)

Total Head

3

H3=P3/w(m)

v3 = Q/A3

v3 2/ 2g (m)

Total Head

4

H4=P4/w(m)

v4 = Q/A4

v42/ 2g (m)

Total Head

5

H5=P5/w(m)

v5 = Q/A5

v5 2/ 2g (m)

Total Head

6

H6=P6/w(m)

v6 = Q/A6

v6 2/ 2g (m)

Total Head

7

H7=P7/w(m)

v7 = Q/ A7

v7 2/ 2g (m)

Total Head

8

H8=P8/w(m)

v8 = Q/ A8

v82/ 2g (m)

Total Head

Calculations:

Length of the test section = 0.4 m

Piezo tube 1 at 0.05 m c/s area A1 = 0.000961 m2








Cross Sectional Area of Measuring Tank (Am) = 0.09 m2

Height Difference in measuring tank (H) = H1- H2

= m

Volume Collected (V) = H * Am

= m3

Volumetric Flow rate (Q) = V / T

= m3/s

Linear Point Velocity (vi) = Q/ Ai

= m/ sec


26

Velocity head at point – 1 = [vi2 / (2*g)]

= m

Total Head at point –1 = Static Head + Velocity Head

= Hi + vi2 / (2*g)

= m

Check for all the points that sum of velocity head and pressure head is almost constant. Here

potential energy is not being considered as the test section lies in one horizontal plane only.

7. Result:

Draw graph of static head, velocity head, and total head Vs distance of the piezometric tubes.

8. Conclusions

9. Precautions

References

(a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical

Engineering”, 7th


(b) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”,

26th

reprint, McGraw Hill, NY, 2011.

(c) C.J. Geankoplis., “Transport Processes and Separation Process

Principles”, 4th

ed., PHI, 2009.


27

EXPERIMENT NO. 3(b)

VENTURIMETER, ORIFICEMETER & ROTAMETER CALIBRATION

1. Aim & Objective

a) To determine the co-efficient of discharge through Venturimeter & Orificemeter.

b) To calibrate the Rotameter

2. Theory

VENTURIMETER:

A Venturimeter consists of;

1. An inlet section followed by a convergent cone.

2. A Cylindrical Throat.

3. A gradually divergent cone.

The inlet section of the Venturimeter is of the same diameter as that of the pipe,

which is followed by a convergent cone. The convergent cone is a short pipe, which tapers

from the original size of the pipe to that of the Throat of the Venturimeter. The Throat of the

Venturimeter is a short parallel side tube having its cross-sectional area smaller than that of

the pipe. The divergent cone of the Venturimeter is gradually diverging pipe with its cross-

sectional area increasing from that of the Throat to the original size of the pipe. At inlet

section & Throat of the Venturimeter, pressure taps are provided.

ORIFICEMETER:

An Orificemeter consists of a flat circular plate with a circular hole called Orifice,

which is concentric with the pipe axis. An orifice plate is a thin plate with a hole in the

middle. It is usually placed in a pipe in which fluid flows. When the fluid reaches the orifice

plate, the fluid is forced to converge to go through the small hole; the point of maximum

convergence actually occurs shortly downstream of the physical orifice, at the so-called vena

contracta point. As it does so, the velocity and the pressure changes. Beyond the vena

contracta, the fluid expands and the velocity and pressure change once again. By measuring

the difference in fluid pressure between the normal pipe section and at the vena contracta, the

volumetric and mass flow rates can be obtained from Bernoulli's equation.

ROTAMETER:

The Rotameter is a variable – area meter that consists of an enlarging transparent tube

and a metering “float” (actually heavier than the liquid) that is displaced upward by the

upward flow fluid through the tube. The tube is graduated to read the flow directly. Notches

in the float cause it to rotate and thus maintain a central position in the tube. The float is an

indicating element, and greater the flow rate, the higher the float rides in the tube.

3. Experimental set up


28

Schematic Diagram for Venturimeter, Orificemeter & Rotameter Apparatus

Flow Diagram of Venturimeter Flow Diagram of Orificemeter


VENTURIMETER & ORIFICEMETER:

Starting Procedure:

a) Clean the apparatus and make All Tanks free from Dust.

b) Close the drain valves provided.

c) Fill Sump tank ¾ with Clean Water and ensure that no foreign particles are there.

d) Close all Flow Control Valves given on the water line and open By-Pass Valve.

e) Close all Pressure Taps of Manometer connected to Venturimeter & Orificemeter.

f) Ensure that On/Off Switch given on the Panel is at OFF position.

g) Now switch on the Main Power Supply (220 Volts AC, 50 Hz).

h) Switch on the Pump.

i) Operate the Flow Control Valve to regulate the flow of water in the desired Test

Section.

j) Open the Pressure Taps of Manometer of related Test section, very slowly to avoid

the blow of water on manometer fluid.


29

k) Now open the Air release Valve provided on the Manometer, slowly to release the air

in manometer.

l) When there is no air in the manometer, close the Air release valves.

m) Adjust water flow rate in desired section with the help of Control Valve.

n) Record the Manometer reading.

o) Measure the flow of water, discharged through desired test section, using Stop Watch

and Measuring Tank.

p) Repeat Steps the same procedure for different flow rates of water, operating Control

Valve and By-Pass valve.

q) When experiment is over for one desired test section, open the By-Pass Valve fully.



CALIBRATION OF ROTAMETER:

a) Close the ball valves provided in the Venturimeter and Orificemeter pipelines.

b) Open the ball valve provided in the Rotameter pipeline.

c) Now switch on the main power supply and switch on the pump.

d) Set the flow rate with the help of by pass and flow control valves provided in

Rotameter pipeline.

e) Measure the discharge with the help of measuring tank and stopwatch.

f) The actual discharge, verify the set value of Rotameter.

g) Repeat the same procedure for different flow rates.

Closing Procedure:

a) When experiment is over, close all Manometers Pressure Taps first.

b) Switch off Pump.

c) Switch off Power Supply to Panel.

5. Observation

For venturimeter & orificemeter

OBSERVATION TABLE:

S.No. Pressure

difference, h

(cm)

Rise of Water level in

Measuring Tank, R

(cm)

Time taken for R, t

(sec.)

Average Co-efficient of Discharge:

For rotameter

OBSERVATION TABLE:

S.No. Rotameter

Reading in

LPH, Qth

Rise of Water level in

Measuring Tank, R

(cm)

Time taken for R, t

(sec.)


30

FORMULAE:

For both Venturimeter & Orificemeter:

1. Head losses,

1

100 W

mhH

, m of water ------------------- (1)

2. Theoretical discharge (Qt):

2

2

2

1

21 2

aa

gHaaQt

, m

3/s ------------------- (2)

3. Actual discharge (Qa):

100*t

AxRQa , m

3/s ------------------- (3)

4. Co-efficient of discharge (Cd):

Cd = Qa/Qt ------------------- (4)

For Rotameter:

5. Actual discharge:

10003600100*

xxt

AxRQa , m

3/s ------------------- (5)

DATA:-

H= Head loss

h = Pressure difference (m)

A = 0.1 m2

ρm = Density of manometric fluid = 13600 kg/ m3

ρW = Density of working fluid = 1000 kg/ m3

g = Acceleration due to Gravity = 9.81 m/sec2

For Venturimeter:

d1 = Dia. at inlet of the venturimeter = 28mm = 0.028m

d2 = Dia. at throat of the Venturimeter = 14mm = 0.014m

a1 = d1 2/4 cm

2 Area at inlet of Venturimeter = 6.157 * 10

-4 m

2

a2 = d2 2/4 cm

2 Area at throat of Venturimeter = 1.539* 10

-4 m

2

For Orificemeter:

d1 = Dia. at inlet of Orificemeter = 28mm. = 0.028m

d2 = Dia. of Orifice Plate = 14mm. = 0.014m

a1 = d12/4 Area at inlet of Orificemeter = 6.157 * 10

-4m

2

a2 = d22/4 m

2 Area of Orifice Plate = 1.539* 10

-4 m

2

6. Result and discussion

For Venturimeter and Orificemeter

S. No. Actual discharge Qa

, m3/s

Theo. Discharge Qt ,

m3/s

Cd = Qa/Qt


31

For Rotameter

S. No. Rotameter Reading in Qth, m3/s Theo. Discharge Qt,

m3/s

7. Conclusion

8. Precaution






32

EXPERIMENT NO. 3(c)

FLOW THROUGH HELICAL COIL

1. Aim

To determine the critical Reynolds number of a fluid flowing through the helical coil.

To determine the friction factor for flow of water through helical coil.

2. Apparatus

Helical coil experiment set up, Stop watch, Vernier Caliper

3. Theory

Helical coils are used widely in processing industries for cooling and heating applications

since the centrifugal forces experienced by the fluid acts to promote contact with the channel

wall, thereby tending to insure good contact with the wall, and an enhanced heat transfer

Some of their main advantages over straight tubes are high heat and mass transfer

coefficients, and space economy in terms of area per unit volume.

Helical coils are used for heating or cooling in process tanks. When a fluid flows through a

curved tube, centrifugal force acting upon the various elements of fluid moving with different

velocities causes secondary circulation. Secondary flow results in higher heat transfer

characteristics. Further, secondary flow stabilizes the laminar flow leading of a higher critical

Reynolds number for transition from laminar to turbulent flow.

The use of helically coiled exchangers continues to increase. Applications include liquid

heating/cooling, steam heaters, vaporizers, cryogenic cooling and vent condensing. Listed

below are the details for standard services in which helical exchangers warrant consideration.

Seal Coolers, Condensers, Cryogenic Vaporizers, Compressor Inter- and After-Coolers.

4. Experimental Procedure

a) With the discharge valve closed, switch ON the pump.

b) Slowly open the discharge valve and note down the manometer reading.

c) Note down the time taken for 1 cm raise in the discharge tank.

d) Repeat the same procedure for different flow rates of water.

5. Observation

d = Inner diameter of helical pipe = 0.932×10-2

m

v = velocity of water in pipe (m/sec)

µ = dynamic viscosity of water at 25 oC (0.890 ×10

-3 Ns/m

2)

A = cross-sectional area of coil = m2

H1 = one side manometer reading.

H2=other side manometer reading.

H2 – H1 = difference in manometer reading.

V= volume of water collected.

T= time in seconds

OBSERVATION TABLE:

S. No H1 H2

H1-H2

V(m3)

T(sec)

Q (m3/s)


33

1

2

3

4

6. Calculation

Volume collected is calculated by measuring tank, V = l * b * height

(Units in meter) = m3

Flow rate Q = V/T = m3 /sec

Velocity v = Q/A = m/sec

On the basis of velocity we can find Reynolds number,

Re = ρ * v * d

ρ = density of water = 1000 kg/m3

v= velocity

d = inner diameter of coil

= dynamic viscosity of water

Calculation for fanning friction factor

F = (∆p/ ρ g) * (D/L) * (2g/v2)

∆p=pressure difference in manometer

g =9.81 m/s2

D = overall diameter of coil = 0.235 m

H = overall height of coil = 0.35 m

N = No. of turns = 14

R =D/2

LENGTH OF COIL L = N2 * R

2

9R + 10 H

Put the value of L in the above equation and find out value of friction factor of helical coil.

7. Result and Conclusion The critical Reynolds number of the fluid flowing through the coil was found to be

_______.

The friction factor for flow of water through helical coil was found to be ______

8. Precaution


34

EXPERIMENT NO. 4(a)

STUDY OF PITOT TUBE – AIR

1. Aim

To determine the velocity profile for the flow of air through a circular pipe using Pitot Tube.

2. Objective

To measure the velocity at different points in the flow.

3. Theory

A pitot tube is a device to measure the point velocities in the flow field of a fluid. It generally

consists two concentric tubes arranged parallel to the flow .The outer tube is perforated at one

end with small holes in its wall, which are perpendicular to the flow direction and lead to the

annular space. The annular space is otherwise sealed except for a manometer lead on the

other end .The inner tube has a small opening at one end (the same end in which holes are

there in the outer tube) parallel to the flow and pointed in the direction opposite to the flow

direction. The other end of the tube has a manometer lead. The two tubes are connected to

two limbs of a manometer through the manometer leads. At equilibrium there is no fluid

within the pitot tube. The annular space serves to transmit the static pressure. The flowing

fluid is brought to rest at the entrance of the inner tube, and this tube transmits the impact

pressure (or stagnation pressure) equivalent to the kinetic energy of the flowing fluid at that

position. If and are the stagnation pressure measured by the inner and outer tubes

respectively, the point velocity, , at a given position can be obtained from the Bernoulli

equation, by neglecting the frictional losses, as

u0 = √ {2(Ps-P0)/ℓf} = √ {2gh(ℓm-ℓf)/ℓf}

Where ℓf and ℓm are the densities of the process fluid and the manometric fluid respectively,

h is the difference in the level of the manometric fluid in the two limbs of the manometer,

and g is the acceleration due to gravity. Thus by changing the position of the pitot tube the

local velocities in the flow field can be obtained. The velocity profile at a given cross section

of the pipe is obtained by moving the pitot along the diameter of the pipe. By knowing the

point velocities at a given cross section, the volumetric flow rate, Q, can be determined

Q = 0∫D/2

2ᴫru0dr

Where D is the pipe diameter, and r is the radius at which u0 is measured.

4. Requirements

Pipe line with provision for supply of air, pitot tube, inclined manometer.

5. Procedure

1. Start the flow of air by switching on the blower.

2. Divide the pipe diameter in 9 equal parts to fix the radial position at which the point

velocities are to be determined.

3. Keep the pitot tube at the radial position.

4. Once the flow steadies (indicated by unchanging or slightly fluctuating level

difference in the manometer), record the level difference in the manometer.

5. Record the ambient temperature (taken same as the air temperature).

6. Repeat step 4 by moving the pitot tube at the radial determined in step 2.

6. Observations


35

Ambient temperature (T) =

Air density at T = kg/m³

Density of manometric liquid at T = kg/ m³

S.No. r,mm h,mm

7. Formula

1)Point velocity, u0

u0 = √ {2gh(ℓm-ℓf)/ℓf}

2)Volumetric flow rate,Q

Q = 0∫D/2

2ᴫru0dr

8. Results and discussion

1. Calculate the point velocities at different radial positions.

2. Plot the point velocity versus the radial position.

3. Calculate the volumetric flow rate from the point velocities .

9. Conclusions


36

STUDY OF PITOT TUBE - WATER

1. Aim

To determine the velocity profile across the cross section of pipe for the flow of water

using Pitot tube and thereby determine the co-efficient of Pitot tube for different flow rates.

2. Objective

To measure the velocity at different points in a pipe for different flow rates.

3. Theory

It is a device used for measuring the velocity of flow at any point in a pipe. It is based on the

principle that if the velocity of flow at a point becomes zero, there is increase in pressure due

to the conversion of the kinetic energy into pressure energy.

The Pitot tube consists of a capillary tube, bent at right angle. The lower end, which is bent

through 90, is directed in the upstream direction. The liquid rises up in the tube due to

conversion of kinetic energy into pressure energy. The velocity is determined by measuring

the rise of liquid in the tube.

When a Pitot tube is used for measuring the velocity of flow in a pipe or other closed conduit

the Pitot tube may be inserted in the pipe as shown in figure. Since a Pitot tube measures the

stagnation pressure head (or the total head) at its dipped end. The pressure head may be

determined directly by using piezometeric readings between the Pitot tube and pressure

taping at the pipe surface. Consider two points (1) and (2) at the same level in such a way that

point (2) is just at the inlet of the Pitot -tube and point (1) is far away from the tube. At point

(1) the pressure is p1 and the velocity of the stream is v1. However, at point (2), is called

stagnation point, the fluid is brought to rest and the energy has been converted to pressure

energy. Therefore the pressure at (2) is p2, the velocity v2 is zero and since (1) and (2) are in

the same horizontal plane, so z1 = z2.

Applying Bernoulli’s equation at points (1) and (2)

g

v

w

p

g

v

w

p

22

22

2

2

11

v2 = 0

w

p

w

p

g

v 1212

2


37

w

p

w

pgv 12

1 2

gHv 21

This is theoretical velocity.

Actual velocity gHCv vact21

4. Requirements

Pitot tube : Material SS/copper of compatible size fitted

with scale.

Test Section : Material Clear Acrylic, compatible to 1” dia

Pipe.


Flow Measurement : Using Measuring Tank,

Capacity 40 Ltrs.



Control Panel Comprises of: Standard make On/Off Switch, Mains Indicator, etc.

5. Experimental set-up The apparatus consists of a Pitot tube made of SS and fixed below a pointer gauge. The

pointer gauge is capable to measure the position of Pitot tube in transparent test section. The

pipe has a flow control valve to regulate the flow. Piezometric tubes are provided to

determine the velocity head. A pump is provided to circulate the water. Discharge is

measured with the help of measuring tank and stopwatch


Starting Procedure:

1. Clean the apparatus and make Tank free from Dust.

2. Close the drain valves provided.

3. Fill Sump tank ¾ with Clean Water and ensure that no foreign particles are there.

4. Close all Flow Control Valves given on the water line and open By-Pass Valve.

5. Ensure that On/Off Switch given on the Panel is at OFF position.

6. Now switch on the Main Power Supply (220 Volts AC, 50 Hz).

7. Switch on the Pump.

8. Operate the Flow Control Valve to regulate the flow of water through orifice.

9. Adjust water flow rate to desired rate with the help of flow Control Valve.

10. Set the Pitot tube at the center of test section

11. Record the piezometric reading and measure the discharge with the help of measuring

tank and stop watch.

12. Now move the Pitot tube up and down at the same flow rate and note the piezometric

readings to find out the velocity at different points in pipe.

13. Calculate the co – efficient of Pitot tube from actual and theoretical velocities and plot

the velocities at different points inside the pipe.

14. Repeat the same procedure for different flow rates of water, operating Control Valve,

and By-Pass valve.


38

Closing Procedure:

1. When experiment is over Switch off Pump.

2. Switch off Power Supply to Panel.

3. Drain water from all tanks with the help of given drain valves.

FORMULAE

1. Discharge,

t

RAQ

* ----------------------- (1)

2. Actual Velocity,

a

Qvact ----------------------- (2)

3. Theoretical velocity

ghvth 2 ------------------------ (3)

4. Co – efficient of Pitot tube,

th

actv

v

vC ------------------------- (4)

5. Velocity at any point,

ghCv v 2

7. Observation table

S.No. Pressure head at

different points on up

side

Pressure

head at

center

Pressure head at

different points on down

side

R(cm) t(sec)

8 mm 6 mm 4 mm 0 4 mm 6 mm 8 mm

1.

2.

3.

CALCULATION TABLE

S.

No.

Cv v8 v6 v4 v0 v4 v6 v8

1.

2.

3.

DATA


39

A = 0.1 m2

a = 0.0006157 m2

g = 9.81 m/ s2

NOMENCLATURE

A = Area of measuring tank.

a = Cross section area of test section

R = Rise of water level in measuring tank.

h = Piezometric difference

CV = Co efficient of Pitot tube

g = Acceleration due to gravity

va = actual velocity of fluid.

Q = discharge at outlet.

t = time for R.


Calculate the point velocities at different radial positions.

Calculate the co-efficient of Pitot Tube for different flow rates.

Plot the point velocity versus the radial position.

9. Conclusion

10. Precautions






6. Always use clean water.

7. If apparatus will not in use for more than one month, drain the apparatus completely,



40

EXPERIMENT NO. 4(b)

STUDY OF REYNOLDS APPRATUS

1. Aim

To study the flow regime for liquid flow in a pipe, and to find the critical Reynold's number.

2. Objective

To find the volumetric flow rate corresponding to various flow regimes.

3. Theory

Osborne Reynolds (1883) demonstrated different flow regimes when a fluid flows through a

pipe. These regimes are: laminar, turbulent and transition or intermediate regimes. For a

given pippe and fluid, laminar flow exists at low velocity characterized by ordered sliding

past of liquid layer over one another without lateral mixing; while turbulent flow is found at

high velocity in which the flow pattern is random and there is intense lateral mixing in the

fluid. In between these two extremes, a transition flow regime is observed in which the flow

alters between laminar and turbulent flows. Reynolds’s found that the nature of flow depends

on the diameter of the pipe, and viscosity, density and average velocity of the fluid; and these

factors can be combined into one dimensionless group, called the Reynolds’s number, the

magnitude of which would indicate the flow regime. The Reynolds number, Re is defined as

Re = Du/µ = Du/v

Where D = pipe diameter, u= fluid velocity, ρ= fluid density, µ = fluid viscosity, and v =

kinematic viscosity.

Physically, Reynolds number is interpreted as the ratio of the inertial force to viscous force

acting on the fluid. In general, D and v represent some characteristic dimension of the flow

domain and characteristic velocity of the fluid. Thus laminar flow exists at small Reynolds

number, and turbulent flow at large Reynolds number.

The minimum Reynolds number at which laminar flow disappears is called the critical

Reynolds number. The value of critical Reynolds number depends on the geometry of the

flow domain (circular or rectangular pipe, open or closed channel, flat plate etc.) and flow

configuration (flow around a bluff body, in packed bed etc.)

4. Requirements Reynolds apparatus, water source, dye, measuring cylinder, stop watch.

5. PROCEDURE

1. Fill in the tank with water, and the dye- chamber with dye.

2. Note the water temperature.

3. Start the water flow and maintain a small flow rate, enough to fill the whole pipe

cross section.

4. Once the flow stabilizes, start the dye injection. The injection rate should be just

enough to give a clear visible streak of the dye.

5. Observe the pattern of the dye streak. The dye should flow in a straight line.

6. Increase the water in small and equal increments, and observe the dye streak.

7. Repeat step 6 until some undulations commence in the streak. Note the corresponding

volumetric flow rate of water, which is the critical Reynolds number. Appearance of

the undulations signifies the initiation of the intermediate or transition flow.


41

Note At this point the undulations will be unstable so that there will be some portion

of the dye streak which will be undulating and some portion which will

not.

8. Keep increasing the flow rate of the liquid further until at one point there is found a

complete dispersion of the dye (indicated by the liquid getting colour through the

cross section) just as it comes out of the injection needle. This point shows the

conversion to a fully turbulent regime.

9. Note the corresponding the volumetric flow rate.

6. Observations Temperature of the liquid =

Pipe diameter = m

S.No Flow Regime Volume of water collected Time taken

1 Laminar

2 Transition

3 Fully Turbulent

Volumetric flow rate corresponding to start of the Laminar flow = m³/s

Volumetric flow rate corresponding to start of the transition flow = m³/s

Volumetric flow rate corresponding to fully turbulent flow = m³/s

Liquid density at observed temperature = kg/m³

Liquid viscosity at the observed temperature = kg/m.

7. Results and discussion 1. Calculate the Reynolds number corresponding to the transition flow and turbulent

flow.

The liquid velocity is calculate as u = 4Q/ π D², Where Q is the volumetric flow

rate.

2. Compare the critical Reynolds number and the Reynolds number for transition to fully

turbulent flow observed with those reported in the literature. Discuss the possible

sources of discrepancy, if any.

8. Conclusion


42

EXPERIMENT NO. 5 (a)

CENTRIFUGAL PUMP CHARACTERISTICS

1. AIM

Study and analysis of centrifugal pump characteristics.

2. Objectives 1. To study the operating characteristics of a centrifugal pump.

2. To study the operating characteristics of two centrifugal pumps operated in series and in

parallel.

3. Theory Pumps are the fluid moving machineries which increase the mechanical energy of the fluids

to be displaced. The energy increase may be used to increase the velocity, the pressure or the

elevation of the fluids. A large number of pumps, differing widely in principle and

mechanical construction, have been developed to meet a wide variety of operating conditions.

For selection of pumps for a specific application requires the knowledge of operating

conditions of the system and applicability of different available pumps.

The mechanical energy of the liquid is increased by centrifugal action. Centrifugal pumps

are classified as single suction and double suction pumps depending upon the suction from

either one side or from both sides respectively.

