CHEMICAL ENGINEERING SERIES
EVAPORATION
Compilation of Lectures and Solved Problems
CHEMICAL ENGINEERING SERIES 2
EVAPORATION
EVAPORATION
A unit operation that involves the concentration of a solution consisting of a non-volatile solute
and a volatile solvent
It is conducted by vaporizing a portion of the solvent to produce a concentrated solution of thick
liquor.
It differs with other unit operations in such a way that:
1. Distillation: in evaporation vapor is usually a single component
2. Drying: in evaporation, residue is liquid, sometimes a highly viscous one
3. Crystallization: focus is on concentrating a solution rather than forming crystals
Calculations for the Different Methods of Operations of Evaporators:
1. Single Effect Evaporators – used when the required capacity of operation is relatively small
and/or cost of steam is relatively cheap compared to the evaporator cost.
Vapor, V
where:
TV
- Mass flow rates of feed,
HV
PV
vapor, and steam
respectively
- Temperatures of feed,
product and vapor,
respectively
P
Steam, S
I
TS
- Liquid enthalpy of feed
λS
and product, respectively
Condensate
- Vapor enthalpy
TS2
TI
- Operating temperature
Feed, F
- Operating pressure
xF
TF
- Mass fraction of solute in
hF
Product, P
feed and product
CP, F
xP
respectively
TP
hP
Over-all Material Balance:
Solute Balance:
Enthalpy Balance:
Heat Balance:
(
)
must be evaluated at
or
If vacuum pressure is given,
CHEMICAL ENGINEERING SERIES 3
EVAPORATION
2. Forward Feed Multiple Effect Evaporators – fresh feed is added to the first effect and flows
to the next in the same direction as the vapor flow. This is used when the feed is hot or when
the final concentrated product might be damaged at high temperatures
VI
Steam, S
VII
VIII
P1, T1
P2 ,T2
P 3 , T3
TI
TII
TIII
F
LII
LI
P
3. Backward Feed Multiple Effect Evaporators – fresh feed enters the last and coldest effect
and continues until the concentrated product leaves the first effect. This is used when the
fresh feed is cold. This type of evaporation would requires liquid pump for each effect since
flow is from low to high pressure
VI
Steam, S
VII
VIII
P1, T1
P2 ,T2
P 3 , T3
TI
TII
TIII
LI
LII
F
P
4. Mixed Feed Multiple Effect Evaporators – fresh feed enters any of the available effects and
continues not necessarily to the effect next to it.
5. Parallel Feed Multiple Effect Evaporators – involves the adding of fresh feed and the
withdrawal of concentrated product from each effect. The vapor from each effect is still used
to heat the next effect. This method is used mainly when the feed is almost saturated and
solid crystals are the product, as in the evaporation of brine to make salt.
CHEMICAL ENGINEERING SERIES 4
EVAPORATION
Performance Evaluation of Steam-Heated Evaporators
1. Capacity – number of kilograms of water vaporized per hour
Evaporator Capacity
Where:
-
Rate of heat transfer through the heating
surface of an evaporator
Over-all heat transfer coefficient
Heat transfer surface area
Over-all temperature drop
2. Steam Economy – number of kilograms vaporized per kilogram of steam fed to the unit
Boiling point Evaluation (BPE) of a solution is the increase in boiling point over that of water
1. Small for dilute solutions and organic colloids solution
2. Large enough for concentrated solutions of inorganic salts; BPE can be estimated using
th
Figure 11-124 (CHE HB 8 edition)
Dϋhring’s Rule – the boiling point of a given solution is a linear function of the boiling point of
th
pure water at the same pressure. Figure 16.3 (Unit Operations 7 edition by McCabe and
Smith)
For solutions with BPE:
(
)
∑
CHEMICAL ENGINEERING SERIES 5
EVAPORATION
PROBLEM # 01:
A triple effect forward feed evaporator is being used to evaporate a sugar solution containing 5 wt
% solids to a concentrated solution of 80 %. The boiling point rise of the solutions (independent
2
of pressure) can be estimated from BPR °C = 1.78x + 6.22x , where x is wt fraction of sugar in
solution. Saturated steam at 205.5 kPa (121.1°C saturation temperature) is being used. The
pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 10,000 kg/h at
26.7°C. The heat capacity of the liquid solutions is 4.19 – 2.35x, kJ/kg·K. The heat of solution is
considered to be negligible. The coefficients of heat transfer have been estimated as U 1 = 3123,
2
U2 = 1987, and U3 = 1136 W/m ·K. if each effect has the same surface area, calculate the area,
the steam rate used and the steam economy.
