CHEMICAL ENGINEERING SERIES
Compilation of Lectures and Solved Problems
CHEMICAL ENGINEERING SERIES
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. 2. 3.
Distillation: in evaporation vapor is usually usually a single component Drying: in evaporation, residue is liquid, sometimes a highly highly viscous one Crystallization: focus is on concentrating a solution rather than forming crystals crystals
Calculations for the Different Methods of Operations of Evaporators: 1.
Single Effect Evaporators – Evaporators – used 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,
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CHEMICAL ENGINEERING SERIES
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. 2. 3.
Distillation: in evaporation vapor is usually usually a single component Drying: in evaporation, residue is liquid, sometimes a highly highly viscous one Crystallization: focus is on concentrating a solution rather than forming crystals crystals
Calculations for the Different Methods of Operations of Evaporators: 1.
Single Effect Evaporators – Evaporators – used 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,
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CHEMICAL ENGINEERING SERIES
EVAPORATION
2.
Forward Feed Multiple Effect Evaporators – Evaporators – fresh 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
3.
P
Backward Feed Multiple Effect Evaporators – fresh – fresh feed enters the last and coldest effect and continues until the concentrated product leaves the first effect. This is used when 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
P3 , T 3
TI
TII
TIII
LI
LII
F
P
4.
Mixed Feed Multiple Effect Evaporators – Evaporators – fresh 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 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.
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Performance Evaluation of Steam-Heated Evaporators 1.
Capacity – Capacity – number number of kilograms of water vaporized per hour Evaporator Capacity
Where:
2.
-
Rate of heat transfer through the heating surface of an evaporator Over-all heat transfer coefficient Heat transfer surface area Over-all temperature drop
Steam Economy – Economy – number number of kilograms vaporized per kilogram of steam fed to the unit
Boiling point Evaluation (BPE) of (BPE) of a solution is the increase in boiling point over that of water 1. 2.
Small for dilute solutions and organic colloids solution Large enough for concentrated solutions of inorganic salts; BPE can be estimated using th Figure 11-124 (CHE HB 8 edition)
Dϋhring’s Dϋhring’s Rule – Rule – the 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:
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CHEMICAL ENGINEERING SERIES
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 205.5 205.5 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 – 4.19 – 2.35x, 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, 2 U2 = 1987, and U 3 = 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
P3 , T3
TI
TII
TIII
F
L2
L1
Step 1: rd From steam table, at 13.4 kPa (pressure of the vapor space 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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Where:
Assume , initially equal rate of evaporation in each effect
st
For the 1 effect:
nd
For the 2 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
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CHEMICAL ENGINEERING SERIES
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
nd
For 2 effect:
From steam table, at T V1 = 109°C
From steam table, H(vapor enthalpy) at
rd
For 3 effect:
From steam table, at TV2 = 89.89°C
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CHEMICAL ENGINEERING SERIES
EVAPORATION
From steam table, H(vapor enthalpy) at
Equate
and
Step 5: Solve for heat transfer area for each effect: st For 1 effect:
nd
For 2 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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
∑ ∑ ∑ st
nd
To solve for BPR for the 1 and 2 effects: st 1 effect:
nd
2 effect:
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:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
From steam table, steam at 121.1°C
From steam table, H(vapor enthalpy) at
nd
For 2 effect:
From steam table, at T V1 = 103.94°C
From steam table, H(vapor enthalpy) at
rd
For 3 effect:
From steam table, at T V2 = 87.61°C
From steam table, H(vapor enthalpy) at
Equate
and
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Step 5: Solve for heat transfer area for each effect: st For 1 effect:
nd
For 2 effect:
rd
For 3 effect:
Areas are almost equal, therefor e new assumptions are valid: ANSWERS: Area of each effect:
Steam Requirement
Steam Economy:
PROBLEM # 02:
A solution with a negligi ble boiling point rise is being evaporated in a tripl e effect evaporator using saturated steam at 121.1°C. The pressure in the vapor of the last effect is 25.6 kPa abs. The
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CHEMICAL ENGINEERING SERIES
EVAPORATION 2
heat transfer coefficients are U 1 = 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
VII
VIII
T1
T2
T3
TI
TII
TIII
p3 = 25.6 kPaabs
F
LI
LII
P
∑ ∑ Using the heat transfer area equation:
Assume equal heat transfer flux f or each effect
From steam table,
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CHEMICAL ENGINEERING SERIES
EVAPORATION
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CHEMICAL ENGINEERING SERIES
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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
At 3 atm, from steam table,
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CHEMICAL ENGINEERING SERIES
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 surf ace 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
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CHEMICAL ENGINEERING SERIES
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:
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CHEMICAL ENGINEERING SERIES
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 st
1 Effect Steam Feed to E-I Liquor from E-III Vapor to E-II Product n
2 Effect Feed from H- c Vapor to E-III Liquor to E-III rd
3 Effect Vapor to condenser Liquor to E-I Feed Feed to “b” Feed to “c”
DESIGNATION
TEMPERATURE, °F 230 106 172 187
162 144 145
101 106 60 91 134
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CHEMICAL ENGINEERING SERIES
EVAPORATION
st
Consider heat balance around 1 effect:
