at2251 atht notes.pdf

8/10/2019 AT2251 ATHT Notes.pdf

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UNIT I GAS POWER CYCLES

1. Define: Thermodynamic cycles.

Ans: Thermodynamic cycle is defined as the series of processes performed on the system, so

that the system attains to its original state.

2. Define the term compression ratio.

Ans: Compression ratio is the ratio between total cylinder volume to clearance volume. It isdenoted by the letter r

3. What is the range of compression ratio for SI and diesel engine?

Ans: For petrol of SI engine 6 to 8 : For diesel engine 12 to 18.

4. Write the expression for efficiency of the otto cycle?

Ans:1

Efficiency n =1 - ---------(r)r-1

5. What is meant by cutoff ratio?

Ans: Cutoff ratio is defined as the ratio of volume after the heat addition to before the heat

addition. It is denoted by the letter p

6. What are the assumptions made for air standard cycle.

Ans:1. Air is the working substance.

2. Throughout the cycle, air behaves as a perfect gas and obeys all the gas laws.

3. No chemical reaction takes place in the cylinder

4.

Both expansion and compression are strictly isentropic5. The values of specific heats of the air remain constant throughout the cycle.

7. What is the difference between otto and Diesel cycle.

Otto Cycle Diesel Cycle

1. Otto cycle consist of two adiabatic andtwo constant volume process.

1. It consists of two adiabatic, one constantvolume and one constant pressure

processes.

2. Compression ratio is equal to expansion

ratio

2. Compression ratio is greater than

expansion ratio.

3. Heat addition takes place at constantvolume.

3. Heat addition takes place at constantpressure

4. Compression ratio is less. It is variesfrom 6 to 8.

4. Compression ratio is more. It variesfrom 12 to 18.

8. Define: Mean effective pressure of an I.C. engine.

Ans: Mean effective pressure is defined as the constant pressure acting on the piston during the

working stroke. It is also defined as the ratio of work done to the stroke volume or pistondisplacement volume.

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9. Define: Specific fuel consumption.

Ans: SFC is defined as the amount of fuel consumed per brake power hour of work.

10. What is meant by calorific value of a fuel.

Ans: Calorific value of a fuel is defined as the amount of heat liberated by the compete

combustion of unit quantity of a fuel.

11. Give the expression for efficiency of the Dual cycle.

Ans:

1 Kpr - 1

Efficiency n = 1 - ------- -----------------(r)r-1 (K-1) + yK(p-1)

where,

rCompression ratiokpressure or Expassion ratio

pcut off ratio and

yadiabatic index

11. What are the factors influencing of the Dual cycle?

Ans: 1. Compression ratio 2.cut off ratio 3. pressure ratio and 4. heat supplied at constantvolume and constant pressure.

12. Give the expression for efficiency of the Brayton cycle.

Ans:1

Efficiency n = 1 - ---------- where Rppressure ratio.

(Rp)y-1

13. What is a Gas turbine? How do you classify.

Ans: Gas turbine is an axial flow rotary turbine in which working medium is gas.

Classification of gas turbine.1. According to the cycle of operation

a) open cycle b) closed cycle and c) semiclosed cycle.

2. According to the process

a) constant volume and b) constant pressure process.

14. What is meant by open cycle gas turbine?Ans: In open cycle gas turbine, the exhaust gas form turbine is exhausted to the atmosphereand fresh air is taken in compressor for every cycle.

15. Differentiate open and closed cycle gas turbines.

Open cycle gas turbine Closed cycle gas turbine

1. Working substance is exhausted to the

atmosphere after one cycle.

1. The same working substance is

recirculated again and again.

2. Pre-cooler is not required 2. Pre-cooler is required to cool the exhaust

gas to the original temperature.




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3. High quality fuels are used 3. Low quality fuels are used

4. For the same power developed size and

weight of the plant is small

4. Size and weight are bigger.

16. What is the function of intercooler in gas turbines? Where it is placed?Ans: The intercooler is placed between L.P. and H.P. compressors. It is used to cool the gas

coming form L.P. compressor to its original temperature.

17. Why re-heater is necessary in gas turbine? What are its effects?

Ans: The expansion process is very often performed in two sperate turbine stages. The re-

heater is placed between the H.P. and L.P. turbines to increase the enthalpy of the exhaust gascoming from H.P. turbine.

Effects:

1. Turbine output is increased for the same compression ratio

2. Thermal efficiency is less.

18. What is the function of regenerator in gas turbine?

Ans: The main function of heat regenerator is to exchange the heat from exhaust gas to thecompressed air for preheating before combustion chamber. It increases fuel economy and

increase thermal efficiency.

1. Derive the air standard efficiency of Otto cycle with neat sketch




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2. Problem




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3.Derive the air standard efficiency of diesel cycle with neat sketch.




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4. PROBLEM




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5 PROBLEM




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6. PROBLEM




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7 PROBLEM




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8 PROBLEM




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UNIT II REFRIGERATION CYCLES AND RECIPROCATING CYCLES

1. What is meant by single acting compressor?

In single acting compressor, the suction, compression and delivery of air take place on one side

of the piston.

2. What is meant by double acting compressor?

In double acting reciprocating compressor, the suction compressin and delivery of air take

place on both side of the piston.

3. What is meant by single stage compressor?

In single stage compressor, the compression of air from the initial pressure to the final

pressure is carried out in one cylinder only.

4. Define clearance ratio

Clearance ratio is defined as the ratio of clearance volume to swept volume (or) stroke volume.

Vc Vc clearance volume

C = -------- Vs swept volumeVs

5. What is compression ratio?

Compression ratio is defined as the ratio between total volume and clearance volume.

Total volumeCompression ratio = -------------------

Clearance Volume

6. What are the factors that effect the volumetric efficiency of a reciprocating

compressor?

1) Clearance volume 2) Compression ratio.

7. What is the difference between complete (or) perfect inter cooling and incomplelte (or)

imperfect inter cooling.

Perfect Inter cooling

When the temperature of air leaving the intercooler (T3) is equal to the original atmospheric air

temperature (T1), then the inter cooling is known as perfect inter cooling.

Imperfect Inter coolingWhen the temperature of air leaving the inter cooler (T3) is more than original atmospheric airtemperature (T1), then the inter cooling is known as Imperfect inter cooling.

8. Power requirement of a refrigerator is _________

Ans: Inversely proportional to cop

9. In SI Units, one ton of refrigeration is equal to __________

Ans: 210KJ/min

10. The capacity of a domestic refrigerator is in the range of __________

Ans: 0.1 to 0.3 tonnes.




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11. COP of a refrigerator working on a reversed carnot code is _________

Ans: T2

-----------

T1 T2

12. The vapour compression refrigerator employs the __________cycle

Ans: Reversed carnot

13. In vapour compression cycle the condition of refrigerant is dry saturated vapour

________

Ans: Before entering the compressor.

14. Name four important properties of a good refrigerant

Ans: 1. Low boiling point2. High critical temperature & pressure

3. Low sp.heat of liquid

4. Non flammable and non explosive.

15. Name some of the equipments used in air conditioning system

Ans: 1. Filter

2. Cooling coil3. Heating coil

4. Compressor

5. Condeser6. Evaporator

16. Name any four commonly used refrigerants

Ans; 1. Ammonia (NH3)2. Carbon di oxide (CO2)3. Sulphur di oxide (SO2)

4. Freon 12.

17. What are the factors to be considered in air conditioning a room?

Ans: 1. Temperature of air

2. Humidity of air

3. Purity of air4. Motion of air.

1. PROBLEM




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2. PROBLEM




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3.PROBLEM

4.What are the difference between vapour compression and absorbtion system?