SOLUTION:
V1
Steam, S
V2
V3
P1, T1
P2 ,T2
P 3 , T3
TI
TII
TIII
F
L1
L2
Step 1:
rd
From steam table, at 13.4 kPa (pressure of the vapor space at 3 effect)
From the given:
rd
For the 3 effect with x = 0.80
(
)
(
)
(
(
)
)
Step 2:
Consider solute balance around the system:
(
)(
*
Over-all material balance around the system
L3
CHEMICAL ENGINEERING SERIES 6
EVAPORATION
Where:
Assume , initially equal rate of evaporation in each effect
(
)
st
For the 1 effect:
(
)(
For the 2
nd
(
)
effect:
)(
)
Step 3:
st
nd
To solve for BPR for the 1 and 2 effects:
st
1 effect:
(
)
(
)
(
)
nd
2 effect:
(
)
(
)
(
)
∑
∑
(
∑
)
Estimate ΔT for each effect using equation 8.5-6
∑
(
)*
+
∑
(
)*
+
∑
(
)*
+
Calculate actual boiling point of solution for each effect using the estimated ΔT
CHEMICAL ENGINEERING SERIES 7
EVAPORATION
Step 4: Heat Balance with 0°C as datum
st
For 1 effect:
(
)
(
)
(
)
From steam table, steam at 121.1°C
(
)
From steam table, H(vapor enthalpy) at
(
(
For 2
)(
nd
(
)(
)
effect:
)
)
(
(
)
(
(
)
)(
)(
)
)
From steam table, at TV1 = 109°C
(
)
From steam table, H(vapor enthalpy) at
(
(
)(
)(
)
(
( )(
)(
)
)(
)
rd
For 3 effect:
(
)
(
)
(
From steam table, at TV2 = 89.89°C
)
)
(
)
(
)
CHEMICAL ENGINEERING SERIES 8
EVAPORATION
(
)
From steam table, H(vapor enthalpy) at
(
(
)(
(
Equate
)
)(
)(
(
)
)(
( )(
) (
)
)
)(
and
Step 5:
Solve for heat transfer area for each effect:
st
For 1 effect:
(
*(
*(
(
For 2
nd
*
)
effect:
(
*(
*
*(
(
)
rd
For 3 effect:
(
*(
(
*
*(
)
Since areas are not close, then another trial should be done
TRIAL 2:
Conduct new material balance using the computed L values
)
CHEMICAL ENGINEERING SERIES 9
EVAPORATION
(
)(
(
)
)(
)
st
To solve for BPR for the 1 and 2
st
1 effect:
(
)
(
)
(
nd
2 effect:
(
)
(
)
(
∑
nd
effects:
)
)
∑
(
∑
)
Average Area from trial 1:
For the new
(
)(
*
(
)(
*
(
)(
*
Adjust to attain total ΔT of 63.53
(
)(
*
(
)(
*
Calculate actual boiling point of solution for each effect using the estimated ΔT
Step 4: Heat Balance with 0°C as datum
st
For 1 effect:
(
)
(
)
CHEMICAL ENGINEERING SERIES 10
EVAPORATION
(
)
From steam table, steam at 121.1°C
(
)
From steam table, H(vapor enthalpy) at
(
(
For 2
)(
nd
(
)(
)
)
(
effect:
)
(
)
(
(
)
)(
)(
)
)
From steam table, at TV1 = 103.94°C
(
)
From steam table, H(vapor enthalpy) at
(
(
)(
)(
)
(
( )(
)(
)
)
)(
)
(
rd
For 3 effect:
(
)
(
)
(
From steam table, at TV2 = 87.61°C
(
)
)
From steam table, H(vapor enthalpy) at
(
)
(
)
)
(
)
CHEMICAL ENGINEERING SERIES 11
EVAPORATION
(
)(
(
Equate
)(
)(
)
)(
( )(
) (
)
)(
and
Step 5:
Solve for heat transfer area for each effect:
st
For 1 effect:
(
*(
*
*(
(
For 2
nd
)
effect:
(
*(
*
*(
(
)
rd
For 3 effect:
(
*(
(
*
*(
)
Areas are almost equal, therefore new assumptions are valid:
ANSWERS:
Area of each effect:
Steam Requirement
Steam Economy:
)
CHEMICAL ENGINEERING SERIES 12
EVAPORATION
PROBLEM # 02:
A solution with a negligible boiling point rise is being evaporated in a triple effect evaporator using
saturated steam at 121.1°C. The pressure in the vapor of the last effect is 25.6 kPa abs. The
2
heat transfer coefficients are U1 = 2840, U2 = 1988, and U3 = 1420 W/m ·K and the areas are
equal. Estimate the boiling point in each of the evaporators.
