prof. r. shanthini 28 may 2012 content of lectures 13 to 17: evaporation: -factors affecting...

Prof. R. Shanthini 28 May 2012

Content of Lectures 13 to 17:

Evaporation: - Factors affecting evaporation- Evaporators - Film evaporators- Single effect and multiple effect evaporators- Mathematical problems on evaporation

PM3125: Lectures 13 to 15

Principal reference: Chapter 8 in C.J. Geankoplis, Transport Processes and Unit Operations, 3rd Edition, Prentice-Hall of India


EvaporationEvaporation

- Suppose that we have a dilute solution of a solute (say, sugar) dissolved in a solvent (say, water)

- We need to remove part of the solvent (water) to produce a concentrated solution

- This can be achieved by heating the solution so as to evaporate the solvent (water)

- This process is known as evaporation.

- The industrial equipment used for this purpose in known as an evaporator.


Types of evaporators

Open kettle or pan evaporator:




Concentrate

Condensate

Pan

Boiler

Jacket

Pressure gauge

Steam




- simplest form of evaporators

- inexpensive

- simple to operate

- very poor heat economy

- in some cases paddles and scrapers for agitation are used



Horizontal-tube evaporator:

Dilute feed

Condensate

Concentratedproduct

Vapour

Steam inlet



Horizontal-tube evaporator:- relatively cheap

- used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales

- poor liquid circulation (and therefore unsuitable for viscous liquids)



Vertical-type short-tube evaporator: - Liquid is inside the tubes

- Steam condenses outside the tubes

- used for non-viscous liquids having high heat-transfer coefficients and liquids that do not deposit scales



Vertical-type short-tube evaporator:

Dilute feed

Condensate

Concentratedproduct

Vapour

Steam inlet



Falling-film-type evaporator:



More types are given in additional handouts uploaded at the course website.


Factors effecting evaporation:

Concentration in the liquid:

- Liquid feed to an evaporator is relatively dilute.

- So its viscosity is low, and heat-transfer coefficient high.

- As evaporation proceeds, the solution becomes concentrated.

- So viscosity increases and heat-transfer coefficient drops.

- Density and the boiling point of solution also increase.



Solubility:

- As solution is heated, concentration of the solute in the solution increases.

- In case the solubility limit of the solute in solution is exceeded, then crystals may form.

- Solubility of the solute therefore determines the maximum concentration of the solute in the product stream.

- In most cases, the solubility of the solute increases with temperature. This means when a hot concentrated solution from an evaporator is cooled to room temperature, crystallization may occur.



Temperature sensitivity of materials:

- Pharmaceuticals products, fine chemicals and foods are damaged when heated to moderate temperatures for relatively short times.

- So special techniques are employed to reduce temperature of the liquid and time of heating during evaporation



Foaming and frothing:

- Solutions like organic compounds tend to foam and froth during vaporization.

- The foam is carried away along with vapor leaving the evaporator.

- Entrainment losses occur.



Pressure and temperature:

- The boiling point of the solution is related to the pressure of the system.

- The higher the operating pressure of the evaporator, the higher the temperature at boiling.

- Also, as the concentration of the dissolved material in solution increases by evaporation, the temperature of boiling may rise (a phenomenon known as boiling point rise/elevation).

- To keep the temperatures low in heat-sensitive materials, it is often necessary to operate under atmospheric pressure (that is, under vacuum).



Scale deposition:

- Some solutions deposit solid materials (called scale) on the heating surfaces.

- The result is that the overall heat-transfer coefficient (U) may drastically decrease, leading to shut down of the evaporators for cleaning purposes.



Materials of construction:

- Evaporators are made of some kind of steel.

- However many solutions attack ferrous metals and are contaminated by them.

- Copper, nickel, stainless steels can also be used.


Method of operation of evaporators

- When a single evaporator is used ,the vapor from the boiling liquid is condensed and discarded. This is called single effect evaporation.

- It is simple but utilizes steam ineffectively.