In a single suction centrifugal pump the liquid enters through a suction connection concentric

with the axis of a high-speed rotary element called the impeller, which carries radial vanes

integrally cast in it. Liquid flows outward in the spaces between the vanes and leaves the

impeller at a considerably greater velocity with respect to the ground than at the entrance to

the impeller. In a properly functioning pump the space between the vanes is completely filled

with liquid flowing without cavitation. The liquid leaving the outer periphery of the impeller

is collected in a spiral casing called the volute and leaves the pump through a tangential

discharge connection. In the volute, the velocity head of the liquid from the impeller is

converted into pressure head. The power is supplied to the fluid by the impeller and is

transmitted to the impeller by the torque of the drive shaft, which usually is driven by a

direct-connected motor at constant speed, commonly at 1750 rpm. Another common type

uses a double-suction impeller, which accepts liquid from both sides. Also, the impeller itself

may be a simple open spider, or it may be enclosed or shrouded.

4. Experimental Set up

4.1. Requirements Centrifugal pump test rig, bucket, watch.

4.2. Schematic Diagram of Experimental Setup The schematic diagram of the experimental setup is shown below:


43

V8

V-Notch

Measuring Tank

Sump Tank

P1

P2

P3

P4

V1

V2

V4

V3 C1 C2

V5

OPERATION OPEN CLOSE

SINGLE PUMP : V1, V4, V5 V2, V3, V8, V9

SERIES : V1, V2, V5 V3, V4, V8, V9

PARALLEL : V1, V3, V4, V5 V2,V8,V9

5. Experimental Procedure First, prime the pump by pouring water through valve V8. Subsequently, the following

sequence of operations are to be carried out:

5.1. Calibration of V-Notch

1. Measure the width of the V-notch at the top and depth of the V-notch.

2. Fill the storage tank with water.

3. Open valves V1, V2 and V5, and close valves V3, V4 & V9.

4. Start running the pumps.

5. Fill the channel with water until water starts spilling over the notch to the outlet.

6. Stop the water supply by closing the bench supply valve (V5) with the pump in running

condition.

7. Allow the water above the crest height to spill over the notch.

8. When the water level is at the level of the crest of the notch, bring the point gauge exactly

to the water surface, and note the reading (say R1) on the longer scale which coincides

with the “zero” of the shorter scale. (This is the datum level for subsequent height-

measurements when the water flows in the channel.)

9. Open valve V5 completely and wait for the level (or head) of water in the channel to

stabilize.

10. Bring the point gauge to the surface of the water by rotating the knob attached to the

scale, and note the reading (say R2) on the longer scale that coincides with the “zero” of

the shorter scale.


44

11. Collect the water in the bucket for an arbitrary but known duration, and determine the

flow rate of water by dividing the volume (or mass) of water collected by the time of

collection.

12. The difference (R2-R1) is called the head which is related to the volumetric flow rate by

the following equation:

2

5

2tan15

8hgCQ d

; Where, h = R2-R1

13. Adjust the water flow rate at seven different values from the maximum to zero flow rate

in the channel by manipulating valve V5 and determine the water flow rate and the

corresponding head.

14. Calculate the discharge coefficient (Cd) in each case, and take the arithmetic average of

these values as the Cd of the notch.

5.2. Pump Characteristics (Single Pump)

1. After carrying out the steps 1 to 3 mentioned in “Calibration of V-Notch”, open the valve

V5 fully.

2. Note the RPM of the pump from the control panel.

3. Note the gage pressure on the outlet line (P4 in the figure) and on the inlet line (P1 in the

figure). The difference of the readings in P4 and P1 is the head delivered by the pump.

4. Note the power supplied to the pump from the display on the control panel by pressing

the appropriate button (labeled W).

5. Note the head of the V-notch.

6. Repeat steps 2 to 5 for seven different flow rates of water spanning the whole range of

flow rate up to zero flow by manipulating V5.

5.3. Pump Characteristics (Pump in Series)

1. Fully open the valves V1, V2 and V5, and close valves V3 and V4.



figure).






7. Repeat steps 1 to 6 for three different RPMs.

5.4. Pump Characteristics (Pump in Parallel)

1. Fully open the valves V1, V3, V4 and V5, and close valve V2.



figure).







6. Observations


45

6.1. Calibration of V-Notch Width across the top of the V-notch =

Depth of the V-notch =

Initial point gauge reading at zero flow rate (R1) =

S No Point gauge

reading, R2

Head of

the V-notch

(R2 – R1)

Amount of

Water collected

Duration of

collecting water

6.2 Pump Characteristics*

1. RPM =

2. Initial point gauge reading at zero flow rate (R1) =

3. Length of Pipe in single pump operation =

4. Length of Pipe in series operation =

5. Length of Pipe in parallel operation =

6. Elevation =

7. Diameter of the suction pipe =

8. Diameter of the discharge pipe =

S No Point gauge

Reading, R2

Head of

The V-notch

Gauge pressure reading Power to the

pump P1 P4

* Make separate tables for different configurations (single/series/parallel) of the pumps and for different RPMs.

7. Model Calculations

For steady incompressible flow of a liquid of density , the developed head is given as

fhZZg

VV

g

PPH

)(

212

2

1

2

212

(1)

where subscripts 1 and 2 signify the values at the suction and delivery ports of the pump; P

is the pressure; V is the velocity; Z is the elevation; and hf is the friction head (that is, the

frictional losses) in the line between the suction and delivery ports. hf is given by

g

VK

D

Lfh ff

24

2

(2)

where f is the friction factor, L is length of the pipe, D is the diameter of the pipe, Kf is the

loss factor for fittings. Friction factor (f) depends on roughness and type of flow


46

(laminar/turbulent) and read from friction factor chart. Alternatively, it can be calculated

from the equations:

Re

16

Nf ; for laminar flow

The friction factor for turbulent flow can be calculated by using Colebrook equations (either

implicit form or explicit form) given below:

7.1. Implicit Forms of Colebrook Equation

There are at least three forms of the Colebrook Equation that can be found in current

literature on hydraulics. These are:

fNDf Re

1051.2

7.3log2

1 (3)

fNDf Re

107.182

log274.11

(4)

fD

N

D

f Re

10103.9

1log2log214.11

(5)

where,

f is the Friction Factor and is dimensionless

ε is the Absolute Roughness and is in units of length

D is the Inside Diameter and, as these formulas are written, is in the same units as ε.

NRe is the Reynolds Number and is dimensionless.

Note that ε/D is the Relative Roughness and is dimensionless.

These three equations are referred to as “Implicit” Equations. “Implicit” means that “f”, the

Friction Factor, is “Implied or understood though not directly expressed”2. Simply stated,

the equations ARE NOT in the form of “f = ………”. These are sometimes referred to as

“equivalent” but the results will vary when calculated to the fourth significant digit.

These equations can be solved for “f” given the Relative Roughness (ε/D) and the Reynolds

Number, (NRe), by iteration. Such iterations can be performed using an electronic

spreadsheet. A spreadsheet, “Friction Factor Formulas for Cheresource.xls” is available at

http://www.cheresources.com/colebrook1.shtml presented for demonstration. The

spreadsheet contains four worksheets. The first “Tab” is labeled “Iterations”. The Iterative

solutions are generated by breaking the formulas in two parts, that which is left of the equal

sign and that which is right of the equal sign (See row 20 as an example). The Iteration then

tests values of “f” that will result in the difference between the two sides to be zero or very

close to zero. (A complete explanation was published in the ASHRAE Journal of September,

2002: see Reference 4 for details)

7.2. Explicit Forms of Colebrook Equation

http://www.cheresources.com/colebrook1.shtml


47

There are four Explicit forms of Colebrook Equations reported in literature. (Ref: available

on internet at http://www.cheresources.com/colebrook1.shtml). Out of these the Serghide’s

solution is the most accurate one, which is given below:

Serghide’s Solution (see Reference 3 for details).

22 )))2(/()(( ABCABAf (6)

where

)]/12()7.3/[(log2 Re10 NDA

)]/51.2()7.3/[(log2 Re10 NADB

)]/51.2()7.3/[(log2 Re10 NBDC

Serghide can be used across the entire range of the Moody Diagram. Its accuracy is

unparalleled amongst the Explicit Equations evaluated here. It appears to be based on Eqn. 3,

as do all the Explicit Equations presented. There is less deviation between Serghide and Eqn.

3 then there is between Eqn. 3 and either Eqn. 4 or Eqn. 5.

6.0log07.41

Re fNf

; (7)

Eqn. 7 is the von Karman equation for turbulent flow which can be used only if the pipe is

smooth. In our case, the pipe is not smooth and hence this equation cannot be used.

Colebrook equation has to be used.

Power developed by the pump (or the power delivered to the fluid) Pf is given as

gQHPf (8)

where Q is the volumetric flow rate of the liquid.

The mechanical efficiency of the pump is given as

B

f

P

P (9)

where PB is the total power supplied to the pump drive from an external source.

Operating (or performance) characteristics of a pump is commonly illustrated by plots of

actual head developed H, power consumption Pf, and efficiency versus the volumetric flow

rate.

8. Results & Discussion

8.1. Calibration of V-notch

S No Volumetric flow rate Head of the V-notch Cd

http://www.cheresources.com/colebrook1.shtml


48

Average Cd =

8.2. Pump Characteristics (RPM = )

S No Volumetric flow

rate

Head developed by

the pump

Power delivered by

the pump Efficiency

1. Plot head, power delivered to the pump, power delivered to the liquid, and efficiency

versus the capacity of the pump (i.e., flow rate of water).

2. Suggest the operating point from the efficiency versus flow rate plot.

9. Conclusions

10. Precautions

Do not run the pump with the delivery valve closed for a long time to avoid damage of the

pump.

References


Engineering, 6th


2. Babu, B.V.,"Pumps: Selection & Trouble Shooting", IPT (Indian Plumbing Today), Vol.

2005 (No. 6), pp. 41-49, November-December, 2005.

3. T.K.Serghide’s implementation of Steffenson’s accelerated convergence technique,

reportedly to have appeared in Chemical Engineering March 5, 1984.

4. Lester, T. “Calculating Pressure Drops in Piping Systems.” ASHRAE Journal Sept.

2002.

EXPERIMENT NO. 5 (b)

RECIPROCATING PUMP CHARACTERISTICS

1. AIM

Study and analysis of operating characteristics of double acting reciprocating pump.

2. Objectives To study the operating characteristics of a double acting piston-type reciprocating pump.

3. Theory A piston pump consists of a cylinder with a reciprocating piston. The fluid is drawn through

an inlet check valve into the cylinder by the withdrawal of a piston and is then forced out

through a discharge check valve on the return stroke. Most piston pumps are double acting

with the liquid admitted alternately on each side of the piston so that one part of the cylinder

is being filled while the other is being emptied. The energy added per unit mass of liquid by




49

the pump is termed as the head developed by the pump, which is expressed in length

dimension.


4.1. Requirements Reciprocating pump test rig, watch.

4.2. Experimental Setup

P2

P1

V1

V2

V3

LL

V4

To Drain

Measuring

Tank

Sump

Tank

5. Experimental Procedure 1. Fill the storage tank with water and ensure that the measuring tank is empty.

2. Open the valve on the delivery line, and allow the flow from the delivery line to drain

directly into the sump tank with the help of the drain valve.

3. Adjust the RPM of the motor to an arbitrary but known value.

4. Note the gauge pressures at the suction and delivery line of the pump.

5. Note the power to the pump from the display on the control panel by pressing appropriate

button.

6. Collect water in the measuring tank for a given duration of time and note the change in

water level in this tank during this period.


flow rate (excluding zero flow rate) by manipulating the delivery valve.


6. Observations Technical specifications of the pump:

i. Type: Double acting

ii. Power: Connected to 1 HP, 3 phase, 440V AC motor.

iii. Pump bore (diameter): 44.5 mm

iv. Pump stroke: 35 mm


50

Inside diameter of the suction pipe =

Inside diameter of the delivery pipe =

RPM =

Length of the suction pipe =

Length of the delivery pipe =

Elevation =

S No

Time of collection

of water in

measuring tank

Change in water

level in the

measuring tank

Gauge pressure

reading Power to

the pump Suction

(P1)

Delivery

(P2)

Note: Make separate tables for different RPMs.


For steady incompressible flow of a liquid of density , the developed head is given as

fhZZg

VV

g

PPH

)(

212

2

1

2

212

(1)

where subscripts 1 and 2 signify the values at the delivery and suction ports of the pump; P is

the pressure; V is the velocity; Z is the elevation; and hf is the friction head (that is, the

frictional losses) in the line between the suction and delivery ports. hf is to be calculated as

described in previous experiment, i.e., centrifugal pump characteristics.

Power developed by the pump (or the power delivered to the fluid) Pf is given as

gQHPf (2)

where Q is the volumetric flow rate of the liquid.

The mechanical efficiency of the pump is given as

B

f

P

P (3)

where PB is the total power supplied to the pump drive from an external source.

Volumetric efficiency (V) of a reciprocating pump is defined as the ratio of the actual

discharge (Q) to the theoretical discharge (Qth). The theoretical discharge is the volume swept

by the piston of the pump per unit time. The difference between the theoretical and actual

discharge of the pump is called the slip.

th

VQ

Q (4)


51

where Qth for a double acting pump is given by

NLdQ Bth

60

2

42 2

(5)

where dB and L are the bore diameter and length of the stroke of the pump, and N is the speed

of rotation of the crank (attached to the motor) in RPM.

Operating (or performance) characteristics of a pump is commonly illustrated by plots of

actual head developed H, power consumption PB and Pf and efficiency versus the capacity

of the pump (i.e., the volumetric flow rate of liquid).


RPM =

Theoretical discharge of the pump =

S No Volumetric flow

rate

Head developed

by the pump

Power delivered

by the pump Η ηV

1. Plot head, power delivered to pump, power delivered to the liquid, and mechanical

efficiency versus the capacity of the pump (flow rate of water).

2. Suggest the operating point from the efficiency versus flow rate plot.

9. Conclusions

10. Precautions

Never run the pump with the delivery valve closed.

References


Engineering, 6th


4. Babu, B.V.,"Pumps: Selection & Trouble Shooting", IPT (Indian Plumbing Today), Vol.

2005 (No. 6), pp. 41-49, November-December, 2005.




52


HEAT PIPE DEMONSTRATOR

1. Aim

To study the performance of heating of heat pipe and compare its working with the best

conductor.

2. Objective

To calculate the thermal conductivity of the given heat pipes.

3. Theory

A heat pipe or heat pin is a heat-transfer device that combines the principles of both thermal

conductivity and phase transition to efficiently manage the transfer of heat between two solid

interfaces. At the hot interface of a heat pipe a liquid in contact with a thermally conductive

solid surface turns into a vapor by absorbing heat from that surface. The vapor then travels

along the heat pipe to the cold interface and condenses back into a liquid - releasing the latent

heat. The liquid then returns to the hot interface through either as capillary action, centrifugal

force, or gravity, and the cycle repeats. Heat pipes are thermal superconductors, due to the

very high heat transfer coefficients for boiling and condensation.

3.1 Working Principle

The heat pipe is a heat-transfer element with extremely high thermal conductivity and large

heat transfer capacity, which accomplishes the heat transfer through the latent heat of phase

change of the working fluid inside the sealed vacuum tube. For the typical gravity heat pipe,

the up end of the heat pipe releases heat, while the working media is condensed into the

liquid. The condensed liquid, with the pull of gravity, will return to the hot side along the

inner wall of the heat pipe and will be heated and vaporized again. In this way, the heat will

be transferred from one end to the other successively by the heat transfer of phase change, the

thermal resistance inside the heat pipe is very small, which results in a larger heat transfer

capacity at a smaller temperature difference.

In terms of the gravity heat pipe, due to its simple structure, single-direction heat

conductivity, and special heat transfer mechanism (the heat exchange between the hot fluid

and the cold one is done outside the heat pipe), the heat transfer can be easily enhanced, the

working fluid is readily evaporated and boiled and the heat pipe is quick to start up the heat-

transfer capability of the heat pipe can rival and even exceed that of stainless steel and

copper. The heat pipe can be used individually or as combinedly. The heat pipe heat

exchanger consisting of heat pipes is characterized by such merits as high heat transfer

efficiency, low flow resistance, compact structure, high reliability, and excellent maintenance

economy, and is widely applied in the space technology, electronic, metallurgical, motive

power, and petroleum and chemical trades.

https://en.wikipedia.org/wiki/Thermal_conductivity

https://en.wikipedia.org/wiki/Thermal_conductivity

https://en.wikipedia.org/wiki/Phase_transition

https://en.wikipedia.org/wiki/Interface_(chemistry)

https://en.wikipedia.org/wiki/Interface_(chemistry)

https://en.wikipedia.org/wiki/Liquid

https://en.wikipedia.org/wiki/Vapor

https://en.wikipedia.org/wiki/Latent_heat

https://en.wikipedia.org/wiki/Latent_heat

https://en.wikipedia.org/wiki/Liquid

https://en.wikipedia.org/wiki/Capillary_action

https://en.wikipedia.org/wiki/Centrifugal_force

https://en.wikipedia.org/wiki/Centrifugal_force

https://en.wikipedia.org/wiki/Boiling

https://en.wikipedia.org/wiki/Condensation


53

Fig. Liquid and Vapor pressure distribution along the heat pipe


4.1. Requirements

The apparatus consist of three different test specimen viz. stainless steel, copper & heat pipe

(Al) which is banded with water cut at the top and heaters at the bottom.

4.2. Schematic Diagram of Experimental Setup

The schematic diagram of the experimental setup is shown below

:

5. Procedure

1. Plug-in socket of control panel & set up instrument in proper position.

2. Fill up the cup provided on each of the pipe with water.

3. Start the main switch of control panel.


54

4. Increase slowly the input to heater by the dimmerstat starting from 0 volts position

and ampere I changes accordingly.

5. Adjust input so that it correspond out to certain Watts to be calculated (i.e. Q = V *

I) out as maximum by help of dimmerstat by varying Voltage.

6. See that both inputs remain constant throughout the experiment.

7. The steady state condition can be checked by rectal temperature of thermocouple of

1 to 8 zone temperature indicator provided on Control panel out of which 1 to 6

channel temperature Indicator are actually working with 1,2 – Stainless Steel , 3,4 –

Heat Pipe (Al) and 5,6 – Copper material heat pipe.

8. Note down the reading in the all the tube with Time up to which it reaches before

and after steady state as given in observation table.

6. Observations & Calculations

Data given:

1. Length of SS, Copper and Heat pipe (L) :- 300 mm

2. Diameter of pipe (Dp) :- 250 mm

3. Area of pipe (Ap) :- 0.04908 m2

6.1 Observation table for S.S. pipe.

Thermocouple

No.

Thermocouple

Position (mm)

(Before Steady State)

Temperature

(K)

(Before Steady

State)

Temperature

(K)

(At Steady State)

1 50

2 200

6.2 Observation table for Heat pipe.

Thermocouple

No.

Thermocouple

Position (mm)


Temperature

(K)

(Before Steady

State)

Temperature

(K)

(At Steady State)

3 50

4 200

6.3 Observation table for Copper pipe.

Thermocouple

No.

Thermocouple

Position (mm)


Temperature

(K)

(Before Steady

State)

Temperature

(K)

(At Steady State)

5 50

6 200

6.4 Formulae & Calculations

QA = V * I


55

QA = - KA * AP *

QA – Power, Watts

V – Voltage Potential, Volts

I – Ampere Current, Ampere

QA - Heat Transfer Rate, Watts

Ap - Area of Pipe, m2

T - Change in temperature from Start to Steady state ,K

x - Change in position of Thermocouple from Start to Steady state (i.e. 50 & 200 mm)

KA - Thermal Conductivity of Material A, W / m . K

For Copper, Heat pipe and Stainless Steel pipe :

KA =

= W / m .K

7. Results

Thermocouple

Position

(mm)

Thermal Conductivity

( W/m K)

SS pipe Heat pipe Copper pipe

50

200

8. Precautions

1. Ensure that the Dimmerstat is at 0 position at before switching on and off the

equipment.

2. Increase the energy input gradually to the heater during initial set-up experimentation.

3. Never use the heater at full wattage for longer period of time.


THERMAL CONDUCTIVITY OF SOLIDS

1. Aim

To measure the thermal conductivity of solids.

2. Objective

(i) To Determination the Overall Thermal Conductivity of Composite wall.

(ii) To check that the Thermal Resistances in Composite wall are connected in series.


56

3. Theory

Thermal conductivity is defined as the fundamental property of material which gives the

measure of the effectivity of material in transmitting heat through it.

For the measurement of the thermal conductivity K what is required is to have one

dimensional heat flow through the flat specimen, an arrangement for maintaining its faces at

the constant temperature and metering method to measure the heat flow through a known

area.

Knowing the heat input to the central plate heater, the temperature difference across the each

specimen, its thickness and the area, one can calculate the K by the following formula.

K = q * L / [2 * A* ( - )]

Where,

K Thermal Conductivity of the sample, W / m

q Heat flow rate in the specimen, W

A Area of the specimen, m²

Hot side average temperature,

Cold side average temperature

L Thickness of the specimen, m


4.1. Requirements

Thermal Conductivity Apparatus with composite wall (Mild steel, Hylam (Paper Grade) &

Wood), C clamps


Two sections of composite walls are positioned on either side of the plate heater (Ni-Cr wire

packed in upper and lower mica sheets, 1000 W)

Two thermocouples (2&3) are used to measure the hot face temperature at the upper and

lower heater plate, 4 & 5 are used to measure the temperature at the other end of mild steel

plate (25mm), 6 & 7 to measure temperature at far end of hylam sheet (20 mm) and 8 & 9 for

far end of the wooden plate (12 mm). (see figure). Diameter of the plates is 300 mm

Specimens are held in position by the help of C clamps. The whole assembly is enclosed in

wooden box with one side transparent of visualization.

Voltmeter and Ammeter are used to measure the energy input to the heater. This energy input

to the heater can be varied using Dimmer stat. Digital temperature indicator with selector

switch on the control panel indicates the temperature at different positions in the composite


57

wall. Indicator lamp Indicates ON/OFF position of the heater. MCB has been provided to

switch ON/OFF the power to the equipment.

5. Procedure

The specimens are placed on either side of the heating plate assembly uniformly touching

each other. Then predetermined heat input is given to the heater using the Dimmer stat. The

input of the heater (current & voltage) and the thermocouple readings are observed every 5

minutes till a reasonably steady state condition is reached (steady temperature gradient). The

readings are recorded in the observation table. The final steady state values are taken for

calculations.

6.Observations and calculations:

6.1 Observation Table

S.No. Volt.

(V)

Amp.

(A)

T2

( )

T3

( )

T4

( )

T5

( )

T6

( )

T7

( )

T8

( )

T9

( )

6.2 Calculations

Heat transfer Area Perpendicular to Heat Flow,

A = (π /4) * D² = ________ m²

Heat Input Q = V * I = ________ W

Thermal Conductivity for individual specimen,

= (Q * ) / [2 * * ( – ], W/m0C

i = 1, 2 and 3 for MS, Hylam and Wooden Plate respectively.

Where,

= ( + ) / 2 for MS plate, ( + ) / 2 for Hylam Plate, ( + ) / 2 for Wooden

Plate,

= ( + ) / 2 for MS Plate, ( + ) / 2 for Hylam Plate, ( + ) / 2 for Wooden

Plate,

Overall Thermal Conductivity,

= (Q * L) / [2 * A *((( + ) / 2) – (( + ) / 2)))], W / m

Indiviual Thermal Resistance,

Ri= L/KiAi, 0C/W

In Series, Thermal Resistance Rt of the composite wall is given by

RT= ∑ Ri , i = 1, 2 and 3 for MS, Hylam and Wooden Plate respectively.

7. Results Overall thermal conductivity of the material is……. W/m

0C

8. Precautions 1. Ensure that the Dimmer state is at 0 position at before switching on and off the

equipment.

2. Increase the energy input gradually to the heater during initial set-up experimentation.


58

3. Never use the heater at full wattage for longer period of time.

EXPERIMENT NO. 6 (c)

THERMAL CONDUCTIVITY OF LIQUIDS

1. Aim

To measure of the effectivity of the liquid in transmitting heat through it.

2. Objective

To determine the Thermal conductivity of liquid.