(Source: Transport Processes and Separation Processes, Geankoplis)
SOLUTION:
VI
Steam, S
121.1 C
T2
T3
TI
TII
TIII
LI
Using the heat transfer area equation:
Assume equal heat transfer flux for each effect
( )
( )
∑
From steam table,
∑
VIII
T1
F
( )
VII
LII
p3 = 25.6
kPa abs
P
CHEMICAL ENGINEERING SERIES 13
EVAPORATION
(
)(
(
)
(
)(
(
)
)
)
(
)(
(
)
)
CHEMICAL ENGINEERING SERIES 14
EVAPORATION
PROBLEM # 03:
A
forced-circulation
evaporator
is
to
concentrate 60,000 kg/h of 44 percent NaOH to
65 percent using steam at 3 atm pressure. The
feed temperature and the condensing
temperature are both 40°C. The density of the
3
feed solution is 1,450 kg/m . If the over-all
2
heat transfer coefficient is 2,000 W/m ·°C,
calculate (a) the steam requirement, in kg/h; (b)
the heat transfer area required.
(Source: Unit Operations of Chemical
th
Engineering, 7 edition, McCabe and Smith)
SOLUTION:
Consider solute balance (NaOH balance)
(
)(
*
Consider Over-all material Balance:
Consider enthalpy balance:
From figure 16.6
(
)
From figure 16.3 (McCabe and Smith)
For
the corresponding boiling point of solution at 65% conc,
From figure 16.6, at 65% concentration and 101.67°C
For the superheated vapor, assume Cp of steam is 0.45 BTU/lb·°F
(
)
From steam table at 40°C, HV, To = 1106.72 BTU/lb
CHEMICAL ENGINEERING SERIES 15
EVAPORATION
[
*(
(
(
*(
*
(
*(
(
)]
*(
*
At 3 atm, from steam table,
( )
(
(
)
)(
)
*
CHEMICAL ENGINEERING SERIES 16
EVAPORATION
PROBLEM # 04:
A triple-effect evaporator of the long-tube is to be used to concentrate 35,000 gal/h of a 17%
solution of dissolved solids to 38% dissolved solids. The feed enters at 60°Fand passes through
three tube-and-shell heaters, a, b, and c, in series and then through the three effects in order II,
III,I. Heater “a” is heated by vapor taken from the vapor line between the third effect and the
condenser, heater “b” with vapor from the vapor line between the second and the third effects,
and heater “c” with vapor from the line between the first and the second effect. In each heater the
warm end temperature approach is 10°F. Other data are given below:
Steam to I
230°F, dry and saturated
Vacuum on III
28 in (referred to a 30-in barometer)
Condensates leave steam chests at condensing temperatures
Boiling point elevations
1°F in II, 5°F in III, 15°F in I
2
Coefficients, in BTU/h·ft ·°F, corrected for BPE - 450 in I, 700 in II, 500 in III
All effects have equal areas of heating surface
Concentration,
% solids
Specific
gravity
Specific heat,
BTU/lb·°F
10
20
30
35
40
1.02
1.05
1.10
1.16
1.25
0.98
0.94
0.87
0.82
0.75
Calculate (a) the steam required in lb/h; (b) heating surface per effect; (c) economy in lb per lb of
steam; and (d) the latent heat to be removed in the condenser
th
(Source: Unit Operations of Chemical Engineering, 7 edition, McCabe and Smith)
SOLUTION:
I
II
c
P
xP= 0.38
III
b
CONDENSER
a
F= 35,000 gal/h
xF = 0.17
TF = 60 F
CHEMICAL ENGINEERING SERIES 17
EVAPORATION
Consider solute balance:
At 17% concentration, sp. gr = 1.041
(
)(
)
Consider over-all material balance:
Assume equal evaporation in each effect:
(
*
Material Balance for each effect:
For second effect:
Over-all material balance:
Solute balance:
(
)(
)
For third effect:
Over-all material balance:
Solute balance:
(
)(
For first effect:
To check:
Over-all material balance:
)
CHEMICAL ENGINEERING SERIES 18
EVAPORATION
Solute balance:
(
)(
)
TEMPERATURE DISTRIBUTION
Δ
rd
rd
Temperature of vapor leaving the 3 effect corresponds to the pressure in the 3 effect
From steam table, at 2 in Hg
There’s a need to assume values of ΔT 1, ΔT2, ΔT3
(
)(
(
)(
)
(
)
)(
(
)(
)
)
( )
( )
STREAM