- To evaporate 1 kg of water from the solution we require 1-1.3 kg of steam.

- Increasing the evaporation per kg of steam by using a series of evaporators between the steam supply and condenser is called multiple effect evaporation

Single-effect evaporation:

Multiple-effect evaporation:


Calculation methods for single-effect evaporators

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1

Feed: F – mass flow ratexF – mass fraction of solute in feed

TF – temperature of feed

hF – enthalpy of feed

Vapour leaving the evaporator: V – mass flow rateyV – mass fraction of solute in vapour

T1 – temperature of vapour

HV – enthalpy of vapour

Concentrate leaving the evaporator: L – mass flow ratexL – mass fraction of solute in concentrate

T1 – temperature of concentrate

hL – enthalpy of concentrate

Steam: S – mass flow ratePS – steam pressure

TS – steam temperature

HS – enthalpy of steam

hS – enthalpy of condensate

P – pressure in the evaporatorT1 – temperature in the evaporator



Overall material balance:

F = L + V

Solute balance:

F xF = L xL + V yV

If the vapour is free of solute:

F xF = L xL

Heat balance:

F hF + S HS = L hL + V HV + S hS

Rewriting:

F hF + S (HS - hS) = L hL + V HV

F hF + S λ = L hL + V HV

where λ = HS - hS

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1



Energy lost by the steam

q = S λ = S (HS – hS)

In case of no energy loss to the environment, q amount of energy gets transferred from steam to the solution through the tube wall of area A and overall heat transfer coefficient U.

Therefore,

q = U A ΔT = U A (TS – T1)

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1


Example 1:

A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt solution entering at 38ºC to a final concentration of 1.5 wt %.

The vapor space of the evaporator is at 101.325 kPa (1.0 atm abs) and the steam supplied is saturated at 150 kPa. The overall coefficient U = 1704 W/m2.K.

Calculate the amounts of vapor and liquid products and the heat-transfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water.



Data provided:

F = 9072 kg/h

xF = 1 wt % = 0.01 kg solute / kg feed

TF = 38ºC

xL = 1.5 wt %

= 0.015 kg solute / kg liquid product

P = 101.325 kPa (1.0 atm abs)

PS = 150 kPa

U = 1704 W/m2.K

T1 = saturated temperature at P (= 101.325 kPa) = 100ºC

TS = saturated temperature at 150 kPa = 111.4ºC

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1



Overall material balance:

F = L + V

Solute balance:

F xF = L xL (no solute in the vapour)

Heat balance:


where λ = HS – hS

Data provided:

F = 9072 kg/h

xF = 0.01 kg solute / kg feed

TF = 38ºC

xL = 0.015 kg solute / kg liquid product

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K q = S λ = U A ΔT = U A (TS – T1)

Amounts of vapor and liquid products = ?

F, xF and xL are known, and therefore

L = 6048 kg/h and V = 3024 kg/h

Available equations:



Heat balance:


where λ = HS – hS

Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K

q = S λ = U A ΔT = U A (TS – T1)

Heat transfer area A = S λ / U (TS – T1) = ?

Available equations:

S λ = L hL + V HV – F hF

= (F – V) hL + V HV – F hF

= F (hL – hF) + V (HV – hL )

= F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa )



Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K


S λ = F Cp (T1 - TF) + V (Latent heat of vapourization at 101.325 kPa)

F, T1 , TF and V are already known.

Cp = 4.14 kJ/kg.K (assumed)

Latent heat of vapourization at 101.325 kPa = 2256.7 kJ/kg

Therefore,

S λ = (9072) (4.14) (100 – 38) +(3024) (2256.7) kJ/h

= 9152862 kJ/h



Data known:

F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h

TF = 38ºC

P = 101.325 kPa; T1 = 100ºC

PS = 150 kPa; TS = 111.4ºC

U = 1704 W/m2.K


S λ = 9152862 kJ/h = 9152862 * 1000 / 3600 W

T1 and TS are known

U = 1704 W/m2.K

Therefore,

A = S λ / U (TS – T1)

= [9152862 * 1000 / 3600] / [1704 * (111.4 – 100)]

= 130. 9 m2


Effects of processing variables on evaporator operation:

Effect of feed temperature:

- The inlet temperature of the feed has a large effect on the evaporator operation.