3. Theory

Thermal conductivity is the property of a material to conduct heat. Heat transfers occurs

across materials of high thermal conductivity than across materials of low thermal

conductivity. Thermal conductivity of materials is temperature dependent. For the

measurement of the thermal conductivity K what is required is to have one dimensional heat

flow through the flat specimen, an arrangement for maintaining its faces at the constant

temperature and metering method to measure the heat flow through a known area.

http://en.wikipedia.org/wiki/List_of_materials_properties

http://en.wikipedia.org/wiki/Heat_conduction

http://en.wikipedia.org/wiki/Heat


59

Knowing the heat input to the central plate heater, the temperature difference across the each

specimen, its thickness and the area, one can calculate the K by the following formula.

K = q * L / [A * ( Theater avg – Tsample liquid temp ) ]

where,

K Thermal Conductivity of the sample, W / m C

q Heat flow rate in the specimen, W

A Area of the specimen, m2

TH,av Hot side average temperature, C

TC,av Cold side average temperature, C

L Thickness of the test section, m

Thermal conductivity is important in building insulation and related fields. However,

materials used in such trades are rarely subjected to chemical purity standards.Several

construction materials' ''k'' values are listed below. These should be considered approximate

due to the uncertainties related to material definitions.

The following table is meant as a small sample of data to illustrate the thermal conductivity

of various types of substances.

Some typical thermal conductivity (k values)

Material Thermal

Conductivity

(W·m−1

·K−1

)

Temperature

(K)

Ice 1.6 - 2.2 293

Water 0.32 293

Glycerol 0.29 293

Rubber (92%) 0.16a 303

Alcohols OR Oils 0.1- 0.21 293

Air 0.024 - 0.0262 273-300

Oxygen (O2) 0.0238 293

Nitrogen (N2) 0.0234 - 0.026 293 - 300


4.1. Requirements

Thermal Conductivity Apparatus with composite wall (Mild steel, Hylam (Paper Grade) &

Wood), C clamps

.

4.2 Description

Test Section (SS 304, 5 mm width, 150 mm diameter) is resting on a heater (500 W, Ni-Cr)

.The heat passing across the test section is removed by the cooling water jacket.

The housing made of Mild Steel has been filled with glass wool to ensure minimum heat loss

to the surroundings.

The total assembly of test section, , main heater and cooling jacket are held in position by the

help of bolts.

Voltmeter and Ammeter are used to measure the energy input to the heater. This energy input

to the heater can be varied using Dimmerstat. Digital Temperature indicator with selector

switch on the control panel indicates the temperature at different positions. Indicator Lamp

indicates ON/ OFF position of the heater. MCB has been provided to switch ON/ OFF the

power to the equipment.

http://en.wikipedia.org/wiki/Watt

http://en.wikipedia.org/wiki/Metre

http://en.wikipedia.org/wiki/Kelvin

http://en.wikipedia.org/wiki/Ice

http://en.wikipedia.org/wiki/Water

http://en.wikipedia.org/wiki/Glycerol

http://en.wikipedia.org/wiki/Rubber

http://en.wikipedia.org/wiki/Alcohol

http://en.wikipedia.org/wiki/Oil

http://en.wikipedia.org/wiki/Earth's_atmosphere

http://en.wikipedia.org/wiki/Oxygen

http://en.wikipedia.org/wiki/Nitrogen


60

5. Procedure

1. Fill the test section with the liquid whose thermal conductivity is to be measured.

2. Switch ON the Main heater and set the desired heat input through the test section

using Dimmerstat.

3. Start the cooling water supply to the cooling water jacket.

4. Now observe the temperatures T1 to T6 after 15 minutes and note down their values

once they become reasonably constant .



Sr.

No

Volt

(V)

Amp.

(A)

T1

(C)

T2

(C)

T3

(C)

T4

(C)

T5

(C)

T6

(C)

1

2

3

4

5

T1= heater temperature

T2= cold water inlet temp to upper plate

T3= hot water outlet temp of upper plate

T4= cold water inlet to lower plate

T5= hot water outlet of lower plate

T6= sample liquid inner temp

T Heater = Heater temperature= T1

Tsample = sample liquid temperature

Diameter of test section = 150 mm

Thickness of test section = 20 mm

6.2 Calculations Heat Transfer Area Perpendicular to Heat Flow,

A = (2 * * r) = ------------------ m

2

Heat Input Q = V * I = ------------ Watt

Thermal Conductivity , K = (Q * L) / [A * (Theater – T 6)] = ----------- W/ m C.

7. Results Thermal conductivity of the material is……. W/m

0C

8. Precautions 1. Ensure that the Dimmerstat is at 0 position at before switching on and off the

equipment

2. Increase the energy input gradually to the heater during initial set-up experimentation

3. Never use the heater at full wattage for longer period of time


61


DROPWISE AND FILWISE CONDENSATION

1. Aim:

To determine the inside and outside heat transfer coefficient of Filmwise and

Dropwise condenser.

To study the Dropwise and Filmwise condensation phenomena

2. Apparatus:

Filmwise and Dropwise Condensers enclosed in a Borosilicate Glass Tube with flow control

valves, Steam generator with heating elements, Digital Temperature Indicator with selector

switch, Rotameter.

3. Theory:

Condensation Heat Transfer:

The process of condensation is the reverse of boiling. Whenever a saturated vapor comes in

contact with a surface at a lower temperature, condensation occurs. There are two modes of

condensation; filmwise, in which the condensate wets the surface forming a continuous film

which covers the entire surface and dropwise in which the vapor condenses into small liquid

droplets of various sizes which fall down the surface in a random fashion.

Filmwise condensation generally occurs on clean uncontaminated surfaces. In this type of

condensation the film covering the entire surface grows in thickness as it moves down the

surface by gravity. There exists a thermal gradient in the film and so it acts as a resistance to

heat transfer. In dropwise condensation a large portion of the area of the plate is directly

exposed to the vapor, making heat transfer rates much larger (5 to10 times) than those in

filmwise condensation.


62

4. Experimental Setup

T4

Apparatus description: The apparatus consist of:

Steam Generator: (8 liter capacity) equipped with 2 kW heater and Pressure Gauge, Manual

Release Valve, Feed Line and Steam Line.

Dropwise Condenser:

MOC: Copper with chrome plating

Dimensions: ID (di) = 16 mm

OD (do) = 19 mm

Length (L) = 170 mm

Filmwise Condenser:

MOC: Copper with Natural finish

Dimensions: ID (di) = 16 mm

OD (do) = 19 mm

Length (L) = 170 mm

Temperature Indicator with Selector Switch measures the Temperature of :

T1 Steam Chamber Temp.

T2 Cooling Water Inet to dropwise condenser

T3 Cooling Water Inet to filmwise condenser

T4 Dropwise Condenser outlet Temp

T5 Filmwise Condenser outlet Temp

T6 Dropwise condensation surface Temp

T7 Filmwise condensation surface Temp

DROPWISE – FILMWISE CONDENSATION APPARATUS

T2-3

T1

T6 T7

T5

STEAM

INLET

WATER

INLET


63

5. EXPEIRMENTAL PROCEDURE

a) Fill the steam generator with about 10 – 15 liter of water (preferably soft).

b) Connect supply socket to the mains and switch on the heater.

c) Switch on the heater while keeping the steam line and feed line valves in closed position.

d) Adjust the temperature of the steam near to the 105 – 110 OC.

e) Allow the steam generation to take place.

f) This may take 30 – 40 minutes depending on the initial temperature of the feed water.

g) The pressure of the generated steam will be indicated on the pressure gauge. Note down

the Pressure reading of the steam inlet.

h) Now select the condenser to be tested first and open the ball valve of the same for cooling

water supply.

i) Now start the supply of cooling water in the selected condenser (Dropwise or Filmwise)

j) Depending upon the type of condenser under test Dropwise or Filmwise condensation can

be visualized.

k) If water flow rate is low than steam pressure in chamber will rise and pressure gauge will

read the pressure.

l) If the water flow rate is matched than condensation will occur at more or less at

atmospheric pressure.

m) Process of Dropwise and Filmwise condensation can be easily viewed through the front

glass window of main unit.

n) Note down the inlet temperature of the cooling water, Outlet Cooling water Temperature

as indicated by the DTI.

o) Slowly open the steam line valve and allow the steam to enter the steam chamber.

p) Observe the condensation phenomena and also note down the condenser temperature,

steam inlet temperature.

q) Measure and note the cooling water flow rate through Rotameter provided.

r) Repeat the above procedure for the second type of condenser.

s) Use the wiper provided if the fog/ mist restricts the visualization of the glass vessel

t) At the time of steady state ,the outlet temperature of the water flow inside the tube

becomes constant.

PRECAUTIONS:

a) Do not start heater supply unless water is filled in the test unit.

b) Operate gently the selector switch of temperature indicator to read various temperatures.

c) Increase the temperature gradually of the heater during initial set-up experimentation.

d) Never use the heater at full wattage for longer period of time.

e) Use the proper range of Rotameter.

f) Operate the change over switch of temperature indicator gently from one position to

other, i.e. from 1 to 4 position

6. Observations:

Filmwise Condensation:

Cooling Water Flow Rate (mw): LPM

Temperature:

T1 Steam Temp:

T3 Cooling Water In Filmwise Condenser:

T7 Filmwise Condenser Surface Temp:

T5 Cooling water outlet from film wise condensation


64

Dropwise Condensation:

Cooling Water Flow Rate (mw): LPM

Temperature:

T1 Steam Temp:

T2 Cooling Water in Drop wise Condenser:

T6 Drop wise Condenser Surface Temp:

T4 Cooling water outlet from drop wise condensation surface

Calculations: Normally steam will not be pressurized, but the pressure gauge reads some pressure than

properties of steam should be taken at that pressure or otherwise atmospheric pressure will be

taken.

First calculate the heat transfer coefficient inside the condenser under test. For this properties of water are taken at bulk mean temperature of water

i.e. for dropwise condensation mean temp = (T2 + T4) / 2.

Film-wise condensation mean temp = (T3+T5)/2

Following properties are required:

Density of water kg / m3

Kinematic Viscosity m2/ sec

Thermal Conductivity ‘k’ kcal / hr m C

Prandtl Number Pr

Reynolds Number NRe = v di /

Where

v = (mw*4/ * di2) = ( Flow rate/Area)

di = Inner Diameter of Condenser

If this value of NRe > 2100 then flow is turbulent, below this value flow is laminar.

Normally flow will be turbulent in the tube.

Nusselt Number NuD = 0.023 (ReD)0.8

(Pr)0.4

Inside heat transfer coefficient ( hi ) = NuD * k/ di kcal / hr m2 C

Calculate heat transfer coefficient on outer surface of the condenser HO:

For this properties of water are taken at bulk mean temperature of condensate

(Ts + Tw) / 2.

Density of water kg / m3

Kinematic Viscosity m2/ sec

Thermal Conductivity ‘k’ kcal / hr m C

Prandtl Number Pr

Reynolds Number NRe = v di /

Where, Ts Temperature of steam,

T6-7 Temperature of condenser wall (DROPWISE AND FILMWISE)

do outside diameter of condenser

1. ho = 0.725 * 2 * g * k3 / ( Ts - Tw) * do

From these values overall heat transfer coefficient (U) can be calculated,

1/ U = 1 / hi + (di / do) (1 / ho)

U = kcal / hr m2 C

The same procedure can be repeated for another condenser.


65

Except for some exceptional cases overall heat transfer coefficient for dropwise condensation

will be higher than that of filmwise condensation. Results may vary from theory to some

degree due to unavoidable heat losses from the glass tube walls.

7. Result:

GIVEN DATA:

First calculate the heat transfer coefficient inside the condenser under test. For this properties

of water are taken at bulk mean temperature of water

i.e. (T2 + T4) / 2. Following properties are required:

PHYSICAL PROPERTIES AT WATER MEAN TEMPERATURE

Physical Properties Dropwise Condenser

Filmwise Condenser

Steam Temperature OC Ts 99.3

99.4

Mean Temperature oC Tm(T2+T4 / 2) 35

35

Density of Water kg/m3 993.95

993.95

Viscosity

Kg/ m s µ 0.00073 0.00073

Kinematics Viscosity

m2/ s

0.732*10-6

0.732*10-6

Cooling Water

Flowrate LPM Q 2 2

Vol. Flowrate

m3/s

m 0.000033 0.000033

Velocity

m/s V 0.164 0.164


kcal / hr m C K 0.537 0.537

Specific Heat

CP 0.997 0.997

Now calculate the heat transfer coefficient outside the condenser under test. For this

properties of condensate are taken at bulk mean temperature of Condenser wall and the Steam

Inlet Temperature i.e. (Tw + Ts) / 2. Following properties are required:

PHYSICAL PROPERTIES AT CONDENSATE MEAN TEMPERATURE

Physical Properties Dropwise Condenser Filmwise Condenser


66

Steam Temperature OC T1 99.3 99.4

Mean Tamp 0C

Tm 78.45 82.2

Density of Water kg/m3 971.8 971.8

Viscosity

Kg/ m s µ 0.000355 0.000355

Kinematics Viscosity

m2/ s

0.365*10-6

0.365*10-6

Cooling Water

Flowrate LPM Q 2

Vol. Flowrate

m3/s

m 0.000033

Velocity

m/s V 0.164


kcal / hr m C K 0.579

Specific Heat

CP 0.997


67


UNSTEADY STATE HEAT TRANSFER UNIT

AIM

(1) To study the unsteady state temperature response of finite geometric shapes.

(2) To calculate the value of surface conductance (h).

1. THEORY:

When the temperature at any given point in a system changes with time, the heat transfer is at

unsteady state. The phenomenon of unsteady state heat transfer is widely utilized in industrial

processes, such as the cooling of metal ball bearings. The calculation of heat transfer

coefficients makes it possible, among other things, to predict the amount of time a system

takes to reach steady state conditions.

In this experiment, we want to use a lumped capacity analysis to simplify our

calculations, in which only the effects of convection are significant. This occurs when the

internal resistance is negligible, and the Biot Number, Bi, is less than 0.1 as described in

Equation (1).1

k

hxBi (1)

where k is the thermal conductivity and x is a characteristic dimension of the body obtained

from the volume to area ratio.

Derivation of Lumped-Sum Analysis:

tpxC

h

TT

TT

0

ln (1*)

where x is the relative radius, described by:

A

Vx (2*)

V is the volume of the cylinder and A is the surface area. For a long cylinder this becomes:

2

rx (3*)

Plugging into equation (1*)

trC

h

TT

TT

p

2ln

0

(4*)


68

trC

h

TT

TT

p

2ln

0

(2)

where Cp is the heat capacity and is the density of the material. T is the temperature of a

large bath, T is the center temperature of the cylinder at time, t, and T0 is the initial center

temperature.

This lumped sum model makes several approximations, including that of an infinite

tank. As discussed in previous reports, this approximation is not valid for the tank in the

Rothfus lab using materials with high internal resistance, such as Plexiglas. However, using

materials with low internal resistance such as those in this experiment, the lumped sum model

is a close approximation. In addition to the approximation of an infinite bath, the h value is

the average value over the entire surface of the cylinder. This introduces a problem of

conduction through the ends of the cylinder, which must be accounted for by using an

appropriate length (L) to diameter (D) ratio2:

2D

L (3)

Using Plexiglas end caps helps to eliminate heat flow through the ends of the cylinders,

making the infinite cylinder approximation more accurate. Using various L/D ratios, it is

possible to test the accuracy of this assumption, which is addressed in this experiment.

In general, the average heat-transfer coefficient on immersed bodies is correlated in terms of

the Nusselt number1:

3/1

PrReNCNN m

Nu (4)

where NPr is the Prandtl number which is constant for this experiment.

The variables C and m are constants based on the Reynolds number. From all previous

reports using similar cylinders, the Reynolds number was found to be within the range of 4 x

103 to 4 x 10

4 corresponding to C=0.683 and m=0.618

3.

DvN Re (5)

where v, , and are properties of water which are constant throughout the experiment.

k

hDN Nu (6)


69

where k is the thermal conductivity of water, and D is the diameter of the cylinder. From

these equations it can be seen that h will decrease with increased diameter.

382.01

D

Ch

(7)

where C1 is a proportionality constant. This relationship will hold true for infinite cylinders,

but with increasing diameter and constant, finite length, the end effects will become more

apparent, and therefore the Biot number will increase.

While equation (6) contains a relation between the k value of the surrounding water

and the convective heat transfer coefficient, it does not indicate any relationship between the

h value and the thermal conductivity of the material itself. In fact, this equation would imply

that the material has no affect on the heat transfer coefficient, as long as the lumped sum

approximation has been justified, meaning the internal resistance is relatively low. However,

in previous experiments it has been noted that the heat transfer coefficient varies with

changing material properties.4

When the effects of internal resistance are not negligible, the lumped-sum analysis

described above is invalid, and Heisler charts must be used to calculate the convective heat

transfer coefficients. To use these charts, dimensionless numbers are calculated and applied

to a chart with data for objects of varying geometry. In this case, we will use the following

described values and apply them to a Heisler chart (see figure A 4.1 below) to determine the

heat transfer coefficient of a long cylinder.5

21

x

tX (8)

0TT

TTY

(9)

xh

km

(10)

1x

xn (11)

pC

k

(12)

X is a dimensionless number which accounts for the effects of the shape and material, where

k is the solid thermal conductivity, and x1 is the radius of the cylinder. Y is a dimensionless

temperature constant, m is the inverse of the Biot number, relating conductive effects and

convective effects, and n is a relative radius, in which x is the location of the thermocouple.

The results gathered from Heisler charts will be accurate even with significant internal

resistance and are recommended anytime the Biot number is greater than 0.1.


70

Figure A4.1: Heisler chart.

2. Experimental setup:

In this experiment, the effects of material selection and radius on the heat transfer coefficient,

h, of a solid cylinder were measured.

For this experiment the materials were ordered and were prepared by cutting the appropriate

material into its desired length. Then Sphere & Cylinder were drilled in the centre to a depth

that corresponded to half its length (See Figure A2.1 for an example of how this was done).

A Type J, iron/ constantan, thermocouple was glued into the centre of each cylinder using an

epoxy with a high thermal conductivity. This particular epoxy was chosen because it would

not have any affect on the temperature readings that were obtained from the thermocouple.

In essence, the presence of the epoxy could be ignored. (For more accurate results, the ends

of the cylinders would be sealed with Plexiglas end caps and a two-part epoxy. The epoxy

and Plexiglas acted as an insulator to prevent heat loss from the ends of the cylinders.)

The apparatus consist of a 20 litre SS 304 tank equipped with 1.5 kW heater and a drain

valve. The tank is well insulated from outside to prevent heat losses to the surroundings. The

temperature of the fluid inside the tank can be controlled by the regulator provided on the

tank and is measured using the digital temperature indicator. Two shapes of known geometry

and metals are provided with thermocouples to study the unsteady state heat transfer.

T1 Hot Oil/ Water Bath Temperature

T2 Mild Steel Sphere

T3 Copper Cylinder

Test Body Shape 1:

Geometry = Cylinder

MOC = Copper


71

Diameter = 50 mm

Length = 150 mm

Density () = 8900 kg/m3

Specific Heat (Cp) = 0.38 kJ/ kg oC

Test Body Shape 2:

Geometry = Sphere

MOC = Mild Steel

Diameter = 50 mm

Density () =

Specific Heat (Cp) =


72

3. Schematic diagram:

4. Procedure:

A hot water bath is used . This water bath provides a constant temperature environment for

the experiment to be conducted .The test body was placed in the ambient temperature and is

allowed to achive this temperature . Once this temperature is achieved ,as shown by the

reference thermocouple , the test body to be tested is then transferred to the hot water bath at

particular temperature (about 65 degree c), which was kept constant using thermostat

provided. The cylinder is allowed to equilibrate to this temperature .digital temperature

indicator and stopwatch is used to record the temperature read by thermocouple every 5

seconds.

I. Fill the heating tank with fluid upto 75% level so that test specimen can be

dipped into the fluid.

II. Switch on the heater and maintain the desired temperature of the water /oil

using the regulator provided on the tank .

III. Note down this temperature (Ta) and intial temperature of the body as

indicated by the DTI.

IV. Now dip the selected body into this hot fluid with the help of stand provided.

V. With selector switch of DTI at proper thermocouple number and start noting

down the unsteady state temperature response (T) for a regular time interval

(about 5 or 10 secs interval).

Stirrer Reference

Thermocouple

Hot Water Bath:

Constantly stirred. Kept at

65 C


73

VI. After the temperature reaches the steady state take the body out of the heating

tank and note down the unsteady state temperature response of the body while

cooling .Or dip the body into the cold water bath and note down the unsteady

state temperature response (T) for a regular time interval for cooling.

Repeat the above procedure with second body.

5. OBSERVATION TABLE:

Body Shape : Copper Cylinder

Diameter : 50 MM

Length :150 mm

Volume : 2.9437 * 10-4

MM2

Surface Area : 0.0235 M

Thermal Conductivity : 206 w/m.k

Density :

Specific Heat :

Cycle : Heating / Cooling

Surrounding Temperature (Ta) : C

Initial body Temperature (Ti) : C

Sr.

No.

Time

(sec)

Temperature

Response

(C)

(T - Ta/

Ti – Ta)

Unit Surface

Conductance

(w/ m2*

oC)

Biot

Number

Fourier

Number

(t) T = T3 h Bi Fo

GRAPHS: Plot [(T-Ta)/ (Ti-Ta)] vs. t for heating as well as cooling cycle.

OBSERVATION TABLE:

Body Shape : MS Sphere

Diameter : 50 MM

Volume :2.9437 * 10-4

MM2

Surface Area :0.0235 M

Thermal Conductivity : 206 w/m.k

Density :8900 kg/m3

Specific Heat : 0.38 kJ/ kg oC

Cycle : Heating / Cooling

Surrounding Temperature (Ta) : C

Initial body Temperature (Ti) : C

Sr.

No.

Time

(sec)

Temperature

Response

(C)

(T - Ta/

Ti – Ta)

Unit Surface

Conductance

(w/ m2*

oC)

Biot

Number

Fourier

Number

(t) T = T3 h Bi Fo


74

GRAPHS: Plot [(T-Ta)/ (Ti-Ta)] vs. t for heating as well as cooling cycle.

SAMPLE CALCULATION:

Sample calculation of h using lumped-sum model

(For Aluminium D=0.05 M, L=0.150 M)

First, obtain the equation of the line.where x=(v/As),X=

Next, use the slope of that equation in the following equation:

2

pCrslopeh

SAMPLE CALCULATION OF H FROM HEISLER CHART

First, choose a point on the graph of Y vs. t. Then substitute those values into the following

equations:

21

x

tX

0TT

TTY

1xh

km

Using the Heisler chart, find the point, which corresponds to a value for m. Rearranging the

equation for m yields the following:

1xm

kh

Sample calculation of C1 value:

For 3.0-inch diameter aluminum cylinder, from Table 2 the average experimental

h is 1335.

h = 1335 Km

inW

2

D = 3.0 in.