DESIGNATION
TEMPERATURE,
°F
st
1 Effect
Steam Feed to E-I
Liquor from E-III
Vapor to E-II
Product
230
106
172
187
nd
2 Effect
Feed from H- c
Vapor to E-III
Liquor to E-III
162
144
145
rd
3 Effect
Vapor to condenser
Liquor to E-I
101
106
Feed
Feed to “b”
Feed to “c”
60
91
134
CHEMICAL ENGINEERING SERIES 19
EVAPORATION
st
Consider heat balance around 1 effect:
(
)
For x = 0.2692, Cp = 0.89156 BTU/lb·°F
From steam table, at T1 = 172 °F, λ = 995 BTU/lb; at Ts = 230°F, λ = 958.8 BTU/lb
)(
)(
)
(
)
(
)
(
Consider heat balance around the second effect and heater “c”
st
nd
Note that the vapor coming from 1 effect will be used to heat the heater and the 2 effect
(
)
(
)
(
)
For x = 0.17, Cp = 0.952 BTU/lb·°F
From steam table, at T2 = 144 °F, λ = 1011.64 BTU/lb
(
) (
)(
)(
)
Consider heat balance around the third effect and heater “b”
nd
rd
Note that the vapor coming from 2 effect will be used to heat the heater and the 3 effect
(
)
(
)
(
For x = 0.2084, Cp = 0.9341 BTU/lb·°F
From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb
)
(
) (
)(
)(
(
)(
)(
)
Equate
Substitute
(
and
and
)
(
)
)
CHEMICAL ENGINEERING SERIES 20
EVAPORATION
From equation
(
)
(
)
(
)
(
*(
Δ
*
*(
(
(
)
)
(
*(
*(
(
*(
Δ
*
*(
(
(
(
*
)
)
(
*(
*(
(
)
*(
*
*(
(
(
)
*
)
Since surface of each effect is not the same, therefore, previous assumptions need to be readjusted
To adjust ΔT:
Assume constant q and U
(
)
(
)
(
*
(
)(
*
(
)(
*
(
)(
*
CHEMICAL ENGINEERING SERIES 21
EVAPORATION
SECOND TRIAL:
Recompute using the adjusted ΔT:
STREAM
DESIGNATION
TEMPERATURE,
°F
st
1 Effect
Steam Feed to E-I
Liquor from E-III
Vapor to E-II
Product
230
106
170
185
nd
2 Effect
Feed from H- c
Vapor to E-III
Liquor to E-III
160
143
144
rd
3 Effect
Vapor to condenser
Liquor to E-I
101
106
Feed
Feed to “b”
Feed to “c”
60
91
133
st
Consider heat balance around 1 effect:
(
)
(
For x = 0.2692, Cp = 0.89156 BTU/lb·°F
From steam table, at T1 = 170 °F, λ = 996.2 BTU/lb; at T s = 230°F, λ = 958.8 BTU/lb
)(
)(
)
(
)
(
)
Consider heat balance around the second effect and heater “c”
st
nd
Note that the vapor coming from 1 effect will be used to heat the heater and the 2 effect
(
)
(
)
(
)
For x = 0.17, Cp = 0.952 BTU/lb·°F
From steam table, at T2 = 143 °F, λ = 1012.23 BTU/lb
(
) (
)(
)(
)
CHEMICAL ENGINEERING SERIES 22
EVAPORATION
Consider heat balance around the third effect and heater “b”
nd
rd
Note that the vapor coming from 2 effect will be used to heat the heater and the 3 effect
(
)
(
)
For x = 0.2084, Cp = 0.9341 BTU/lb·°F
From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb
)
(
) (
)(
)(
(
)(
)(
)
(
Equate
and
Substitute
(
and
)
(
)
From equation
(
)
(
)
(
)
(
*(
*(
(
(
)
)
(
*(
*(
(
*(
*
(
(
*
)
)
*(
*(
*(
)
(
*(
(
(
*
*
(
)
*
)
CHEMICAL ENGINEERING SERIES 23
EVAPORATION
*(
(
)
Since surface of each effect is not the same, therefore, previous assumptions need to be readjusted
To adjust ΔT:
Assume constant q and U
(
)
(
)
(
*
(
)(
*
(
)(
*
(
)(
*
Since there will be no change in ΔT’s, therefore, assumptions are correct:
( )
( )
ECONOMY:
( )
(
(
)
*(
*(
)
From steam table, at T3 = 101 °F, λ = 1036.44 BTU/lb
(
*(
*
( )
CHEMICAL ENGINEERING SERIES 24
EVAPORATION
PROBLEM # 05:
A single effect evaporator concentrates 1 MT of
10% wt sucrose solution to 50%. The feed
enters the evaporator at 20°C and has a
specific heat of 1.0.
The evaporator is
maintained at a vacuum of 800 mm Hg against
a barometric reading of 760 mm Hg. The heat
2
is provided by saturated steam at 8.8 kg/cm
gage. Assuming that no sensible heat is
recovered in the evaporator, calculate the
weight of heating steam, in kg, needed for
concentrating the sucrose solution.