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5.Explain with neat sketch vapour absorbtion




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6 Problem




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RECIPROCATING AIR COMPRESSOR

1. Explain with neat sketch the Operation of an Air Compressor

Operation:

To understand the operation of an air compressor, let us assume the cycle and indicator diagram for a

simple single stage reciprocating air compressor, as shown below. (Click the image to enlarge.)

The simple reciprocating air compressor has a piston which reciprocates inside the

cylinder wall and cylinder head. The piston is attached to the crankshaft with the help of

a connecting rod and thus the rotation of the crankshaft causes the piston to move up

and down inside the cylinder. The crankshaft is mounted on the crank case. The

cylinder head contains valve pockets where the suction and delivery valve are fixed.

These suction and delivery valves are of simple pressure differential types. They openand close, due to the pressure difference on either side of the valve plates.

1. When the compressor stops or idles for some time, it is always assumed that there is

some residual compressed air left in the cylinder space. This residual air expands when

the piston moves down. The pressure drops in the cylinder space at a particular point as

the piston moves down, where the pressure inside the cylinder becomes lesser than the

atmospheric pressure. Thus this difference in pressure makes the suction or inlet valve

open.




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2. This opening of inlet valve allows fresh air to be drawn inside the cylinder space as

the piston still continues to move in the downward direction. The inlet valve will remain

open till there is pressure difference between the atmosphere and inside of the cylinder

space. As the pressure difference starts to reduce, the inlet valve starts slowly closing.

The inlet valve closes completely when there is no pressure difference and then the

piston reaches bottom dead center (BDC), and it starts to travel in an upward direction.

At this position, both the inlet and delivery valve remains closed. Thus as the piston

moves up, the pressure starts to build inside the cylinder space.

3. The delivery valve starts to open when there is a pressure difference between the

cylinder space and air receiver. Let us assume the air receiver is at a pressure of 7 bar.

The deliveryvalve will not open until the pressure inside the cylinder space is slightly

above 7 bar. As the piston moves in upward direction, the pressure increases and at

some point the pressure grows beyond 7 bar making the deliveryvalve open. Thus the

compressed air is delivered into the air receiver.


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4. As the piston reaches top, the pressure starts to fall and the delivery valve starts to

close. The residual compressed air remaining in the space again starts to expand as the

piston moves down continuing the next cycle.

2. Explain with neat sketch theoretical P-V diagram of air compressor:

Referring to the diagram, the theoretical air compressor P-V diagram can be

understood.

4-1:

The air compressor draws in air from the atmosphere. The atmospheric pressure is P1.

The initial volume when the piston is at top is zero (Assuming there is no bumping

clearance). Thus as the piston moves from top to bottom, a volume of air V1 is drawn

into the compressor. The temperature of air is T1.




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1-2:

As the piston moves up, the air is compressed polytropically (PV^n= C). The pressure of

air increases from P1 to P2. The volume decreases from V1 to V2. The temperature

increases from T1 to T2.

2-3:

The compressed air at the pressure P2, volume V2 and temperature T2 is delivered out

of the compressor to the air receiver.

3. Explain with neat sketch the Practical P-V diagram of air

compressor:

In my last article, we have seen the P-V diagram and understood the operation of an air

compressor. But in practice, the diagram is not so perfect. The figure shown here

represents the actual practical P-V diagram of an air compressor. The points, 1234

represents the theoretical diagram. But there are some shaded portions above and

below the work done area. It is necessary to give certain explanation for these

additional areas which add up to the work done by the compressor.

Referring to the diagram, at point 4, when the clearance air has reduced to the

atmospheric pressure, the inlet valve will not open immediately. The pressure drops

lower than the atmospheric pressure and the inertia of the valves are overcome by the

pressure difference. Thus the valve is forced open by the atmospheric air and it rushes

into the cylinder chamber. There is a valve bounce and the pressure does not remain

constant inside the cylinder. The pressure slightly increases and then decreases after

which reaches somewhat steady intake of air. This negative pressure difference is

called as the Intake Depression.




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The same occurs at the point 2, where the delivery valve delays to open. The

compressed air pressure inside the cylinder of compressor reaches a pressure slightly

more than the air receiver pressure. The delivery valve then opens causing a slight

decrease in pressure but always above the air receiver pressure enabling the

compressor to deliver air. Then it reaches a stable point after some Valve Bounce and

then reaches point 3 where the delivery valve closes to continue its cycle. Thus in

practice, the compressor requires more power to compensate for the additional work

done due to the depressions in delayed opening of the valves.

4. What are the Limi tations of a Single-Stage Air Compressor?

Refer to the enclosed diagram, the single stage air-compressor is compressing from

pressure P1 to Pressure P2, completing the cycle 1234, where 3-4 is the clearance air

expansion. Also V1-V4 is the effective swept volume or the effective volume where the

fresh air from atmosphere is sucked in. The mass of air flowing through the compressoris controlled by this effective swept volume V1-V4.

If any restriction is placed on the delivery of the air compressor, for example: the

dischargevalve throttled, then the delivery pressure of the air compressor increases.

From the diagram, let us say the new delivery pressure is P5. Then the operating cycle

will be 1567, where 6-7 is the clearance expansion of air and the effective swept volume

is V1-V7. Thus it is evident that the effective swept volume (V1-V4) is more than (V1-

V7). Thus when the delivery pressure of the single-stage air compressor is increased,

the effective swept volume is reduced.

If the delivery pressure is further increased (assuming the compressor is so strong to

work), the delivery pressure reaches P8, and the compression follows the curve 1-8,

where there will be no delivery of compressed air. Thus when the delivery pressure of a

single-stage compressor is increased, the mass flow rate also increases.


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Since the delivery pressure increases, the associated temperature also increases. Thus

the temperature of the air after compression is so high as to cause mechanical

problems and the amount of heat is actually theenergy loss.

If a single-stage machine is required to deliver a high-pressure compressed air, then itrequires

1. Heavy moving/working components to compress air to such a high pressure,

2. There might be some balancing problems due to heavy moving parts,

3. The power requirement for such heavy parts movement is too high,

4. There will high torque fluctuations,

5. To compensate for the torque fluctuations, a heavy flywheel is required.

6. Better cooling arrangements are required, and

7. Lubricating oil which does not get vaporized at such high temperatures.

A multi-stage compressor is one in which there are several cylinders of different

diameters. The intake of air in the first stage gets compressed and then it is passed

over a cooler to achieve a temperature very close to ambient air. This cooled air is

passed to the intermediate stage where it is again getting compressed and heated. This

air is again passed over a cooler to achieve a temperature as close to ambient as

possible. Then this compressed air is passed to the final or the third stage of the air

compressor where it is compressed to the required pressure and delivered to the air

receiver after cooling sufficiently in an after-cooler.

Advant ages of Mult i-s tage compress ion:

1. The work done in compressing the air is reduced, thus power can be saved

2. Prevents mechanical problems as the air temperature is controlled


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3. The suction and delivery valves remain in cleaner condition as the temperature

and vaporization of lubricating oil is less

4. The machine is smaller and better balanced

5. Effects from moisture can be handled better, by draining at each stage

6. Compression approaches near isothermal

7. Compression ratio at each stage is lower when compared to a single-stage

machine

8. Light moving parts usually made of aluminum, thus less cost and better

maintenance

Refer to the diagram of a multi-stage compressor, where it is evident that the work done

by the compressor is less when compared to a single-stage machine for same deliverypressure.

Practical Understanding of Multi -stage Compressor

We know that PV^n = C, where n is the polytropic compression index.

If we want to compress air from atmospheric pressure to a pressure of 30 bar, and say

the ambient temperature is 27 degree Celsius:

The compression index n = 1.35 and the compression ratio for single-stage compressor

would be 30:1.

Also we know that T1/T2 = (P1/P2) ^ ((n-1)/n).

Thus when calculated using the above expression, T2= 450 degree Celsius. Thus it is

evident that the delivery temperature of compressed air is 450 degree Celsius.




