Post on 08-Jul-2015
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PIECENAMP
Module 9 – Steady state simulation 1
First Example: A Single Effect Evaporator
(to be done in Excel)
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Module 9 – Steady state simulation 2
Evaporation
Function is to concentrate solution
What affects evaporation?
• Rate at which heat is transferred to the liquid
• Quantity of heat required to evaporate mass of water
• Maximum allowable temperature of liquid
• Pressure which evaporation takes place
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Module 9 – Steady state simulation 3
Single Effect Vertical Evaporator
Three functional sections
• Heat exchanger
• Evaporation section
• liquid boils and evaporates
• Separator
• vapor leaves liquid and passes off to other equipment
Three sections contained in a vertical cylinder
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Module 9 – Steady state simulation 4
• In the heat exchanger section (calandria), steam condenses in the outer jacket
• Liquid being evaporated boils on inside of the tubes and in the space above the upper tube stack
• As evaporation proceeds, the remaining liquors become more concentrated
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Module 9 – Steady state simulation 5
Tf, xf, hf, ṁf
TL, xL, hL, ṁL
Ts, Hs, ṁs
Tv, yv, Hv, ṁV
Ts, hs, ṁs
P = kPa
U = J/m2 s oC
A = ? m2
Condensate S
Vapor V
Concentrated liquid L
Steam S
Feed F
Diagram of Single Effect Evaporator
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Module 9 – Steady state simulation 6
Material and Heat Balances
ṁF = ṁL + ṁV
ṁFxF = ṁLxL + ṁVyV
q = UAΔT
ΔT = Ts – TL
Heat given off by vaporλ = Hs – hs
ṁFhF + ṁsHs = ṁLhL + ṁVHV+ ṁshs
ṁFhF + ṁsλ = ṁLhL + ṁVHV
q = ṁs(Hs-hs) = ṁsλ
ṁsλ – ideal heat transferred in evaporator
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Module 9 – Steady state simulation 7
Finding the Latent Heat of Evaporation of Solution and the Enthalpies
• Using the temperature of the boiling solution TL, the latent heat of evaporation can be found;
• The heat capacities of the liquid feed (CpF) and product (CpL) are used to calculate the enthalpies of the solution.
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Module 9 – Steady state simulation 8
Property Effects on the Evaporator
• Feed Temperature– Large effect
– Preheating can reduce heat transfer area requirements
• Pressure– Reduction
• Reduction in boiling point of solution
• Increased temperature gradient
• Lower heating surface area requirements
• Effect of Steam Pressure– Increased temperature gradient when higher pressure
steam is used.
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Module 9 – Steady state simulation 9
Boiling-Point Rise of Solutions
• Increase in boiling point over that of water is known as the boiling point elevation (BPE) of solution
• BPE is found using Duhring’s Rule– Boiling point of a given solution is a linear
function of the boiling point of pure water at the same pressure
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Module 9 – Steady state simulation 10
Duhring lines (sodium chloride)
http://www.nzifst.org.nz/unitoperations/evaporation4.htm
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Module 9 – Steady state simulation 11
Problem Statement
(McCabe 16.1 modified)
A single-effect evaporator is used to concentrate 9070 kg/h of a 5% solution of sodium chloride to 20% solids. The gauge pressure of the steam is 1.37 atm; the absolute pressure in the vapor space is 100 mm Hg. The overall heat transfer coefficient is estimated to be 1400 W/m2 oC. The feed temperature is 0oC. Calculate the amount of steam consumed, the economy, and required heating surface.
First Example Excel Spreadsheet
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Module 9 – Steady state simulation 12
1. Draw Diagram and Label Streams
9070 kg/h feed, 0oC, 5% solids,
hF
TL, 20% solids, hL, ṁL
Ts, Hs, 1.37 atm gauge, ṁs
Tv, 0% solids, Hv, ṁv
Ts, hs, ṁs
P= 100 mm Hg
U = 1400 W/m2 oC
A=?
Condensate S
Vapor V
Liquor L
Steam S
Feed F
q=?
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Module 9 – Steady state simulation 13
2. Perform Mass BalancesṁF = ṁL + ṁV
[9070 kg/h = ṁL kg/h+ ṁV kg/h]
ṁFxF = ṁLxL + ṁVyV (note that yv is zero because only
vapor is present, no solids)
[0.05 * 9070 kg/h = 0.2 * ṁL kg/h + 0]
• Can solve for ṁv and ṁL
ṁV = 6802.5 kg/h, ṁL = 2267.5 kg/h
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Module 9 – Steady state simulation 14
3. Perform Heat Balances to find the Economy
ṁFhF + ṁSHS = ṁLhL + ṁVHV+ ṁShS
ṁFhF + ṁSλ = ṁLhL + ṁVHV
q = ṁS(HS- hS) = ṁSλ
The economy is defined as the mass of water evaporated per mass of steam supplied.
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Module 9 – Steady state simulation 15
Needed Data
• Boiling point of water at 100 mm Hg = 51oC (from steam tables)
www.nzifst.org.nz/unitoperations/appendix8.htm
• Boiling point of solution = 88oC (from Duhring lines)
http://www.nzifst.org.nz/unitoperations/evaporation4.htm
• Boiling point elevation = 88 – 51 = 37oC
• Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88oC and 100 mm Hg [.133 bar]) = 2664 kJ/kg(F&R, p.650) – also called the latent heat of evaporation
• Heat of vaporization of steam (Hs-hs = λ ) at 1.37 atm gauge [20 lbf/in2] = 939 Btu/lb = 2182 kJ/kg (McCabe, App.7, p.1073)
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Module 9 – Steady state simulation 16
Finding the enthalpy of the feed
1. Find the heat capacity of the liquid feed
feed is 5% sodium chloride, 95% water
p, i pimixall mixturecomponents
C = x C
yNaCl=0.05
ywater=0.95
Cp,water=4.18 kJ/kgoC
Cp,NaCl=0.85 kJ/kgoC
(Cp)F = 0.05*0.85 + 0.95*4.18 = 4.01 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
F p F ref,Fh = C (T - T )
hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kg
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Module 9 – Steady state simulation 17
Finding the enthalpy of the liquor
1. Find the heat capacity of the liquor
feed is 20% sodium chloride, 80% water
yNaCl=0.20
ywater=0.80
Cp,water=4.18 kJ/kgoC
Cp,NaCl=0.85 kJ/kgoC
Cp,L = 0.20*0.85 + 0.80*4.18 = 3.51 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
L p, L refLh = C (T - T )
hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kg
p, i pimixall mixturecomponents
C = x C
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Module 9 – Steady state simulation 18
ṁLhL + ṁVHV - ṁFhF = ṁSHS - ṁShS = ṁS(HS- hS) = ṁSλ
λ = (HS-hS) = 2182 kJ/kg
(2267.5 kg/h *309.23 kJ/kg) + (6802.5 kg/h * 2664 kJ/kg) – (0) = ṁS (HS-hS)
q = ṁS (2182 kJ/kg)
Heat Balances
ṁs=8626.5 kg/h
q = 8626.5 kg/h*2182 kJ/kg = 1.88x107 kJ/h = 5228621 W = 5.23 MW
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Module 9 – Steady state simulation 19
Find the Economy
= ṁV/ṁS
6802.5 kg/hEconomy = = 0.788
8626.5 kg/h
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Module 9 – Steady state simulation 20
4. Calculate Required Heating Surface
Condensing temperature of steam (1.37 atm gauge = 126.1oC
q = UAΔT
A = q/UΔT
2
o
2 o
5228621 WA = = 98.02 m
W1400 (126.1 - 88) C
m C