Vapor, V
Steam, S
8.8 kg/cm2 gage
Sucrose Soln
F= 1 MT
xP = 0.10
TF= 20 C
Sp ht = 1.0
T1
TI
(Source: CHE BP May 1990)
SOLUTION:
Consider sucrose balance
(
)(
)
Consider Over-all material Balance:
Consider heat balance:
Since system involves solution of non-electrolytes, assume negligible BPE
(
)
From the steam table, @ 160 mm Hg,
(
)(
*(
)
From steam table at 160 mm Hg,
(
)(
*
2
From steam table at 8.8 kg/cm gage
(
)
*
+
pvacuum = 600 mm Hg
Pbarometric= 760 mm Hg
P
xP = 0.50
CHEMICAL ENGINEERING SERIES 25
EVAPORATION
PROBLEM # 06:
A solution of organic colloids is to be concentrated
from 20 to 65% wt solids in an evaporator.
Saturated steam is available at 172 kPa absolute
and the pressure in the condenser is 61.07
vacuum. The feed enters at 25°C and its specific
heat is 4.0 J/g·°C. The solution has a negligible
2
elevation in boiling point. OHTC is 1,000 W/m ·°C
and the evaporator must evaporate 9,000 kg/h.
a) Determine the steam consumption, kg/h
b) How many square meters of heating surface
are required?
c) What is the steam economy?
Vapor,
V = 9,000 kg/h
pvacuum = 61.07 cm
Steam, S
172 kPa abs
Organic Colloid
F
xF = 0.20
TF= 25 C
sp ht = 4.0 J/g· C
SOLUTION:
Consider sucrose balance
(
)
Consider Over-all material Balance:
Consider heat balance:
(
)
(
*
From the steam table, @ 460 mm Hg,
(
*(
*(
)
From steam table at 460 mm Hg,
(
)(
*
T1
TI
P
xP = 0.65
CHEMICAL ENGINEERING SERIES 26
EVAPORATION
From steam table at 172 kPa abs
( )
Δ
(
)
)(
(
( )
)
CHEMICAL ENGINEERING SERIES 27
EVAPORATION
PROBLEM # 07:
An evaporator is to concentrate 10% wt caustic
soda solution to 50% wt. Feed enters at 100°F.
Steam is available saturated at 50 psig and the
evaporator can be operated at 9.96 psi vacuum.
2
OHTC of the evaporator is 500 BTU/h·ft ·°F.
d) Determine the heating area required for the
production of 10,000 lb/h of the 50% wt NaOH.
e) What is the steam economy?
Vapor,
V
pvacuum = 9.96 psi
Steam, S
50 psig
NaOH soln
F
xF = 0.100
TF= 100 F
T1
TI
P = 10,000 lb/h
xP = 0.50
SOLUTION:
Consider NaOH balance
(
)(
)
Consider Over-all material Balance:
Consider enthalpy balance:
Since solution is an electrolyte, it can be expected that there will be BPE
Solve for TI for a vacuum pressure of 9.96 psi vacuum (evaporator pressure)
From steam table,
From figure 16.3 (McCabe and Smith), for 50% NaOH solution and T I of 159.95°F
⁄
From steam table at 87°F,
(
)
(
)(
*
From figure 16.6 (McCabe and Smith),
At 100°F and 10% NaOH
At 159.95 °F and 50% NaOH
(
*(
*
(
*(
*
(
*(
*
CHEMICAL ENGINEERING SERIES 28
EVAPORATION
Δ
(
(
)
For steam at 50 psig
*(
)
( )
( )
CHEMICAL ENGINEERING SERIES 29
EVAPORATION
PROBLEM # 08:
A 10% wt NaOH at 80 °F is to be concentrated in a
single effect evaporator to 40% wt. Steam is
supplied at 20 psig and the vacuum pressure of the
barometric condenser is 26 in Hg. 100 gpm of water
is fed to the condenser and the water leaving the
condenser (including the condensate) is at 100 °F.
2
OHTC of evaporator is 200 BTU/h·ft ·°F. Calculate
the heating surface required for the evaporator.
(Source: CHE BP May 1984)
tB =T =
100 F
Vapor,
V
pvacuum = 26”Hg
Steam, S
20 psig
NaOH soln
F
xF = 0.10
TF= 80 F
T1
tA = 70 F
100 gpm
TI
P
xP = 0.40
SOLUTION:
Consider heat balance around the condenser:
(
)
(
)(
*(
)
(
)
Assume barometric pressure of 1 atm or 29.921 in Hg
From steam table, at 3.921 in Hg,
The vapor will be condensed first before lowering to 100°F, thus, Cp of the liquid water
should be used
(
(
)
)
(
)(
For the evaporator
Consider NaOH balance
(
)
Consider Over-all material Balance:
)
CHEMICAL ENGINEERING SERIES 30
EVAPORATION
Consider enthalpy balance:
Since solution is an electrolyte, it can be expected that there will be BPE
From figure 16.3 (McCabe and Smith), for 40% NaOH solution and T 1 of 124.37°F
⁄
From steam table at 124.37°F,
(
)
(
)(
*
From figure 16.6 (McCabe and Smith),
At 80°F and 10% NaOH
At 170 °F and 40% NaOH
(
*(
*
(
Δ
(
(
)
For steam at 20 psig
*(
)
*(
*
(
*(
*
CHEMICAL ENGINEERING SERIES 31
EVAPORATION
PROBLEM # 09:
4,500 kg/h of a 10% wt sugar solution is to be concentrated to 30% wt. Feed enters at 21°C.