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Do you think of any harm to compressor will happen at this temperature?

The main issue is the lubricating oil mist and associated explosion. The lubricating oil at

this temperature will ignite and cause very severe problem. Thus if multi-staging is

used, the delivery air temperature is controlled very close to the ambient air and there isno possibility of lubricating oil and associated problems.

Why Cylinder Diameter reduces as Pressure Increases

Now let us understand a very practical issue.

Have you ever noticed in a multi-stage compressor, the diameter of cylinder liners of

each stage is different? To be more elaborate, the first stage cylinder diameter is

biggest and the intermediate stage is the second biggest and the third or final stage isthe smallest of all. Do you know the reason for this?

As the pressure of compression increases, the cylinder diameter decreases. The mass

flow rate or the amount of air passing through each stage is same. Thus as the pressure

increases with the same mass flow rate, the volume occupied by air must reduce. This

is accomplished by reducing the diameter and thus the volume of the cylinder.




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UNIT- III CONDUCTION

1. What is thermal diffusivity?

It is the ration of thermal conductivity to the thermal capacity

2. Define overall heat transfer coefficient

The overall heat transfer U gives the heat transmitted per unit area per unit time perdegree temperature difference between the bulk fluids on each side to the metal or solid.

3. Define the Fourier Law of heat conduction

The rate of flow of heat in a simple homogeneous solid is directly proposional to thearea of the section at right angle to the direction of heat flow and change of temperature with

respect to the length of the path of the heat flow.

4. What are the mechanisms in heat transfer through solids?

Latice Vibration

Transport of Free Electrons

5. Define thermal conductivity of metals

Thermal conductivity is defined as the amount of energy conducted through a body of

unit area and unit thickness in unit time when the difference in unit temperature between thefaces causing the heat flow in unit temperature difference.

6. What is thermal contact resistance?

Due to the reduced area and presence of voids, a large resistance to heat flow occurs atthe interface. This resistance is known as thermal contact resistance.

7. What ate the purpose of thermal insulation?

Prevent the heat flow from the system to surroundings

Prevent the heat flow from the surroundings to system

8. Name few common types of fin

Uniform straight fin, Tapered straight fin, Annular fin, Pin fins

9. Define fin efficiency

The efficiency of fin is defined as the ratio of the actual hat transferred by the fin to themaximum heat transferable by fin, if entire fin area were at a base temperature.

10. What are the conditions that make the fin effective?




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Thermal conductivity should be large

Convective heat transfer coefficient should be small

Thickness of the fin should be small

11. Define thermal diffusivity

Thermal diffusivity = Thermal conductivity/Thermal capacity

What are the various mode of heat transfer?

Conduction, Convection, Radiation

12. What are the factors influence the thermal conductivity?

Material structure, Moisture content, Density, Pressure and Temperature, Homogeneousmixture of components

13. Define critical thickness of insulation

The thickness up to which the heat flow increases and after which heat flow decreases istermed as critical thickness.

14. Define thermal insulation

A material, which retards the flow of heat with reasonable effectiveness, is known as

insulation.

15. Name few applications of finned surfaces

Radiator of automobiles, Air-cooled engine cylinder heads, Condenser and evaporator inrefrigeration and AC systems, Electric motor bodies and Electronic chip heat sinks

16. What are the assumptions made in the analysis of heat flow through fins?

Steady state heat conduction

No heat generation within the fin

Uniform convective heat transfer coefficient

One dimensional heat conduction

Homogeneous properties of materials

17. Define fin effectiveness

Fine effectiveness is the ratio of the fin heat transfer rate to the heat transfer rate thatwould exist without a fin.

18. What are the conditions that make the fin effective?




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Thermal conductivity should be large

Convective heat transfer coefficient should be small

Thickness of the fin should be small

PART

B

1. Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick Wall whose thermal conductivityis k =0.8 W/m C. On a Certain day, the temperatures of the inner and the outer surfaces Ofthe wall are measured to be 14C and 6C, respectively. Determine the rate of heat lossthrough the wall on that day.Assumptions1 Heat transfer through the wall is steady since the surface temperatures remainconstant at the specified values. 2 Heat transfer is one-dimensional since any significanttemperature gradients will exist in the direction from the indoors to the outdoors. 3. Thermalconductivity is constant.

Properties The thermal conductivity is given to be k= 0.8 W/mC.

AnalysisThe surface area of the wall and the rate of heat loss through the wall are

A ( (4 6 24m) m) m2

( . )( )

.Q kA

T T

L

1 2 08

14 6

03W/ m. C)(24 m

C

m

2 512 W

2. Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick wall whose thermal conductivity isk =0.8 W/m C. On a certain day, the temperatures of the inner and the outer surfaces of thewall are measured to be 14C and 6C, respectively. Determine the rate of heat loss throughthe wall on that day.assuming the space between the two glass layers is evacuated.

Assumptions 1 Heat transfer through the window is steady since the indoor and outdoortemperatures remain constant at the specified values. 2 Heat transfer is one-dimensional sinceany significant temperature gradients will exist in the direction from the indoors to theoutdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer byradiation is negligible.

Properties The thermal conductivity of the glass and air are given to be kglass= 0.78 W/mC

and kair= 0.026 W/mC.

Analysis The area of the window and the individual resistances are

A ( . ( .12 2 2 4m) m) m2

6C14C

L=0.3 m

Q

Wall

Air




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C/W2539.0

0167.01923.0)0016.0(20417.02

C/W0167.0)m4.2(C).W/m25(

11

C/W1923.0)m4.2(C)W/m.026.0(

m012.0

C/W0016.0)m4.2(C)W/m.78.0(

m003.0

C/W0417.0)m4.2(C).W/m10(

11

2,211,

o

2o22

2,o

22

2

2

21

1glass31

221

1,i

convconvtotal

conv

air

conv

RRRRR

AhRR

Ak

LRR

Ak

LRRR

AhRR

The steady rate of heat transfer through window glass then becomes

W114

C/W2539.0

C)]5(24[21

totalR

TTQ

The inner surface temperature of the window glass can be determined from

C19.2 =C/W)W)(0.0417114(C24

o1,11

1,

11conv

conv

RQTTR

TTQ

3. A cylindrical resistor element on a circuit board dissipates 0.15 W of power in anenvironment at 40C. The resistor is 1.2 cm long, and has a diameter of 0.3 cm. Assumingheat to be transferred uniformly from all surfaces, determine (a) the amount of heat thisresistor dissipates during a 24-h period, (b) the heat flux on the surface of the resistor, inW/m2, and(c) the surface temperature of the resistor for a combined convection and radiation heattransfer coefficient of 9 W/m2 C.Assumptions1Steady operating conditions exist. 2 Heat is transferred uniformly from all

surfaces of the resistor.Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is

Wh3.6 h)W)(2415.0(tQQ (b) The heat flux on the surface of the resistor is

222

m000127.0m)m)(0.012003.0(4

m)003.0(2

42

DL

DAs

2

W/m11792

m000127.0

W15.0

sA

Qq

(c) The surface temperature of the resistor can be determined from

C171

)m7C)(0.00012.W/m(1179

W15.0)(

22

s

sss

hA

QTTTThAQ

4. Heat is to be conducted along a circuit board that has a copper layer on one side. Thecircuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxylayers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces,determine the percentages of heat conduction along the copper (k =386 W/m C) and epoxy(k = 0.26 W/m C) layers. Also determine the effective thermal Conductivity of the board.

Assumptions1Steady operating conditions exist. 2 Heat transfer is one-dimensional sinceheat transfer from the side surfaces is disregarded 3 Thermal conductivities are constant.

PropertiesThe thermal conductivities are given to be k= 386 W/mC for copper and 0.26W/mC for epoxy layers.