75

0.382D

1Ch (equation (9))

Therefore, C1 = 2031 Km

inW

2

h values for all trials

Material trials: L/D trials: Radius Trials:

D=1.5 in. L=6 in. aluminum D=2 in. aluminum L=6 in.

h(W/m^2*K) h(heisler) h(W/m^2*K) h(heisler) h(W/m^2*K) h(heisler)

steel 1 974.771 945 L=4 in 1 1801.954 1622 D=1 in 1 2086.937

steel 2 1039.984 945 L=4 in 2 1823.516 1352 D=1 in 2 2083.856

steel 3 985.068 859 L=4 in. 3 1829.677 1352 D=1 in 3 2048.432

AVERAGE 999.941 916 AVERAGE 1818.382 1442 AVERAGE 2073.075

stdev 35.05835762 49.65212315 stdev 14.55702862 155.8845727 stdev 21.39699107

copper1 2136.796 L=6 in. 1 1943.6467 1474 D=1.5 in 1 1266.024

copper2 2042.045 L=6 in. 2 1919.0046 1622 D=1.5 in 2 1520.152

copper3 1979.967 L=6 in. 3 1885.1217 1622 D=1.5 in 3 1240.691

AVERAGE 2052.936 AVERAGE 1915.924333 1573 AVERAGE 1342.289

stdev 78.97970765 stdev 29.38383805 85.44783984 stdev 154.5537942

brass1 1809.447 1365 L=7.5 in. 1 2008.3322 1622 D=2 in 1 1697.273

brass2 1753.195 1365 L=7.5 in. 2 1959.048 1707 D=2 in 2 1152.051

brass3 1906.326 1560 L=7.5 in. 3 1949.8072 1622 D=2 in 3 1497.05

AVERAGE 1822.989333 1430 AVERAGE 1972.3958 1650 AVERAGE 1448.791333

stdev 77.45851718 112.5833025 stdev 31.46294217 49.07477288 stdev 275.7959962

aluminum1 1266.024 L=9 in. 1 1774.232 1622 D=2.5 in 1 1478.568 1298

aluminum2 1520.152 L=9 in. 2 1811.1953 1622 D=2.5 in 2 1259.093 1442

aluminum3 1240.691 L=9 in. 3 1801.9545 1622 D=2.5 in 3 1517.072 1442

AVERAGE 1342.289 AVERAGE 1795.793933 1622 AVERAGE 1418.244333 1394

stdev 154.5537942 stdev 19.23631521 0 stdev 139.1671645 83.13843876

D=3 in 1 1270.644 1298

D=3 in 2 1270.644 1298

D=3 in 3 1464.706 1442

AVERAGE 1335.331333 1346

stdev 112.0417479 83.13843876

6. Results

7. Discussion

8. Conclusion

9. PRECAUTIONS

1. Wait till steady state is achieved.

2. Use stabilized power supply.


76


HEAT TRANSFER IN AGITATED VESSEL (JACKET & COIL)

1. Aim & Objective To study heat transfer phenomena in agitated vessel with jacket & helical coil system and

comparison experimental overall heat transfer coefficients in both jacket & coil.

2. Apparatus

2 nos. pump driven setup consisting of agitated vessel with helical coil, 2 sump tanks, PID

temperature controller, temperature indicators, rotameters.

3. Theory

Whenever the fluid motion is varied by external means the heat is transferred by forced

convection. Most of the time fluid is agitated by circulating the hot and cold fluids at rapid

rates on the opposite sides of pipes or tubes. The rate of heat transfer by forced convection to

an incompressible fluid traveling in turbulent flow in a pipe of uniform diameter at constant

mass rate has been found to be influenced by the velocity, density, specific heat, thermal

conductivity, viscosity of fluid as well as the inside diameter of the pipe.

The velocity, viscosity, density and diameter affect the thickness of fluid film at the pipe wall

through which the heat must first be conducted and they also influence the extent of fluid

mixing. The thermal conductivity of the fluid and the specific heat reflects the variation of the

average fluid temperature as a result of unit heat absorption. A simple jacketed pan or kettle

is very commonly used in the chemical industries as a reaction vessel. In many cases, such as

in nitration or sulphonation reactions, heat has to be removed or added to the mixture in order

either to control the rate of reaction or to bring it to completion.

The addition or removal of heat is conveniently arranged by passing steam or water through a

jacket fitted to the outside of the vessel or through a helical coil fitted to the inside. In either

case some form of agitator is used to obtain even distribution in the vessel. This may be of

the anchor type for very thick mixes or a propeller or turbine if the contents are too viscous.

4. Experimental Setup:

Agitated vessel consist impeller mounted on the shaft, which is driven by a motor. The

formation of swirling water in an agitated vessel can be prevented by introduction of baffles.

The schematic is shown in below figure


77

Fig 1: Schematic diagram of Heat Transfer in Agitated Vessel experiment

5. Experimental Procedure:

Open the drain cock of the vessel and remove all the water from inside.

Fill the agitated vessel with about 4-5 liter of water (about 70-75% height of the vessel)

for a batch operation.

Switch on the heater and heat the water to the desired degree of hotness (about 70-80 C).

Intermittently switch ON the pump with bypass line valve fully open and supply valve

fully closed to ensure thorough mixing of water in the tank to ensure uniform

temperature.

Measure the temperature of bulk water periodically with the help of the thermocouple

placed in the thermowell.

Now start the hot water supply in the coil or jacket depending upon the experiment. And

measure its inlet and outlet temperatures periodically with the thermocouples placed in

the thermowells.

Note down the values of cooling water inlet and outlet temperatures and cooling water

flow rate, hot water inlet and outlet temperatures and hot water flow rate.


Thermal conductivity of SS (K) = 16.86 W/ m.K

Diameter of agitated vessel (Da) = 0.73 m

Height of agitated vessel (h) = 0.265 m

Diameter of coil (Dc) = 0.11m

Length of coil (Lc) = 5.115 m

6.1 Properties of water at 331 K (58 0C)

Density of water 984.1 Kg/cm3


78

Specific heat of water 4187 J/Kg.K

Viscosity of water 485 * 10-6

Pa.sec

6.2 Observation table

For hot water in coil:

S.No Hot water

flow rate

(litres/min)

Cold water

flow rate

(litres/min)

T1

(° C)

T2

(° C)

T3

(° C)

T4

(° C)

For hot water in jacket:

S.No Hot water

flow rate

(litres/min)

Cold water

flow rate

(litres/min)

T1

(° C)

T2

(° C)

T5

(° C)

T6

(° C)

6.3 Calculations:

6.3.1 Experimental Value calculations:

For hot water in coil:

Heat Gained by the Cooling Water (Q) = mw Cp (T4– T3)/60

= .................J/ sec

= Uexpt Acoil [ T2 - ( T3 + T4 ) / 2 ]

For hot water in jacket:

Heat Gained by the Cooling Water (Q) = mw Cp (T6 – T5)/60

= .................. J/ sec

= Uexpt Ajacket [ T2 - ( T5 + T6) / 2 ]

Calculate the value of Uexpt from the above equations for both jacket and coil.

Where,

QC = Cold water flow rate, LPM.

QH = Hot water flow rate, LPM.

T1 = Bulk water inlet temperature of vessel.

T2 = Bulk water outlet temperature of vessel.

T3 = Coil inlet temperature, oC.

T4 = Coil outlet temperature, oC.

T5 = Jacket inlet temperature, oC.

T6 = Jacket outlet temperature, oC.

Ajacket = 0.11 m2

7. Results & Conclusions

Experimental heat transfer coefficients for hot water flow through coil are _______ W/m2 °C


79

Experimental heat transfer coefficients for hot water flow through jacket are _______ W/m2

°C

8. PRECAUTIONS:

Make sure during the test period Hot Water Tank should not be emptied totally and the heater

must not be exposed to air if the Heater is ON, otherwise it will be damaged.


80


FLUIDIZED BED HEAT TRANSFER UNIT

1. AIM

To study the performance of the heat transfer in Fluidized bed

To study the heat transfer coefficient for heat transfer in fluidized bed.

2. Apparatus:

2 nos. pump driven setup consisting of fluidized bed, 2 sump tanks, PID temperature

controller, temperature indicators, rotameters.

3. THEORY:

When a liquid or a gas is passed at very low velocity up through a bed of solid particles, the

particles do not move ,and the pressure drop is given by the Ergun equation. If the fluid

velocity is steadily increased ,the pressure drop and the drag on individual particles

increase,and eventually the particles start to move and become suspended in the fluid. The

terms “fluidization” and “fluidized bed” are used to describe the condition of fully suspended

particles,since the suspension behaves like a dense fluid.

Fluidized beds are used extensively in the chemical process industries, particularly for the

cracking of high-molecular-weight petroleum fractions.Such beds inherently possess

excellent heat transfer and mixing characteristics.These are devices in which a large surface

area of contact between a liquid and a gas ,or a solid and a gas or liquid is obtained for

achieving rapid mass and heat transfer and for chemical reactions.

The fluidized bed is one of the best known contacting methods used in the processing

industry, for instance in oil refinery plants. Among its chief advantages are that the particles

are well mixed leading to low temperature gradients, they are suitable for both small and

large scale operations and they allow continuous processing. There are many well established

operations that utilize this technology, including cracking and reforming of hydrocarbons,

coal carbonization and gasification, ore roasting, Fisher-Tropsch synthesis, coking, aluminum

production, melamine production, and coating preparations. Nowadays, you will find

fluidized beds used in catalyst regeneration, solid-gas reactors, combustion of coal, roasting

of ores, drying, and gas adsorption operations The application of fluidization is also well

recognized in nuclear engineering as a unit operation for example, in uranium extraction,

nuclear fuel fabrication, reprocessing of fuel and waste disposal.

4. EXPERIMENTAL PROCEDURE:

Fill the water in cold water supply tank with about 100 litre of clean tap water. After

filling tank upto 50% start water pump and fill the overhead heater tank by

controlling bypass and supply valve.

As the heater tank fills with water to a level of heater, Switch on the immersion type

heater provided in the hot water tank and heat the water to the desired temperature

(about 60-70 C). Intermittently switch ON the pump with bypass line valve fully

open and supply valve fully closed to ensure thorough mixing of water in the tank to

ensure uniform temperature.

After achieving the desired temperature of water in the hot water tank, allow the hot

water to flow through inner pipe side and adjust the flow rate to the desired value

using the valve of rotameter. Drain the exit of the hot water to the drainage.

Start the cold water supply on the outer pipe side and adjust the flow rate to the

desired value-using valve of rotameter. Now place the outlet of outer tube side in to

the drain line.


81

Observe the inlet and outlet temperature of both cold and hot water streams and note

down them after they achieve steady state.

Also note down the Flow rates of hot water and cold water with the help of

Rotameters.

Repeat the above procedure either by changing the Flow rates or by changing the

inlet temperature of the hot water.

5. OBSERVATION TABLE AND CALCULATION :

Density of water = 1 kg/l

Diameter of bed do = 2 inch

Length of bed L = 12 inch

Specific heat of water = CpH = CpC = 4187 J/Kg.K

Hot Water Inner Tube Side Cold Water Outer Tube Side

Flow Rate

mh

LPM

Inlet

Temp.

T1

(C)

Outlet

Temp.

T2

(C)

Flow Rate

mc

LPM

Inlet

Temp.

T3

(C)

Outlet

Temp.

T4

(C)

Flow rate of hot water in kg/ s mH = mh * / 60

=____________ kg/ s

Heat Transferred by the Hot Water to the Cold Water

QH = mH * CpH * (T1 – T2)

=____________ W

Flow rate of cold water in kg/ s mC = mc * / 60

= ___________ kg/ s

Heat Gained by the Cold Water from the Hot Water

QC = mC * CpC * (T4 – T3)

= ____________ W

Average Heat Q = (QH + QC) / 2

= ____________ W

Area for the Heat Transfer, A = * do * L

= ____________ m2

For Parallel flow

( T1 – T3 ) - ( T2 – T4 )

LMTD = --------------------------------

ln ( ( T1 – T3 ) / ( T2 – T4 ) )

= _____________ 0C

Overall Heat transfer coefficient can be calculated using,


82

U = Q / (A * LMTD)

= ____________ W/ m2 C

6. RESULT AND DISCUSSIONS:

FOR PARALLEL FLOW

OVERALL HEAT TRANSFER COEFFICIENT = ---------------- W/ m2.C

7. PRECAUTIONS:

Make sure during the test period Hot Water Tank should not be emptied totally and the

heater must not be exposed to air if the Heater is ON, otherwise it will be damaged.


83


PARALLEL FLOW AND COUNTER FLOW HEAT EXCHANGER

1. Aim

a) To calculate the overall Heat transfer coefficient of the heat exchanger.

b) To calculate the local heat transfer coefficient of the heat exchanger using Seider

Tate Equation and Dittus-Boelter Equation.

2. Objective

To study the heat transfer phenomena in parallel and counter flow arrangements.

3. Theory

The double-pipe heat exchanger is one of the simplest types of heat exchangers. It is called a

double-pipe exchanger because one fluid flows inside a pipe and the other fluid flows

between that pipe and another pipe that surrounds the first. This is a concentric tube

construction. Flow in a double-pipe heat exchanger can be co-current or counter-current.

There are two flow configurations: co-current is when the flow of the two streams is in the

same direction, counter current is when the flow of the streams is in opposite directions.

As conditions in the pipes change: inlet temperatures, flow rates, fluid properties, fluid

composition, etc., the amount of heat transferred also changes. This transient behaviour

leads to change in process temperatures, which will lead to a point where the temperature

distribution becomes steady. When heat is beginning to be transferred, this changes the

temperature of the fluids. Until these temperatures reach a steady state their behaviour is

dependent on time.

In this double-pipe heat exchanger a hot process fluid flowing through the inner pipe transfers

its heat to cooling water flowing in the outer pipe. The system is in steady state until

conditions change, such as flow rate or inlet temperature. These changes in conditions cause

the temperature distribution to change with time until a new steady state is reached. The new

steady state will be observed once the inlet and outlet temperatures for the process and

coolant fluid become stable. In reality, the temperatures will never be completely stable, but

with large enough changes in inlet temperatures or flow rates a relative steady state can be

experimentally observed.

As with any process the analysis of a heat exchanger begins with an energy and material

balance. Before doing a complete energy balance a few assumptions can be made. The first

assumption is that the energy lost to the surroundings from the cooling water or from the U-

bends in the inner pipe to the surroundings is negligible. We also assume negligible potential

or kinetic energy changes and constant physical properties such as specific heats and density.

These assumptions also simplify the basic heat-exchanger equations.


84


85

4. Calculation

Assumption

Steady state conditions exist.

Fluid properties remain constant and are evaluated at a temperature of fluid.

Nomenclature

T refers to the temperature of the warmer fluid.

t refers to the temperature of the cooler fluid.

w subscript refers to the warmer fluid.

h subscript refers to hydraulic diameter

c subscript refers to the cooler fluid.

a subscript refers to the annular flow area or dimension.

p subscript refers to the tubular flow area or dimension.

1 subscript refers to an inlet condition.

2 subscript refers to an outlet condition.

e subscript refers to equivalent diameter.

Fluid Properties

Counter current:

S.no

Flow rate

mh

(Kg/m)

Flow rate

mc

(Kg/m)

Hot water in

(ºC)

Hot water out

(ºC)

Cold water in

(ºC)

Cold water out

(ºC)


86

Co-current:

Tubing Sizes

IDa =

IDp = ODp =

Flow Area

Area of pipe A= π*D*L =

Area of annulus A = π*De*L =

Fluid Velocity

5. Results

The Over All heat transfer coefficient =

The local heat transfer coefficient =

E


SHELL AND TUBE HEAT EXCHANGER

1. Aim

Study of Shell and Tube Heat Exchanger

2. Objective

To calculate overall heat transfer coefficient for shell & tube heat exchanger.

S.no

Flow rate

mh

(Kg/m)

Flow rate

mc

(Kg/m)

Hot water in

(ºC)

Hot water out

(ºC)

Cold water in

(ºC)

Cold water out

(ºC)


87

3. Theory Heat Exchanger is device in which heat is transferred from one fluid to another. The

necessity for doing this arises in a multitude of industrial applications. Common examples of

heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator

and the steam boiler of a thermal power plant.

Heat Exchangers are classified in three categories:

1) Transfer Type.

2) Storage Type.

3) Direct Contact Type.

A transfer type of heat exchanger is one on which both fluids pass simultaneously through the

device and heat is transferred through separating walls. In practice, most of the heat

exchangers used are transfer type ones.

Shell and tube heat exchangers are the most widely used in chemical process industries. A

shell and tube exchanger consists of a bundle of tubes enclosed in a cylindrical shell. The

ends of the tubes are fitted into tube sheets, which separate the shell-side and tube-side fluids.

Baffles are provided in the shell to direct the fluid flow and support the tubes. The tubes are

arranged in triangular or square pitch. One fluid flows inside the tube and is called the tube

side fluid, the other fluid flows outside the tubes and is called the shell side fluid.

The shell and tube heat exchangers are further classified according to flow arrangement as -

1. Single Pass

2. Multiple Pass

The simplest shell and tube heat exchanger is 1-1 heat exchanger (one shell pass and one tube

pass), the other types are 1-2 heat exchangers and 2-4 heat exchanger.

The transfer of heat from the hot fluid to the wall or tube surface is accompanied by

convection through the tube wall or plate by conduction, and then by convection to the cold

fluid. The flow arrangement in a shell and tube heat exchanger could either be co-current or

countercurrent. The heat transfer coefficients outside the tube bundles are referred as shell

side coefficients. Baffles also increase the convection coefficient of the shell side fluid by

inducing turbulence and a cross flow velocity component.

Any heat exchanger design requires rigorous analysis. One of the most essential parts of the

heat exchanger analysis is determination of the overall heat transfer coefficient. The overall

heat transfer coefficient is defined in terms of the total thermal resistance to heat transfer

between two fluids.

The heat lost by the hot fluid can be calculated

hq = Heat Transfer rate to the hot water.

KCal/hr TTCmq hohiPhhh

Heat taken by the cold fluid can also be calculated

cq = Heat Transfer rate to the cold water.

KCal/hr TTCmq cicoPccc

2

qqQ hc

avg

mo

avg

oTA

QU

4. Experimental setup

4.1. DESCRIPTION


88

The apparatus consists of 1-2 Pass Shell and Tube heat exchanger. The hot fluid is hot

water, which is attained from an insulating water bath using a magnetic drive pump and it

flow through the inner tube while the cold water flowing through the annuals. For flow

measurement rotameters are provided at inlet of cold water and outlet of hot water line.

The hot water bath is of recycled type with Digital Temperature Controller 0 to 100 oC.

4.2. REQUIREMENTS

Water supply 20 lit/min (approx.), Drain, Electricity Supply: 1 Phase, 220 V AC, and 4 kW,

Floor area of 1.5 m x 0.75 m

4.3. SPECIFICATIONS

1. Shell

Material = S.S.

Dia. = 220 mm

Length = 500 mm

25% cut baffles at 100 mm distance 4 Nos.

2. Tube

Material = S.S

OD = 16 mm

ID = 13 mm

Length of tubes = 500 mm

Nos. of tubes = 24

3. Temperature Controller = Digital 0 – 199.9oC

4. Temperature Sensors = RTD PT-100 type (5 nos.)

5. Temperature Indicator = Digital 0 to 200oC with multi-channel switch.

6. Electric Heater = 230 V AC 2 kW (2 Nos.)

7. Flow measurement = Rotameter (2 No.)

8. Water Bath = Material: SS insulated with ceramic wool and powder

coated MS outer Shell fitted with heating elements.

9. Pump = FHP magnetic drive pump (max. operating temp

85oC).

5. Experimental procedure:

Starting Procedure:

1. Clean the apparatus and make Water Bath free from Dust.

2. Close all the drain valves provided.

3. Fill Water Bath ¾ with Clean Water and ensure that no foreign particles are there.

4. Connect Cold water supply to the inlet of Cold water Rotameter Line.

5. Connect Outlet of Cold water from Shell to Drain.

6. Ensure that all On/Off Switches given on the Panel are at OFF position.

7. Now switch on the Main Power Supply (220 V AC, 50 Hz).

8. Switch on Heater by operating Rotary Switch given on the Panel.

9. Set Temperature of the Water Bath with the help of Digital Temperature Controller.

10. Open Flow Control Valve and By-Pass Valve for Hot Water Supply.

11. Switch on Magnetic Pump for Hot Water supply.

12. Adjust Hot water flow rate with the help of Flow Control Valve and Rotameter.

13. Record the temperatures of Hot and Cold water Inlet & Outlet when steady state is

achieved.

Closing Procedure:

1. When experiment is over, Switch off heater first.

2. Switch of Magnetic Pump for Hot Water supply.


89

3. Switch off Power Supply to Panel.

4. Stop cold water supply with the help of Flow Control Valve.

5. Stop Hot water supply with the help of Flow Control Valve.

6. Drain Cold and Hot water from the shell with the help of given drain valves.

7. Drain Water Bath with the help of Drain valve.

6. Observation & calculation:

DATA:

Inside heat transfer area, Ai = 3.187 x 10-3 m2

Outside heat transfer area, Ao = 4.827 x 10-3 m2

OBSERVATION TABLE:

S.

No.

Hot water side Cold water side

Flow rate mh Kg/hr Thi oC Tho

oC Flow rate mc kg/hr Tci

oC Tco

oC

1

2

3

4

CALCULATIONS:

1. Rate of heat transfer from hot water,

hohiphhh TTCMQ , Watt

2. Rate of heat transfer to cold water,

cicopccc TTCMQ , Watt

3. Average heat transfer,

2

QQQ ch , Watt

4. LMTD,

1

2

12m

T

Tln

TTT

Where: T1 = Thi - TCi (for parallel flow)

= Thi - TCo (for counter flow)

and T2 = Tho - TCo (for parallel flow)

= Tho- TCi (for counter flow)

Note that in a special case of Counter Flow Exchanger exists when the heat capacity rates

Cc & Ch are equal, then Th i - Tc o = T h o - T c i thereby making Ti = To . In this case

LMTD is of the form 0/0 and so undefined. But it is obvious that since T is constant

throughout the exchanger, hence

Tm = Ti = To

(acc. to ref. Fundamental of Engineering Heat & Mass Transfer by R.C. Sachdeva, Pg. 499)

5. Overall heat transfer coefficient,

mi

iTFA

QU

, W/m

2-OC


90

mO

OTFA

QU

, W/m

2-OC

Heat Transfer rate, is calculated as

Qh = -------- W

Qc = -------- W

Q = 2

ch QQ W

(Assume Cph = Cpc = 4.176 J /kg- ºC)

LMTD - logarithmic mean temperature difference that can be calculated as per the following

formula:

LMTD = Tm = oi

oi

TTln

TT

Where Ti = T h i - T c i (for parallel flow)

= T h i - Tc o (for counter flow)

and To = T ho -T c o (for parallel flow)

= T ho - T c i (for counter flow)

Note that in a special case of Counter Flow Exchanger exists when the heat capacity rates Cc

& Ch are equal, then Th i - Tc o = T h o - T c i thereby making Ti = To. In this case, LMTD is

of the form 0/0 and so undefined. But it is obvious that since T is constant throughout the

exchanger, hence

Tm = Ti = To

Overall heat transfer coefficient can be calculated by using.

Q = U ATm

Ui = Q/(Ai Tm) W/m² ºC

Uo = Q/(Ao Tm) W/m² ºC

NOMENCLATURE:

Mc = Cold water flow rate, Kg/s

Mh = Hot water flow rate, Kg/s

Tci = Cold water inlet temp.

Thi = Hot water inlet temp.

Tco = Cold water outlet temp.

Tho = Hot water outlet temp.

Tc = Mean temp. of cold water

Th = Mean temp. of hot water ,

c = Density of Cold fluid, kg/m3

Cpc = Specific heat of cold fluid, J/kg-C

pc = Thermal Conductivity of cold fluid, W/mC

h = Density of Hot fluid, kg/m3

Cph = Specific heat of Hot fluid, J/kg oC

Ph = Thermal Conductivity of Hot fluid, W/mC

Qh = Heat lost by hot water, W

Qc = Heat gained by cold water, W

Q = Average heat transfer, W

LMTD = Logarithm mean temp. difference

A = Area of Heat Transfer, m2

Do = Outer dia of S.S tube, m

L = length of the tube, m


91

U = Overall heat transfer coefficient, W / m2 o

C

7. Results & discussion

8. Conclusion

9. Precautions & maintenance instructions

References:

1. Holman, J.P., “Heat Transfer”, 8th

ed.,McGraw Hill, NY, 1976.

2. Kern, D.Q., “Process Heat Transfer”, 1st ed.,McGraw Hill, NY, 1965.

3. McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”,

4th

ed.McGraw Hill, NY, 1985.

4. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol.

- 1”, 5th

ed.Asian Books ltd., ND, 1996.


92


PLATE TYPE HEAT EXCHANGE

1. Aim:

To analyze the performance of an existing plate type Heat Exchanger.

2. Objectives:

1. To calculate Overall Heat Transfer Coefficient & effectiveness for plate type Heat

Exchanger.

2. To analyze effects of changing the flow rate for hot water & cold water fluids.

3. Theory:

A heat exchanger is a device that facilitates the transfer of heat between two fluids at

different temperatures. The main mechanisms for heat transfer in heat exchangers are

conduction and convection. There are many types of heat exchangers. An example of one is a

plate and frame heat exchanger. A plate and frame heat exchanger is a device that utilizes

corrugated pressed plates for heat by counter-current flow.

Other types of heat exchangers include the miniature, brazed heat exchanger, traditional

gasket plate heat exchangers, welded heat exchangers, plate and shell heat exchanger, and the

spiral heat exchanger.

