Evaporation.ppt

BKC3413: Chapter 7 FKKSA, KUKTEM
1
Chapter 2
EVAPORATION

2
Content
• Type of Evaporation equipment and
Methods
• Overall Heat Transfer Coefficient in
Evaporators
• Calculation Methods for Single Effect
Evaporators
• Calculation Methods for Multiple Effects
Evaporators
• Condenser for Evaporator
• Evaporation using Vapor Recompression

3
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.
• 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.

4
Processing Factors
Concentration in
the liquid
solubility
Temperature sensitivity
of materials
Foaming or frothing
Pressure and temperature
Scale deposition
Materials of construction

5
Processing Factors
• Concentration
dilute feed, viscosity , heat transfer coefficient, h
concentrated solution/products, , and h .
• Solubility
concentration , solubility  , crystal formed.
solubility  with temperature .
• Temperature.
heat sensitive material degrade at higher temperature &
prolonged heating.

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

7
Effect of Processing Variables on Evaporator Operation.
• 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.
Is the feed is under pressure & TF > Tbp, additional
vaporization obtained by flashing of feed.
• P1
desirable T  [Q = UA(TS – T1)],
A  & cost .
T1 depends on P1 will  T1.

8
• PS
 PS will  TS but high-pressure is costly.
optimum TS by overall economic balances.
• BPR
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.
• Enthalpy–concentration of solution.
for large heat of solution of the aqueous solution.
to get values for hF and hL.

9

10

11
feed, F
TF , xF , hF.
steam, S
TS , HS
concentrated liquid, L
T1 , xL , hL
condensate, S
TS , hS
vapor,V to condenser
T1 , yV , HV
P1
T1
heat-exchanger
tubes
Simplified Diagram of single-effect evaporator

12
• Single-effect evaporators;
• 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.
• 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.

13
Calculation Methods for Single-effect Evaporator.
• Objectives: to calculate
- vapor, V and liquid, L flowrates.
- heat transfer area, A
- overall heat-transfer coefficient, U.
- Fraction of solid content, xL.
• 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

14
• To calculate A or U,
- no boiling point rise and negligible heat of
solution:
calculate hF, hL, Hv and .
where,  = (HS – Hs)
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  = L hL + V HV
solve for A and U:
q = S  = U A T = UA (TS – T1)

15
• To get BPR and the heat of solution:
- calculate T1 = Tsat + BPR
- get hF and hL from Figure 8.4-3.
- get S & HV from steam tables for superheated
vapor or
HV = Hsat + 1.884 (BPR)
- solve for S:
- solve for A and U:
q = S  = U A T = UA (TS – T1)

16
Example 8.4-1: Heat-Transfer Area in Single-Effect
Evaporator.
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. 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.

17
U = 1704 W/m2
T1 A = ?
P1 = 101.325 kPa
F = 9072 kg/h
TF = 311 K
xF = 0.01
hF.
S , TS , HS
PS = 143.3 kPa
L = ?
T1 , hL
xL = 0.015
S, TS , hS
V = ?
T1 , yV , HV
Figure 8.4-1: Flow Diagram for Example 8.4-1

18
Solution;
Refer to Fig. 8.4-1 for flow diagram for this solution.
For the total balance,
F = L + V
9072 = L + V
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

19
Since we assumed the solution is dilute as water;
cpF = 4.14 kJ/kg. K (Table A.2-5)
From steam table, (A.2-9)
At P1 = 101.325 kPa, T1 = 373.2 K (100 ºC).
HV = 2257 kJ/kg.
At PS = 143.3 kPa, TS = 383.2 K (110 ºC).
 = 2230 kJ/kg.
The enthalpy of the feed can be calculated from,
hF = cpF (TF – T1)
hF = 4.14 (311.0 – 372.2)
= -257.508 kJ/kg.

20
Substituting into heat balance equation;
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
The heat q transferred through the heating surface
area, A is
q = S ()
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.

21
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.
calculate the steam used, the steam economy in kg
vaporized/kg steam used, and the heating surface
area in m2.

22
U = 1560 W/m2
T1 A = ?
P1 = 11.7 kPa
F = 4536 kg/h
TF = 60 ºC
xF = 0.2
hF.
S = ?
TS , HS
PS = 172.4 kPa
L, T1 , hL
xL = 0.5
S, TS , hS
V, T1 , yV , HV
Figure 8.4-4: Flow Diagram for Example 8.4-3

23
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,
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

24
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,
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.

25
For saturated steam at 172.4 kPa, from steam table, we get
TS = 115.6 ºC and  = 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 HV = Hsat + cP BPR
= 2590 + 1.884 (40.6) = 2667 kJ/kg.
Substituting into heat balance equation and solving for S,
4535 (214) + S (2214) = 1814 (505) + 2722 (2667)
S = 3255 kg steam /h.

26
The heat q transferred through the heating surface area, A is
q = S ()
q = 3255 (2214) (1000 / 3600) = 2 002 000 W
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

27
EVAPORATION
steam, TS
feed, TF
concentrate
from first
effect.
vapor T1
(1)
T1
(2)
T2
(3)
T3
concentrate
from second
effect.
concentrated
product
condensate
vapor T2 vapor T3
to vacuum
condenser
Simplified diagram of forward-feed triple-effect evaporator

28
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 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 .

29
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 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.

30
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
- 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,
 T = T1 + T2 + T3 = TS – T3

31
- hence we know that  T are approximately
inversely proportional to the values of U,
- 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  T
3
2
1
1
1
1
1
1
1
U
U
U
U
T
T






32
EVAPORATION
Simplified diagram of backward-feed triple-effect evaporator
steam, TS
feed, TF
vapor T1
(1)
T1
(2)
T2
(3)
T3
concentrated
product
condensate
vapor T2 vapor T3
to vacuum
condenser

33
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
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.

34
EVAPORATION
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 & 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.

35
EVAPORATION
Ex. 8.5-1 : Evaporation of Sugar Solution in a Triple-Effect
Evaporator.
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 22 680 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.

36
EVAPORATION
S = ?
TS1 = 121.1 ºC
PS1 = 205.5 kPa
(2)
T3
T1
T2
F = 22680
xF = 0.1
TF = 26.7 ºC
T1 , L1 , x1
V1 = 22,680 – L1
(1) (3)
V2 = L1 – L2
V3 = L2 - 4536
TS1 TS3
TS2
T2 , L2 , x2
T3
L3 = 4536
x3 = 0.5
P3 = 13.7 kPa
Fig. 8.5-1: Flow diagram for example 8.5-1

37
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,
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

38
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.
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)

39
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,
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





 
)
1136
1
(
)
1987
1
(
)
3123
1
(
)
3123
1
(
97
.
65




40
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
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.
The above data T1, T2 and T3 are getting from iteration-s

41
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
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

42
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.
S1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.
Effect 2:
H2 = HS3 + 1.884 BPR2= 2654 + 1.884(0.65) = 2655 kJ/kg.
S2 = H1 – hS2 = 2685 – 441 = 2244 kJ/kg.
Effect 3:
H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.
S3 = H2 – hS3 = 2655– 361 = 2294 kJ/kg.

43
Write the heat balance on each effect. Use 0ºC as a datum.
FcPF (TF –0) + SS1 = L1cP1 (T1 –0) + V1H1 ,, ………(1)
22680(3.955)(26.7-0)+2200S = 3.869L1(105.54-0)+(22680-L1)2685
L1cP1 (T1 –0) + V1S2 = L2cP2 (T2 –0) + V2H2 ………(2)
3.869L1(105.54-0)+(22680-L1)2244=3.684L2(86.84-0)+(L1-L2)2655
L2cP2 (T2 –0) + V2S3 = 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 L1&L2 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

44
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








 
  W
x
x
V
q S
6
2
1
2 10
492
.
3
1000
2244
3600
5602








 
  W
x
x
V
q S
6
3
2
3 10
830
.
3
1000
2294
3600
6010








 
 
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 




45
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.
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)

46
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,
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 




 97
.
65
34
.
32
86
.
16
77
.
16

47
These T’ values are readjusted so that T 1`= 16.77,
T 2`= 16.87, T 3` = 32.36, and 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
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

48
Then the new values of the enthalpy are,
(1) H1 = HS2 + 1.884 BPR1 = 2682 + 1.884(0.35) = 2683 kJ/kg.
S1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.
(2) H2 = HS3 + 1.884 BPR2 = 2654 + 1.884(0.63) = 2655 kJ/kg.
S2 = H1 – hS2 = 2683 – 440 = 2243 kJ/kg.
(3) H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.
S3 = H2 – hS3 = 2655– 362 = 2293 kJ/kg.
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

49
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








 
  W
x
x
V
q S
6
2
1
2 10
539
.
3
1000
2243
3600
5675








 
  W
x
x
V
q S
6
3
2
3 10
855
.
3
1000
2293
3600
6053








 
 
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 




50
EVAPORATION
The average area Am = 105.0 m2 to use in
each effect.
steam economy = ???? [Q/Vapor Flowrate]
025
.
2
8960
6416
6053
5675
3
2
1






S
V
V
V

51
THANK
YOU

Evaporation.ppt

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