- When feed is not at its boiling point, steam is needed first to heat the feed to its boiling post and then to evaporate it.

- Preheating the feed can reduce the size of evaporator heat-transfer area.



Effect of pressure:

- Pressure in the evaporator sets the boiling point of the solution (T1).

- Steam pressure determines the steam temperature (Ts)

- Since q = U A (TS – T1), larger values of (TS – T1) will help reduce the heat-transfer area needed and hence the cost of evaporator.

- Vacuum can be maintained in the solution side using a vacuum pump.

- For example, if the pressure in Example 1 is reduced to 41.4 kPa, boiling point of water reduces to 349.9 K and that would increase the (TS – T1) from 10 K to 33.3 K. A large decrease in heat-transfer area

would be obtained.



Effect of steam pressure:

- High pressure provides high Ts values, and hence TS – T1 will increase.

- High pressure steam is however more costly.

- Therefore, overall economic balances must be considered to determine the optimum steam pressure.


Boiling point rise of solutions:

- In example 1, the solution is assumed to be dilute enough to be considered to have the same thermal properties as water. It is not true always.

- For concentrated solutions, heat capacity and boiling point are quire different from that of water.

- Duhring’s rule is an empirical law that relates the boiling point of a solution to the boiling point of the solvent at different pressures for a solution of given concentration.


Boiling point rise of solutions (an example):

-

Duhring plot for boiling point of sodium chloride solutions


Enthalpy-concentration charts of solutions:

-

See the handout.


Example 2:

An evaporator is used to concentrate 4536 kg/h of a 20% NaOH solution entering at 60ºC to a product of 50% solids. The pressure of the saturated steam used is 170 kPa and the vapor space pressure of the evaporator is at 12 kPa. The overall coefficient U is 1560 W/m2.K.

Calculate the steam used, the steam economy (in kg vapourized / kg steam used) and the heating surface area.



Data provided:

F = 4536 kg/h


TF = 60ºC

xL = 50 wt %


P = 12 kPa = 0.12 bar

PS = 170 kPa = 1.7 bar

U = 1560 W/m2.K

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1

T1 ≠ saturated temperature at P (= 0.12 bar) = 49.4oC

TS = saturated steam temperature at 1.7 bar = 115.2oC


Calculation methods for single-effect evaporators Amounts of vapor and liquid products = ?

F, xF and xL are known, and therefore L = 1814 kg/h and V = 2722 kg/h

Steam used = ?

S λ = L hL + V HV - F hF

hF = enthalpy of 20% solution at 60oC = 214 kJ/kg

(using the enthalpy–concentration chart) hL = enthalpy of 50% solution at T1 = ?

(using the enthalpy–concentration and boiling-point rise charts)

Saturated temperature at P (= 0.12 bar) = 49.4oCUsing the boiling-point rise chart, we get 89.5oC ( read against 49.4oC and 50 wt%) as the boiling point of the solution. That is T1 = 89.5oC

hL = enthalpy of 50% solution at 89.5oC = 505 kJ/kg

(using the enthalpy–concentration chart)HV = enthalpy of superheated steam at 89.5oC and 0.12 bar = 2667 kJ/kg

(using the superheated steam table)




Steam used = S = ?

S λ = 1814 x 505 + 2722 x 2667 – 4536 x 214 = 7204940 kJ/h

λ = latent heat of vapourization of water at 1.7 bar and 115.2oC

= 2216 kJ/kg (using the steam table)

Therefore S = 7204940 / 2216 kg/h = 3251 kg/h

Steam economy = kg vapourized / kg steam used = V / S = ?