Resistor

0.15 W

Q




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Analysis We take the length in the direction of heat transfer to beL and the width of theboard to be w. Then heat conduction along this two-layer board can be expressed as

L

Twktkt

L

TkA

L

TkAQQQ

epoxycopper

epoxycopper

epoxycopper )()(

Heat conduction along an equivalent board of thickness t = tcopper+ tepoxy and thermalconductivity keffcan be expressed as

L

Twttk

L

TkAQ

)( epoxycoppereff

board

Setting the two relations above equal to each other and solving for the effective conductivitygives

k t t kt kt k kt kt

t teff eff ( ) ( ) ( )

( ) ( )copper epoxy copper epoxy

copper epoxy

copper epoxy

Note that heat conduction is proportional to kt. Substituting, the fractions of heat conductedalong the copper and epoxy layers as well as the effective thermal conductivity of the boardare determined to be

( ) ( .

( ) ( . .

( ) ( ) ( ) . . .

( )

( )

.

..

( )

( )

.

..

kt

kt

kt kt kt

fkt

kt

fkt

kt

copper

epoxy

total copper epoxy

epoxy

epoxy

total

copper

copper

total

386 0 0386

0 26 0 000312

00386 0000312 0038912

0 000312

0 0389120008

00386

00389120992

W / m. C)(0.0001 m) W/ C

W/ m. C)(0.0012 m) W/ C

W/ C

0.8%

99.2%

andkeff

( . . . )

( . . )

386 0 0001 0 26 0 0012

00001 00012

W/ C

m29.9 W /m. C

5. A1-mm-thick copper plate (k = 386 W/m C) is sandwiched between two 5-mm-thickepoxy boards (k = 0.26 W/m C) that are 15 cm 20 cm in size. If the thermal contactconductance on both sides of the copper plate is estimated to be 6000 W/m C, determinethe error involved in the total thermal resistance of the plate if the thermal contactconductance are ignored.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional sincethe plate is large. 3 Thermal conductivities are constant.

Properties The thermal conductivities are given to be k= 386 W/mC for copper plates and

k= 0.26 W/mC for epoxy boards. The contact conductance at the interface of copper-epoxy

layers is given to be hc= 6000 W/m2C.

Analysis The thermal resistances of different layers for unitsurface area of 1 m2are

C/W00017.0)mC)(1.W/m6000(

1122

c

contact

cAh

R

Epoxy

Q

Copper

Tstepoxytcopper




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R L

kAplate 2

m

(386 W/ m. C)(1 m ) C / W

00012 6 10

6..

R L

kAepoxy 2

m

(0.26 W/ m. C)(1 m ) C / W

0005001923

..

The total thermal resistance is

C/W03914.001923.02106.200017.02

22

6

epoxyplatecontacttotal

RRRR

Then the percent error involved in the total thermal resistance of the plate if the thermalcontact resistances are ignored is determined to be

%87.0

10003914.0

00017.02100

2Error%

total

contact

R

R

which is negligible.

E ox

5 mm

E ox

5 mm

Copp

er

Q

R

R T T

R Rplate




37/56

6.A 4-m-high and 6-m-wide wall consists of a long 18-cm 30-cm cross section of horizontalbricks (k =0.72 W/m C) separated by 3-cm-thick plaster layers (k = 0.22 W/m C). Thereare also 2-cm-thick plaster layers on each side of the wall, and a 2-cm-thick rigid foam ( k=0.026 W/m C) on the inner side of the wall. The indoor and the outdoor temperatures are22C and =4C, and the convection heat transfer coefficients on the inner and the outer sides

are h1 = 10 W/m2 C and h2 =20 W/m2 C, respectively. Assuming one-dimensional heattransfer and disregarding radiation, determine the rate of heat transfer through the wall.

Assumptions1Heat transfer is steady since there is no indication of change with time. 2 Heattransfer through the wall is one-dimensional.3 Thermal conductivities are constant.4 Heattransfer by radiation is disregarded.

PropertiesThe thermal conductivities are given to be k= 0.72 W/mC for bricks, k= 0.22

W/mC for plaster layers, and k= 0.026 W/mC for the rigid foam.

Analysis We consider 1 m deep and 0.33 m high portion of wall which is representative of

the entire wall. The thermal resistance network and individual resistances are

R Rh A

R R L

kA

R R R LkA

R R R L

h A

R R

i conv

foam

plasterside

plastercenter o

,( ( . )

.

.

( . ( . ).

.( . ( . )

.

.

( . ( . ).

1

1

1

6

5

1 1

10 0 33 10303

0 02

0 026 0 33 12 33

0 020 22 0 30 1

0303

018

0 22 0 015 15455

W / m . C) m C / W

m

W / m. C) m C / W

mW / m. C) m

C / W

m

W / m. C) m C / W

2 2

2

2 2

3 2

4 brick

conv

mid

mid

total i mid o

L

kA

R Rh A

R R R RR

R R R R R R

018

0 72 0 30 10833

1 1

20 033 10152

1 1 1 1 1

5455

1

0833

1

5455081

2 0 303 2 33 2 0 303 081 01524 201

2

2

3 4 5

1 2

.

( . ( . ).

( ( . ).

. . ..

. . ( . ) . ..

,

m

W / m. C) m C / W

W / m. C) m C / W

C / W

C / W

2

o 2

The steady rate of heat transfer through the wall per 033. m2is

[( ( )]

..Q

T T

Rtotal

1 2

22 4

4 201619

C

C / WW

Then steady rate of heat transfer through the entire wall becomes

( .( )

.Qtotal

619

4 6

033W)

m

m

2

2450 W

7.A50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open

space at 15C. The average temperature of the outer surface of the pipe is measured to be150C. If the combined heat transfer coefficient on the outer surface of the pipe is 20 W/m2

R2

R3

R4R5

R6

R7

T2

R

Ri

T1




38/56

C, determine (a) the rate of heat loss from the steam pipe, (c) the thickness of fiberglassinsulation (k =0.035 W/m C) needed in order to save 90 percent of the heat lost. Assumethe pipeTemperature to remain constant at 150C.

Assumptions1Heat transfer is steady since there is no indication of any change with time. 2Heat transfer is one-dimensional since there is thermal symmetry about the center line and novariation in the axial direction. 3 Thermal conductivity is constant. 4 The thermal contactresistance at the interface is negligible. 5 The pipe temperature remains constant at about 150

C with or without insulation. 6 The combined heat transfer coefficient on the outer surfaceremains constant even after the pipe is insulated.

PropertiesThe thermal conductivity of fiberglass insulation is given to be k= 0.035 W/mC.Analysis (a) The rate of heat loss from the steam pipe is

2m71.15m)50(m)1.0( DLAo

W42,412=C)15150)(m71.15(C).W/m20()(22 airsobare TTAhQ

(c) In order to save 90% of the heat loss and thus to reduce it to 0.142,412 = 4241 W, thethickness of insulation needed is determined from

kL

rr

Ah

TT

RR

TTQ

oo

airs

insulationo

airsinsulated

2

)/ln(1 12

Substituting and solving for r2, we get

m0692.0

)m50(C)W/m.035.0(2

)05.0/ln(

)]m50(2(C)[.W/m20(

1

C)15150(W4241 2

2

22

r

r

r Then the thickness of insulation becomes

cm1.92 592.612 rrtinsulation

8.Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and maintainsthe hot water at 55C. The tank is located in a small room whose average temperature i 27C,and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and 12W/m2 C, respectively. The tank is placed in another 46-cm-diameter sheet metal tank ofnegligible thickness, and the space between the two tanks is filled with foam insulation ( k=0.03 W/m C). The thermal resistances of the water tank and the outer thin sheet metalshell are very small and can be neglected. The price of electricity is $0.08/kWh, and the homeowner pays $280 a year for water heating. Determine the fraction of the hot water energy cost

of this household that is due to the heat loss from the tank. Hot water tank insulation kitsconsisting of 3-cm-thick fiberglass insulation (k = 0.035 W/m C) large enough to wrap theentire tank are available in the market for about $30. If such an insulation is installed on thiswater tank by the home ownerhimself, how long will it take for this additional insulation to pay for itself?s

Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2Heat transfer is one-dimensional since there is thermal symmetry about the centerline and novariation in the axial direction. 3 Thermal properties are constant. 4 The thermal contactresistance at the interface is negligible. 5 Heat transfer coefficient accounts for the radiation

effects, if any.

