1. NAMP PIECE
First Example: A
Single Effect
Evaporator
(to be done in Excel)
Module 9 – Steady state simulation 1

2. NAMP PIECE
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
Module 9 – Steady state simulation 2

3. NAMP PIECE
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
Module 9 – Steady state simulation 3

4. NAMP PIECE
• 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
Module 9 – Steady state simulation 4

5. NAMP PIECE
Diagram of Single Effect Evaporator
Vapor V
Tf, xf, hf, ṁf Tv, yv, Hv, ṁV
U = J/m2 s oC
Feed F
P = kPa
Ts, Hs, ṁs
A = ? m2
Condensate S
Ts, hs, ṁs
Steam S
Concentrated
liquid L TL, xL, hL, ṁL
Module 9 – Steady state simulation 5

6. NAMP PIECE
Material and Heat Balances
q = UAΔT
ΔT = Ts – TL
Heat given off by vapor
ṁF = ṁL + ṁV λ = H s – hs
ṁFxF = ṁLxL + ṁVyV ṁFhF + ṁsHs = ṁLhL + ṁVHV+ ṁshs
ṁFhF + ