PM3125: Lectures 13 to 15
Content of Lectures 13 to 17:
Evaporation:
-
Factors affecting evaporation
Evaporators
Film evaporators
Single effect and multiple effect evaporators
Mathematical problems on evaporation
Principal reference:
Chapter 8 in C.J. Geankoplis, Transport Processes and Unit Operations,
3rd Edition, Prentice-Hall of India
Prof. R. Shanthini
28 May 2012
Evaporation
- 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.
Prof. R. Shanthini
28 May 2012
Types of evaporators
Open kettle or pan evaporator:
Prof. R. Shanthini
28 May 2012
Types of evaporators
Open kettle or pan evaporator:
Pan
Pressure gauge
Steam
Boiler
Jacket
Condensate
Concentrate
Prof. R. Shanthini
28 May 2012
Types of evaporators
Open kettle or pan evaporator:
- simplest form of evaporators
- inexpensive
- simple to operate
- very poor heat economy
- in some cases paddles and scrapers for agitation are used
Prof. R. Shanthini
28 May 2012
Types of evaporators
Horizontal-tube evaporator:
Vapour
Dilute feed
Steam
inlet
Condensate
Concentrated
product
Prof. R. Shanthini
28 May 2012
Types of evaporators
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)
Prof. R. Shanthini
28 May 2012
Types of evaporators
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
Prof. R. Shanthini
28 May 2012
Types of evaporators
Vertical-type short-tube evaporator:
Vapour
Dilute feed
Steam
inlet
Condensate
Concentrated
product
Prof. R. Shanthini
28 May 2012
Types of evaporators
Falling-film-type evaporator:
Prof. R. Shanthini
28 May 2012
Types of evaporators
More types are given in additional handouts
uploaded at the course website.
Prof. R. Shanthini
28 May 2012
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.
Prof. R. Shanthini
28 May 2012
Factors effecting evaporation:
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.
Prof. R. Shanthini
28 May 2012
Factors effecting evaporation:
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
Prof. R. Shanthini
28 May 2012
Factors effecting 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.
Prof. R. Shanthini
28 May 2012
Factors effecting evaporation:
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).
Prof. R. Shanthini
28 May 2012
Factors effecting evaporation:
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.
Prof. R. Shanthini
28 May 2012
Factors effecting evaporation:
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.
Prof. R. Shanthini
28 May 2012
Method of operation of evaporators
Single-effect evaporation:
- 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.
Multiple-effect evaporation:
- Increasing the evaporation per kg of steam by using a series of
evaporators between the steam supply and condenser is called
multiple effect evaporation
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
Feed:
F – mass flow rate
xF – mass fraction of solute in feed
TF – temperature of feed
hF – enthalpy of feed
P
Vapour leaving the evaporator:
V – mass flow rate
Condensate, S yV – mass fraction of solute in vapour
T1 – temperature of vapour
PS, TS, hS
HV – enthalpy of vapour
T1
Concentrate, L
xL, T1, hL
Steam:
S – mass flow rate
PS – steam pressure
TS – steam temperature
HSProf.
– enthalpy
of steam
R. Shanthini
hS28
– May
enthalpy
2012 of condensate
Concentrate leaving the evaporator:
L – mass flow rate
xL – mass fraction of solute in concentrate
T1 – temperature of concentrate
hL – enthalpy of concentrate
P – pressure in the evaporator
T1 – temperature in the evaporator
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
Overall material balance:
F=L+V
P
Solute balance:
F xF = L xL + V yV
T1
If the vapour is free of solute:
F xF = L xL
Condensate, S
PS, TS, hS
Concentrate, L
xL, T1, hL
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
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
Energy lost by the steam
q = S λ = S (HS – hS)
P
T1
Condensate, S
PS, TS, 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.
Concentrate, L
xL, T1, hL
Prof. R. Shanthini
28 May 2012
Therefore,
q = U A ΔT = U A (TS – 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.
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Data provided:
F = 9072 kg/h
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
xF = 1 wt % = 0.01 kg solute / kg feed
P
TF = 38ºC
xL = 1.5 wt %
T1
Condensate, S
PS, TS, hS
= 0.015 kg solute / kg liquid product
P = 101.325 kPa (1.0 atm abs)
PS = 150 kPa
Concentrate, L
U = 1704 W/m2.K
xL, T1, hL
T1 = saturated temperature at P (= 101.325 kPa) = 100ºC
Prof. R. Shanthini
28 May 2012
TS = saturated temperature at 150 kPa = 111.4ºC
Calculation methods for single-effect evaporators
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
Available equations:
Overall material balance:
F=L+V
Solute balance:
F xF = L xL (no solute in the vapour)
Heat balance:
F hF + S λ = L hL + V HV
where λ = HS – hS
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
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Data known:
Available equations:
F = 9072 kg/h; L = 6048 kg/h, V = 3024 kg/h
P = 101.325 kPa; T1 = 100ºC
Heat balance:
F hF + S λ = L hL + V HV
where λ = HS – hS
PS = 150 kPa; TS = 111.4ºC
q = S λ = U A ΔT = U A (TS – T1)
TF = 38ºC
U = 1704 W/m2.K
Heat transfer area A = S λ / U (TS – T1) = ?
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 )
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
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
Heat transfer area A = S λ / U (TS – T1) = ?
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
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
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
Heat transfer area A = S λ / U (TS – T1) = ?
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
Prof. R. Shanthini
28 May 2012
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 heattransfer area.
Prof. R. Shanthini
28 May 2012
Effects of processing variables on evaporator operation:
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.
Prof. R. Shanthini
28 May 2012
Effects of processing variables on evaporator operation:
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.
Prof. R. Shanthini
28 May 2012
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.
Prof. R. Shanthini
28 May 2012
Boiling point rise of solutions (an example):
-
Prof. R. Shanthini
28 May 2012
Duhring plot for boiling point of sodium chloride solutions
Enthalpy-concentration charts of solutions:
-
See the handout.
Prof. R. Shanthini
28 May 2012
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.
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Data provided:
F = 4536 kg/h
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
xF = 20 wt % = 0.2 kg solute / kg feed
P
TF = 60ºC
xL = 50 wt %
T1
Condensate, S
PS, TS, hS
= 0.5 kg solute / kg liquid product
P = 12 kPa = 0.12 bar
PS = 170 kPa = 1.7 bar
Concentrate, L
U = 1560 W/m2.K
xL, T1, hL
T1 ≠ saturated temperature at P (= 0.12 bar) = 49.4oC
Prof. R. Shanthini
28 May 2012
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) = 49.4oC
Using 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)
Prof. R. Shanthini
28 May 2012
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 = ?
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
Prof. R. Shanthini
28 May 2012
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.
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Vapour, V
yV, T1, HV
Data provided:
F = 4536 kg/h
Feed, F
xF, TF, hF
Steam, S
PS, TS, HS
xF = 20 wt % = 0.2 kg solute / kg feed
P
TF = 60ºC
xL = 50 wt %
T1
Condensate, S
PS, TS, hS
= 0.5 kg solute / kg liquid product
P = 12 kPa = 0.12 bar
PS = 170 kPa = 1.7 bar
Concentrate, L
U = 1560 W/m2.K
xL, T1, hL
T1 = saturated temperature at P (= 0.12 bar) = 49.4oC
Prof. R. Shanthini
28 May 2012
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 = (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
Prof. R. Shanthini
28 May 2012
Calculation methods for single-effect evaporators
Compare the solutions of Examples 2 and 3 and
discuss the importance of considering the boiling pint
rise and enthalpy change of concentrated solution.
Prof. R. Shanthini
28 May 2012
Double-effect evaporators
Prof. R. Shanthini
28 May 2012
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
Prof. R. Shanthini
28 May 2012
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.
Prof. R. Shanthini
28 May 2012
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