Saturated steam at 110°C is available and the temperature in the condenser is 43°C. Specific
heat of the solutions may be taken as constant at 4 J/g·°C. Determine (a) heating surface
required, (b) steam consumption, and (c) steam economy, for each of the following cases:
2
I.
II.
III.
Single effect, OHTC = 2,840 W/m ·°C
2
Double effect, forward feed; U1 = 2,270, U2 = 1,700 W/m ·°C
2
Double effect, backward feed; U1 = 2,270, U2 = 1,700 W/m ·°C
SOLUTION:
Consider sugar balance
(
)(
*
Consider Over-all material Balance:
CASE I: SINGLE EFFECT
Vapor,
V
T1 = 43 C
Steam, S
110 C
Sugar soln
F=4,500 kg/h
xF = 0.10
TF= 21 C
T1
TI
P
xP = 0.30
Consider heat balance:
Since solution is non electrolyte, BPE is negligible
Temperature of vapor leaving the evaporator is the same as the condenser temperature
(
(
)
*(
*(
)
CHEMICAL ENGINEERING SERIES 32
EVAPORATION
From steam table, at 43°C
(
*(
Δ
(
(
*
)
)(
)
(
)
From steam table, at 110°C,
(
(
)
)
CASE II: DOUBLE EFFECT, FORWARD FEED
CHEMICAL ENGINEERING SERIES 33
EVAPORATION
VI
VII
T2 = 43 C
Steam, S
110 C
T1
T2
TI
TII
Sugar soln
F=4,500 kg/h
xF = 0.10
TF= 21 C
PI
Initial Assumptions:
Assume equal evaporation rates
(
*
st
Consider material balance around the 1 effect
Solve for the temperature distribution; to assume ΔT’s
Assume equal heat flux
( )
Δ
( )
( Δ )
( Δ )
P
xP = 0.30
CHEMICAL ENGINEERING SERIES 34
EVAPORATION
Δ
Δ (
)
Δ (
*
Ts = 110 C
Δ
ΔTI
Δ
(
)(
Δ
Δ (
TI
*
)
Δ (
*
ΔTII
Δ
Δ
TII = 43 C
(
)(
(
nd
Enthalpy balance around the 2
*
)
effect:
st
Enthalpy balance around the 1 effect:
(
Equate
)
and
(
(
)
)
(
)
)
(
(
)
(
)(
)
(
)
(
)(
)
From steam table;
CHEMICAL ENGINEERING SERIES 35
EVAPORATION
(
)(
)
(
)(
(
)
)
(
(
)(
( )
Equate
and
(
*(
*
Δ
(
)(
)
)(
)
Δ
(
)
)
CHEMICAL ENGINEERING SERIES 36
EVAPORATION
Since AI = AII, therefore original assumptions are correct
(
)
(
)
(
)
CASE III: DOUBLE EFFECT, BACKWARD FEED
VI
VII
T2 = 43 C
Steam, S
110 C
T1
T2
TI
TII
PI
P
xP = 0.30
Initial Assumptions:
Assume equal evaporation rates
(
*
st
Consider material balance around the 1 effect
Sugar soln
F=4,500 kg/h
xF = 0.10
TF= 21 C
CHEMICAL ENGINEERING SERIES 37
EVAPORATION
Solve for the temperature distribution; to assume ΔT’s
Assume equal heat flux
( )
( )
( Δ )
Δ
( Δ )
Δ
Ts = 110 C
Δ (
ΔTI
)
Δ (
*
Δ
TI
ΔTII
TII = 43 C
Δ
(
Δ
Δ (
Δ
(
nd
st
effect:
Enthalpy balance around the 1 effect:
and
(
(
)
)
*
)
)(
(
Enthalpy balance around the 2
Equate
)(
)
Δ (
*
*
CHEMICAL ENGINEERING SERIES 38
EVAPORATION
(
)
(
)(
(
)
(
)(
)
)
From steam table;
(
)
(
)
(
)(
)
( )
Equate
and
(
*(
(
*
)(
)
(
)(
)
(
)(
)
CHEMICAL ENGINEERING SERIES 39
EVAPORATION
)(
(
)
Since AI = AII, therefore original assumptions are correct
(
)
(
)
(
)
CHEMICAL ENGINEERING SERIES 40
EVAPORATION
PROBLEM # 10:
Glycerine is to be concentrated from 12% to 72% in a single-effect evaporator. The inlet steam
used is at 25 psig and comes out at 170°F. The vapor space in the evaporator has 25 inches Hg
vacuum. Ten metric tons of glycerine per hour are fed at 85°F. The concentrated product is at
125°F. Calculate the amount of water evaporated in kg/h.
(Source: MRII Reviewer)
SOLUTION:
Vapor,
V
P = 25” Hg vac
Steam, S
25 psig
F=10 MT/h
xF = 0.12
TF= 85 F
TI
170 F
P
xP = 0.72
TF= 125 F
Consider solute balance:
(
)(
*
Consider over-all material balance:
CHEMICAL ENGINEERING SERIES 41
EVAPORATION
PROBLEM # 11:
A feed of 4,535 kg/h of a 2.0 % wt salt solution at 311 K enters continuously a single effect
evaporator and is being concentrated to 3%. The evaporation is at atmospheric pressure and the
2
area of the evaporator is 69.7 m . Saturated steam at 383.2 K is supplied for heating. Since the
solution is dilute, it can be assumed to have the same boiling point as water. The heat capacity
of the feed can be taken as Cp = 4.10 kJ/kg·K. Calculate the amounts of vapor and liquid product
and the over-all heat transfer coefficient.
(Source:
Principles of Transport Processes and Separation Processes, 4
Geankoplis)
SOLUTION:
Vapor,
V
P = 1 atm
Steam, S
383.2 K
F=4535 kg/h
xF = 0.02
TF= 311 K
TI
P
xP = 0.03
Consider solute balance:
(
)(
*
Consider over-all material balance:
Consider heat balance:
th
edition, by
CHEMICAL ENGINEERING SERIES 42
EVAPORATION
Since the evaporator operates at 1 atm, operating temperature will be the the
temperature corresponding to 1 atm or water boiling point (373 K)
(
(
)
*(
*(
)
From the steam table at 373 K,
(
(
*(
*
)
(
)(
)
CHEMICAL ENGINEERING SERIES 43
EVAPORATION
PROBLEM # 12:
2
An evaporator having an area of 83.6 m and U = 2270
2
W/m ·K is used to produce distilled water for a boiler feed.
Tap water having 400 ppm dissolved solids at 15.6 °C is fed
to the evaporator operating at 1 atm pressure abs.
Saturated steam at 115.6°C is available for use. Calculate
the amount of distilled water produced per hour if the outlet
liquid contains 800 ppm solids.
Vapor,
V
xF = 0.04
TF= 15.6 C
(Source: Principles of Transport Processes and Separation
th
Processes, 4 edition, by Geankoplis)
SOLUTION:
Assume that the boiling point of the solution is 100°C (1 atm operating pressure)
(
)
)(
(
)(
)
Consider solute balance:
(
)
Consider over-all material balance:
Consider heat balance:
(
)
For water, Cp = 4.1868 kJ/kg·°C
(
)(
*(
From the steam table at 100°C,
( )(
(
)
*
)
P = 1 atm
Steam, S
115.6 C
TI
P
xP = 0.08
CHEMICAL ENGINEERING SERIES 44
EVAPORATION
PROBLEM # 13:
A single effect evaporator is concentrating a feed of 9,072 kg/h
of a 10 wt % solution of NaOH in water to a product of 50 %
solids. The pressure of saturated steam used is 42 kPa (gage)
and the pressure in the vapor space of the evaporator is 20
kPa (abs). The over-all heat transfer coefficient is 1,988
2
W/m ·K. Calculate the steam used, the steam economy in kg
vaporized/kg steam, and the area for the following feed
conditions:
a) Feed temperature of 288.8 K
b) Feed temperature of 322.1 K
Vapor,
V
PV = 20 kPa (abs)
Steam, S
42 kPa (gage)
NaOH soln
F=9,072 kg/h
xF = 0.10
TF
TI
(Source: Principles of Transport Processes and Separation
th
Processes, 4 edition, by Geankoplis)
SOLUTION:
Consider NaOH balance:
(
)(
*
Consider over-all material balance:
For 20 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)
Since system is NaOH, it is expected to have BPE
From figure 8.4-2 (Geankoplis), for 50% NaOH concentration and T V = 333.06 K (60.06°C)
[
(
)]
From steam table, at 60.06°C,
⁄
,
⁄
P
xP = 0.50
CHEMICAL ENGINEERING SERIES 45
EVAPORATION
(
*[
(
)(
⁄
)]
A. For feed temperature of 288.8 K
Consider enthalpy balance:
⁄
From steam table, at 42 kPa gage,
,
From figure 8.4-3 (Geankoplis)
(
*(
(
*
(
)
(
)
(
*(
(
*(
)(
)(
)
)
*(
*
CHEMICAL ENGINEERING SERIES 46
EVAPORATION
B. For feed temperature of 322.1 K
Consider enthalpy balance:
⁄
From steam table, at 42 kPa gage,
,
From figure 8.4-3 (Geankoplis)
(
*(
(
*
(
)
(
)
(
*(
(
*(
)(
)(
)
)
*(
*
CHEMICAL ENGINEERING SERIES 47
EVAPORATION
PROBLEM # 14:
In order to concentrate 4,536 kg/h of an NaOH solution
containing 10 wt % NaOH to a 20 wt % solution, a single
2
effect evaporator is being used with an area of 37.6 m .
The feed enters at 21.1 °C. Saturated steam at 110 °C is
used for heating and the pressure in the vapor space of
the evaporator is 51.7 kPa abs. Calculate the kg/h of
steam used and the over-all heat transfer coefficient.
(Source: Principles of Transport Processes and
th
Separation Processes, 4 edition, by Geankoplis)
Vapor,
V
PV = 51.7 kPa (abs)
Steam, S
110 C
NaOH soln
F=4,536 kg/h
xF = 0.10
TF = 21.1 C
TI
P
xP = 0.20
SOLUTION:
Consider NaOH balance:
(
)(
*
Consider over-all material balance:
For 51.7 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)
Since system is NaOH, it is expected to have BPE
From figure 8.4-2 (Geankoplis), for 20% NaOH concentration and T V = 82.06°C
(
[
)]
⁄
From steam table, at 82.06°C,
(
*[
(
⁄
)(
⁄
,
)]
CHEMICAL ENGINEERING SERIES 48
EVAPORATION
Consider enthalpy balance:
⁄
From steam table, at 110°C,
From figure 8.4-3 (Geankoplis)
(
*(
(
(
*
(
)
)
(
*(
(
*(
)(
)(
)
)
*(
*
CHEMICAL ENGINEERING SERIES 49
EVAPORATION
PROBLEM # 15:
An evaporator is concentrating F kg/h at 311 K of a 20 wt %
solution of NaOH to 50%. The saturated steam for heating
is at 399.3 K. The pressure in the vapor space of the
evaporator is 13.3 kPa abs. The over-all heat transfer
2
2
coefficient is 1,420 W/m ·K and the area is 86.4 m .
Calculate the feed rate F of the evaporator.
Vapor,
V
PV = 13.3 kPa (abs)
Steam, S
399.3 K
NaOH soln
F
xF = 0.20
TF = 311 K
(Source: Principles of Transport Processes and Separation
th
Processes, 4 edition, by Geankoplis)
TI
P
xP = 0.50
SOLUTION:
Consider NaOH balance:
(
)( )
Consider over-all material balance:
For 13.3 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)
Since system is NaOH, it is expected to have BPE
From figure 8.4-2 (Geankoplis), for 50% NaOH concentration and T V = 51.39°C
(
[
)]
⁄
From steam table, at 51.39°C,
(
)[
(
Consider enthalpy balance:
⁄
)(
,
)]
⁄
CHEMICAL ENGINEERING SERIES 50
EVAPORATION
From steam table, at 399.3 K,
⁄
From figure 8.4-3 (Geankoplis)
(
(
)(
(
*
)
( )(
*
)
(
)(
)(
)
CHEMICAL ENGINEERING SERIES 51
EVAPORATION
PROBLEM # 16:
A single effect evaporator is concentrating a feed
solution of organic colloids from 5 to 50 wt %. The
solution has a negligible boiling point elevation. The
heat capacity of the feed is Cp= 4.06 kJ/kg·K and the
feed enters at 15.6°C. Saturated steam at 101.32 kPa
is available for heating, and the pressure in the vapor
space of the evaporator is 15t.3 kPa. A total of 4,536
kg/h of water is to be evaporated. The over-all heat
2
transfer coefficient is 1,988 W/m ·K. What is the
2
required surface area in m
and the steam
consumption?
Vapor,
V = 4,536 kg/h
PV = 15.3 kPa (abs)
Steam, S
101.32 kPa
F
xF = 0.05
TF = 15.6 C
TI
P
xP = 0.50
(Source: Principles of Transport Processes and
th
Separation Processes, 4 edition, by Geankoplis)
SOLUTION:
Consider NaOH balance:
(
)( )
Consider over-all material balance:
For 15.3 kPa evaporator vapor space pressure, the temperature of the vapor (from steam table)
For steam at 101.32 kPa
Since no BPE
Consider heat balance:
(
)
CHEMICAL ENGINEERING SERIES 52
EVAPORATION
(
(
(
*(
*(
(
*(
*
)
)
)(
)