Plate and frame heat exchangers, are used in many different processes at a broad range of

temperatures and with a wide variety of substances. Plate and frame heat exchangers have

been used in industry since 1930. During recent years, research into Plate and frame heat

exchangers has increased considerably and there is now a state-of-the-art compilation of

knowledge on this topic.

During liquid-liquid heat exchange, it is important to maintain relatively low temperatures

and pressures. The heat exchange process should not exceed temperatures and pressures of

250 degrees Celsius and 25 atmospheres respectively. Plate and frame heat exchangers are

not exclusive for liquid-liquid heat exchange, but also have some usage in liquid-gas and

liquid-condensing-vapor combination heat exchange.

Plate and frame heat exchangers are devices that are made up of many plates stacked together

on a frame, and either bolted or welded together. Each plate, consists of tiny little

passageways, through which, the fluids are passed. The cold and hot fluids travel through two

different passage ways in a counter current flow, which maximizes the heat transfer by

creating a temperature gradient that is constantly from the hot fluid to the cold fluid, as

illustrated below.


93

FIG 1: Single-pass countercurrent flow.

In Fig 1, it can be seen how two fluids move within the plates. The black arrows, for

example, represent the hot fluid and the gray arrows represent the cold fluid. As can be seen

in the above diagram, the fluids move in opposing directions as they cross the plates.

FIG 2: Parts of a typical plate and frame heat exchanger

FIG 2 above shows the different parts of a Plate and frame heat exchangers. The fixed head is

where the two fluids enter. One fluid enters through the bottom, and the other fluid enters

through the top. The movable follower is the back of Plate and frame heat exchangers. It can

be adjusted back and forth for the addition or removal of plates in the plate pack. The guiding


94

and carrying bar provide support for the plate pack and the follower runs along these bars.

The clamping bolts secure the stacked structure.

FIG 3(a) Intermitting FIG 3(b) Chevron

Fig 3 above shows the two main types of plates. The first employs the intermitting

corrugations. These corrugations are at right angles with the fluid flows. The fluid goes down

the plate and hits the corrugation line, runs to the other end of the plate, and goes down

through the slit at the end of the line, after which, the fluid hits another corrugation plate and

continues this pattern throughout the plate. The maximum gap size between the corrugations

is 3 to 5 mm maximum. With this style of plate, turbulence in the flow is increased by

constantly hitting the plate. The size of the corrugation gaps is directly proportional to the

amount of turbulent increase in the fluid.

The chevron corrugations are slanted downward at an angle of beta. The fluid flows steadily

down the plate crisscrossing back forth. The plate in FIG 3(b) has the troughs going in the

opposite direction for the other fluids. This maximizes the crossing points of the two fluids,

thus, increasing the heat transfer. If the angle is about 80 degrees, the heat transfer increases

due to the swirling flow of the two fluids in opposite directions in opposing plates.

FIG 4: Flow patern of fluids in plate type heat exchanger

3.1. Advantages:


95

Flexibility in the efficiency: The number of plates can be adjusted for more or less

heat transfer

Compactness: The numerous amounts of little corrugations increase the surface area

for heat transfer in a small volume.

Low Cost: The plates are inexpensive and easy to produce

Less Corrosion: the plates are typically made of stainless steel

Low Maintenance: very easy to clean

Temperature Control: can maintain a relatively low temperature difference in fluid

because of the small diameters inside the corrugations.

High thermal efficiency

Reduced Liquid Volume: due to narrow flow channels

Due to their efficiency, they can be selected to use less coolant

3.2. Disadvantages:

Temperature and Pressure Limitations: Temperature cannot go above 250 degrees

Celsius and pressure cannot go exceed 25 atmospheres.

Leakage: Corrosive materials can cause leakage

Pressure Increase: The small diameter of the corrugations can cause a heavy

increase in the pressure, so pumping costs must to taken into account.

3.3. Applications: Some Plate Heat Exchangers are designed specifically for use in the food, dairy and brewing

industries. A major feature of Plate and frame heat exchangers used for this process is a

sanitary plate designed to achieve optimum distribution of the product over the entire plate

surface. These liquid food plants are designed for the concentration of malt, beer, yeast, fruit

juices, pulp and other liquid food products to remove water and stabilize enzymes, prior to

product storage.

The Plate and Frame Heat Exchanger, can also be used in the cooling and heating of fibrous

materials, such as, fruit juices and fluids containing pulp, fruit purees, turbid fruit juices,

dairy mixes, citrus pulp, and highly viscous liquids.

The unique capabilities of the PHE makes it suitable for a wide range of applications that

extend beyond refrigeration such as: refrigerant evaporating & condensing, heat pumps,

steam heating, engine or hydraulic oil cooling, swimming pool heating, and heat recovery for

industrial applications e.g. waste water, dye works, paper manufacture etc.

When choosing a heat exchanger, the design engineer has to take into account many factors.

Some of these factors include, fluid characteristics, operating pressure, operating temperature

and the range of possible flow rates.


The miniature exchanger supplied for the Lab project consists of a pack of 11 plates, 9 of

which have water on both sides that contributes to heat transfer. The plates have sealing

gaskets and are held together in a frame between a fixed end plate and moving end plate. Two

nuts/bolts passing through the end plates compress the plates and gaskets together. Hot and

cold fluids flow between channels on alternate sides of the plates to promote heat transfer.

The plate heat exchanger with 11 plates is configured for 5 passes in series. Although the

overall flow arrangement may be either countercurrent or Co-current, the flow arrangements

on either side of each individual plate alternates between countercurrent and Co-current

patterns.


96

Each end plate of the heat exchanger incorporates tapings for the hot and cold fluids to

enter/leave the exchanger, and thermocouples in each of the tapings allow the temperatures of

the fluids to be measured. The four type K thermocouple temperature sensors are labeled T1

to T4 for identification and each lead is terminated with a miniature thermocouple plug for

connection to the appropriate socket on the service unit. Flexible tubing attached to each fluid

inlet/outlet is terminated with a ferrule to allow rapid connection to the appropriate quick

release fittings on the HT30X service unit.

In normal countercurrent operation the flexible connections are hot water inlet adjacent to

temperature sensor T1, hot water outlet adjacent to temperature sensor T2, cold-water inlet

adjacent to temperature sensor T3, and cold-water outlet adjacent to temperature sensor T4.

The pattern of holes in the plates and the shape of the gaskets determine the direction of low

through the exchanger. The pates are made of SS-316 and incorporate a locating groove for

the gasket. Each plate has a pressed chevron pattern to promote turbulence and provide

multiple support points. As a result of the turbulence promoters, which cause flow separation

around protuberances, turbulent-like behavior can begin at Reynolds numbers as low as

several hundred, thereby enhancing heat transfer performance.

Silicone rubber gasket on each plate ensures that the adjacent flow channels are sealed from

each other.

5. Experimental Procedure:

1. Fill the thermic fluid tank with about 75 liter of thermic fluid (say water - here).

2. Switch on the immersion type heater (6 kW) provided in the thermic fluid tank and heat

the thermic fluid to the desired temperature (about 50-60 C). Intermittently switch ON

the pump with bypass line valve fully open and supply valve fully closed to ensure

through mixing of thermic fluid in the tank to ensure uniform temperature.

3. After achieving the desired temperature of thermic fluid in the thermic fluid tank, switch

ON the pump (0.5 HP) and allow the hot thermic fluid to flow through plates and adjust

the flow rate to the desired value using the valve for about five minutes. Recycle the exit

of the hot thermic fluid to the thermic fluid tank.

4. Start the cold water supply into plates and adjust the flow rate to the desired value.

5. Monitor the hot thermic fluid inlet temperature and maintain it at the constant value by

switching the heater either on/ off with the help of thermostat provided on the control

panel of the tank.

6. Observe the inlet and outlet temperature of both cold water and hot thermic fluid streams

and note down them after they achieve steady state.

7. Also note down the flow rates of hot thermic fluid and cold water with the help of

Rotameters.

8. Repeat the above procedure either by changing the flow rates or by changing the inlet

temperature of the hot thermic fluid.

6. Observations:

S. No.

Hot Water

Cold Water

Flow rate

mh

(LPM)

Inlet

Temp

T1

(C)

Outlet

Temp

T2

(C)

Flow rate

mc

(LPM)

Inlet

Temp.

T3

(C)

Outlet

Temp.

T4

(C)


97

7. Calculations: The actual area of heat transfer is computed using the following equation:

A = N*a = N * L * W

Where,

N= number of plates

a= projected area of a single plate

L= the length component of the projected area of the plate

W= the width component of the projected area of the plate

Flow rate of hot water in kg/ s

mH = mh * / 60

= __________ kg/ s


QH = mH * CpH * (T1 – T2)

= ___________ W

Flow rate of cold water in kg/ s

mC = mc * / 60

= __________ kg/ s


QC = mC * CpC * (T4 – T3)

= ___________ W

True Temperature Difference

442

31

4231

42

31ln

)42(31

TT

TT

TTTTT

= ______________0C

Now, average heat transfer

Q = (QH + QC )/ 2

= ______________ W

Designed Overall Heat Transfer Co-efficient

Uc = TmA

Q

*

= __________ W/m2 0C


98

E= EFFECTIVENESS OF THE EXCHANGER.

=MCp)cold/(MCp)min * ∆ Tcold / ∆ Thot

= ∆ Tcold / ∆ Thot

where,

∆ Tcold= the change in temperature of the cold fluid

∆ Thot= the change in temperature of the hot fluid

7.1. Model Calculations:

S. No.

Hot Water

Cold Water

Flow rate

mh

(LPM)

Inlet

Temp

T1

(C)

Outlet

Temp

T2

(C)

Flow rate

mc

(LPM)

Inlet

Temp.

T3

(C)

Outlet

Temp.

T4

(C)

1 2 41 32 3 32 36

2 4 42 36 6 34 37

3 6 47 40 6 35 42

4 8 48 44 4 38 46

5 4 48 43 8 42 44

The actual area of heat transfer is computed using the following equation:

A = N*a = N * L * W

Where,

N= number of plates =8

L= the length component of the projected area of the plate = 550 mm

W= the width component of the projected area of the plate = 125 mm

A = N*a = N * L * W

= 8 * 0.550 * 0.125

= 0.55 m2

Sample calculations for reading no -2:

Flow rate of hot water in kg/ s

mH = mh * / 60

= 4 * 1000/(1000*60)

= 0.066 kg/sec


QH = mH * CpH * (T1 – T2)

= 0.066 * 4.187 * (42 – 36)

= 1.6580 W(J/SEC)

Flow rate of cold water in kg/ s


99

mC = mc * / 60

= 6 * 1000/(1000*60)

= 0.1 kg/sec


QC = mC * CpC * (T4 – T3)

= 0.1 * 4.187 * (37 – 34)

= 1.2561 W(J/SEC)

True Temperature Difference

32

41ln

)32(41

TT

TT

TTTTTlm

=3/(ln 2.5)

= 3.27 0C

Now, average heat transfer

Q = (QH + QC )/ 2

= 2.9141 W

Designed Overall Heat Transfer Co-efficient

TmA

Q

*

= 2.9141/(0.55 * 3.27)

= 1.6202 W/m2 0C

Effectiveness of heat exchanger:

=MCp)cold/(MCp)min * ∆ Tcold / ∆ Thot

= ∆ Tcold / ∆ Thot

= (37–34 )/ (42 – 36)

= 3/6 * 100

=50 %

where,

∆ Tcold= the change in temperature of the cold fluid

∆ Thot= the change in temperature of the hot fluid

8. Results & Discussion:

Hence, the performance of the plate type Heat Exchanger has been analysed and the Overall

Heat Transfer Coefficient & effectiveness for the plate type Heat Exchanger is calculated.

1. Overall Heat Transfer Coefficient of the plate type Heat Exchanger =

2. Effectiveness of the plate type Heat Exchanger =


100

9. Conclusions:

10. Precautions:

1. Make sure during the test period Hot Thermic Tank should not be emptied totally and the

heater must not be exposed to air if the Heater is ON, otherwise it will be damaged.

2. Take all observations within 5 to 10 minitues before cold water temperaturrises

significantly.

11. References:

1. http://en.wikipedia.org/wiki/Plate_heat_exchanger

2. Lawry, F.J., “Plate type heat exchangers”, chem. Eng., June 29, 1959, 89-94.

3. www.tv-me.com/download.php?a=apv.pdf


FINNED TUBE HEAT EXCHANGER

1. Aim:

To analyze the performance of an existing finned tube Heat Exchanger.

2. Objective:

To calculate overall heat transfer coefficient, logarithmic mean temperature difference and

effectiveness of finned tube heat exchanger.

3.Theory The heat which is conducted through a body must frequently be removed by some convection

process. For example, the heat lost by conduction through a furnace wall must be dissipated

to the surroundings through convection. In heat exchanger application a finned tube

arrangement might be used to remove heat from a hot fluid.

Consider the one-dimensional fin exposed to a surrounding fluid at a temperature T. The

temperature of the base of the fin To. Equivalent diameter of annular, de = 4 x Net free flow

area / wetted or heated perimeter

Total Wetted Perimeter, Pw = (Ds + do) + 2YNF

Heat Transfer Perimeter, PHT = do + 2YNF

Net free area in the annular of a longitudinally finned exchanger is

ANF = [(/4) (Ds2 + do

2)] - (XYNF)

http://en.wikipedia.org/wiki/Plate_heat_exchanger

http://www.tv-me.com/download.php?a=apv.pdf


101

Fig1: Crossection of pipe and finned tube

Equivalent Diameter Based on wetted perimeter (hydraulic diameter)

(de)w = 4ANF / Pw

Based on heat transfer perimeter

(de)HT = 4 ANF / PHT

Use (de)w for the calculation of Re-No, Gz-No, D/L for Annular fluid and (de)HT for the

calculation of HT coefficient on annular side and also in evaluating the Gr-No.

Total outside heat transfer surface: (At)

Ao = 2 ( do L - NF L X)

Af = 2 NF L (2Y + X)

At = Ao + Af = 2L (do + 2NFY)

For bare tube Af = At = Ao

Heat Transfer co-relations

For turbulent flow (Re>10,000, L/D>60, 0-67<Pr < 120 )

Nu = 0.023 Re0.8

Pr0.33

= (b / w )0.14

For liquids

= (Tb / Tw)0.5

For gases (heating Tb/Tw < 1 for cooling =1, Tb /Tw>1)

For laminar flow : (Re < 2100)

Nu = 1.86 (Re Pr D/L )0.33

For Gz > 100

Nu = 1.86 (Re Pr D/L )0.33

+ 0.87 [1 + 0.015 Gr0.33

]

For liquids = (b / w )0.14

For gases = (Tb / Tw)n

For temperature ratio of 0.5 - 2, n=0, =1

For transition flow:

Nu = 0.116 (Re0.67

- 125) Pr0..33

[1+ (D/L)0.67

]

Fin Performance

Fin Effectiveness (f) is defined as the actual heat transferred by the fin divided by the heat

that ideally would be transferred if the entire fin were at base temperature.

f = mY

mY)tanh(

m = (hfPf / hfaf)1/2

= (2hf / kf X)0.5

for rectangular fin


102

af = 2(L + X) 2L

Fin effectiveness f is defined as the ratio of heat transferred through the fin to heat

transferred through the same bare surface having no fins.

f = f (surface area of fin / cross sectional area of fin)

Efficiency of the total heat transfer surface is expressed as weighted for efficiency:

(f)wtd = f (Af /At) + Ao /At

Over all heat transfer coefficient - U,

Q = U A (LMTD)


Fins increase heat transfer surface per unit length and reduce the size of heat exchanger

required for a given service. Fins are usually welded to the tube but may also be integrally

formed. Heat exchangers with fins are made up of copper. Consider a horizontal finned tube

(Longitudinal) double pipe heat exchanger (water to air). Water flow rate can be measured by

rotameter and air flow rate can be measured by using orifice provided.

5. Procedure:

a) Clean the apparatus and make water bath free from dust.

b) Close all the drain valves provided.

c) Fill water bath ¾ with clean water and ensure that no foreign particles are there.

d) Ensure that all ON/OFF switches given on the panel are at OFF position.

e) Select the parallel or counter flow of air and operate the ball valves given on the air

stream line.

f) Now switch on the main power supply (220 V AC, 50 Hz).

g) Switch on heater by operating switch given on the panel.

h) Set temperature of the water bath with the help of digital temperature controller.

i) Open flow control valve for hot water supply.

j) Switch on magnetic pump for hot water supply.

k) Adjust hot water flow rate with the help of flow control valve provided before

rotameter.

l) Record the hot water flow rate with the help of rotameter and cold air flow rate with

the help of orifice-meter provided.

m) Record the temperatures of hot and cold fluid inlet & outlet when steady state is

achieved.

n) Take readings at different flow rates of hot water and cold air.

o) For second run again fix the valve position for parallel or counter flow of air stream.

p) Repeat the steps 8 to 14 of experiment for different flow rates of fluids.

q) Repeat the steps 5 to 16 for another flow of air stream into the heat exchanger.

6. Observations:

Flow arrangement : Counter flow / Parallel flow.

System given : copper pipe with fins

Tube side fluid : Hot water.

Annular fluid : Cold air stream.

Inner Tube : Dia 9 mm, Length 1 m. Material Copper.

Longitudinal Fins : Width 12mm, Length 1m, 4 Nos. Material: Copper.

Outer tube : Dia. 42 mm. Length 1 m. Material G.I.

Water Flow Measurement : Rotameter.


103

Air stream Flow Measurement: Orifice-meter

Hot Water Tank : made of Stainless steel, Double Wall, insulated

with ceramic wool

Hot Water Circulation : Magnetic Pump made of Polypropylene to circulate Hot

Water. Maximum working temperature is 85°C.

Heaters : 2 kW Nichrome wire heater (1 Nos.)

Digital Temperature Controller : 0–199.9°C.

(For Hot Water Tank)

Digital Temperature Indicator: 0-199.9oC, with multi-channel switch

Temperature sensors : RTD PT-100 type 5 Nos.

With Standard make On/Off switch, Mains Indicator etc.

A good quality painted rigid MS Structure is provided to support all the parts

Data:

Copper tube outside dia ,do = 12mm

Copper tube inner dia, di = 9 mm

Fin length or tube length, L = 1000 mm

Fin height, Y = 12 mm

Fin thickness, X = 2 mm

No. of fins, Nf = 04

I D of outer shell, Ds = 42 mm

I D of orificemeter = 6.25mm

Observation Table:

Sr no. Water flow rate Ww

(kg/h) Thi

oC Tho

oC Tci

oC Tco

oC

Hot water inlet, (T1) = Thi

Hot water outlet, (T2) = Tho

Cold air stream inlet, (T3) = Tci

Cold air stream outlet (parallel), (T4) = TCo

Cold air stream outlet (counter), (T5) = TCo

Cold fluid i.e. air mean temp, (Tc) = 2

TTiO CC

OC

Hot fluid i.e water mean temp, (Th) = 2

TTiO hh

OC


104

From data hand book:

Thermal conductivity of fin material (Copper), Kf = 386.0 W/m-oC

Properties of water at temp T = ------- oC

Density = 1000 kg/m3

Specific heat Cp = -------J/kgoC

Thermal Conductivity k = ------- W/m -oC

7. Calculations:

Total heat transfer area = At

At = 2L (do + 2NFY)

= ---------- m2

Qc = Wc Cpc ( Tco – Tci)

Qh = Wh Cph (Tho - Thi)

Qavg = ( Qh + Qc) /2

Logarithmic Mean Temp Difference (LMTD)

LMTD =

)ln(1

2

12

T

T

TT

Over all Heat Transfer Coefficient

U = AxLMTD

Q

Use NTU method to determine the effectiveness of the heat exchanger.

For shell side fluid (air)

Cs = Wc Cpc

For tube side fluid (hot water)

CT = Wh Cph

Identify the minimum either Cmin = Cs & Cmax = CT

Or Cmin = CT & Cmax = Cs

Cmin = --------

Cmax = -----------

CR = max

min

C

C

NTU = minC

UA

For counter flow:

Effectiveness, = )]CNTU(1exp[C1

)]CNTU(1exp[1

RR

R

For parallel flow:

= R

R

C1

)]CNTU(1exp[1

For each run calculate NTU & (Try to keep CR fixed)

Plot % vs NTU for counter flow and for co-counter flow.

Nomenclature:

Tho

Tci

Thi

Tco

0

L

%

NTU = UA/Cmin

CR

CR


105

Ds = inner diameter of shell (outer pipe), m

do = the outer diameter of inner pipe (bare tube), m

de = Equivalent dia. of annular

NF = No. of fins per tube

X = Fin thickness, m

Y = Fin height, m

L = Nominal length of heat exchanger and length of longitudinal fin (in meters)

ANF = Net free area, m2

Qf = Cross Sectional area of fin, m2

Cp = Specific heat, KJ/kg-OC

G = Mean mass velocity, kg/m2-s

h = Heat transfer coefficient, w/m2-OC

k = Thermal conductivity, w/mOC

W = Water flow rate Kg / s.

U = Over all heat transfer co-efficient; w/m2k

Q = Heat load, w

8. Result & Discussions: Hence, the performance of the finned tube Heat Exchanger has been analysed and the Overall

Heat Transfer Coefficient, logarithmic mean temperature difference, & effectiveness for the

finned tube Heat Exchanger is calculated.

1. Logarithmic mean temperature difference of the finned heat exchanger is=

2. Overall Heat Transfer Coefficient of the finned tube Heat Exchanger =

3. Effectiveness of the finned tube Heat Exchanger =

9. Conclusion:

10. Precautions:

1. Never switch on main power supply before ensuring that all the on/off switches given on

the panel are at off position.

2. Never Switch on Heaters before filling water bath ¾ with clean water. It may damage

heaters.

3. Never run the Pump at low Voltage i.e. less than 180 Volts.

4. Never fully close the Delivery and By-pass line Valves simultaneously.


6. To prevent clogging of moving parts, run Pump at least once in a fortnight.


Always use clean water.

References:

1. Sinnott, R.K., Coulson & Richardson’s Chemical Engineering, Vol. 6, Butterworth-

Hienemann, Oxford, 1996.

2. http://www.enggcyclopedia.com/2012/03/finned-tube-type-heat-exchangers/


106

EXPERIMENT NO. 11

STEFAN-BOLTZMANN APPARATUS

1. AIM

To find out the Stefan-Boltzmann constant.

2. Objective

To study radiation heat transfer of black body.

3. Theory

All substances at all temperature (above absolute Kelvin) emit thermal radiation. Thermal

radiation is an electromagnetic wave and does not require any material medium for

propagation. All bodies can emit radiation and have also the capacity to absorb all of a part

of the radiation coming from the surrounding towards it.

The most commonly used law of thermal radiation is the Stefan Boltzmann law which states

that thermal radiation heat flux or emissive power of a black surface is proportional to the

fourth power of absolute temperature of the surface and is given by,

The constant of proportionally is called the Stefan Boltzmann constant and has the value of

5.67 x 10-8

W/m² K4

. The Stefan Boltzmann law can be derived by integrating the Planck’s

law over the entire spectrum of wavelength from 0 to infinity. The objective of this

experimental set up is to measure the value of this constant fairly closely, by an easy

experimental arrangement.

4. Experimental Description

The apparatus is centered on a flanged copper hemisphere B fixed on a flat non-conducting

plate A. The outer surface of B is enclosed in a metal water jacket used to heat B to some

suitable constant temperature.

One RTD PT-100 type temperature sensor is attached to the inner wall of hemisphere B to

measure its temperature and to be read by a temperature indicator.

The disc D, which is mounted in an insulating Bakelite sleeve is fitted in a hole drilled in the

center of the base plate A. An RTD PT-100 temperature sensor is used to measure the

temperature of D i.e. TD. The Temperature Sensor is mounted on the disc to study the rise of

its temperature.

5. Utilities Required Electricity Supply: 1 Phase, 220 V AC, 2 kW.

Table for set-up support


a) Heat the water in the tank by the immersion heater up to a temperature of about below

90 C.

b) The disc D is removed before pouring the hot water in the jacket.

c) The hot water is poured in the water jacket.


