The document discusses evaporation processes and evaporator equipment. It defines evaporation as the removal of water from an aqueous solution through the application of heat. It then describes the types of evaporators, including single effect, multiple effect, forced circulation, open pan, and falling film evaporators. The document also covers important processing factors in evaporation like concentration, solubility, temperature, foaming, and scale deposition. It provides examples of calculating heat transfer area and steam usage for single effect evaporators concentrating different solutions.

1. EVAPORATION
1

2. CONTENT
1. Type of Evaporation equipment and Methods
2. Overall Heat Transfer Coefficient in Evaporators
3. Calculation Methods for Single Effect Evaporators
2
3. Calculation Methods for Single Effect Evaporators
4. Calculation Methods for Multiple Effects Evaporators
5. Condenser for Evaporator
6. Evaporation using Vapor Recompression

3. Topic Outcomes
Define Evaporation process
Describe and determine the processing
factors that involved in evaporation process
identify the types of evaporators and
identify the types of evaporators and
distinguish the suitable evaporator for
certain operations
Describe other specific types of evaporators
that are used in certain industries
3

4. EVAPORATION
Heat is added to a solution to vaporize the solvent, which is usually water.
Case of heat transfer to a boiling liquid.
Vapor from a boiling liquid solution is removed and a more concentrated
solution remains.
4
solution remains.
Refers to the removal of water from an aqueous solution.
Example: concentration of aqueous solutions of sugar. In these cases the
crystal is the desired product and the evaporated water is discarded.

5. Foaming or frothing
Pressure and temperature
Scale deposition
Materials of construction
5
Processing Factors
Concentration in
the liquid
solubility
Temperature sensitivity
of materials
Scale deposition

6. PROCESSING FACTORS
(1) Concentration
dilute feed, viscosity µ↓, heat transfer coefficient, h↑
concentrated solution/products, µ↑, and h ↓.
6
(2) Solubility
concentration ↑, solubility ↓ , crystal formed.
solubility ↑ with temperature ↑.
(3) Temperature.
heat sensitive material degrade at higher temperature prolonged
heating.

7. (4) Foaming/frothing.
caustic solutions, food solutions, fatty acid solutions form
foam/froth during boiling.
entrainment loss as foam accompany vapor.
7
(5) Pressure and Temperature
pressure ↑, boiling point ↑.
concentration ↑, boiling point↑.
heat-sensitive material operate under vacuum.
(6) Material of construction
minimize corrosion.

8. TRAY
Growing PACKING
Growing foam
leaves tower
Lean soln
Sweet gas Sweet gas
8
LI
TRAY
Growing
foam slows/
stops liquid
from
flowing
down
LI
PACKING
Foam effects
less severe, but
can cause some
channeling
Sour gas
Sour gas
Rich soln Rich soln
Lean soln
Forming Effects

9. Type of Evaporation
Equipment
Horizontal tube type
Vertical tube type
Long tube vertical type
Forced-circulation type
9
Forced-circulation type
Open kettle or pan
Open-pan solar evaporator
Falling-film-type evaporator
Agitated-film evaporator

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Wiped film evaporator
10
Operating parameters Operating window
Feed rate kg/h 20 - 100.000
Evaporation rate kg/h up to 40.000
Heating temperature °
°
°
°C up to 380 1)
Pressure on process side barg -1 to 30
Product viscosity at operating temperature mPas up to 70.000
Residence time min 1 2)
Evaporation ratio (concentrate/feed) up to 1:50 3)
Through put % 20 - 100 3)
1) evaporators with electrical inductive heating up to 600 °
°
°
°C
2) in horizontal evaporators longer residence times possible
3) depending on evaporator type

11. Effect of Processing Variables on Evaporator Operation
(1) TF
TF Tbp, some of latent heat of steam will be used to heat up the cold
feed, only the rest of the latent heat of steam will be used to vaporize
the feed.
11
Is the feed is under pressure TF Tbp, additional vaporization
obtained by flashing of feed.
(2) P1
desirable ∆T ↑ [Q = UA(TS – T1)]
A ↓ cost ↓.
T1 depends on P1 will ↓ T1.

12. (3) PS
↑ PS will ↑ TS but high-pressure is costly.
optimum TS by overall economic balances.
(4) BPR
The concentration of the solution are high enough so that the
12
The concentration of the solution are high enough so that the
cP and Tbp are quite different from water.
BPR can be predict from Duhring chart for each solution such
as NaOH and sugar solution.
(5) Enthalpy–concentration of solution.
for large heat of solution of the aqueous solution.
to get values for hF and hL.

13. 13

14. 14

15. feed, F
TF , xF , hF.
vapor,V to condenser
T1 , yV , HV
P1
T1
heat-exchanger
tubes
15
steam, S
TS , HS
concentrated liquid, L
T1 , xL , hL
condensate, S
TS , hS
T1
Simplified Diagram of single-effect evaporator

16. steam, T
feed, TF
vapor T1
(1)
T1
(2)
T2
(3)
T3
vapor T2 vapor T3
to vacuum
condenser
16
steam, TS
concentrate
from first
effect.
concentrate
from second
effect.
concentrated
product
condensate
Simplified diagram of forward-feed triple-effect evaporator

17. steam, TS
feed, TF
vapor T1
(1)
T1
(2)
T2
(3)
T3
vapor T2 vapor T3
to vacuum
condenser
17
Simplified diagram of backward-feed triple-effect evaporator
concentrated
product
condensate

18. The feed (usually dilute) enters at TF and saturated steam at TS
enters the heat-exchange section.
Condensed leaves as condensate or drips.
The solution in the evaporator is assumed to be completely
mixed and have the same composition at T1.
SINGLE EFFECT EVAPORATORS
18
mixed and have the same composition at T1.
The pressure is P1, which is the vapor pressure of the solution at
T1.
Wasteful of energy since the latent heat of the vapor leaving is
not used but is discarded.
Are often used when the required capacity of operation is
relatively small, but it will wasteful of steam cost.

19. CALCULATIONS
a) vapor, V and liquid, L flowrates.
b) heat transfer area, A
c) overall heat-transfer coefficient, U.
CALCULATION METHODS FOR SINGLE-EFFECT
EVAPORAT
19
c) overall heat-transfer coefficient, U.
d) Fraction of solid content, xL.
(1) To calculate V L and xL,
- solve simultaneously total material balance
solute/solid balance.
F = L + V total material balance
F (xF) = L (xL) solute/solid balance

20. (2) To calculate A or U,
- No boiling point rise and negligible heat of solution:
calculate hF, hL, Hv and λ.
where, λ = (HS – hs)
h = cP(T – Tref)
20
h = cP(T – Tref)
where, Tref = T1 = (as datum)
cPF = heat capacity (dilute as water)
HV = latent heat at T1
solve for S:
F hF + S λ = L hL + V HV
solve for A and U:
q = S λ = U A ∆T = UA (TS – T1)

21. (3) To get BPR and the heat of solution:
1) Calculate T1 = Tsat + BPR
2) Get hF and hL from Figure 8.4-3.
3) Get λS HV from steam tables for superheated vapor or
H = H + 1.884 (BPR)
21
HV = Hsat + 1.884 (BPR)
4) Solve for S:
F hF + S λ = L hL + V HV
5) Solve for A and U:
q = S λ = U A ∆T = UA (TS – T1)

22. A continuous single-effect evaporator concentrates 9072 kg/h of a
1.0 wt % salt solution entering at 311.0 K (37.8 ºC) 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
143.3 kPa. The overall coefficient U = 1704 W/m2 .K.
Example 8.4-1: Heat-Transfer Area in Single-Effect Evaporator
22
Calculate the amounts of vapor and liquid product and the heat-
transfer area required. Assumed that, since it its dilute, the solution
has the same boiling point as water.

23. U = 1704 W/m2
P1 = 101.325 kPa
F = 9072 kg/h
TF = 311 K
xF = 0.01
hF.
V = ?
T1 , yV , HV
23
U = 1704 W/m2
T1 A = ?
S , TS , HS
PS = 143.3 kPa
L = ?
T1 , hL
xL = 0.015
S, TS , hS
Figure 8.4-1: Flow Diagram for Example 8.4-1

24. Solution
Refer to Fig. 8.4-1 for flow diagram for this solution.
For the total balance,
F = L + V
9072 = L + V
24
For the balance on the solute alone,
F xF = L xL
9072 (0.01) = L (0.015)
L = 6048 kg/h of liquid
Substituting into total balance and solving,
V = 3024 kg/h of vapor

25. Since we assumed the solution is dilute as water;
cpF = 4.14 kJ/kg. K
From steam table, (A.2-9)
At P1 = 101.325 kPa, T1 = 373.2 K (100 ºC).
HV = 2257 kJ/kg.
25
HV = 2257 kJ/kg.
At PS = 143.3 kPa, TS = 383.2 K (110 ºC).
λ = 2230 kJ/kg.
The enthalpy of the feed can be calculated from,
hF = cpF (TF – T1)
hF = 4.14 (311.0 – 373.2)
= -257.508 kJ/kg.

26. Substituting into heat balance equation;
F hF + S λ = L hL + V HV
with hL = 0, since it is at datum of 373.2 K.
9072 (-257.508) + S (2230) = 6048 (0) + 3024 (2257)
S = 4108 kg steam /h
26
S = 4108 kg steam /h
The heat q transferred through the heating surface area, A is
q = S (λ)
q = 4108 (2230) (1000 / 3600) = 2 544 000 W
Solving for capacity single-effect evaporator equation;
q = U A ∆T = U A (TS – T1)
2 544 000 = 1704 A (383.2 – 373.2)
Solving, A = 149.3 m2.

28. Example 8.4-3: Evaporation of an NaOH Solution.
An evaporator is used to concentrate 4536 kg/h of a 20 % solution of
NaOH in water entering at 60 ºC to a product of 50 % solid. The pressure
of the saturated steam used is 172.4 kPa and the pressure in the vapor
space of the evaporator is 11.7 kPa. The overall heat-transfer coefficient
is 1560 W/m2.K.
28
is 1560 W/m .K.
Calculate the steam used, the steam economy in (kg vaporized/kg steam)
used, and the heating surface area in m2.

29. U = 1560 W/m2
P1 = 11.7 kPa
F = 4536 kg/h
TF = 60 ºC
xF = 0.2
hF.
S = ?
V, T1 , yV , HV
29
U = 1560 W/m2
T1 A = ?
S = ?
TS , HS
PS = 172.4 kPa
L, T1 , hL
xL = 0.5
S, TS , hS
Figure 8.4-4: Flow Diagram for Example 8.4-3

30. Solution
Refer to Fig. 8.4-4, for flow diagram for this solution.
For the total balance,
F = 4536 = L + V
For the balance on the solute alone,
30
For the balance on the solute alone,
F xF = L xL
4536 (0.2) = L (0.5)
L = 1814 kg/h of liquid
Substituting into total balance and solving,
V = 2722 kg/h of vapor

31. To determine T1 = Tsat + BPR of the 50 % concentrate product, first
we obtain Tsat of pure water from steam table. At 11.7 kPa, Tsat =
48.9 ºC.
From Duhring chart (Fig. 8.4-2), for a Tsat = 48.9 ºC and 50 %
NaOH , the boiling point of the solution is T1 = 89.5 ºC. hence,
31
NaOH , the boiling point of the solution is T1 = 89.5 ºC. hence,
BPR = T1 - Tsat = 89.5-48.9 = 40.6 ºC
From the enthalpy-concentration chart (Fig.8.4-3), for
TF = 60 ºC and xF = 0.2 get hF = 214 kJ/kg.
T1 = 89.5 ºC and xL = 0.5 get hL = 505 kJ/kg.

32. For saturated steam at 172.4 kPa, from steam table, we get
TS = 115.6 ºC and λ = 2214 kJ/kg.
To get HV for superheated vapor, first we obtain the enthalpy at Tsat =
48.9 ºC and P1 = 11.7 kPa, get Hsat = 2590 kJ/kg. Then using heat
capacity of 1.884 kJ/kg.K for superheated steam. So H = H + c BPR
32
1 sat
capacity of 1.884 kJ/kg.K for superheated steam. So HV = Hsat + cP BPR
= 2590 + 1.884 (40.6) = 2667 kJ/kg.
Substituting into heat balance equation and solving for S,
F hF + S λ = L hL + V HV
4535 (214) + S (2214) = 1814 (505) + 2722 (2667)
S = 3255 kg steam /h.

33. The heat q transferred through the heating surface area, A is
q = S (λ)
q = 3255 (2214) (1000 / 3600) = 2 002 000 W
Solving for capacity single-effect evaporator equation;
33
Solving for capacity single-effect evaporator equation;
q = U A ∆T = U A (TS – T1)
2 002 000 = 1560 A (115.6 – 89.5)
Solving, A = 49.2 m2.
Steam economy = 2722/3255
= 0.836

34. EVAPORATION
feed, TF
vapor T1
(1)
T
(2)
T
(3)
T
vapor T2 vapor T3
to vacuum
condenser
34
steam, TS
concentrate
from first
effect.
T1 T2 T3
concentrate
from second
effect.
concentrated
product
condensate
Simplified diagram of forward-feed triple-effect evaporator

35. EVAPORATION
Forward-feed multiple/triple-effect evaporators;
- the fresh feed is added to the first effect and flows to
the next in the same direction as the vapor flow.
- operated when the feed hot or when the final concentrated
35
- operated when the feed hot or when the final concentrated
product might be damaged at high temperature.
- at steady-state operation, the flowrates and the rate of
evaporation in each effect are constant.
- the latent heat from first effect can be recovered and reuse.
The steam economy ↑, and reduce steam cost.
- the Tbp ↓ from effect to effect, cause P1 ↓.

36. EVAPORATION
Calculation Methods for Multiple-effect Evaporators.
Objective to calculate;
- temperature drops and the heat capacity of
evaporator.
- the area of heating surface and amount of vapor leaving
36
- the area of heating surface and amount of vapor leaving
the last effect.
Assumption made in operation;
- no boiling point rise.
- no heat of solution.
- neglecting the sensible heat necessary to heat
the feed to the boiling point.

37. EVAPORATION
Heat balances for multiple/triple-effect evaporator.
- the amount of heat transferred in the first effect is
approximately same with amount of heat in the second
effect,
q = U1 A1 ∆T1 = U2 A2 ∆T2 = U3 A3 ∆T3
37
q = U1 A1 ∆T1 = U2 A2 ∆T2 = U3 A3 ∆T3
- usually in commercial practice the areas in all effects are
equal,
q/A = U1 ∆T1 = U2 ∆T2 = U3 ∆T3
- to calculate the temperature drops in evaporator,
Σ ∆T = ∆T1 + ∆T2 + ∆T3 = TS – T3

38. - hence we know that Σ ∆T are approximately inversely
proportional to the values of U,
3
2
1
1
1
1
1
1
1
U
U
U
U
T
T
+
+
Σ∆
=
∆
38
- similar eq. can be written for ∆T2 ∆T3
- if we assumed that the value of U is the
same in each effect, the capacity equation,
q = U A (∆T1 + ∆T2 + ∆T3 ) = UA Σ ∆T

39. EVAPORATION
steam, T
feed, TF
vapor T1
(1)
T1
(2)
T2
(3)
T3
vapor T2 vapor T3
to vacuum
condenser
39
Simplified diagram of backward-feed triple-effect evaporator
steam, TS
T1 T2 T3
concentrated
product
condensate

40. EVAPORATION
Backward-feed multiple/triple-effect evaporators;
- fresh feed enters the last and coldest effect and continues
on until the concentrated product
leaves the first effect.
- advantageous when the fresh feed is cold or
40
- advantageous when the fresh feed is cold or
when concentrated product is highly viscous.
- working a liquid pump since the flow is from low
to high pressure.
- the high temperature in the first effect reduce the
viscosity and give reasonable heat-transfer
coefficient.

41. Step-by-step Calculation Method for Triple-effect Evaporator (Forward Feed)
For the given x3 and P3 and BPR3
From an overall material balance, determine VT = V1 + V2 + V3
(1st trial – assumption)
Calculate the amount of concentrated solutions their concentrations in each effect using material
balances.
Find BPR ∆T in each effect Σ∆T.
41
Find BPR ∆T in each effect Σ∆T.
If the feed is very cold, the portions may be modified appropriately, calculate the boiling point in
each effect.
Calculate the amount vaporized and concentrated liquid in each effect through energy material
balances.
If the amounts differ significantly from the assumed values in step 2, step 2, and 4 must be repeated with
the amounts just calculated.
Using heat transfer equations for each effect, calculate the surface required for each effect
If the surfaces calculated are not equal, revise the ∆TS . Repeat step 4 onward until the areas are distributed satisfactorily.

42. A triple-effect forward-feed evaporator is being used to evaporate a sugar
solution containing 10 wt% solids to a concentrated solution of 50 %. The boiling-
point rise of the solutions (independent of pressure) can be estimated from (BPR
ºC = 1.78x + 6.22 x2 ), where x is wt fraction of sugar in solution. Saturated
steam at 205.5 kPa and 121.1ºC saturation temperature is being used. The
pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 22
EXAMPLE. 8.5-1 : Evaporation of Sugar Solution in a
Triple- Effect Evaporator
42
pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 22
680 kg/h at 26.7 ºC. the heat capacity of the liquid solutions is cP = 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 U1 = 3123, U2 = 1987, and U3 = 1136
W/m2.K. If each effect has the same surface area, calculate the area, the steam
rate used, and the steam economy.

43. EVAPORATION
S = ?
(2)
T3
T1
T2
F = 22680
xF = 0.1
TF = 26.7 ºC
V1 = 22,680 – L1
(1) (3)
V2 = L1 – L2
V3 = L2 - 4536
P3 = 13.7 kPa
43
S = ?
TS1 = 121.1 ºC
PS1 = 205.5 kPa
T1 , L1 , x1
TS1 TS3
TS2
T2 , L2 , x2
T3
L3 = 4536
x3 = 0.5
Fig. 8.5-1: Flow diagram for example 8.5-1

44. Solution,
The process flow diagram is given in Fig. 8.5-1..
Step 1,
From steam table, at P3 = 13.4 kPa, get Tsat = 51.67 ºC. Using the
BPR equation for third effect with xL = 0.5,
44
BPR3 = 1.78 (0.5) + 6.22 (0.52) =2.45 ºC.
T3 = 51.67 + 2.45 = 54.12 ºC. (BPR = T – Ts)
Step 2,
Making an overall and a solids balance.
F = 22 680 = L3 + (V1 + V2 + V3)
FxF = 22 680 (0.1) = L3 (0.5) + (V1 + V2 + V3) (0)
L3 = 4536 kg/h
Total vaporized = (V1 + V2 + V3) = 18 144 kg/h

45. Assuming equal amount vaporized in each effect,
V1 = V2 = V3 = 18 144 / 3 = 6048 kg/h
Making a total material balance on effects 1, 2, and 3, solving
F = 22 680 = V1 + L1 = 6048 + L1, L1 = 16 632 kg/h.
45
F = 22 680 = V1 + L1 = 6048 + L1, L1 = 16 632 kg/h.
L1 = 16 632 = V2 + L2 = 6048 + L2, L2 = 10 584 kg/h.
L2 = 10 584 = V3 + L3 = 6048 + L3, L3 = 4536 kg/h.
Making a solids balance on each effect, and solving for x,
22 680 (0.1) = L1 x1 = 16 632 (x1), x1 = 0.136
16 632 (0.136) = L2 x2 = 10 584 (x2), x2 = 0.214
10 584 (0.214) = L3 x3 = 4536 (x3), x3 = 0.5 (check)

46. Step 3, The BPR in each effect is calculated as follows:
BPR1 = 1.78x1 + 6.22x1
2 = 1.78(0.136) + 6.22(0.136)2
= 0.36ºC.
BPR2 = 1.78(0.214) + 6.22(0.214)2 =0.65ºC.
BPR3 = 1.78(0.5) + 6.22(0.5)2 =2.45ºC. then,
Σ∆T available = T – T (sat) – (BPR + BPR + BPR )
46
Σ∆T available = TS1 – T3 (sat) – (BPR1 + BPR2 + BPR3 )
= 121.1 – 51.67 – (0.36+0.65+2.45) = 65.97ºC.
Using Eq.(8.5-6) for ∆T1 , ∆T2 , and ∆T3
∆T1 = 12.40 ºC ∆T2 = 19.50 ºC ∆T3 = 34.07 ºC
3
2
1
1
1
1
1
1
1
U
U
U
U
T
T
+
+
Σ∆
=
∆
( )
)
1136
1
(
)
1987
1
(
)
3123
1
(
)
3123
1
(
97
.
65
+
+
=

47. However, since a cold feed enters effect number 1, this effect requires
more heat. Increasing ∆T1 and lowering ∆T2 and ∆T3 proportionately
as a first estimate, so
∆T1 = 15.56ºC ∆T2 = 18.34 ºC ∆T3 = 32.07 ºC
The above data ∆T1, ∆T2 and ∆T3 are getting from iteration-s
47
To calculate the actual boiling point of the solution in each effect,
T1 = TS1 - ∆T1 = 121.1 – 15.56 = 105.54 ºC.
T2 = T1 - BPR1 - ∆T2 = 105.54 – 0.36 – 18.34 = 86.84 ºC.
TS2 = T1 –BPR1 = 105.54 – 0.36 = 105.18 ºC.
T3 = T2 - BPR2 - ∆T3= 86.84 – 0.65 – 32.07 = 54.12 ºC.
TS3 = T2 –BPR2 = 86.84 – 0.65 = 86.19 ºC.

48. The temperatures in the three effects are as follows:
Effect 1 Effect 2 Effect 3 Condenser
TS1 = 121.1ºC TS2 = 105.18 TS3 = 86.19 TS4 = 51.67
T1 = 105.54 T2 = 86.84 T3 = 54.12
48
Step 4,
The heat capacity of the liquid in each effect is calculated from the
equation cP = 4.19 – 2.35x.
F: cPF = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.K
L1: cP1 = 4.19 – 2.35 (0.136) = 3.869 kJ/kg.K
L2: cP2 = 4.19 – 2.35 (0.214) = 3.684 kJ/kg.K
L3: cP3 = 4.19 – 2.35 (0.5) = 3.015 kJ/kg.K

49. The values of the enthalpy H of the various vapor streams relative to
water at 0 ºC as a datum are obtained from the steam table as follows:
Effect 1:
H1 = HS2 + 1.884 BPR1 = 2684 + 1.884(0.36) 2685 kJ/kg.
λ = H – h = 2708 – 508 = 2200 kJ/kg.
49
λS1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.
Effect 2:
H2 = HS3 + 1.884 BPR2= 2654 + 1.884(0.65) = 2655 kJ/kg.
λS2 = H1 – hS2 = 2685 – 441 = 2244 kJ/kg.
Effect 3:
H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.
λS3 = H2 – hS3 = 2655– 361 = 2294 kJ/kg.

50. Write the heat balance on each effect. Use 0ºC as a datum.
FcPF (TF –0) + SλS1 = L1cP1 (T1 –0) + V1H1 ,, ………(1)
22680(3.955)(26.7-0)+2200S = 3.869L1(105.54-0)+(22680-L1)2685
L1cP1 (T1 –0) + V1λS2 = L2cP2 (T2 –0) + V2H2 ………(2)
3.869L (105.54-0)+(22680-L )2244=3.684L (86.84-0)+(L -L )2655
50
3.869L1(105.54-0)+(22680-L1)2244=3.684L2(86.84-0)+(L1-L2)2655
L2cP2 (T2 –0) + V2λS3 = L3cP3 (T3 –0) + V3H3 ………(3)
3.68L2(86.84-0)+(L1-L2)2294=4536(3.015)(54.1-0)+(L2-4536)2600
Solving (2) and (3) simultaneously for L1L2 and substituting into(1)
L1 = 17078 kg/h L2 = 11068 kg/h L3 = 4536 kg/h
S = 8936kg/h V1 = 5602kg/h V2 = 6010kg/h
V3 = 6532kg/h

51. EVAPORATION
Step 5, Solving for the values of q in each effect and area,
( ) W
x
x
S
q S
6
1
1 10
460
.
5
1000
2200
3600
8936
=






=
= λ
( ) W
x
x
V
q S
6
2
1
2 10
492
.
3
1000
2244
3600
5602
=






=
= λ
( )
6010 

51
( ) W
x
x
V
q S
6
3
2
3 10
830
.
3
1000
2294
3600
6010
=






=
= λ
( )
2
6
1
1
1
1 4
.
112
65
.
15
3123
10
460
.
5
m
x
T
U
q
A =
=
∆
=
( )
2
6
2
2
2
2 8
.
95
34
.
18
1987
10
492
.
3
m
x
T
U
q
A =
=
∆
=
( )
2
6
3
3
3
3 1
.
105
07
.
32
1136
10
830
.
3
m
x
T
U
q
A =
=
∆
= 2
3
2
1
4
.
104
3
)
(
m
A
A
A
Am =
+
+
=

52. EVAPORATION
Am = 104.4 m2, the areas differ from the average value by less than
10 % and a second trial is really not necessary. However, a second
trial will be made starting with step 6 to indicate the calculation
methods used.
52
Step 6,
Making a new solids balance by using the new L1 = 17078, L2 =
11068, and L3 = 4536, and solving for x,
22 680 (0.1) = L1 x1 = 17 078 (x1), x1 = 0.133
17 078 (0.130) = L2 x2 = 11 068 (x2), x2 = 0.205
11 068 (0.205) = L3 x3 = 4536 (x3), x3 = 0.5 (check)

53. EVAPORATION
Step 7. The new BPR in each effect is then,
BPR1 = 1.78(0.133) + 6.22(0.13)2 =0.35ºC.
BPR2 = 1.78(0.205) + 6.22(0.205)2 =0.63ºC.
BPR3 = 1.78(0.5) + 6.22(0.5)2 =2.45ºC. then,
Σ∆T available = 121.1 – 51.67 – (0.35+0.63+2.45) = 66.0 ºC.
53
Σ∆T available = 121.1 – 51.67 – (0.35+0.63+2.45) = 66.0 ºC.
The new ∆T are obtained using Eq.(8.5-11),
( ) C
A
A
T
T
m
°
=
=
∆
=
∆ 77
.
16
4
.
104
4
.
112
56
.
15
1
1
'
1
( ) C
A
A
T
T
m
°
=
=
∆
=
∆ 86
.
16
4
.
104
8
.
95
34
.
18
2
2
'
2
( ) C
A
A
T
T
m
°
=
=
∆
=
∆ 34
.
32
4
.
104
1
.
105
07
.
32
3
3
'
3
C
T °
=
+
+
=
Σ∆ 97
.
65
34
.
32
86
.
16
77
.
16

54. These ∆T’ values are readjusted so that ∆T 1`= 16.77,
∆T 2`= 16.87, ∆T 3` = 32.36, and Σ∆T = 66.0 ºC. To
calculate the actual boiling point of the solution in each
effect,
(1) T1 = TS1 + ∆T 1` = 121.1 – 16.77 = 104.33ºC
(2) T2 = T1 – BPR1 - ∆T 2` = 104.33 – 0.35 – 16.87 = 87.11 ºC
TS2 = T1 – BPR1 = 104.33 – 0.35 = 103.98ºC
(3) T3 = T2 – BPR2 - ∆T 3` = 87.11 – 0.63 – 32.36 = 54.12 ºC
54
(3) T3 = T2 – BPR2 - ∆T 3` = 87.11 – 0.63 – 32.36 = 54.12 ºC
TS3 = T2 – BPR2 = 87.11 – 0.63 = 86.48 ºC.
Step 8;
Following step 4 to get cP = 4.19 – 2.35x,
F: cPF = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.K
L1: cP1 = 4.19 – 2.35 (0.133) = 3.877 kJ/kg.K
L2: cP2 = 4.19 – 2.35 (0.205) = 3.705 kJ/kg.K
L3: cP3 = 4.19 – 2.35 (0.5) = 3.015 kJ/kg.K

55. Then the new values of the enthalpy are,
(1) H1 = HS2 + 1.884 BPR1 = 2682 + 1.884(0.35) = 2683 kJ/kg.
λS1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.
(2) H2 = HS3 + 1.884 BPR2 = 2654 + 1.884(0.63) = 2655 kJ/kg.
λS2 = H1 – hS2 = 2683 – 440 = 2243 kJ/kg.
(3) H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.
λS3 = H2 – hS3 = 2655– 362 = 2293 kJ/kg.
55
Writing a heat balance on each effect,and solving,
(1) 22680(3.955)(26.7-0)+2200S = 3.877L1(104.33-0)+(22680-L1)2683
(2) 3.877L1(104.33-0)+(22680-L1)2243=3.708L2(87.11-0)+(L1-L2)2655
(3) 3.708L2(87.11-0)+(L1-L2)2293=4536(3.015)(54.1-0)+(L2-4536)2600
L1 = 17005 kg/h L2 = 10952 L3 = 4536 S = 8960
V1 = 5675 V2 = 6053 V3 = 6416

56. EVAPORATION
Solving for q and A in each effect,
( ) W
x
x
S
q S
6
1
1 10
476
.
5
1000
2200
3600
8960
=






=
= λ
( ) W
x
x
V
q S
6
2
1
2 10
539
.
3
1000
2243
3600
5675
=






=
= λ
( ) W
x
x
V
q S
6
3
2
3 10
855
.
3
1000
2293
3600
6053
=






=
= λ
56
( ) W
x
x
V
q S3
2
3 10
855
.
3
1000
2293
3600
=




=
= λ
( )
2
6
'
1
1
1
1 6
.
104
77
.
16
3123
10
476
.
5
m
x
T
U
q
A =
=
∆
=
( )
2
6
'
2
2
2
2 6
.
105
87
.
16
1987
10
539
.
3
m
x
T
U
q
A =
=
∆
=
( )
2
6
'
3
3
3
3 9
.
104
36
.
32
1136
10
855
.
3
m
x
T
U
q
A =
=
∆
=

57. EVAPORATION
The average area Am = 105.0 m2 to
use in each effect.
57
steam economy = 025
.
2
8960
6416
6053
5675
3
2
1
=
+
+
=
+
+
S
V
V
V