RT

R

T




39/56

Properties The thermal conductivity of plastic cover is given to be k= 0.15 W/mC.Analysis In steady operation, the rate of heat transfer from the wire is equal to the heatgenerated within the wire,

W80)A10)(V8( VIWQ e The total thermal resistance is

C/W4051.00735.03316.0

C/W0735.0)m10(C)W/m.15.0(2

)1/2ln(

2

/ln(

C/W3316.0m)]m)(10(0.004C)[.W/m24(

11

plasticconvtotal

12plastic

2conv

RRR

kL

rrR

AhR

oo

Then the interface temperature becomes

C62.4

)C/W4051.0)(W80(C30total1

total

21 RQTTR

TTQ

The critical radius of plastic insulation is

mm25.6m00625.0C.W/m24

CW/m.15.0

2

h

k

rcr Doubling the thickness of the plastic cover will increase the outer radius of the wire to 3 mm,which is less than the critical radius of insulation. Therefore, doubling the thickness of plasticcover will increase the rate of heat loss and decrease the interface temperature.

9. Obtain a relation for the fin efficiency for a fin of constant cross-sectional area Ac,perimeter p, length L, and thermal conductivity k exposed to convection to a medium at Twith aheat transfer coefficient h. Assume the fins are sufficiently long so that the temperature of thefin at the tip is nearly T. Take the temperature of the fin at the base to be Tb and neglect heattransfer from the fin tips. Simplify the relation for (a) a circular fin of diameter D and (b)rectangular fins of thickness t..Assumptions 1 The fins are sufficiently long so that the temperature of the fin at the tip is

nearly T . 2 Heat transfer from the fin tips is negligible.

AnalysisTaking the temperature of the fin at the base to be Tb and using the heat transferrelation for a long fin, fin efficiency for long fins can be expressed as

finActual heat transfer rate from the fin

Ideal heat transfer rate from the fin

if the entire fin were at base temperature

hpkA T T

hA T T

hpkA

hpL L

kA

ph

c b

fin b

c c( )

( )

1

This relation can be simplified for a circular fin of diameter D and rectangular fin ofthickness t and width w to be

fin,circular

fin,rectangular

1 1 4 1

2

1 1

2

1

2

1

2

2

L

kA

ph L

k D

D h L

kD

h

L

kA

ph L

k wt

w t h L

k wt

wh L

kt

h

c

c

( / )

( )

( )

( )

( )

Rconv

T2

Rplastic

T1

h, T

D

Tb




40/56

10.A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heattransfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to bedetermined.

Assumptions1 Steady operating conditions exist. 2 The temperature along the fins varies inone direction only (normal to the plate). 3 Heat transfer from the fin tips is negligible. 4 Theheat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal

properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect ofradiation from the fins.

Properties The thermal conductivity of the aluminium plate and fins is given to be k= 237

W/mC.

Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of thecircular fins can be determined to be

a hp

kA

h D

k D

h

kDc

2

4

4 4 35

237 000251537

/

( )

( )( . ).

W/ m . C

W/ m. C mm

2-1

fin

-1

-1

m m

m m

tanh tanh( . . )

. ..

aL

aL

1537 003

1537 0030935

The number of fins, finned and unfinned surface areas, and heat transfer rates from thoseareas are

n 1

0006 000627 777

m

m) m)

2

( . ( .,

W2107

C)30100)(m86.0)(C.W/m35()(

W300,15

C)30100)(m68.6)(C.W/m35(935.0

)(

m86.04

)0025.0(

2777714277771

m68.64

)0025.0()03.0)(0025.0(27777

427777

22unfinnedunfinned

22

finfinmaxfin,finfinned

222

unfinned

222

fin

TThAQ

TThAQQ

D

A

DDLA

b

b

Then the total heat transfer from the finned plate becomes

kW17.4 W1074.12107300,15 4unfinnedfinnedfintotal, QQQ

The rate of heat transfer if there were no fin attached to the plate would beA

Q hA T T b

no fin2

no fin no fin2 2

m m m

W / m C m C W

( )( )

( ) ( . )( )( )

1 1 1

35 1 100 30 2450

Then the fin effectiveness becomes

7.102450

400,17

finno

finfin

Q

Q

3 cm

0.6 cmD=0.25 cm




41/56

UNIT- IV CONVECTION

1. What is dimensional analysis?

Dimensional analysis is a matehematical technique which make use of the study of

the dimensions for solving seferal engineering problems.

2. Define free convection

If the fluid motion is produced due to change in density resulting from temperaturegradient, the mode is said to be natural convection.

3. What is forced convection?

If the fluid motion is artificially created by means of external force like a blower orfan that type of heat transfer is know as forced convection.

4. What are the advantages of dimensional analysis?

It express the functional relationship between the variables in dimensional termsIt enable getting up a theoretical solution in a simplified dimensional form

5.What are the factors changes the thickness of boundary layer?

Turbulence in the ambient flowSurface roughness, Pressure gradientViscosity of the fluid and temperature difference between the surface and fluid

6. What is meant by Newtonian and Non-Newtonian flow?

The fluid, which obeys Newtons law of viscosity, is called Newtonian flow and other

type of fluid is called Non-Newtonian flow.

7. Define hydrodynamic and thermal boundary layer.

In hydrodynamic boundary layer the velocity of flow is less than 99% of free streamvelocity.

In thermal boundary layer, temperature of the fluid is less than 99% of free streamtemperature.

8. What are the limitations of dimensional analysis?

The complete information is not provided by this analysisNo information about the internal mechanism of physical phenomenonDes not give any clue regarding the selection of variables.

9.Define lower critical Reynolds number

It defines the limit below which all turbulence, no matter how severe, entering theflow from any source will eventually be damped out by viscous action.




42/56

10.Define Nusselt Number.

It is the ratio of heat transfer rate, Q to the rate at which heat would be conductedwithin the fluid under a temperature gradient.

11. Define Grashoff number

It is related with natural convection heat transfer. It is defined as the ratio of theproduct of inertia force and buoyancy force to the square of viscous force.

12. What is boundary layer?

A layer adjacent to the boundary is called boundary layer.

13. What are the factors that change the boundary layer from laminar to turbulent?

Turbulence in ambient flow, surface roughness, pressure gradient, plate curvature andtemperature difference between fluid and boundary.

14. Define skin friction coefficient.

It is defined as the ration of shear stress at the plate to the dynamic head caused byfree stream velocity.

15. Define thermal boundary layer.

In a boundary layer wherein the temperature variation exists is called the thermalboundary layer.

16. Define free convection.

Free or natural convection is the process of heat transfer which occurs due tomovement of the fluid particles by density changes associated with temperature differential ina fluid.

17. Give few examples of free convection

The cooling of transmission lines, electrical transformers and rectifiersThe heating of rooms by use of radiatorsThe heat transfer from hot pipes and ovens surrounded by cooler airCooling of reactor core.

PARTB

1. Hot engine oil flows over a flat plate. The temperature and velocity of the oil are 30 C&3

m/s respectively. The temperature of the plate is 30C. compute the total drag force and therate of heat transfer per unit width of the plate.




43/56

Assumptions1 Steady operating condition exists. 2 The critical Reynolds number is Recr=

5105. 3 Radiation effects are negligible.

PropertiesThe properties of engine oil at the film temperature of (Ts + T)/2 = (80+30)/2

=55C = 328 K are (Table A-13)

1505PrCW/m.141.0

/sm10123kg/m867 263

k

AnalysisNoting thatL = 6 m, the Reynolds number at the end of the plate is

Re(

.LL

V

3

123 10146 10

6

5m/ s)(6 m)

m / s2

which is less than the critical Reynolds number. Thus we havelaminar flow over the entire plate. The average frictioncoefficient and the drag force per unit width are determinedfrom

N3.812

m/s))(3kg/m867(

)m16)(00347.0(2

00347.0)1046.1(328.1Re328.1

232

2

5.055.0

VsfD

Lf

ACF

C

Similarly, the average Nusselt number and the heat transfercoefficient are determined using the laminar flow relations fora flat plate,

Nu hL

k

h k

LNu

L

0664 0664 146 10 1505 2908

0141

62908 683

0 5 1 3 5 0 5 1 3. Re Pr . ( . ) ( )

.( ) .

. / . /

W / m. C

mW/ m . C

2

The rate of heat transfer is determined from Newton's law of cooling

3. Wind is blowing parallel to the wall of a house. The temperature and velocity of the air is5oC& 55 km/hr. Calculate the rate of heat loss from the wall.

Assumptions 1 Steady operating condition exists. 2 The critical Reynolds number is Recr=

5105. 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties.

Properties The properties of air at 1 atm and the film

temperature of (Ts+ T)/2 = (12+5)/2 = 8.5C

7340.0Pr

/sm10413.1

CW/m.02428.0

25-

k

AnalysisAir flows parallel to the 10 m side:

Ts= 30C

Oil

V

= 3 m/s

T

= 30C

L = 6 m

Ts= 12

C

Air

V

= 55 km/h

T

= 5C

L




44/56

The Reynolds number in this case is

7

2510081.1

/sm10413.1

m)m/s](10)3600/100055[(Re

L

L

V

which is greater than the critical Reynolds number. Thus we have combined laminar andturbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient andthen heat transfer rate are determined to be

C.W/m43.32)10336.1(m10

CW/m.02428.0

10336.1)7340.0](871)10081.1(037.0[Pr)871Re037.0(

24

43/18.073/18.0

NuL

kh

k

hLNu L

kW9.08

W9081C5)-)(12mC)(40.W/m43.32()(

m40=m)m)(104(

22

2

ss

s

TThAQ

wLA

If the wind velocity is doubled:

7

2510163.2

/sm10413.1

m)m/s](10)3600/1000110[(Re

LL

V

which is greater than the critical Reynolds number. Thus we have combined laminar andturbulent flow. Using the proper relation for Nusselt number, the average heat transfercoefficient and the heat transfer rate are determined to be

C.W/m88.57)10384.2(m10

CW/m.02428.0

10384.2)7340.0](871)10163.2(037.0[Pr)871Re037.0(

24

43/18.073/18.0

NuL

kh

k

hLNu L

kW16.21

W206,16C5)-)(12mC)(40.W/m88.57()(

m40=m)m)(410(

22

2

ss

s

TThAQ

wLA

4. Air is flowing over the steam pipe having steam temperature of 90 oC .The velocityand temperature of the air are 70Cand 50 km/ hr respectively. Calculate rate of heatloss by the air on the steam pipe

Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air isan ideal gas with constant properties.

PropertiesThe properties of air at 1 atm and the film temperature of (T s+ T)/2 = (90+7)/2

= 48.5C are (Table A-15)

7232.0Pr

/sm10784.1

CW/m.02724.0

25-

k

Analysis The Reynolds number is

Air

V

= 50 km/h

T

= 7C

Pipe

D = 8 cm

T= 90C




45/56

425

10228.6/sm10784.1

m)(0.08]s/h)0m/km)/(3601000(km/h)(50[Re

DV

The Nusselt number corresponding to this Reynolds number is

1.159

000,282

10228.61

7232.0/4.01

)7232.0()10228.6(62.03.0

000,282

Re1

Pr/4.01

PrRe62.03.0

5/48/5

4

4/13/2

3/15.04

5/48/5

4/13/2

3/15.0

k

hDNu

The heat transfer coefficient and the heat transfer rate become

C.W/m17.54)1.159(m08.0

CW/m.02724.0 2

NuD

kh

length)m(per=C7)-)(90mC)(0.2513.W/m17.54()(

m0.2513=m)m)(108.0(

22

2

W1130

TThAQ

DLA

ssconv

s

5. The components of an electronic system located in a horizontal square duct (20cm20 cm)is cooled by air flowing over the duct. The velocity and temperature of the air are 200 m/min& 300C.Determine the total power rating of the electronic device.

Assumptions 1 Steady operating condition exists. 2 Radiation effects are negligible. 3 Air isan ideal gas with constant properties. 4 The local atmospheric pressure is 1 atm.


temperature of (Ts+ T)/2 = (65+30)/2 = 47.5C are

7235.0Pr/sm10774.1

CW/m.02717.0

25-

k

AnalysisThe Reynolds number is

425

10758.3/sm10774.1

m)(0.2m/s(200/60)Re

DV

Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be

2.112)7235.0()10758.3(102.0PrRe102.03/1675.043/1675.0

k

hDNu

The heat transfer coefficient is

C.W/m24.15)2.112(m2.0

CW/m.02717.0 2 NuD

kh

6. Water at 15C is to be heated to 65C by passing it over a bundle of 4-m-long 1-cm-diameter resistance heater rods maintained at 90C. Water approaches the heater rod bundlein normal direction at a mean velocity of 0.8 m/s. The rods arc arranged in-line withlongitudinal and transverse pitches of SL= 4 cm and ST= 3 cm. Determine the number of tuberowsNL in the flow direction needed to achieve the indicated temperature rise.

Assumptions 1 steady operating condition exists. 2 The surface temperature of the rods is

constant.PropertiesThe properties of water at the mean temperature of (15C +65C)/2=40C

Air

30C

200 m/min

20 cm

65C




46/56

k = 0.631 W/m-K = 992.1 kg/m3

Cp=4.179 kJ/kg-K Pr = 4.32

= 0.65310-3kg/m-s Prs= Pr@ Ts= 1.96

Also, the density of water at the inlet temperature of 15C (for use in the mass flow rate

calculation at the inlet) is i=999.1 kg/m

3

.

AnalysisIt is given thatD= 0.01 m, SL= 0.04 m and ST= 0.03 m, and V= 0.8 m/s.

Then the maximum velocity and the Reynolds number

based on the maximum velocity become

m/s20.1m/s)8.0(01.003.0

03.0max

VVDS

S

T

T

232,18skg/m10653.0

m)m/s)(0.0120.1)(kg/m1.992(Re

3

3max

DD

V

The average Nusselt number is determined using

the proper relation from Table 7-2 to be

3.269)96.1/32.4()32.4()232,18(27.0

)Pr(Pr/PrRe27.0Nu

25.036.063.0

25.036.063.0

sDD

Assuming that NL> 16, the average Nusselt number and heat transfer coefficient for all thetubes in the tube bank become

3.269NuNu , DND L

CW/m994,16m0.01

C)W/m631.0(3.269 2,

D

kNuh L

ND

Consider one-row of tubes in the transpose direction (normal to flow), and thus take NT=1.Then the heat transfer surface area becomes

LLtubes NNDLNA 1257.0m)m)(401.0()1( Then the log mean temperature difference, and the expression for the rate of heat transfer

become

C51.45)]6590/()1590ln[(

)6590()1590(

)]/()ln[(

)()(ln

esis

esis

TTTT

TTTTT

LLs NNThAQ 220,97)C51.45()C)(0.1257W/m994,16(2

ln The mass flow rate of water through a cross-section corresponding to NT=1 and the rate ofheat transfer are

kg/s91.95m/s))(0.8m0.03)(4kg/m1.999( 23 VAm c W10004.2C)1565(J/kg.C)9kg/s)(41791.95()( 7 iep TTCmQ

Substituting this result into the heat transfer expression above we find th e number of tuberows

206 LLs NNThAQ 220,97W10004.27

ln

7.Cooling water available at 10C is used to condense steam at 30C in the condenser of apower plant at a rate of 0.15 kg/s by circulating the cooling water through a bank of 5-m-long

1.2-cm-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of 4m/s, and leaves at a temperature of 24C. The tubes are nearly isothermal at 30C.Determine

SL

ST

V=0.8 m/s

Ti=15C

Ts=90C




47/56

the average heat transfer coefficient between the water and the tubes, and the number of tubesneeded to achieve the indicated heat transfer rate in the condenser.

Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe isconstant. 3 The thermal resistance of the pipe is negligible.

Properties The properties of water at the average temperature of (10+24)/2=17C are

CJ/kg.5.4184

kg/m7.998 3

pC

Also, the heat of vaporization of water at 30C iskJ/kg2431fgh .

Analysis The mass flow rate of water and the surface area are

kg/s0.4518=m/s)(44

m)(0.012)kg/m7.998(

4

23

2

mmc

DAm VV

The rate of heat transfer for one tube is

W468,26)C1024)(CJ/kg.5.4184)(kg/s4518.0()( iep TTCmQ The logarithmic mean temperature difference and the surface area are

C63.11

1030

2430ln

1024

ln

ln

is

es

ie

TT

TT

TTT

2

m0.1885=m)m)(5012.0( DLAs The average heat transfer coefficient is determined from

C.kW/m12.12

W1000

kW1

)C63.11)(m1885.0(

W468,262

lnln

TA

QhThAQ

ss

The total rate of heat transfer is determined from

kW65.364kJ/kg)2431)(kg/s15.0( fgcondtotal hmQ Then the number of tubes becomes

13.8W468,26

W650,364

Q

QN totaltube

9. Water is to be heated from 10C to 80C as it flows through a 2-cm-internal-diameter, 7-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform

heating throughout the surface of the tube. The outer surface of the heater is well insulated, sothat in steady operation all the heat generated in the heater is transferred to the water in the

Steam, 30

C

L = 5 m

D = 1.2 cm

Water

10

C

4 m/s

24

C




48/56

tube. If the system is to provide hot water at a rate of 8 L/min, determine the power rating ofthe resistance heater. Also, estimate the inner surface temperature of the pipe at the exit .

Assumptions1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The innersurfaces of the tube are smooth.

PropertiesThe properties of water at the average

temperature of (80+10) / 2 = 45C are (Table A-9)

91.3Pr

CJ/kg.4180

/sm10602.0/

CW/m.637.0

kg/m1.990

26-

3

pC

k

AnalysisThe power rating of the resistance heater is

kg/s132.0kg/min921.7)/minm008.0)(kg/m1.990( 33 Vm

W38,627 C)1080)(CJ/kg.4180)(kg/s132.0()( iep TTCmQ

The velocity of water and the Reynolds number are

Vmc

V

A

(8 / )

( . /.

10 60

002 404244

3m / s

m)m/ s

3

2

101,14/sm10602.0

m)m/s)(0.02(0.4244Re

26

hmDV

which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in thiscase are roughly

m20.0m)02.0(1010 DLL th

which is much shorter than the total length of the duct. Therefore, we can assume fully

developed turbulent flow in the entire duct, and determine the Nusselt number from79.82)91.3()101,14(023.0PrRe023.0 4.08.04.08.0

k

hDNu h

Heat transfer coefficient is

C.W/m2637)79.82(m02.0

CW/m.637.0 2

NuD

kh

h

Then the inner surface temperature of the pipe at the exit becomes

C113.3

es

s

eess

T

T

TThAQ

,

2

,

C)80)](m7)(m02.0()[C.W/m2637(W627,38

)(

Water

10

C3 m/s 80

C

(Resistance heater)

L

D = 2 cm




49/56

10. Water is boiling in a 12-cm-deep pan with an outer diameter of 25 cm that is placed ontop of a stove. The ambient air and the surrounding surfaces are at a temperature of 25C, andthe emissivity of the outer surface of the pan is 0.95. Assuming the entire pan to be at anaverage temperature of 98C, determine the rate of heat loss from the cylindrical side surfaceof the pan to the surroundings by (a) natural convection and (b) radiation. (c) If water is

boiling at a rate of 2 kg/h at 100C, determine the ratio of the heat lost from the side surfacesof the pan to that by evaporation of water. The heat of vaporization of water at 100C is 2257kJ/kg.

Assumptions 1 Steady operating conditions exist. 2 Air is anideal gas with constant properties. 3 The local atmospheric

pressure is 1 atm.

PropertiesThe properties of air at 1 atm and the film temperature

of (Ts+T)/2 = (98+25)/2 = 61.5C are (Table A-15)

1-

25

K00299.0K)2735.61(

11

7198.0Pr

/sm10910.1

CW/m.02819.0

fT

k

Analysis (a) The characteristic length in this case is the height of the pan, m.12.0LLc

Then,

6

225

3-12

2

3

10299.7)7198.0()/sm10910.1(

)m12.0)(K2598)(K00299.0)(m/s81.9(Pr

)(

LTTgRa

s

We can treat this vertical cylinder as a vertical plate since

4/14/164/1

35and thus0.25


50/56

2

2

m09425.0)m12.0)(m25.0(

C.W/m720.6)60.28(m12.0

CW/m.02819.0

DLA

NuL

kh

s

and

W46.2 C)2598)(m09425.0)(C.W/m720.6()(22

TThAQ ss

(b) The radiation heat loss from the pan is

W5.9

444282

44

)K27325()K27398().KW/m1067.5)(m09425.0)(10.0(

)( surrssrad TTAQ

(c) The heat loss by the evaporation of water is

W1254kW254.1)kJ/kg2257)(kg/s3600/2( fghmQ

Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation ofwater then becomes

4.2%

042.01254

9.52.46f

11. Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermalconductivity k =0.78 W/m C, and emissivity 0.9. The room and the walls that face thewindow are maintained at 25C, and the average temperature of the inner surface of thewindow is measured to be 5C. If the temperature of the outdoors is 5C, determine (a) theconvection heat transfer coefficient on the inner surface of the window, (b) the rate of totalheat transfer through the window, and (c) the combined natural convection and radiation heattransfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermalresistance of the glass in this case?

Assumptions 1 Steady operating conditions exist. 2 Air is anideal gas with constant properties. 3 The local atmospheric

pressure is 1 atm.


temperature of (Ts+T)/2 = (5+25)/2 = 15C are (Table A-15)

Q

Outdoors

-5

C

Glass

Ts= 5C

= 0.9

L = 1.2 m

Room

T

= 25

C




51/56

1-

25

K003472.0K)27315(

11

7323.0Pr

/sm10471.1

CW/m.02476.0

fT

k

Analysis (a) The characteristic length in this case is the height of the window, m.2.1LLc

Then,

9

225

3-12

2

3

10986.3)7323.0()/sm10471.1(

)m2.1)(K525)(K34720.0)(m/s81.9(Pr

)(

0

cs LTTgRa

7.189

7323.0

492.01

)10986.3(387.0825.0

Pr

492.01

Ra387.0825.0

2

27/816/9

6/19

2

27/816/9

6/1

Nu

2m4.2m)m)(22.1(

)7.189(m2.1

CW/m.02476.0

sA

NuL

kh C.W/m3.915 2

(b) The sum of the natural convection and radiation heat transfer from the room to thewindow is

W9.187C)525)(m4.2)(C.W/m915.3()(22

convection ss TThAQ

W3.234])K2735()K27325)[(.KW/m1067.5)(m4.2)(9.0(

)(

444282

44radiation

ssurrs TTAQ

W422.2 3.2349.187radiationconvectiontotal QQQ

(c) The outer surface temperature of the window can be determined from

C65.3)m4.2)(CW/m.78.0(

)m006.0)(W346(C5)(

2

total,,,,total

s

isososiss

kA

tQTTTT

t

kAQ

Then the combined natural convection and radiation heat transfer coefficient on the outerwindow surface becomes

C.W/m20.35 2

C)]5(65.3)[m4.2(

W346

)(or

)(

2,,

totalcombined

,,combinedtotal

ooss

ooss

TTA

Qh

TTAhQ

Note that T QR and thus the thermal resistance R of a layer is proportional to thetemperature drop across that layer. Therefore, the fraction of thermal resistance of the glass isequal to the ratio of the temperature drop across the glass to the overall temperaturedifference,

4.5%)(or045.0)5(25

65.35

total

glass

total

glass

TR

T

R

R

which is low. Thus it is reasonable to neglect the thermal resistance of the glass.




52/56


53/56

10. Define irradiation and radiosity

Irradiation is defined as the total radiation incident upon a surface per unit time per unitareaRadiosity is used to indicate the total radiation leaving a surface per unit time per unit

area.

PART- B

1.Determine the view factorsF13andF23between the rectangular surfaces

Assumptions The surfaces are diffuse emitters and reflectors.

Analysis From Fig.

24.0

5.02

11

5.02

1

31

3

F

W

L

W

L

and

29.0

12

2

5.02

1

)21(321

3

F

W

LL

W

L

We note thatA1=A3. Then the reciprocity and superposition rules gives

0.24 3113313131A FFFAF

05.024.029.0 32323231)21(3 FFFFF

Finally, 0.05 322332 FFAA

2.Consider a 4-m 4-m 4-m cubical furnace whose floor and ceiling are black and whoseside surfaces are reradiating. The floor and the ceiling of the furnace are maintained attemperatures of 550 K and 1100 K, respectively. Determine the net rate of radiation heattransfer between the floor and the ceiling of the furnace.

Assumptions 1 Steady operating condition exist 2 The surfaces are opaque, diffuse, and gray.3 Convection heat transfer is not considered.

PropertiesThe emissivities of all surfaces are = 1 since they are black or reradiating.

Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the sidesurfaces to be surface 3. The furnace can be considered to be three-surface enclosure with aradiation network shown in the figure. We assume that steady-state conditions exist. Sincethe side surfaces are reradiating, there is no heat transfer through them, and the entire heatlost by the ceiling must be gained by the floor. The view factor from the ceiling to the floor ofthe furnace is F12 02 . . Then the rate of heat loss from the ceiling can be determined from

1

231312

211

11

RRR

EEQ

bb

where

W= 2 m

(2)L 2= 1 m

L 1= 1 m

L3= 1 mA3 (3)

A2

A1 (1)

T2= 550 K

2= 1

T1= 1100 K

1= 1

Reradiatingside surfacess

a= 4 m




54/56

244284

22

24428411

W/m5188)K550)(K.W/m1067.5(

W/m015,83)K1100)(K.W/m1067.5(

TE

TE

b

b

and

A A1 22

4 16 ( )m m2

RA F

R RA F

12

1 12

13 23

1 13

1 116 0 2

03125

1 1

16 080 078125

( )( . ).

( )( . ).

mm

mm

2-2

2

-2

Substituting,

kW747

W1047.7

)m078125.0(2

1

m3125.0

1

W/m)5188015,83( 51

2-2-

2

12Q

3.A thin aluminum sheet with an emissivity of 0.15 on both sides is placed between two verylarge parallel plates, which are maintained at uniform temperatures T1=900 K and

T2 =650 K and have emissivities 0.5 and 2 ,0.8, respectively. Determine the net rate ofradiation heat transfer between the two plates per unit surface area of the plates and Comparethe result with that without the shield.

Assumptions1Steady operating conditions exist 2The surfaces are opaque, diffuse, andgray. 3Convection heat transfer is not considered.

PropertiesThe emissivities of surfaces are given to be 1= 0.5, 2= 0.8, and 3= 0.15.AnalysisThe net rate of radiation heat transfer with a thin aluminum shield per unit area ofthe plates is

2W/m1857

115.0

1

15.0

11

8.0

1

5.0

1

])K650()K900)[(KW/m1067.5(

111111

)(

44428

2,31,321

42

41

shieldone,12

TTQ

The net rate of radiation heat transfer between the plates in the case of no shield is

244428

21

42

41

shield,12 W/m035,12

18.0

1

5.0

1

])K650()K900)[(KW/m1067.5(

111

)(

TTQ no

Then the ratio of radiation heat transfer for the two cases becomes

,

,

,

Q

Q

12

12

1857

12 035

one shield

no shield

W

W

1

6

T2= 650 K

2= 0.8

T1= 900 K

1= 0.5

Radiation

shield




55/56

4.Two very large parallel plates are maintained at uniform temperatures of T1=1000 K andT2 = 800 K and have emissivities of 1 ,2 and 0.2, respectively. It is desired to reduce the netrate of radiation heat transfer between the two plates to one-fifth by placing thin aluminumsheets with anemissivity of 0.15 on both sides between the plates. Determine the number ofsheets that need to be inserted.

Assumptions1Steady operating conditions exist 2The surfaces are opaque, diffuse, andgray. 3Convection heat transfer is not considered.

PropertiesThe emissivities of surfaces are given to be 1= 0.2, 2= 0.2, and 3= 0.15.

AnalysisThe net rate of radiation heat transfer between the plates in the case of no shield is

2

44428

21

42

41

shield,12

W/m3720

12.0

1

2.0

1

])K800()K1000)[(KW/m1067.5(

111

)(

TTQ no

The number of sheets that need to be inserted in order to reduce the net rate of heat transfer

between the two plates to onefifth can be determined from

3

92.2

115.0

1

15.0

11

2.0

1

2.0

1

])K800()K1000)[(KW/m1067.5()W/m(3720

5

1

111

111

)(

shield

shield

444282

2,31,3

shield

21

42

41

shields,12

N

N

N

TTQ

5. A2-m-internal-diameter double-walled spherical tank is used to store iced water at 0C.Each wall is 0.5 cm thick, and the 1.5-cm-thick air space between the two walls of the tank is

evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated spaceare polished so that each surface has an emissivity of 0.15. The temperature of the outerwall

T2= 800 K

2= 0.2

T1= 1000 K

1= 0.2

Radiation shields3= 0.15




56/56

of the tank is measured to be 20C. Assuming the inner wall of the steel tank to be at 0C,determine (a) the rate of heat transfer to the iced water in the tank and ( b) the amount of iceat 0C that melts during a 24-h period.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and

gray.PropertiesThe emissivities of both surfaces are given to be 1= 2= 0.15.Analysis(a) Assuming the conduction resistance s of the walls to be negligible, the rate ofheat transfer to the iced water in the tank is determined to be

m m2A D1 12 22 01 12 69 ( . ) .

W107.4

2

444282

2

2

1

2

2

1

41

421

12

04.201.2

15.015.01

15.01

])K2730()K27320)[(KW/m1067.5)(m69.12(

11

)(

D

D

TTAQ

(b) The amount of heat transfer during a 24-hour period is

kJ9275)s360024)(kJ/s1074.0( tQQ The amount of ice that melts during this period then becomes

kg27.8kJ/kg7.333

kJ9275

if

ifh

QmmhQ

D2= 2.04 m

T2= 20C

2= 0.15

D1= 2.01 m

T1= 0

C1= 0.15

Vacuum

Icedwater

0C


at2251 atht notes.pdf

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