107

d) The hemispherical enclosure B and A will come to some Uniform temperature in a

short time after filling the hot water in the jacket. The thermal inertia of hot water is

quite adequate to prevent significant cooling in the time required to conduct the

experiment.

e) The disc D is now inserted in A at a time when its temperature is TD.

f) Start noting the temperature change for every five second for a minute.

7. Specifications Hemispherical enclosure diameter = 200 mm

Base plate, Bakelite diameter = 250 mm.

No. of Temperature Sensor mounted on B = 1

No. of Temperature Sensor mounted on D = 1

Temperature indicator digital (RTD PT-100) = 0-199.90C

Immersion water heater of suitable capacity and tank for hot water = 1.5 KW

The surface of B and A forming the enclosure are black to make their absorptivity to be

approximately unity. The copper surface of the disc D is also blacked.


8.1 Data

Mass of water in the hemispherical tank : 5.1x10-3

kg

Specific heat of water : 4180 J/kg K

Area of the disc, AD : 3.14x10-3

m2


Time t (second) Temperature TD in ºC

5

10

15

20

25

30

8.3 Calculation

The radiation energy falling on D from the enclosure is given by

The emissivity of the disc D is taken as unity, (assuming black disc). The radiant energy disc

D is emitting into enclosure will be

Net heat input to disc D per unit time is given by subtracting both the equations,

If the disc D has a mass m and specific heat s then a short time after D is inserted in A,


108

In this equation

denotes the rate of rise of temperature of the disc D at the instant

when its temperature is TD and will vary with time. It is clearly measured at time t = 0

before heat conducted from A to D begins to have any significant effect. This is obtained

from plot of temperature rise of D with respect to time and obtaining its slope at t = 0 when

temperature is TD. The temperature sensor mounted on disc is to be used for this purpose.

Note that the disc D with its sleeve is placed quickly in position and start recording the

temperature at fixed time intervals. The whole process must be completed in 30 seconds of

time. Longer disc D is left in position the greater is the probability of errors due to heat

convection from A to D.

Temperature of water : ____0C

Temperature of hemispherical enclosure at A, T : ____ 0C

Temperature of disc at the instant when it is inserted at D, TD : ____ 0C

Temperature time response of the disc. Note down the temperature TD at the time interval of

5 second. Plot the graph of TD and time (in seconds).

Obtain from the graph,

Value of can be obtained by using,

Nomenclature:

AD = Area of disc D.

T = Temperature of enclosure

TD = Temperature of disc

m = mass of disc

s = specific heat of the disc material.

9. Precautions & Maintenance Instructions

a) Always use clean water and heater should be completely dipped in the water before

switch ON the heater.

b) Always take the reading for the first minute after fixing the disc.

c) Use the stabilize AC single phase supply.

d) Never switch on mains power supply before ensuring that all the ON/OFF switches

given on the panel are at OFF position.

e) Voltage to heater should be constant.

f) Keep all the assembly undisturbed.

g) Never run the apparatus if power supply is less than 180 V and above than 240 V.

h) Operate selector switch of temperature indicator gently.

i) Always keep the apparatus free from dust.


109

j) Don’t switch ON the heater before filling the water into the bath.

10. Troubleshooting

1. If electric panel is not showing the input on the mains light. Check the fuse and also

check the main supply.

2. If digital temperature indicator displays “1” on the screen check the computer socket

if loose tight it.

3. If temperature of any sensor is not displayed in digital temperature indicator check the

connection and rectify that.

References

5. Holman, J.P., “Heat Transfer”, 8th

ed., pp. 396, McGraw Hill, NY, 1976.

6. Kern, D.Q., “Process Heat Transfer”, 1st ed., pp. 69, 74, McGraw Hill, NY, 1965.

7. Perry, R.H., Green, D.(editors), “Perry’s Chemical Engineers’ Handbook”, 6th

ed., pp.

10/53, McGraw Hill, NY, 1985.

8. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering

Vol. - 1”, 5th

ed., pp. 390, Asian Books ltd., ND, 1996.


110

EXPERIMENT NO. 12

CROSS-CIRCULATION DRYING

1. Aim

Study of drying characteristics of porous and non-porous solids under forced draft condition

with cross flow of air.

2. Objectives 1. To determine the critical moisture content.

2. To calculate the total drying time.

3. To calculate the humidity of inlet and outlet air with respect to drying time.

3. Theory Drying of solid generally means the removal of relatively small amounts of a liquid from the

solid material to reduce the liquid content to an acceptably low value. There are various

modes of drying a material in various types of dryers. In the present experiment, we study the

cross-circulation drying in an adiabatic or direct dryer. In this case, hot gas (air) is blown over

a bed of wet solids (brick particles wetted with water) under constant drying conditions by

maintaining the temperature, humidity and the velocity of the air across the drying surface

constant. A drying curve is then obtained by plotting drying rate against moisture content of

the solid. In drying, it is necessary to remove free moisture from the surface and also

moisture from the interior of the material. Thus there is interplay of the surface evaporation,

inter-particle and intra-particle diffusion processes during the drying of a material. This gives

rise to several distinct periods in the drying curve as the moisture content of the solid is

reduced from the high initial value to its final value, as explained below in terms of different

periods:

I0 (Initial Period): An initial period during which the drying rate may increase or decrease

rapidly from an initial value and the drying conditions are adjusting themselves to the steady

state condition, which is the next period. This period is of relatively short duration and in

some experiments may be unobservable. Therefore, this period is neglected in the

mathematical analysis of the process.

I (Constant Rate Period): An early stage of drying during which the drying rate remains at a

constant value, that is, is independent of the moisture content. During the constant rate

period, it is assumed that drying takes place from a saturated surface of the material by

diffusion of the water vapor through a stationary air film into the air stream. This period may

be absent if the initial moisture content of the solid is less than a certain minimum.

II (Falling Rate Period 1): During this period, there is insufficient water on the surface to

maintain a continuous film of water. The entire surface is no longer wetted, and wetted area

continually decreases in this first falling-rate period until the surface is completely dry.

During this period, the drying rate decreases more or less linearly with continued decrease of

water content.

III (Falling Rate Period 2): This second falling-rate period begins when it may be assumed

that the surface is dry. In this period evaporation will be taking place from within the solid

and the vapor reaches the surface by molecular diffusion through the material. The drying

rate in this zone decreases further, but generally in a non-linear fashion with the moisture

content.

The moisture content at which the drying rate falls for the first time is the first critical

moisture content, and the moisture content at which the drying rate falls for the second time

is the second critical moisture content. In case of non-porous solids, due to the absence of

intra-particle moisture and hence intra-particle diffusion, the constant-rate period is followed


111

by only one falling rate-period so that there is only one critical moisture content (see McCabe

et al., 2001, for detail).

Drying continues until equilibrium moisture content in the solid is attained. Nature of the

drying curve depends on the nature of the solid (porous or non-porous), and the temperature,

humidity and flow rate of the drying medium.

The total time of drying, tT, is determined from the following equation (the derivation is

given in McCabe et al., 2001)

2

1 ln)(X

XXXX

AR

mt c

cc

c

sT (1)

Where,

ms = mass of bone-dry solid

A = area of drying

Rc = rate at first critical point

X = free-moisture content, mass of water per unit mass of dry solid

Xc = free-moisture content at first critical point

X1 = initial free-moisture content

X2 = final free-moisture content

Free moisture content, X, is given by

XXX T (2)

Where,

XT = total free-moisture content

X* = equilibrium free-moisture content

The equilibrium moisture content of a non-porous insoluble material is practically zero so

that the whole moisture content is free moisture.


4.1. Requirements Dryer assembly, blower, beaker, water, brick particles (porous), glass beads (non-porous),

heater, physical balance, digital anemometer, thermometers (4 in number).



112

Outlet of air

Valve Blower

Anemometer

Measurement

space

Inlet of air

Wet bulb

temperature

thermometer Dry bulb

temperature

thermometer

Dryer

Heater

5. Experimental Procedure 1. Weigh the empty pan of the dryer.

2. Take out the pan from the dryer, and after filling it with about 250 gm of brick particles

place it back in the drying chamber. Note the weight of the pan with its contents.

3. Take out the water-cups for wet-bulb temperature (WBT) thermometer from the dryer and

keep them aside. Inject water in the water-cups using a 5 ml-pipette after every 30

minute.

4. Switch on the heater and the blower. Adjust the valve on the inlet line to give a constant

air velocity at about 4 m/s. Measure the air velocity with the digital anemometer.

5. Keep blowing the air for about 30 minutes for the system to reach a steady state, which is

indicated by an insignificant change in the weight of the filled pan. Note the dry-bulb

temperature and the weight of the pan at this state.

6. Take out the pan from the dryer and soak the brick particles in water for about 10 minutes

in a beaker. Then spread the wet brick particles over the pan uniformly. Fill the cups of

the WBT thermometers with water and wrap wet cotton around the bulb of these

thermometers.

7. Put one cup in the cup-holder and one WBT thermometer each at the inlet and outlet ports

of the dryer. Ensure that the bulb of the thermometer remains dipped in the cup.

8. Place the pan back in the drying chamber.

9. Note: Step 8 should follow step 7 with very little time gap.

10. Record the dry-bulb and wet-bulb temperatures at the inlet and the outlet of the dryer, and

weight of the pan. These are the reading at time t = 0.

11. Record the dry-bulb and wet-bulb temperatures at the inlet and the outlet of the dryer and

the weight of the pan, initially at intervals of about 2 minutes, and later at intervals of

about 5 minutes as the rate of drying (indicated by the rate of change of the weight of the

pan) decreases.

12. Continue the run until there is no significant change in the weight of the pan (and so the

temperatures; also the pan plus solid would weigh almost the same as the one noted in

step 5 before wetting the particles).

13. Repeat the experiment with the glass beads at the two air velocities of about 2 and 4 m/s.

6. Observations 1. Surface area of pan = 0.0245 m

2

2. Air velocity =

3. Weight of pan, w1 =


113

4. Weight of (pan + dry brick particles), w2 (at step 5) =

Material (glass beads/brick particles):

S No Time Weight of pan and

wet solid, w3

Inlet temperature Outlet Temperature

Dry-bulb Wet-bulb Dry-bulb Wet-bulb


1. Calculate the total moisture content (XT) as (w3 – w2)/(w2 - w1), and plot it with time.

2. Determine the rate of drying by finding the slope of the XT vs. t curve at different times.

(Truly speaking, in case of porous brick particles, one should find the equilibrium

moisture content and then determine the rate of drying using the free moisture content;

since it is difficult to find the equilibrium moisture content [why?], we are working with

the total moisture content).

3. Make the drying-curve by plotting rate of drying vs. moisture content.

4. Find the critical moisture content(s) and the corresponding drying rate(s) from the drying

curve.

5. Check the equilibrium moisture content of glass beads.

6. Using Eq. 1 calculate the total time of drying.

7. Compute the humidity of the air at the inlet and the outlet of the dryer at 5 minutes

interval and plot it against time at inlet and outlet.


(Tables, Graphs, Comparison with those reported in literature)

9. Conclusions

10. Precautions

Reference 1. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical Engineering,

6th


2. Richardson, J F, Harker, J H and Backhurst, J R, Coulson and Richardson's Chemical

Engineering, 5th

edition, Vol-2, Asian Books Private Limited, New Delhi, India, 2002.



http://www.fetchbook.info/search_J.F._Richardson/searchBy_Author.html

http://www.fetchbook.info/search_J.H._Harker/searchBy_Author.html

http://www.fetchbook.info/search_J.R_Backhurst/searchBy_Author.html


114

EXPERIMENT NO. 13(a)

OPEN PAN EVAPORATOR

1. Aim

To determine the overall heat transfer coefficient and the economy of Open Pan Evaporator

when evaporating saturated sodium chloride brine.

2. Theory

Evaporation is a process for concentrating a solution by vaporizing part or all of the solvent-

usually water. The objective of the evaporation is to concentrate a solution consisting of a

nonvolatile solute and a volatile solvent. Essentially, evaporation may be considered to be a

special case of heat transfer in which heat is transferred from condensing vapors, from hot

gases, or directly by radiation to a liquid at constant temperature, usually its boiling point at

the operating pressure. By far the most common evaporators are those in which heat is

transferred through a metal wall from condensing steam to a boiling liquid.

Evaporation differs from the drying in that the residue is a liquid sometimes a highly viscous

one rather than a solid. It differs from the distillation, in that the vapor usually is a mixture,

no attempt is made in the evaporation step to separate the vapor into fractions. It differs from

the crystallization in that emphasis placed on concentrating a solution, e.g. in the evaporation

of brine to produce common salt, the line between evaporation and crystallization is far from

the sharp. Evaporation sometimes produces slurry of crystals in saturated mother liquor.

Normally, in evaporation the thick liquor is the valuable product and the vapor is condensed

and discarded. In one specific situation, however, the reverse is true. Mineral – bearing water

often is evaporated to give a solid – free product for boiler feed, for special process

requirements, or for human consumption. This technique is often called water distillation, but

technically it is evaporation.

The vapor coming out of evaporator can be used as heating media for another evaporator

which will be operating at pressure lower than the pressure in the evaporator from which

vapors are issuing so as to provide sufficient temperature gradient for heat transfer in that

evaporator. When single evaporator is put into service and vapors leaving the evaporator are

condensed and discarded, the method is known as single effect evaporation. The economy of

single effect evaporator is always less than one. Generally for evaporation of one kg of water

from a solution, 1 to 1.3 kg of steam is required.

3. Procedure

a) Prepare approximately 11.6 kg of saturated sodium chloride brine (20 % w/w).

b) Charge this material to the open pan evaporator with bottom drain valve in closed

position.

c) Open the steam supply valve and also the steam to flow to the steam jacket, collect

the condensate in a separate vessel. Also note down the pressure indicated by the

pressure gauge.

d) After an hour stop the steam supply and measure the volume of condensate collected

as well as temperature of the boiling liquor.

e) Open the bottom drain valve of an evaporator and weigh the concentrate of sodium

chloride.

f) Wash the open pan evaporator to remove any traces of sodium chloride.

4. Observation and Calculation

4.1 Observations:


115

Initial weight of saturated sodium chloride brine (20% w/w) taken = 11.6 kg

Final weight of saturated sodium chloride brine = ------kg

Time of operation () = 1hr

Pressure of the steam supplied to the steam jacket (P) = 0.5 kg/ cm2

Temperature of boiling liquid (T2) = -------- C

Temp. of steam inlet (T1) = --------C

Temp. of condensate (T3) = --------C

Amount of condensate collected (w) = ------kg

Latent heat of vaporization of steam at the given pressure from steam table ()

=-------- kcal/ kg

4.2 Calculations

1) Amount of water evaporated (m)

= (Initial wt. of satd. NaCl brine (20% w/w) taken –

Final weight of satd. NaCl brine)

= -------- kg

2) Total energy given up by the steam to the brine solution (Q)

= * w

= -------kcal

3) Overall Heat Transfer Coefficient (U):

Q/ = U * A * (T1 – T2)

Where, Q/ = heat transferred, kcal/ hr

A= area of heat transfer surface, m2

t2 = temperature of condensing steam, C

t1 = temperature of boiling liquor, C

U = ----------- Kcal / Hr m2 OC

4) Economy, kg of water evaporated per kg steam

= (m/ w)* 100

5. Result

1) Overall Heat Transfer Coefficient (U) =--------- Kcal / Hr m2 OC

2) Economy, kg of water evaporated per kg steam =---------

6. Conclusion

7. Precaution

Keep partially open the steam trap valve during the performance of an experiment for

keeping the flow of steam continuous.

Do not increase the pressure above 1.5 kg/cm2 in steam generator.

Circulate the water in pipes after completion of an experiment to avoid the chocking

of solids inside the pipeline.

Drain the water from steam generator and condensate collector after completion of an

experiment.


116

Operate gently the selector switch of temperature indicator to read various

temperatures.

Increase the temperature gradually of the heater during initial set-up experimentation.


117

EXPERIMENT NO. 13(b)

VAPOR IN AIR DIFFUSION

1. Aim

To determine the diffusion coefficient of acetone (liquid) in air by using Arnold's cell.

2. Theory

Diffusion describes the spread of particles through random motion from regions of higher

concentration to regions of lower concentration. The time dependence of the statistical

distribution in space is given by the diffusion equation. The concept of diffusion is tied to that

of mass transfer driven by a concentration gradient, but diffusion can still occur when there is

no concentration gradient (but there will be no net flux). If two gases are inter diffusing with

continual supply of fresh gas and removal of the products of diffusion, this diffusion reaches

an equilibrium state with constant concentration gradients. This is known as steady state

diffusion. If also there is no total flow in either direction the rates of diffusion of A and B, NA

and NB are equal but have opposite sign.

The diffusivity or diffusion coefficient, D is a property of the system dependent upon

temperature, pressure and nature of the components. The dimension is length2/time.

This is the case of pseudo-steady-state diffusion in which one of the boundaries shift with

time with the effect that the length of the diffusion path changes, though only by a small

amount over a long period of exposure.

When this condition exists:

---------- (1)

The molar flux at any instant in the gas phase

Where Z=z1-z2 = the length of the diffusion path in time t.


3.1 Apparatus: Arnold's cell, Thermometer, Scale

3.2 Chemicals: Acetone.


a) Acetone is filled in capillary tube and air bubble, if present, is removed from the tube

carefully.

b) Note down the initial height (H0) of the acetone level in the tube.

c) The tube is placed in a water bath to maintain the constant temperature. Note down

the temperature of the water by thermometer.

d) Switch on blower, which blown air across the opening of capillary tube continuously

to remove the vapors evaporated that rises from the surface of the liquid.

21

,

,

1AA

lmB

AB

ZAPP

PRT

P

Z

D

N

2

1

2,

21

, 2

11zz

MtPP

PRT

PD t

LA

AA

lmB

AB


118

e) At regular interval of time note down the drop in the level of acetone.

f) Do necessary calculations and find out the diffusivity (D) from the equation.

g) Also calculate the diffusivity from the GILLILAND’S equation (as shown below) at

the same temperature using standard data and compare the result with the

experimental one.

h) Repeat the same procedure for two-three different temperatures by increasing and

maintaining constant temperature of the water bath and calculate the diffusivity by

both the methods. State your conclusion precisely at the end.

9. Observation Table

S.No Temp

c

Initial ht. of

liquid (mm)

Final ht. of

liquid(mm)

Time

(sec.)

1

2

3

6. Calculation a) Vapor pressure of acetone at given temp. (Calculate by Antoine Equation)

b) PA1 = ______ mm hg (Convert to N/m2, 1mmhg=133.2 N/m

2 )

c) PB1 = P-PA1

d) PA2 = 0 ( Pure B(air) is flowing)

e) PB2 = P-PA2

f) Z1 = ____________ m , Zt =_______ m

g) Diffusivity is given by,

DAB = __________ m2 /sec

7. Diffusivity by Gilliland’s method

VA = Molar volume of acetone = 0.0074 m3/kmol

VB = Molar volume of air = 0.01386 m3/kmol

MA = Molecular weight of acetone = 58 kg/kmol

1

2

12

,

lnB

B

BB

lmB

P

P

PPP

23/13/1

2/12/34

*

/1/110*3.4

BA

BA

VVP

MMTD

2

1

2,

21

, 2

11zz

MtPP

PRT

PD t

LA

AA

lmB

AB


119

MB = Molecular weight of air = 18 kg/kmol

T = Absolute temperature = __________ 0

K

So,

D = __________ m 2

/sec from Gilliland’s eqn.

8. Data Given

a) P= 1 Std.atm. = 760 mm hg = 101325 N/m2

b) R = 8314 N.m/Kmol. K

c) Room Temp = ______ C

d) Density of acetone = 1540 Kg/m3

e) Mol. of acetone = 58 Kg/Kmol.

Antoine Constant For acetone

f) A = 16.6513

g) B = 2940.46

h) C = - 35.93

9. Result

Temperature (K) DAB DTheo.

10. Conclusion

Observe and state the effect of Temperature on Diffusion.


120

EXPERIMENT NO. 14

DIFFERENTIAL DISTILLATION

1. Aim

To verify the Raleigh’s equation for a differential distillation in a binary system and to

estimate relative volatility of the binary system

2. Introduction

Differential distillation is also known as simple distillation. Usually it is operated in a batch

mode. It is used to separate a liquid mixture whose components have fairly large difference in

their boiling points. In this mode of distillation a charge of liquid is placed in the still and it is

distilled off until the liquid charged is all gone or until the operation is no longer doing any

beneficial separating. In batch distillation operations, the liquid in the original charge is

constantly being depleted of its more volatile components. This mode of distillation may also

be used for analytical evaluation of boiling ranges and vaporization characteristics if mixtures

in laboratories.

3. Theory

Differential distillation refers to a batch distillation in which only one vaporization sate is

involved. Lard Rayleigh developed a mathematical equation which represents differential

distillation. The following are the major assumptions in deriving Rayleigh’s equation

1. Still liquid composition is uniform.

2. The distillation process is carried out slowly so that the vapour leaving the still is in

equilibrium with the liquid in still.

3. There is no entrainment of the liquid during the distillation

4. There is no condensation of the vapour evolved before it reaches the condenser.

3.1 Rayleigh’s equation derivation

Let L be the number of moles of the mixture at a given instant. Let x be the mole fraction of

more volatile component in liquid mixture. Let y be the mole fraction of more volatile

component in vapour that is in equilibrium with liquid that is in still. If dL is the differential

amount of liquid vaporized then by applying the material balance with respect to the more

volatile component we get

Input = Lx (1.1)

Output = dLy (1.2)

Accumulation = dxxdLL (1.3)

Therefore dLydxxdLLLx (1.4)

Neglecting dLdx , we get

LdxxdLdLy (1.5)

Therefore, xy

dx

L

dL

(1.6)

The boundary conditions are:

Initially, L = F and Fxx ; F : moles of feed mixture

Finally, L = W and Wxx ; W : moles of with drawn or bottom mixture

W

F

x

x

W

Fxy

dx

L

dL (1.7)


121

T

W

x

xxy

dx

W

Fln (1.8)

Using the vapour liquid equilibrium data for the binary mixture system, the plot is made of

xy

1 vs x .Area under the curve between the ordinate at Fx and Wx gives the values of

the integral on the R.H.S of Eq.(1.8). From this we obtain the value of W, the amount of with

drawn (or) bottoms, to be left to attain the given composition Wx of the liquid. Alternatively,

if we want to calculate the composition of with drawn, when W is known, then L.H.S of

Eq.(1.8) can be found and value of Wx adjusted to obtain the area under the curve equal to the

value of W

Fln .

3.2 Rayleigh’s equation for an ideal system

For an ideal system, average relative volatility, ave , is constant. Therefore for an ideal

system we can write the following equilibrium relation

x1α1

xαy

ave

ave

(1.9)

Substituting Eq.(1.9) in Eq.(1.8) and upon manipulation we get the following relation

W

Fave

W

F

x1W

x1Flnα

Wx

Fxln (1.10)

Eqs.(1.8) and (1.10) are known as Rayleigh’s equations for differential distillation.

4. Experimental setup

The experimental setup to study differential distillation is shown in Figure P1.1. The

distillation still may be fabricated from steel with substantial thickness recommended. The

vapors of the chemical distilled are collected as distillate is shown in the Figure P1.1.


122

Figure P1.1. Differential distillation experimental setup


The following procedure may be followed while conducting the experiment

1. The still is charged with known composition binary system

2. Ensure water flow in the condenser

3. Heating mantle is switched on and heating is carried out sufficiently and slowly.

When the mixture starts boiling, collect the distillate in a flask.

4. This procedure is continued for sufficient length of time (say 2hours) until

approximately 2/3 of the feed distilled.

5. Meantime prepare a calibration chart using known concentrations for the binary

system. For calibration purpose use specific gravity technique (or) refractive index

(RI) method.

6. Measure the specific gravity of the distillate (or) RI of the distillate. Similarly cool the

bottoms to room temperature and measure specific gravity (or) RI of bottoms

6. System used and calculations

The Following binary Chemical may be taken for simple distillation

Ethanol (A)-water (B)

Data required for the calculations from the literature

1. Molecular weights of compounds A and B

2. Densities of components A and B (or) refractive indices of components A and B

3. Vapour- liquid equilibrium (VLE) of components A and B at atmospheric pressure

4. Plot calibration chart

7. Calculations

1. Calculate the amounts and compositions of distillate for the sample i.e D and Dx .

While doing make use of calibration plot for composition calculations.


