EVAPORATION
Evelyn R. Laurito
Ch.E. 411
LEARNING OUTCOMES
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
Describe when and how evaporation is
used as a separation technique in unit
operations
Determine the boiling point of solutions
Differentiate Single and Multiple effect
evaporation with multiple feeding
To set up the material and energy
balances and heat transfer equation
To solve the area and economy of
evaporation systems
WHAT IS EVAPORATION?
 The
conversion of a dilute
solution to a more concentrated
solution through the removal
of solvent
 Requires the use of a heating
medium, usually steam which
is in indirect contact through
a steam chest
 Boiling temperatures must be
reached for evaporation to take place.
 What is the boiling point of water at 1atm?
 Can you make a solution boil at
temperatures less than 100 oC? Ex. 25oC
 Operating pressure is low, usually in vacuum
units. (Ex. 100 mm Hg vac.)
1
VAPOR
STEAM
THICK
LIQUOR
FEED
CONDENSATE
Rising Film Evaporator
Vapor
Rising Film
Evaporator
Feed
Falling Film
Evaporator
Steam
Thick
Liquor
Steam
Condensate
Vapor
Condensate
Feed
Thick Liquor
2
Horizontal-Tube
Evaporator
Short-tube
Evaporator
WHAT IS BOILING POINT?
Boiling point is the temperature at which the confining
pressure becomes equal to the vapor pressure of the
liquid.
 Boiling Point of Solution (T I) is greater than the
Boiling Point of Solvent (T1)
 Boiling Point Elevation or Rise (BPR) = T I – T1
 Strong electrolytes have significant BPR, but organic
solutions usually have negligible BPR.
 Determination of BPR

Colligative Property Equation: BPR = i Kb m
Duhring’s Rule: The boiling point of solution varies linearly
with the boiling point of pure solvent. (See Geankoplis, Fig.
8.4-2/538)
 Boiling Point Nomograph (See Fig. 11-124/11-112 or 11-115/8th
ed))
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3
Find the BPR of a 2 molal solution of NaCl in
water at 1 atm.
 Find the BPR of a 30% NaOH solution at 660 mm
Hg vacuum (Tsatn=53.97oC).


Find the BPR of 30% KCl at 1 atm solution using
Colligative properties and Nomograph
Duhring’s Chart for NaOH
4
H-x-T diagram for NaOH-H2O
SINGLE EFFECT EVAPORATOR
SINGLE EFFECT EVAPORATOR
E, TI,HI
P1
Vo, P0
Ts, Hs
C, P0
Tc, hc
F, xF
TF, hF
L, xL
TI, hL
F = feed or thin liquor rate
L = thick liquor rate
x = solute fraction
E = evaporation rate or capacity
P1 = operating pressure
P0 = steam and condensate pressure
Vo = steam rate = C = condensate rate
Ts, Tc1 = steam, condensate temperatures
TI = boiling point of L = T of vapor
TF = feed temperature
H, h = enthalpy symbols
5
MATERIAL BALANCES

Overall Balance
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F + Vo = E + C + L (but Vo = C)
F=E+L
Solute Balance
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F xF = L xL = Throughput Rate
Throughput Rate = constant
ENTHALPY BALANCE
General Form
 F hF + Vo Hs = E HI + C hc1 + L hL
 For systems w/ h-x-t Data (Ex. NaOH-H2O, Fp.504)
 Vo (Hs- hc1) = E HI + L hL – F hF = q
 hI, hF based on (xL,TI), (xF, TF)
 Hs, hc based on steam table
 For systems with negligible heat of dilution (no
enthalpy data) with liquid enthalpy as zero at the
boiling point of solution.
 hF = CPF (TF – TI); hL = CPL(TI- TI) = 0
 HI = CPW(T1 – TI) + L1 + CPS (TI – T1)
CPW = 4.187, CPS = 1.884, L1 = latent heat at T1
HI = L1 – 2.303 BPR