For x = 0.2692, Cp = 0.89156 BTU/lb·°F From steam table, at T 1 = 172 °F, λ = 995 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 T 2 = 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 T 3 = 101 °F, λ = 1036.44 BTU/lb
Equate
Substitute
and
and
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CHEMICAL ENGINEERING SERIES
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
SECOND TRIAL:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Recompute using the adjusted ΔT: STREAM st
DESIGNATION
1 Effect Steam Feed to E-I Liquor from E-III Vapor to E-II Product n
2 Effect Feed from H- c Vapor to E-III Liquor to E-III r
3 Effect Vapor to condenser Liquor to E-I Feed Feed to “b” Feed to “c”
TEMPERATURE, °F 230 106 170 185
160 143 144
101 106 60 91 133
st
Consider heat balance around 1 effect:
For x = 0.2692, Cp = 0.89156 BTU/lb·°F From steam table, at T 1 = 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 T 2 = 143 °F, λ = 1012.23 BTU/lb
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CHEMICAL ENGINEERING SERIES
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 T 3 = 101 °F, λ = 1036.44 BTU/lb
Equate
Substitute
and
and
From equation
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CHEMICAL ENGINEERING SERIES
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 T 3 = 101 °F, λ = 1036.44 BTU/lb
PROBLEM # 05:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
A single effect evaporator concentrat es 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
pvacuum = 600 mm Hg Pbarometric= 760 mm Hg
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
PROBLEM # 06:
P xP = 0.50
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CHEMICAL ENGINEERING SERIES
EVAPORATION
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
T1
TI
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,
From steam table at 172 kPa abs
P xP = 0.65
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Δ
PROBLEM # 07:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
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= 1 00 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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Δ
For steam at 50 psig
PROBLEM # 08:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
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 = 10 0 F
Vapor, V
pvacuum = 26”Hg Steam, S 20 psig NaOH soln F xF = 0.10 TF= 80 F
T1
t A = 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:
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CHEMICAL ENGINEERING SERIES
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
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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
heat of the solutions may be taken as constant at 4 J/g·°C. Determi ne (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
From steam table, at 43°C
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Δ
From steam table, at 110°C,
CASE II: DOUBLE EFFECT, FORWARD FEED
VI
VII T2 = 43 C
Steam, S 110 C
Sugar soln F=4,500 kg/h xF = 0.10 TF= 21 C
T1
T2
TI
TII
PI
P xP = 0.30
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CHEMICAL ENGINEERING SERIES
EVAPORATION
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
Ts = 110 C
Δ
Δ
Δ
Δ
ΔTI
TI
Δ Δ
ΔTII
Δ
Δ
Δ
TII = 43 C
Δ
nd
Enthalpy balance around the 2 effect:
st
Enthalpy balance around the 1 effect:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Equate
and
From steam table;
Equate
and
34
CHEMICAL ENGINEERING SERIES
EVAPORATION
Δ
Δ
Since AI = AII, therefore original assumptions are correct
CASE III: DOUBLE EFFECT, BACKWARD FEED
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CHEMICAL ENGINEERING SERIES
EVAPORATION
VI
VII T2 = 43 C
Steam, S 110 C
T1
T2
TI
TII
PI
Sugar soln F=4,500 kg/h xF = 0.10 TF= 21 C
P xP = 0.30
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
Ts = 110 C
ΔTI
Δ
Δ
Δ
Δ
TI
ΔTII
Δ
TII = 43 C
Δ
Δ
Δ
Δ
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CHEMICAL ENGINEERING SERIES
EVAPORATION
nd
Enthalpy balance around the 2 effect:
st
Enthalpy balance around the 1 effect:
Equate
and
From steam table;
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Equate
and
Since AI = AII, therefore original assumptions are correct
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CHEMICAL ENGINEERING SERIES
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:
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Consider over-all material balance:
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. th
(Source: Principles of Transport Processes and Separation Processes, 4 edition, by 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
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CHEMICAL ENGINEERING SERIES
EVAPORATION
Consider solute balance:
Consider over-all material balance:
Consider heat balance:
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,
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CHEMICAL ENGINEERING SERIES
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
P = 1 atm
Steam, S 115.6 C TI
P xP = 0.08
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CHEMICAL ENGINEERING SERIES
43
EVAPORATION
From the steam table at 100°C,
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 (Source: Principles of Transport Processes and Separation th Processes, 4 edition, by Geankoplis) SOLUTION:
Consider NaOH balance:
Consider over-all material balance:
Vapor, V PV = 20 kPa (abs) Steam, S 42 kPa (gage) NaOH soln F=9,072 kg/h xF = 0.10 TF
TI
P xP = 0.50
CHEMICAL ENGINEERING SERIES
EVAPORATION
[ ]
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,
,
A. For feed temperature of 288.8 K
Consider enthalpy balance:
⁄ From steam table, at 42 kPa gage,
,
From figure 8.4-3 (Geankoplis)
44
CHEMICAL ENGINEERING SERIES
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)
45
CHEMICAL ENGINEERING SERIES
46
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
CHEMICAL ENGINEERING SERIES
EVAPORATION
[ ] ⁄ ⁄ ⁄ ⁄
From steam table, at 82.06°C,
,
Consider enthalpy balance:
From steam table, at 110°C,
From figure 8.4-3 (Geankoplis)
47
CHEMICAL ENGINEERING SERIES
EVAPORATION
PROBLEM # 15:
An evaporator is concentrat ing 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
48
CHEMICAL ENGINEERING SERIES
EVAPORATION
⁄ ⁄ ⁄ From steam table, at 51.39°C,
,
Consider enthalpy balance:
⁄ From steam table, at 399.3 K,
From figure 8.4-3 (Geankoplis)
49
CHEMICAL ENGINEERING SERIES
50
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