123

2. Using total material balance and more volatile component balances to calculate

amount of residue (bottom),W and composition of the residue, Wx

3. Plot xy

1

vs. x and measure the area between the curve and x-axis to the left of

the ordinate Fxx ,when the area = F/Wln , final Wx

4. Calculate

100

WFln

xydxF/Wln

F

W

x

x

5. Using

W

Fave

W

F

x1W

x1Flnα

Wx

Fxln estimate aveα

8. Presentation of results

1. Mole fraction of mole volatile component in the feed; mole fraction of more volatile

component in the distillate; mole fraction of more volatile component in the bottoms

2. Amounts of distillate and bottoms and their compositions

3. aveα = average relative volatility of the system between Wx and Fx

4. Report % errors

5. One can also compare estimated aveα value with the value obtain with the help of

equilibrium data. Make use the definition of α as

x1x

y1y

α and then calculate aveα

.

6. Using

W

Fave

W

F

x1W

x1Flnα

Wx

Fxln estimate aveα

VLE DATA ETHANOL WATER SYSTEM

x y* x y* 0.0 0.0 32.73 58.26

1.9 17 39.65 61.22

7.21 38.91 50.79 65.64

9.66 43.75 51.98 65.99

12.38 47.04 57.32 68.41

16.61 50.89 67.63 73.85

23.37 54.45 74.72 78.15

26.08 55.8 89.43 89.43


124

CALIBRATION CHART

References:

1. Max Peters, “Elementary Chemical Engineering”, 2nd

ed, McGraw-Hill Book

Company, New York, 1984, pp. 203-204.

2. V.G.Pangarkar and R.R.bhave, “Mass transfer Operations: A laboratory manual in

Chemical Engineering”, Chemical Engineering Education Development Centre,

Indian Institute of Technology Madras, Madras, 1981, pp.20-26.

3. R..E.Treybal, Mass Transfer Operations, McGraw-Hill Book Company, 3rd

ed, New

York, 1984, pp. 367-370.

4. F.Molyneux, Laboratory exercises in Chemical Engineering, Leonard Hill Books,

London, 1967, p.137.

5. ‘Chemical Engineering Laboratory Manual’ published by IIT Bombay, obtained on

the Webpage http://www.che.iitb.ac.in/courses/uglab/uglabs.html accessed on

25.08.2011

1.33

1.335

1.34

1.345

1.35

1.355

1.36

0 0.1 0.2 0.3 0.4 0.5 0.6

Re

frac

tive

Ind

ex

Mole fraction of Ethanol in Ethanol-Water System

http://www.che.iitb.ac.in/courses/uglab/uglabs.html%20accessed%20on%2025.08.2011

http://www.che.iitb.ac.in/courses/uglab/uglabs.html%20accessed%20on%2025.08.2011


125

EXPERIMENT NO. 15

BATCH CRYSTALLIZATION

1. Aim

Determine the yield of crystals using agitated batch crystallizer.

2. Objective

To study the yield of crystals of Sodium Sulfate (Na2SO4) or Sodium Chloride (NaCl) from

its saturated solution using open tank type agitated batch crystallizer.

3. Theory

Crystallization is an important operation in processing as a method of both purification and of

providing crystalline materials in a desired size range. In a crystal, the constituent molecules,

ions or atoms are arranged in a regular manner with the result that the crystal shape is

independent of size and, if a crystal grows, each of the faces develops in regular manner. The

presence of impurities will, however, usually result in the formation of an irregular crystal.

Generally large regular crystals are a guarantee of the purity of the material, though a number

of pairs of materials form “mixed crystals”. In recent years, techniques have been developed

for growing perfect crystals, which are used in the production of semiconductor devices, laser

beams, and artificial gems.

The crystallization process consists essentially of two stages which generally proceed

simultaneously, but which can to some extent be independently controlled. The first stage is

the formation of small particles or nuclei, which must exist in the solution before

crystallization can start, and the second stage is the growth of nuclei. If the number of nuclei

can be controlled, the size of the crystals ultimately formed may be regulated, and this forms

one of the most important features of the crystallization process.

Yield of crystals: The yield of crystals produced by a given cooling may be estimated from

the concentration of the initial solution and the solubility at the final temperature, allowing

for any evaporation, by making solvent and solute balances as follows:

For the solvent, usually water, the initial solvent present is equal to the sum of the final

solvent in the mother liquor, the water of crystallization within the crystals and any water

evaporated or:

w1 = w2 + y(R-1)/ R + w1E ------------------- (i)

Where,

w1 and w2 are the initial and final masses of solvent in the liquor

y is the yield of crystals

R is the ratio (molecular weight of hydrate/ molecular weight of anhydrous salt)

And E is the ratio (mass of solvent evaporated/ mass of solvent in the initial solution)

For the solute:

w1c1 = w2c2 + y/ R ---------------------- (ii)

Where,

c1 and c2 are the initial and final concentrations of the solution in the terms of mass of

anhydrous salt per unit mass of solvent.

From equation (1):


126

w2 = w1(1 – E) – y(R - 1)/ R ---------------------- (iii)

And in equation (2)

w1c1 = c2 [w1(1 – E) – y(R - 1)/ R] + y/ R ----------------------- (iv)

From which the yield,

y = Rw1[c1 – c2 (1 – E)]/ [1 – c2(R-1)] ------------------------ (v)

Broadly, crystallizers may be classified according to whether they are batch or continuous in

operation, and continuous crystallizers may be divided into linear and stirred types which are

referred to later. Crystallizers may also be classified according to which the super saturation

is achieved. In evaporative crystallizers conditions are approximately isothermal, and super

saturation is achieved as a result of the removal of solvent. In cooling crystallizers, super

saturation results from lowering of the temperature of the solution, and this can be effected

either by means of exchange of sensible heat or by evaporative cooling; in the latter case,

there is a small loss of solvent. Evaporative crystallization must of course be used where the

solubility shows little variation with temperature.

The simplest and cheapest type of crystallizer consists of an open tank, which can be used

either as an evaporative or as a cooling crystallizer. In small scale of batch processing, such

crystallizers are quire convenient because of their low first cost, simplicity of operation and

flexibility. They are too wasteful of labor and give too uneven product to be attractive for

large scale continuous processing


The experimental setup has a 5 liter capacity jacketed vessel made of SS 304, 1.6 mm thick,

provided with a FHP variable speed std. make stirrer (Gear Motor, 150 RPM max., for better

control of stirring and revolving the stirrer at low RPM during crystallization).


a) Take about 3-4 liter of water in the unit. Switch ON the power supply and pump and

start circulating the water from the tank in to the jacket at fixed value using the

rotameter. The exit water from the jacket shall be recycled back to the water supply

tank.

b) Set the temperature of the crystallizer content using the front panel of the

Temperature Indicator-Controller (TIC) to the desired value and slowly start adding

the solid to be crystallized and allow it to dissolve completely.

c) Now set the temperature value of the crystallizer on the TIC to the value

corresponding to the value at which the crystallization is to be carried out.

d) Reduce the rate of circulation of hot water in the jacket and maintain it in the range of

about 1.5-2.0 LPM and start draining the water coming out of the jacket.

e) Add the cooling water available in the laboratory at the same rate to the water supply

tank to make up the loss of water in the water supply tank and start recycling the exit

water from the jacket back to the water supply tank.

f) If ice addition is required to bring down the temperature of the Crystallizer obtained

by the cooling water alone, slowly add finely crushed ice to the water supply tank so

that the temperature of the crystallizer do not go below the set value.

g) After 10-15 minutes stop circulation of the cold water and drain the water from the

crystallizer using drain valve.


127

h) Filter and then weigh the water and crystals collected. Collect the crystals from the

crystallizer. Dry them and weigh.

6. Observations and Calculations

T1 = hot water inlet temperature / cold water inlet temperature

T2= hot water outlet temperature / cold water outlet temperature

T3=sample liquid inner temperature

T4=sample liquid outlet temperature (if required)

Initial weight of water taken (w1) = gm

Initial weight of Na2so4 added (m1) = gm

Temperature at which dissolution takes place(t3) = C

Final weight of Na2so4 collected (m2) = gm

Final Cooling Temperature = C

% Yield of crystals = (m2 /Solubility at exp. Temp.) * 100

Yield of crystal = Wt. of Na2so4 obtained as crystals

Wt. of Na2so4 initially taken in feed

7. Result & Discussions

i) Yield of crystals =

ii) Plot the Solubility Curve of Na2SO4 in Water with Temperature

8. Conclusion

9. Precautions


128

EXPERIMENT NO. 16(a)

STEAM DISTILLATION SETUP

1. Aim

To study the characteristics of simple steam distillation using turpentine oil as a feed stock.

2. Objective

To determine the vaporization efficiency, thermal efficiency and percentage recovery of

simple steam distillation.

3. Theory

Steam Distillation is a term used to describe a distillation process with open steam. Steam

distillation is a method for distilling compounds which are heat-sensitive. It is specifically

used where it is desired to separate substances at a temperature lower than their normal

boiling point. The temperature of the steam is easier to control than the surface of a heating

element, and allows a high rate of heat transfer without heating at a very high temperature.

The vapor mixture is cooled and condensed, usually yielding a layer of oil and a layer of

water.

Steam distillation is commonly used in the following situations:

1. To separate relatively small amounts of volatile impurity from a large amount of

material

2. To separate appreciable quantities of higher-boiling materials

3. To recover high-boiling materials from small amounts of impurity that has a higher

boiling point.

4. Where the material to be distilled is thermally unstable or reacts with other

components associated with it at the boiling temperature

5. Where the material cannot be distilled by indirect heating even under low pressure

because of the high boiling temperature

6. Where direct-fired heaters cannot be used because of fire hazards

In a steam distillation process, the liquid is distilled by feeding open steam to the distillation

still. The steam carries with it vapors of volatile liquid and is then condensed to separate the

liquid from water. The essential requirement for carrying out steam distillation is:

1. Substance does not react with steam.

2. Substance is in-soluble in water (immiscible).

From the Hausbrand vapor pressure diagram for turpentine water system at 101.3 kN/m2

pressure, obtain the distillation temperature, TD oC ( 95°C).

From the Hausbrand diagram, curves for turpentine and water at atmospheric pressure cross

at 95°C.

Distillation: Turpentine oil at 101.3 kN/m2 pressure, (1.03kg/cm²) .The intersection of the

two curves gives the distillation temperature.

Figure 1: Plot of PA Vs T and

(101.3 - PB) Vs T.

Distillation Temp.

VapourPressure

T

(101.3 - P )BPA


129

Figure 2: Hausbrand vapor-pressure diagram for turpentine oil-water system


4.1 Description

The apparatus consists of simple batch distillation using steam as a source of heat. The feed

stock used is turpentine oil. Distillate is collected in a separating funnel for the formation of

organic layer and an aqueous layer. Experimental Setup is shown in Fig 3.

Fig 3: Experimental Setup

4.2. Requirements

(Turpentine Oil + Water)

PG

Jacket

Steam

Live

Steam

Turpentine oil In

Water In

Residue Out

Distillate

Cooling Water Out Cooling Water In

Condenser

Heater


130

Steam supply, electricity, turpentine oil, demineralized water and separating funnel


1. Charge the distillation still with 2kg of turpentine oil.

2. Adjust the jacket steam pressure to 170 kN/m2 (Pg,)/1.73 kg/cm² and start the cooling

water supply to the condenser. Record the still temp. Collect the steam condensed in

the jacket.

3. When the temperature in the still reaches 2°C below the distillation temp. (Td), the

jacket steam is stopped and the flow of live steam is started through the steam sparger.

4. The live steam pressure is adjusted around 150 kN/m2 (Ps) /1.52 kg/cm².

5. Weigh the steam condensed in the jacket (WS), kg.

6. Continue the distillation process for sufficient time so that about 50-70% turpentine

charged is distilled (around 1 hr).

7. Stop the steam supply and collect the distillate in the 2L separating funnel. Allow the

formation of organic layer and an aqueous layer. Separate the two phases and weigh

them, (WAD, WBD), kg.

8. Collect the residue, separate the two layers and weigh WAB, WBB (kg).

9. Stop the water supply to the condenser.

6. Observations and Calculation

Mol. Wt. of turpentine oil MA =136.22

Normal boiling point of turpentine oil = 160oC

Sp. heat of turpentine oil at 20°C = CpA = 1.8 kJ/kg °C

Sp. heat of turpentine oil at 50°C = CpA =1.926 kJ/kg °C

Sp. heat of turpentine oil at 100°C = CpA = 2.093 kJ/kg °C

Latent heat of vaporization of turpentine oil, A = 74 kcal/kg = 309.84 kJ/kg

Mol. wt. of Water = MB = 18

System Pressure, P = 101.3 KN/m2

/1.03kg/cm²

Wt. of turpentine (feed) = 2 kg.

Distillation temp., TD = 95 °C

Wt. of Steam Condensed in jacket = WS, kg

Steam pressure in the jacket, Pg = 170 KN/m2

Pressure of live steam, PS = 150 KN/m2

Steam pressure in the jacket = 172.21 kg/m2 = Pg

Live pressure in the still = 151.95 kg/m2 = Ps

Mass of steam condensed during pre-heating (feed change from 23oC to 95

oC) WS = 410 g

Distillation time =

Initial temperature of feed charge, TR = °C

Observed distillation temp., = °C

Distillate readings (Turpentine oil + water):

Wt. of turpentine oil in distillate WAD =

Wt. of water in distillate WBD=

Residue collected (Turpentine oil + water):

Wt. of turpentine oil in residue WAB =

Wt. of water in residue WBB =

Ambient temperature Tr = 23°C

Distillation time = 15 min approximately

Observed distillation temperature =

Converting Weights to moles:


131

No. of moles of turpentine oil in feed

WAF

NAF = ---------- =

MA

No. of moles of turpentine oil in distillate

WAD

NAD = ---------- =

MA

No. of moles of turpentine oil in Residue

WAB

NAF = ----------- =

MA

No. of moles of water in distillate

WBD

NBD = ------------ =

MB

No. of moles of water in residue

WBB

NBB = ---------- =

MB

Applying Material balance for turpentine oil,

NAF = NAD + NAB

Losses from the still turpentine oil =

Steam Distillation Temp, TD =

PAO= Partial Pressure of turpentine oil during distillation = P [NAD/ (NAD + NBD)]

Where P =total pressure = 101.3 KN/m2

Vaporization efficiency =

PAº

= --------- =

PA

WBD (P - PA) MB

And ---------- = ----------------x ----------

WAD MA

where PA is the pressure of turpentine oil and water system at 95 °C from Hausbrand

diagram.

WBD / WAD in distillate =

% Recovery =

Calculation of Thermal efficiency: T

Total heat output

T = --------------------------------- x 100

Total heat input

Heat Output:

QO = WAF (TD – TR) CpA + A WAD = WS + A WAD

TR = ambient temp

TD = distillation temp. =92oC

Heat Input: Qi

Qi = Heat given by condensing steam in the jacket + Heat given by the condensing live steam

during distillation.

Qi = WSJ + (WBD + WBB) [S + Cp (TS – TD)]

Jacket steam pressure = PJ (KN/m2)


132

J =latent heat of vaporization of steam corresponding to pressure PJ, kJ/kg

PS = Pressure of live steam, KN/m2

TS = saturation temp. of steam at PS, °C corresponding to PS, kJ/kg

Cp = Specific heat of steam, kJ/kg°C.

PJ = 172 KN/m2

J = 2216 kJ/kg

PS = 151.95 KN/m2

S = 2230 kJ/kg

Cp = Sp. heat of steam = 2.0 kJ/kgoK

QO

= -------- x 100 =

Qi

NOMENCLATURE

WAF = Wt. of turpentine in feed (kg)

TD = Distillation temp. °C

TR = Reference temp. (Ambient temp) °C

Cp = Specific heat of turpentine oil (kJ/kgoC)

A = Latent heat of evaporation of turpentine (kJ/kg)

WAD = Wt. of turpentine in distillate, (kg)

WS = Wt. of Steam Condensed in the jacket during pre-heating

or charge (kg)

= Latent heat of steam corresponding to jacket steam

pressure P (kN/m2)

WBD = Wt. of water in distillate (kg)

WBB = Wt. of water in residue, kg

S = Latent heat of steam at live steam pressure PS (kN/m2)

TS = saturation temperature of steam at PS

Cp = Specific heat of steam (kJ/kgoC)


Vaporization efficiency=

Recovery=

Thermal efficiency=

8. Conclusion

References

1. McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical

Engineering”, 7th

ed.McGraw Hill, NY, 2005.

2. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering

Vol. - 1”, 5th


3. Treybal, R.E., “Mass Transfer Operations”, 3rd



133

EXPERIMENT NO. 16(b)

VAPOR LIQUID EQUILIBRIUM SETUP

1. Aim & Objective

Study of vapor-liquid equilibrium of Toluene – CCl4 mixture

2. Apparatus VLE setup consisting of heater, condenser, temperature indicator and refractometer, test

tubes.

3. Chemicals Carbon Tetrachloride, Toluene

4. Theory The design of distillation and other contacting equipment requires reliable VLE data.

Although relatively few ideal solutions are known whose equilibrium relations can be

calculated from vapor pressure - temperature data of the pure components, by far the larger

numbers of systems of industrial importance are non-ideal; and attempt to predict the

equilibrium compositions of such mixtures from theoretical considerations alone have not

proved successful. It has been the practice to determine such data experimentally under

various conditions.

Vapor liquid diagram shows relationship between the composition of the vapor and that of

liquid in equilibrium with the vapor for a binary mixture at constant pressure or constant

temperature. If liquid and vapor behave ideally, such curves are calculated as follows:

From Raoult' law:

p1 = P1 X1 (1)

p2 = P2 X2 (2)

Where p1 and p2 are partial pressures of components 1 and 2 in the mixture, P = total pressure

P1 and P2 are vapor pressure of pure components at the same temperature as mixture.

From Dalton's law of partial pressures:

P1= P y 1 (3)

P2= P y 2 (4)

Where, y1 and y2 are the mole fractions of components in vapor.

From these equations we have:

21

2

1PP

pPx

(5)

P

xPy

21

1 (6)

Theoretical VLE curve can thus be calculated by choosing various boiling points of

themixture and calculate x and y as shown above. Carbon tetrachloride - Toluene system

closely follows ideal behavior.


1. The vapor - liquid equilibrium still containing about 15O ml to 200 ml of known

mixture (toluene – CCl4) is heated gently for at least 45 to 60 minutes (to ensure the

mixture has attained equilibrium in above time)

2. The liquid and vapor are then sampled at noted temperature (T) and analyzed for

molar composition of more volatile component in vapor (y) and liquid (x).


134

3. The heating is stopped and the feed liquid is replaced by another feed mixture

(increasing gradually the amount of high boiling liquid), the liquid and vapor are once

more allowed to come to equilibrium (indicated by constant temperature) and

collecting the samples of liquid from the still and the condensate. The still liquid

being hot, should be collected in an ice cooled test tube to avoid any change in

composition due to vaporization. The above procedure should be repeated for at least

six feed compositions.

4. The samples are viewed under a high precision refractometer, and their composition

determined from the pre-determined calibration chart for CCl4 - toluene mixture,

which converts refractive index to mole fractions. This way T-x-y and x-y diagrams

can be plotted.

5. The calibration curve (nD vs. mole fraction of more volatile component, x) should be

generated at a specific temperature (say 250C) by taking different mixtures of CCl4 -

toluene of known molar composition and recording the refractive index of the

mixture. The samples collected should also be evaluated at the same temperature.

6. Observation & Calculation

Correlation for CCl4 – Toluene system w.r.t refractive index is given by the equation

y = -27.757*x + 41.529 ------------- (7)

For VLE curve data

S.No Liquid

refractive

index

Mole

fraction

(x)

Vapor

refractive

index

Mole

fraction

(y)

Temperature

(°C)


Plot the following graphs

1. y vs. x

2. Temperature vs. x

3. Temperature vs. y

8. Conclusion

9. Precautions 1. The still liquid being hot, should be collected in an ice cooled test tube to avoid any

change in composition due to vaporization.

2. The above procedure should be repeated for at least six feed compositions.

References: (a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”,

7th


(b) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”, 26th

reprint,

McGraw Hill, NY, 2011.


135


136

EXPERIMENT NO. 17

TWO PHASE FLOW IN A HORIZONTAL PIPE

1. Aim To Study Two phase flow in Horizontal Pipe.

2. Objective

To study the two-phase flow in a horizontal pipe and to determine the pressure drop for the

two phase flow.

3. Theory

Concurrent gas-liquid flow is very common in chemical process equipment like reactors, re-

boilers, partial condensers, flash vaporizers gas pipe lines etc. As such it is always important

to have a clear understanding of flow pattern and pressure drop in systems involving a two-

phase mixture.

3.1 Flow Pattern

There are various flow regimes or “patterns” that are common to two phase flow systems,

each having different characteristics and associated pressure drops. Two phase flow cannot

simply be described as laminar, transitional and turbulent. The two-phase patterns vary not

only with flow rates and fluid properties, but also depend upon pipe diameter and inclination.

In horizontal pipe, the two phase flows regimes are (in order of lowest gas velocities to

highest) bubble flow, plug flow, stratified flow, wavy flow, slug flow, annular flow and spray

flow. In vertical pipe there are bubble, slug, churn, ripple, annular flow and mist flow.

Most remediation professionals would recognize the appearance of these four patterns from

conducting pilot tests and from observations gathered during full-scale system operation.

However, most practitioners are unaware of the significance of that the type of flow pattern

has on efficient system effectiveness. (Note: The term” Pattern” refers to visual observations

and the flow “regime” applies to flow behavior that can be described by more quantitative

expressions.

Additionally, general estimations of flow pattern can be made using superficial gas and liquid

velocities (Perry, 1999), Horizontal flow patterns can be roughly determined from gas and

liquid superficial velocities, while vertical patterns can be estimated based on gas velocity

only (Note: Superficial velocity is defined as the velocity of a phase as if it was flowing alone

within the pipe.

In horizontal pipe, flow patterns for fully developed flow have been reported in numerous

studies. Transitions between flow patterns are gradual, and subjective owing to the visual

interpretation of individual investigators. In some cases, statistical analysis of pressure

fluctuations has been used to distinguish flow patterns.

Bubble flow is prevalent at high ratios of liquid to gas flow rates. The gas is dispersed as

bubbles which move at velocity similar to the liquid and tend to concentrate near the top of

the pipe at lower liquid velocities.

Plug flow describes a pattern in which alternate plugs of gas and liquid move along the upper

part of the pipe.

In stratified flow, the liquid flows along the bottom of the pipe and the gas flows over a

smooth liquid/gas interface. Similar to stratified flow, wavy flow occurs at greater gas

velocities and has waves moving in the flow direction. When wave crests are sufficiently

high to bridge the pipe, they form frothy slugs which move at much greater than the average

liquid velocity.


137

Slug flow can cause severe and sometimes dangerous vibrations in equipment because of

impact of the high-velocity slugs against bends or other fittings. Slugs may also flood

gas/liquid separation equipment.

In annular flow, liquid flows as a thin film along the pipe wall and gas flows in the core.

Some liquid is entrained as droplets in the gas core. At very high gas velocities, nearly all the

liquid is entrained as small droplets. This pattern is called spray, dispersed, or mist flow

When a liquid and gas flows through a horizontal pipe, the following types of flow may

occur.

1. Both the phases may be in viscous region.

2. Both the phases may be in turbulent region.

3. One of the phases may be in turbulent region and the other in viscous region.

The type of flow of each phase is determined by the Reynolds number calculated on the basis

of superficial velocity (superficial velocity = total flow rate of single phase/ total cross

sectional area of flow)

The pressure drop in a two-phase flow is the sum of the pressure drops due to friction and

pressure drop due to acceleration. In a two phase flow, where gas is one of the phases, the

pressure drop due to acceleration is important because the gas normally flows faster than the

liquid phases and thus accelerates the liquid phase resulting in transfer of energy. The

estimation the pressure drop is not possible analytically. The most common method to


138

estimate the pressure drop for two-phase flow is the one proposed by Lockhart and Martinelli

(Page 6-26, fluid and particle dynamics, Perry’s chemical engineering). In this, each phase is

considered separately and the combined effect is determined.

Lockhart and Martinelli (ibid.) correlated pressure drop data from pipes 25 mm (1 in) in

diameter or less within about 50 percent.

In general, the predictions are high for stratified, wavy, and slug flows and low for annular

flow. The correlation can be applied to pipe diameters up to about 0.1 m (4 in) with about the

same accuracy.

The pressure drop due to two-phase flow is the pressure drop of either of the two phases if it

alone were present, multiplied by a factor.

(P/Z)tp = YL (P/Z)L

(P/Z)tp = Yg (P/Z)g

where YL and Yg are the functions of dimension less parameter X defined as:

X = [(P/Z)L/(P/Z)g]0.5

The two-phase flow analysis can also be expressed in terms of friction factors.

By definition of pressure drop,

P/Z = (2fV2/(gcD))

Friction factor, f = (P/Z) .g. D/2*V2)

where,

v is the superficial fluid velocity, D is pipe I.D, is the fluid density and P/Z is the

pressure drop per unit length

Thus for gas only:

fg = (P/Z)g .g. D/(2*Vg2g)

for liquid only

fl = (P/Z)l .g. D/(2*Vl2l)

for two phase flow

flpl = (P/Z)tp .g. D/(2*Vl2l) (based on superficial velocity of liquid)

flpg = (P/Z)lp .g. D/(2*Vg2g) (based on superficial velocity of gas)

Defining

g = flpg/fg

l = flpl/fl

X2 = [(P/Z)L / (P/Z)g] = [g / l]2

Two phase pressure drop is then expressed as:

(P/Z)tp = g2 (P/Z)g

(P/Z)tp = g2 (P/Z)l

Friction factors for single component flow (gas or liquid) can be easily estimated from:

f = 16/Re (for Re < 2000)

f = 0.079 (Re)-0.25

(for Re > 3000)

Recently empirical co-relations have been derived from a large data bank for void fraction ()

and g and are given below:-

=__________1_______________

(1+0.0904 X 0.548)2.82

g = (1+ X 2ln

) n/2

and, n is obtained from the table given below:-


139

N Liquid Flow Gas Flow

4.12 Turbulent Turbulent

3.61 Viscous Turbulent

3.56 Turbulent Viscous

2.68 Viscous Viscous If X < 1

3.27 Viscous Viscous If X > 1

Pure phase Reynolds numbers are defined as:

Reg = DVg g / g

Rel = DVl l / l

Vl = Superficial velocity of liquid = volumetric flow rate of liquid/cross-section area of

empty pipe.

Vg = Superficial velocity of Gas = volumetric flow rate of gas/cross- section area of

empty pipe.

D is the I.D. of pipe = 25.4mm= 0.0254m

P is the pressure drop across length = Z = 1 meter approximate.


8.

9. Start the water pump and allow water to flow at a constant flow rate

10. Start the compressor and fix the air pressure in the air tank at 1.5 kg/cm² and allow air

to flow at some constant flow rate.

11. Wait for the steady state and observe the flow pattern. Record it. Note the volumetric

flow rates of water and air (rotameter readings).

12. Record the average reading of the manometer (cm of CCl4). Also record the

temperature of water and air by thermometer.

13. Increase the gas flow rate gradually keeping water flow rate constant at the previous

level and repeat steps 4 and 5.

14. Repeat step 6 for at least 6 gas flow rates.

15. Repeat steps 2 to 7 for at least 5 liquid flow rates

16. Stop the compressor.

17. Stop the water pump.

5. Observations

System : Air-water

Distance between manometer taps = 1000mm = 1Z, m

Manometer fluid = mercury

Water temperature = T ambient

Air Temperature = T ambient

Air Pressure = kg/cm²

Density of manometer fluid = m = 1660 kg/m3

Density of water = l = 1000 kg/m3

Density of air = a = 1.205 kg/m3

Viscosity of water (from data book) l = 0.00089 N-s/m2

Viscosity of air (from data book) g = 0.0000192 N-s/m2


140

Run No

Air rotameter reading,

Qg, LPM

Water rotameter reading, Ql,

LPM

Manometer reading R,

cm of CCl4

6. Calculations

I.D. of pipe = 0.0254 D, m

Cross sectional area of pipe A = ------- D², = 5.0670 x 10⁻⁴ m²

4

Volumetric flow rate of air = Qg LPM

Volumetric flow rate of Water = Ql , LPM

Air superficial velocity Vg = ((Qg/60) *10-03

)/A, m/s

Water superficial velocity Vl = ((Ql/60) *10-03

)/A, m/s

Air Reynolds number Reg = DVg g / g

Water Reynolds number Rel = DVl l / l

Experimental two phase pressure drop:

Two phase pressure drop Ptp = (R/100 [m-w)* g], N-s

Two phase pressure drop per unit length = (P/Z) tp, N-s/m²/m

Two phase flow friction factor = ftpg or ftpl

ftpl = (P/Z)tp .g. D/(2*Vl2 l)

(based on superficial velocity of liquid)

ftpg = (P/Z)tp .g. D/(2*Vg2 g)

(based on superficial velocity of Gas)


8. Conclusion


141

EXPERIMENT NO. 18

MASS TRANSFER WITH CHEMICAL REACTION

1. Aim

1. To study the dissolution of benzoic acid in aqueous NaOH solution.

2. To compare the observe enhancement factor for mass transfer with those predicted by the

film and boundary layer models.

2. Theory

Solid-liquid mass transfer plays an important role in some industrial operations. The

dissolution may occur with or without chemical reaction. In case dissolution is accompanied

by solid-liquid reaction, it is desirable to know the enhancement in the rate of mass transfer

due to chemical reaction. In the present experiment we aim at finding the enhancement in the

rate of dissolution due to simultaneous reaction and compare it with the enhancement

predicted on the basis of the film and boundary layer models. The system is: dissolution of

benzoic acid in aqueous NaOH solution.

The dissolution of a solid in a solution accompanied with instantaneous chemical reaction can

be expressed as:

BB Product

Where, A is the solid and B is the liquid phase reactant, assuming the reaction to be

instantaneous so that A and B don’t coexist. The mechanism of solid dissolution involves

dissolution of A in liquid followed by its reaction with species B diffusing from the bull

liquid phase at a reaction plane.

If the film model is applied to this situation, the enhancement factor, defined as the ratio of

the solid-liquid mass transfer coefficient with reaction, kr to the mass transfer coefficient

without reaction, k, given by:

A

B

D

D

D

D

k

k

B

0

A

B

B

Arfilm

--------- (1)

And for boundary layer model it is:

A

B

D

D

D

D

k

k

B

o

3/2

A

B

3/1

B

Arh

---------- (2)


The Set-up consists of a cylinder of benzoic acid mounted on a SS rod and driven by a D.C.

motor. The operational range of rotation is between 10 to 30 rpm. The cylinder is immersed

in an aqueous solution of sodium hydroxide of known concentration in a 500 ml vessel filled

to 2/3rd

its capacity. The position of the benzoic acid cylinder is so adjusted that the liquid

level rises above the top surface by about 3 cm. The dimensions of the benzoic acid cylinder

may be fixed at diameter: 2 to 3 cm, length: 5 to 7 cm. The cylinder can be prepared by

pouring molten benzoic acid in the mould of desired dimensions with 4 to 5 mm SS rod

located in the center of the mould in a vertical position. User can also use the cylinder


142

provided with the set-up.


1. Record the dimensions of the benzoic acid cylinder (length and OD of SS rod) and then fix

I in the vertical position with the D.C Motor. (Do not immerse it in solution as yet)

2. Fill the vessel with aqueous NaOH solution of known concentration of 2/3rd

of its volume.

Record the volume of aqueous NaOH solution added (V).

3. Start the water bath and fix the dissolution temperature (ambient to 50 0C), wait till the

aqueous solution attains the desired temperature. Record the temperature (T).

4. Now fix the benzoic acid cylinder inside the vessel containing aqueous NaOH solution and

start the motor at a fixed rotational speed. (N, rpm)

5. Run the experiment for 10 minutes.

6. Stop the motor and remove the benzoic acid cylinder.

7. Mix thoroughly the contents of vessel and analyze it for un-reacted NaOH concentration

by titration against standard HCL solution.

8. Measure the benzoic acid cylinder dimensions again.

9. Repeat steps 1 to 8 for different concentrations of aqueous NaOH solutions.

10. Repeat steps 1 to 8 using de-ionized water. This run may be carried for about 45 to 60

minutes duration. During this period small samples (5 ml) should be withdraw at regular

intervals of 10 minutes and analyzed for dissolved benzoic acid by titration against 0.02

kmol/m3 HCL solution.


Reaction:

C6H5COOH + NaOH C6H5COONa + H2O

(For each mole of benzoic acid dissolved, 1 mole of NaOH is consumed)

Thus rate of dissolution can be determined by measuring the fall in NaOH concentration.

1. Recommended concentration range for aqueous NaOH solution: 0.5 kmol/m3 to 1.5

kmol/m3.

2. Use the following standard solutions for titration:

a. 0.02 kmol/m3 NaOH solution for the estimation of dissolved benzoic acid.

b. 0.4 to 1 kmol/m3 standard HCL solution for the estimation of un-reacted NaOH.

3. For each titration in case of estimation of NaOH use 10 ml of solution and titrate against

standard HCL solution using suitable indicator.

4. For each titration in case of estimation of dissolved benzoic acid use 5 ml of solution and

titrate against 0.02 kmol/m3 standard NaOH solution.

Let titer value for sample before the experiment = T1 (m3).

Let titer value for sample after the experiment = T2 (m3).

Let titer value for sample = T3 (m3).

Time of dissolution = t (sec)

Dissolution temp = T oC

Hence,

[NaOH]i = (T1/10) [HCL]

[NaOH]f = (T2/10) [HCL]

[C6H5OOH] = (T3/5) [NaOH]

The rate of dissolution of Benzoic acid:


143

R = ({[NaOH]i - [NaOH]f}) X Vol. of NaOH sol. in tank (V m3).

Let the average dimensions (avg. of start and of expt.) of benzoic acid cylinder be Dav, Lav.

Let S.S. rod diameter = dr.

Average surface area, As =

22

42 ravavav dDLD

Specific rate of dissolution = R / As (kmol/m2-s)

Ak

Rexp

Where,

A is the solubility of benzoic acid in water (0.0276 kmol/m3)

k is obtained from physical dissolution run (i.e. with plain water)

After integrating the material balance equation for a species, we have:

A

AInt

V

Ak bs 1

Where,

[Ab] is the bulk liquid cone of benzoic acid at time, t, in kmol/m3 and

[A] is the solubility of benzoic acid in water, kmol/m3.

After plotting t vs. ln [1 - Ab / A], we record the slope of the line given by:

Slope = kAs / V; knowing As and V, we get k.

The theoretical values for enhancement factor, for the film model and boundary layer

model is obtained from Eq.1 and Eq.2 respectively.

The diffusivity of benzoic acid, DA, may be obtained from the literature = 1.04 x 10-9

m2/ at

30 oC.

Value of DB may be taken as = 4.1 x 10-9

m2/ at 30

oC.

Compare the theoretical values for with the experimental values and observe that

boundary layer model is more close to the experimental value.


7. Conclusion


144

EXPERIMENT NO. 19

ADSORPTION IN PACKED BED

1. Aim To understand the phenomena of adsorption in packed beds.

2. Objective To determine adsorption isotherm of acetic acid on activated charcoal. To determine the

adsorption constant (k) and the maximum adsorbed substance amount of acetic acid per gram

of charcoal (Amax) of Langmuir isotherm.

3. Theory Adsorption is a process that occurs when a gas or liquid solute accumulates on the surface of

a solid or a liquid (adsorbent), forming a molecular or atomic film (the adsorbate). It is

different from absorption, in which a substance diffuses into a liquid or solid to form a

solution. The term absorption encompasses both processes, while desorption is the reverse

process.

Adsorption is operative in most natural physical, biological, and chemical systems, and is

widely used in industrial applications such as activated charcoal, synthetic resins and water

purification. Similar to surface tension, adsorption is a consequence of surface energy. In a

bulk material, all the bonding requirements (be they ionic, covalent or metallic) of the

constituent atoms of the material are filled. But atoms on the (clean) surface experience a

bond deficiency, because they are not wholly surrounded by other atoms. Thus it is

energetically favorable for them to bond with whatever happens to be available. The exact

nature of the bonding depends on the details of the species involved, but the adsorbed

material is generally classified as exhibiting physisorption or chemisorptions.

Physisorption or physical adsorption is a type of adsorption in which the adsorbate adheres to

the surface only through Van der Waals (weak intermolecular) interactions, which are also

responsible for the non-ideal behavior of real gases.

Chemisorption is a type of adsorption whereby a molecule adheres to a surface through the

formation of a chemical bond, as opposed to the Van der Waals forces which cause

physisorption.

Adsorption is usually described through isotherms, that is, functions which connect the

amount of adsorbate on the adsorbent, with its pressure (if gas) or concentration (if liquid).

One can find in literature several models describing process of adsorption, namely Freundlich

isotherm, Langmuir isotherm, BET isotherm, etc. We will deal with Langmuir isotherm in

more details:

Langmuir isotherm

In 1916, Irving Langmuir published an isotherm for gases adsorbed on solids, which retained

his name. It is an empirical isotherm derived from a proposed kinetic mechanism.

It is based on four hypotheses:

i. The surface of the adsorbent is uniform, that is, all the adsorption sites are equal.

ii. Adsorbed molecules do not interact.

iii. All adsorption occurs through the same mechanism.


145

iv. At the maximum adsorption, only a monolayer is formed: molecules of adsorption do

not deposit on other, already adsorbed, molecules of adsorbate, only on the free

surface of the adsorbent.

For liquids (adsorbate) adsorbed on solids (adsorbent), the Langmuir isotherm (Fig. 1) can

be expressed by

------------------------ (1)

Where m is the substance amount of adsorbate adsorbed per gram of adsorbent (or kg) of the

unit of m is mol.g-1

.resp. mol. kg-1

Amax is the maximal substance amount per gram (or kg) of the adsorbent.

The unit of Amax is mol.g-1

, resp. mol.kg -1

k is the adsorption constant (mol-1

.dm3), c is the

concentration of adsorbate in liquid (mol.dm-3

).

In practice, activated carbon is used as an adsorbent for the adsorption of mainly organic

compounds along with some larger molecular weight inorganic compounds such as iodine

and mercury.

Activated Carbon - Activated carbon can be manufactured from carbonaceous material,

including coal (bituminous, sub-bituminous, and lignite), peat, wood, or nutshells (i.e.,

coconut).The manufacturing process consists of two phases, carbonization and activation.

The carbonization process includes drying and then heating to separate by-products,

including tars and other hydrocarbons, from the raw material, as well as to drive off any gases

generated. The carbonization process is completed by heating the material at 400–600°C in

an oxygen-deficient atmosphere that cannot support combustion.

4. Experimental Setup The experimental set up consists of a three Glass Columns having different diameter and

height, filled with activated carbon. A common feed inlet (manifold) is provided at the

bottom of each column. Feed is supplied from feed tank by means of pump and rotameter.

Flow rate can be varied by rotameter. A bye-pass arrangement is also provided to maintain

level in feed tank (optional). Adsorption process can be studied by varying bed height and

feed flow rate in each of the three columns turn-wise. The whole set-up is mounted on a

powder coated sturdy MS frame.


146

Glass Column Borosilicate - ID 28mm height 500 mm, ID 38mm height 500mm and ID

52mm and height 750mm, Rotameter Make Eureka 2-20LPH. Pump make Promivac Model

PMP-15, Max head 2.5 mtrs, RPM 2800, single phase 220 Volts @ 50 Hz, sump tank

capacity 20 ltrs apprx.

Chemicals Required (Not in supply scope):

Acetic acid, NaOH, phenolphthalein indicator, activated charcoal granular form (supplied

with set-up).

5. Experimental Procedure a) First prepare the aqueous solution of acetic acid and water.

b) Switch ON the set up and ensure pump discharge.

c) Set the desired flow rate in Rotameter.

d) Open the feed valve provided at the back of column 1 (a common manifold has been

provided for feed routing to all the three columns).

e) Maintain constant flow rate in Rota meter (in case of fluctuation adjust with valve) after

feed starts to come out from top of column, collect the sample in flask.

f) Add 4-5 drops of phenolphthalein indicator in added in each flask and titrate by

standard NaOH solution.

g) Once the endpoint has been reached, read the burette. The volume of the base 0

iX(ml)

that was required to reach the endpoint.

h) Repeat procedure for different flow rates and bed heights.

i) After completing taking readings of Column 1, start feed supply to column NO 2 and

close feed valve of column NO 1.

j) Repeat procedure from 4-5 as mentioned above for different flow rates and bed heights.

k) After completing taking readings of Column 2, start feed supply to column NO 3 and

close feed valve of column NO 2.

l) After completing experiment switch off pump and disconnect set-up from mains. Drain

the SET-UP.


Table 1

Flask

No.

0

iX (ml) 0

ic

(mol/dm3)

Xi

(ml)

ci

(mol/dm3)

mi

(mmol/g)

1/ci

(dm3

/mol)

l/mi

(g/mmol)

1

2

3

4

5

0

ic - Concentration of acetic acid before the adsorption reaction

ci- Concentration of acetic acid after the adsorption reaction

mi- Amount of acetic acid adsorbed per gram of charcoal 0

iX -The volume of the titrant (NaOH)


147

6

Determination of the concentration of acetic acid before (ci0) and after (ci) Adsorption:

,0

0

V

cXc Ti

i ------------------------- (2)

Where 0

iX is the volume of the titrant (NaOH), cT is the concentration of the titrant, V is the

volume of the analytic (acetic acid according to Tab. 2), i=1-6 is the number of flask.

Calculate the concentration of acetic acid after adsorption (ci), using the Eq. 2 and data form

Tab. 3 after adsorption.

Determination of the substance amount of acetic acid adsorbed per gram of charcoal (mol.g-

1) in individual flask:

,)( 0

g

Vccm Aii

i

--------------------------- (3)

Where 0

ic , ci are the concentrations of acetic acid before and after adsorption, respectively. VA

is the volume of the liquid phase in the mixture charcoal – acetic Acid, g is the mass of the

adsorbent – charcoal (in grams), i=1-6 is the number of flask.

Eq. 3 supposes that VA is the same for i=1-6, and also the mass of the charcoal (g). Write

down the obtained values of mi to the Tab. 3.

Determination of k and Amax: The Eq. 1 one can rearrange into a form:

,1111

maxmax AcAm k

------------------------ (4)

Thus,

cf

m

11 should be a straight line.

Use MS Excel to create the dependence

cf

m

11, where c is the concentration of acetic acid

after adsorption. Fit the experimental points with a linear function. The slope represents the

value ofkAmax

1, and the intercept corresponds to

max

1

A

Calculate Amax and k from the slope and the intercept.

7. Precautions

i. Do not run the pump at low voltage i.e. less than 180 Volts.

ii. Never fully close the Delivery line fully.

iii. Always keep apparatus free from dust.

iv. To prevent clogging of moving parts, Run Pump at least once in a fortnight.

v. Always use clean water (use distill water if available in lab).

vi. If apparatus will not in use for more than one month, drain the apparatus completely.

vii. If pump gets jam, open the back cover of pump and rotate the shaft manually.


148

EXPERIMENT NO. 20

HUMIDIFICATION IN WETTED WALL COLUMN

1. Aim

Humidification & Dehumidification study in wetted wall column.

2. Objective

1. To take, and learn to interpret psychometric (wet and dry bulb) data, and to understand

more

generally what is involved in transferring mass across gas-liquid interfaces.

2. To investigate the effects of air flow rates on the mass transfer coefficient, KG (kg mole of

water transferred /m2.h.atm).

3. Theory

A Wetted Wall Column is essentially a vertical tube with means of admitting liquid at the top

and causing it to flow downwards along the inside wall of the tube, under influence of the

gravity. Gas is admitted to the inside bottom of the tube, where it flows through the tower in

contact with liquid (i.e. water).

The mass transfer coefficient KG can be measured in the equipment in which the area of

contact between the two phases is known and boundary layer separation.

The measured flow of air at a measured humidity is brought in to contact with a film of water

at certain temperature and vapor pressure. Moisture is absorbed in the air from the water film

and the resultant humidity of the air and temperature & vapor pressure of the water at the

entry and exit is measured. The rate of diffusion through the gas film, NA, is given by:

NA = KG A (ΔP)m kg mole/h

Where:

KG = Gas film coefficient, kg mole of water transferred/m2 –h-atm. Partial pressure

difference.

A = wetted surface of column

(ΔP)m = log mean partial pressure driving force across the ends of the column.



The schematic diagram of the experimental setup is shown below:


149


b) Feed water to the column from top at a rate at which complete wetting with

minimum ripple formation takes place.

c) Commence and operation with minimum air flow (H =3 cm) and after 10-15 minute

the wet bulb and dry bulb temperature of the inlet air and outlet air at this flow rate

is noted.

d) Read the corresponding humidity (from psychometric chart) and vapor pressure of

water corresponding to the entry and exit temperatures.

e) Repeat step 2 and 3 for 3-4 air flow rates.

f) Calculate experimental value of KG

6. Observations 1. Column diameter, D = 2.54 cm

2. Column effective length, L = 1.04 m

3. Column pressure, P = 760 mmHg/ 1 atm

4. Area for mass transfer A = Dl

= 3.14 2.54 104 = 829.9 cm2

6.1. Observation Table

Air temperature, °C Column Bottom Column Top

Dry bulb

Wet bulb

Air humidity(Kg water/kg dry air) A. HB A.HT


150

Water temperature, °C

7. Calculations:

Air temperature, °C Column Bottom Column Top

Dry bulb

Wet bulb

Air humidity(Kg water/kg dry air)

Water temperature, °C

Molar fraction of water in air at bottom, y1 = (A. H.B / 18) / (A. H.B /18 + 1/29)

y1 =

Molar fraction of water in air at top, y2 = (A. H.T / 18) / (A. H.T/18 + 1/29)

y2 =

Vapour pressure of water at inlet temperature, (P2) T = 0.0508 atm (Bhatt & Vora-343 pp.)

Vapour pressure of water at outlet temperature, (P1) B =

(ΔP)m = [( P1sat

-P1)-( P2sat

-P2)]/[ln(P1sat

-P1/P2sat

-P2)

yi= (y1+y2) / 2

yi= (0.02+0.026) / 2= 0.023

KG (ΔP)m *A / V = [1 / (1-yi)]. In [{(yi-y1)/y2-yi)} {(1-y2) / (1-y1)}]

KG(ΔP)m *A/V= X

H = Head of water

P = H (water - air) g=

V = molar flow rate of gas = 0.65. [(2 P)0.5

/ ] =

Where; [Co / (1-)4 =0.65]

m = Vo /4 (do)2 =

V = molar flow rate of gas

V=mass flow rate of air / mol. Wt. of air =


151

KG = (X * V) / (ΔP)m * A =

Where,

A. H.B= air humidity at bottom

A. H.T= air humidity at top

V = molar flow rate of gas

V0 = mass flow rate of gas

Molar fraction of water in air at bottom=y1, and at top =y2


Hence, the performance of the wetted wall column has been analysed and mass transfer

coefficient is calculated.

Mass Transfer Coefficient of the wetted wall column at flow rate----- is =

9. Conclusion

10. Precautions

1. Do not run the pump with the delivery valve closed for a long time to avoid damage of the

pump.

2. Make sure during the experiment water tank should not be emptied totally.

3. Uniform wetting of the inner surface of the column ensures more valid data.

4. The column will require several minutes to come to steady state, especially with low water

flow rate. Be sure to verify the actual attainment of this condition.

References: 1. Bird R. Stewart W. and Lightfoot E., “Transport Phenomena”, Wiley (2004)

2. Treybal, R. E., Mass Transfer Operations, McGraw Hill Company, New York, 1980

3. McCabe, W. L., Smith J. C., and Harriott, P., Unit Operations of Chemical Engineering, 6th

ed., McGraw Hill Company, New York, 2001

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