NEGLIGIBLE HEAT OF DILUTION: USING
TI AS REFERENCE TEMPERATURE:
ASSUME THE FEED, THICK
LIQUOR AND EVAPORATED SOLVENT ARE
INITIALLY ALL LIQUID AT TI
hF = CpF (TF - TI)
hL = CpL (TI – TI)
HE = 4.187 (T1-TI) + L1 + 1.884 (TI-T1)
E starts as liquid water at TI, boils at T1 and
becomes superheated vapor at TI
or HE = L1 - 2.303(TI-T1) = L1 - 2.303 BPR
In Engl Units: HE = L1 - 0.55 BPR
Substituting into the Enthalpy balance eqn gives the
following:
Heat Balance Equation:
q = Vo(Hs-hc) = E [L1 - 2.303BPR] + F CpF (TI-TF)
6
EVALUATION
OF
H S - HC
Saturated Steam Condensing without subcooling:
 HS - hc = Lo = latent heat at To = sat’n temp.
 Superheated Steam Condensing without
subcooling:
 HS - hc = 1.884(TS - To) + Lo
 Superheated Steam Condensing with subcooling:
 HS - hc = 1.884(TS - To)+ Lo + 4.187(To - Tc)
 Saturated Steam Condensing with subcooling:
 HS - hc = Lo + 4.187(To - Tc)

HEAT TRANSFER EQUATION
q = U A T
q = heat rate supplied by steam = Vo (Hs – hc1)
U = overall heat transfer coefficient
A = heat transfer area needed
T = effective temperature driving force
= To – TI = To – T1 - BPR
To = saturation temperature of steam
EVAPORATOR ECONOMY (E)
This is the ratio of evaporation rate or capacity to
the steam rate
 e = E/Vo
 Ideally, e must be at least 1 to compensate for the
steam condensation
 Single effect evaporation does not usually result
in a high economy.
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7
EXAMPLE 1
Ex.1] A single effect evaporator will handle
18,000 kg/hr of an organic colloid in water.
The solution will be concentrated from 15% to
60% solids. Steam available is at 275 kPa and
140oC. The feed enters at 24oC with a specific
heat of 3.768 kJ/kgK. A pressure of 660 mm
Hg vacuum shall be maintained in the vapor
space. The solution has a negligible BPR.
Overall heat transfer coefficient is 2840
W/m2K. Calculate the economy and evaporator
area.
Given:
Req'd: a) Economy b) A
Sol'n:
OMB: 18000 = L+E
SB: 18000(.15) = L(.6)
L = 4500
E = 13500
at Po = 275 kPa, To = 130.6oC, Lo = 2171.8 kJ/kg
at 660 mm Hg vac or 100 mm Hg abs (13.33 kPa), T1 = 50.6oC
L1 = 2379.1 kJ/kg
HBE: q = Vo[Lo + 1.884 (Ts-To)] = EL1 + F CpF (T1 - TF)
Vo[2171.8+1.884(140-130.6)] = 13500(2379.1)+18000(3.768)(50.6-24)
solving: Vo = 15492 kg/hr
a) Economy = E/Vo = 13500/15492 = 0.87
HTE: q = UA (T0-T1)
= 2840 (3600/1000) A ( 130.6-50.6)
solving : A = 41.47 m2
8
EX.2

Ex 2] A single effect evaporator is used to
produce a concentrate of 10,000 kg/hr of a 20%
solution of NaCl to 35% solids. Saturated steam
is at 275 kPa and the vapor space at 100 mm Hg
abs. Overall heat transfer coefficient is 1250
W/m2K. The feed enters at 30oC with Cp of 3.5
kJ/kgK. Calculate the economy and evaporator
area.
EX.2

