Chapter 10 - Evaporation
Chapter 10 - Evaporation
Chapter 10 - Evaporation
Evaporation
Feed
(volatile solvent
& non-volatile solute)
Concentrated solution
Case of heat transfer to a boiling liquid.
Vapor from a boiling liquid solution is removed and a more concentrated solution
remains.
Examples:
- Concentration of fruit juice, milk or caustic soda
- Production of high quality water from river/sea
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Production of Milk Powder
Standardization to adjust
the content of fat in milk
Heating to sterilize
Falling-film evaporator Spray dryer
to get milk powder
JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 10-3
Processing Factors
(1)Concentration
- Dilute feed: viscosity , heat transfer coefficient, h also TBPsoln. TBPwater
- Concentrated solution: , and h & TBPsoln. (boiling point rise, BPR)
(2) Solubility
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(5) Foaming/frothing
- Caustic solutions, food solutions, fatty acid solutions form foam/froth
during boiling.
Minimize corrosion.
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Types of Evaporation Equipment
Horizontal tube type
Vertical tube type
Long tube vertical type
Forced-circulation type
Open kettle or pan
Open-pan solar evaporator
Falling-film-type evaporator
Agitated-film evaporator
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Effect of Processing Variables on Evaporator Operation
(1) Feed Temperature (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.
- If the feed is under pressure & TF > TBP, additional vaporization is obtained
by flashing of feed.
- Preheating the feed can reduce the size of the evaporator heat transfer area
(2) Evaporator Pressure (P1): q = U A T = UA (TS T1)
- Desirable: large T A & cost . T1 = TBP
JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-7
Boiling Point Rise of Solutions (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.
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Enthalpy-concentration Charts of Solutions
- For large heat of solution of the aqueous solution
- To get values for hF and hL
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Typical Heat Transfer Coefficients for Various Evaporators
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Methods of Operation of Evaporators
1. Single-effect Evaporators
- Used when the required capacity of operation is small and/or the cost of steam is
cheap compared to the evaporator cost.
- Is wasteful of energy since the latent heat of the vapor leaving is not used.
TS , HS Condensate, S
TS , hS
Concentrated liquid, L
T1 , xL , hL
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2. Forward-feed Multiple-effect Evaporators
Used when the feed is hot or when the final concentrated product might be
damaged at high temperatures
T1 > T2 > T3, P1 (atm.) > P2 > P3 (vac.), C1 < C2 < C3,
T1 T2 T3
steam, TS
condensate
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3. Backward-feed Multiple-effect Evaporators
- Used when the feed is cold since a smaller amount of liquid must be heated
to the high temps in the second and first effects
- Used when the final concentrated product is highly viscous. The high
temperature in the first effect reduce the viscosity and give reasonable heat-
transfer coefficient.
condensate
concentrated
product
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4. Parallel-feed Multiple-effect Evaporators
- Involve the adding of fresh feed and the withdrawal of concentrated product
from each effect
- Used when the feed is almost saturated and solid crystals are the product
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Calculations Methods for Single-effect Evaporators
Vapor,V to condenser
T1 , yV , HV
Feed, F P1
heat-exchanger
TF , xF , hF.
tubes
Steam, S A
TS , HS Condensate, S
TS , hS
Concentrated liquid, L
T1 , xL , hL
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(2) To calculate A or U:
- No boiling point rise and negligible heat of solution:
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(3) To get BPR and the heat of solution:
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Example 8.4-1: Heat-Transfer Area in Single-Effect Evaporator
Calculate the amounts of vapor and liquid product and the heat-
transfer area required. Assume that, since it its dilute, the solution has
the same boiling point as water.
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V=?
F = 9072 kg/h T1 , y V , HV
TF = 311 K
xF = 0.01 P1 = 101.325 kPa
hF.
U = 1704 W/m2
T1 =? A = ?
S , TS , HS
PS = 143.3 kPa
S, TS , hS
L=?
T1 , hL
xL = 0.015
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Solution
Total material balance:
F = L + V 9072 = L + V
Balance on the solute (solids) alone:
F xF = L xL 9072 (0.01) = L (0.015)
L = 6048 kg/h of liquid V = 3024 kg/h of vapor
Since it is assumed that the solution is dilute as water cpF = 4.14 kJ/kg. K
hF = cpF (TF T1) (Tref = T1)
hF = 4.14 (311.0 373.2) = -257.508 kJ/kg
hL = cPL(T1 Tref) = 0, since it is at T1 = Tref = 373.2 K
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Substituting into heat balance equation:
F hF + S = L hL + V H V
A = 149.3 m2
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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 C 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.
F = 4536 kg/h V, T1 , yV , HV
TF = 60 C
xF = 0.2 P1 = 11.7 kPa
hF.
U = 1560 W/m2
T1 A= ?
S=?
TS , HS
PS = 172.4 kPa S, TS , hS
L, T1 , hL
xL = 0.5
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Solution
Total material balance:
F = L + V 4536 = L + V
Balance on the solute (solids) alone:
F xF = L xL 4536 (0.2) = L (0.5)
L = 1814 kg/h of liquid V = 2722 kg/h of vapor
First obtain Tsat of pure water from steam table at 11.7 kPa Tsat = 48.9 C
From Duhring chart (Fig. 8.4-2), for a Tsat = 48.9 C and 50 % NaOH :
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From the enthalpy-concentration chart (Fig.8.4-3):
At TF = 60 C and xF = 0.2 hF = 214 kJ/kg
At T1 = 89.5 C and xL = 0.5 hL = 505 kJ/kg
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For saturated steam at 172.4 kPa (from steam table):
TS = 115.6 C and = 2214 kJ/kg.
To get HV for superheated vapor:
first obtain the enthalpy at Tsat = 48.9 C and P1 = 11.7 kPa:
Hsat = 2590 kJ/kg.
Then using heat capacity of 1.884 kJ/kg.K for superheated steam.
HV = Hsat + cP (BPR) = 2590 + 1.884 (40.6) = 2667 kJ/kg
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The heat q transferred through the heating surface area, A is
q = S ()
q = 3255 (2214) (1000 / 3600) = 2 002 000 W
A = 49.2 m2.
JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 10-28