Steam economy = 2722 / 3251 = 0.837

A = S λ / U (TS – T1) = [7204940 * 1000 / 3600] / [1560 * (115.2 – 89.5)]

= 49. 9 m2


Example 3 (Repeat Example 2 assuming that the thermal properties of

the liquid in the evaporator can be approximated by those of water):

An evaporator is used to concentrate 4536 kg/h of a 20% NaOH solution entering at 60ºC to a product of 50% solids. The pressure of the saturated steam used is 170 kPa and the vapor space pressure of the evaporator is at 12 kPa. The overall coefficient U is 1560 W/m2.K.

Calculate the steam used, the steam economy (in kg vapourized / kg steam used) and the heating surface area.



Data provided:

F = 4536 kg/h


TF = 60ºC

xL = 50 wt %


P = 12 kPa = 0.12 bar

PS = 170 kPa = 1.7 bar

U = 1560 W/m2.K

Steam, SPS, TS, HS

Feed, F

xF, TF, hF

Condensate, S

PS, TS, hS

Vapour, V

yV, T1, HV

Concentrate, L

xL, T1, hL

P

T1

T1 = saturated temperature at P (= 0.12 bar) = 49.4oC

TS = saturated steam temperature at 1.7 bar = 115.2oC




Steam used = ?

S λ = L hL + V HV - F hF = (F – V) hL + V HV – F hF = F (hL – hF) + V (HV – hL )

= F Cp (T1 - TF) + V (Latent heat of vapourization at 0.12 bar)

= (4536) (4.14) (49.4 – 60) +(2722) (2383) kJ/h (Cp = 4.14 kJ/kg.K is assumed)

= 6301078 kJ/h

λ = latent heat of vapourization of water at 1.7 bar = 2216 kJ/kg

Therefore S = 6301078 / 2216 kg/h = 2843.5 kg/h

Steam economy = kg vapourized / kg steam used = V / S = ?

Steam economy = 2722 / 2843.5 = 0.957

A = S λ / U (TS – T1) = [6301078 * 1000 / 3600] / [1560 * (115.2 – 49.4)]

= 17 m2



Compare the solutions of Examples 2 and 3 and discuss the importance of considering the boiling pint rise and enthalpy change of concentrated solution.


Double-effect evaporators


Calculation methods for double-effect evaporators

If liquid is to be evaporated in each effect,

and if the boiling point of this liquid is unaffected by the solute concentration,

then writing a heat balance for the first evaporator:

q1 = U1 A1 ΔT1 = U1 A1 (TS – T1)

Similarly, in the second evaporator,

remembering that the "steam" in the second is the vapour from the first evaporator

and that this will condense at approximately the same temperature as it boiled, since pressure changes are small,

q2 = U2 A2 ΔT2 = U2 A2 (T1 – T2)

If the evaporators are working in balance, then all of the vapours from the first effect are condensing and in their turn evaporating

vapours in the second effect. Also assuming that heat losses can be neglected, there is no appreciable boiling-point elevation of the more concentrated solution, and the feed is supplied at its boiling point,

q1 = q2


Calculation methods for double-effect evaporators If the evaporators are working in balance, then all of the vapours from the first effect are condensing and in their turn evaporating vapours in the second effect. Also assuming that heat losses can be neglected, there is no appreciable boiling-point elevation of the more concentrated solution, and the feed is supplied at its boiling point, q1 = q2

That is, U1 A1 ΔT1 = U2 A2 ΔT2

Further, if the evaporators are so constructed that A1 = A2,

the foregoing equations can be combined.

U2 / U1 = ΔT1 / ΔT2

That is, the temperature differences are inversely proportional to the overall heat transfer coefficients in the two effects. This analysis may be extended to any number of effects operated in series, in the same way.


Example 4:

Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h-1 of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m-2 s-1 °C-1 in the first, second and third effects, respectively.

Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect.

Source: http://www.nzifst.org.nz/unitoperations/evaporation2.htm

prof. r. shanthini 28 may 2012 content of lectures 13 to 17: evaporation: -factors affecting...

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