Ex 2] A single effect evaporator is used to
concentrate 20,000 lb/hr of a 20% solution of
NaOH to 50% solids. Saturated steam is at 20
psig and the vapor space at 100 mm Hg. Overall
heat transfer coefficient is 250 BTU/hr ft2-oF.
The feed enters at 100oF. Calculate the economy
and evaporator area.
9
Duhring’s Chart for NaOH
H-x-T diagram for NaOH-H2O
Given:
Reqd: a)Economy
b) A
Soln:
OMB: 20000 = L+E
SB: 20000(.2) = L(.5)
L = 8000
E = 12000 lb/hr
EB:
q = VoLo = EHE + LhL - FhF
for steam at 20 psig or 34.7 psia, To = 258.7oF
Lo = 939.5 BTU/lb
for vapor: at 100 mm Hg or 1.93 psia, T1 = 124.7oF
H1 = 1115.6 BTU/lb
From Duhring's Chart: TI =(T1=124.7,xL=0.5) = 197oF
BPR = 197 -124.7 = 72.3oF
From Enthalpy- Concentration Diagram:
hF = (TF=100oF, xF=0.2) = 55 BTU/lb
hL = (TI=197oF, xL=0.5) = 221 BTU/lb
Substituting in EBE:
Vo(939.5) = 12000 [1115.6 + .45(72.3)] + 8000(221) -20000(55)
Vo = 15375.8 lb/hr
a) Economy = 12000/15375.8 = 0.78
HTE: q = VoLo = UA(To-TI)
= 250 A (258.7-197)
A = 936 ft2
10
EX.3
Ex 3] A single effect evaporator is used to concentrate
1000 kg/hr of a 10% solution of NaCl to 30% solids.
Saturated steam is at 20 psig and the vapor space at 15
KPa. Overall heat transfer coefficient is 1250 W/m2K.
The feed enters at 25 oC . Calculate the economy and
evaporator area. Cp of NaCl solution as 2.1 kJ/kgK
At 15 kPa: T1=53.97 oC L1= 2372.37 kJ/kg
T1=129.15 oF
At 20 psig or 239.2 kPa T0=125.9 oC L0=2185.3 kJ/kg

Ex 3] A single effect evaporator is used to
produce a concentrate of 1000 kg/hr of a 30%
solution of KCl from a 10% solution. Saturated
steam is at 20 psig and the vapor space at 15
KPa. Overall heat transfer coefficient is 1250
W/m2K. The feed enters at 25 oC . Calculate the
economy and evaporator area. Cp of KCl solution
as 2.1 kJ/kgK
EXAMPLE
5 MT/hr of an aqueous solution with 10% sugar will be concentrated
to 50% in a single evaporator system operating at 694 mm Hg vacuum.
Saturated steam at 113oC will condense without subcooling in the
steam chest. Assume that the feed has a specific heat of 3.5 kJ/kg-K,
and the evaporator has an over-all coefficient of 2500 W/m2-K.
Determine the steam requirement, economy and heat transfer area
Needed for the evaporator if the feed enters at:
at a) 25oC
b) 43oC
c) 70oC
Answers: a)A = 15.7381m2 b)A = 15.2381m2 c)A = 14.4881m2
11
SOLUTION
F  5000
xF  0.1
CpF  3.5
Ts  113
U  2500
Pv  694
P 
( Pb  Pv )
760

1
xL  0.5
Pb  760
P  0.08571
1.01325
bar
Material Balances:
F = E + L F xF = L xL
F  xF
L 
L  1000
xL
E  F  L
E  4000
SOLUTION
Case of Negligible BPR/ No Enthalpy Data
Enthalpy Balance:
Vo (Hs - hc1) = E H I + L h I - F hF
From steam table:
T1  316  273
BPR  0
T1  43
HI  L1
To  Ts  273
L1  2579  179
hI  0
To  386
Lo  2222.6
TI  T1
Lo  2696  473.4
HI  2400
SOLUTION
TF  25
Vo 
e 
hF  CpF   TF  TI
E  HI  L  hI  F  hF
Lo
E
Vo
Vo  4460.99163
e  0.89666
 T  Ts  T1
A 
hF  63
Vo  Lo
U   T  3.6
 T  70
A  15.7381
12
SOLUTION
hF  CpF   TF  TI
TF  43
Vo 
e 
E  HI  L  hI  F  hF
Lo
E
Vo  4319.26572
e  0.92608
Vo
 T  Ts  T1
A 
hF  0
 T  70
Vo  Lo
A  15.2381
U   T  3.6
SOLUTION
TF  70
Vo 
e 
hF  CpF   TF  TI
E  HI  L  hI  F  hF
Lo
E
Vo
hF  94.5
Vo  4106.67686
e  0.97402
 T  Ts  T1
Vo  Lo
A 
U   T  3.6
 T  70
A  14.4881
13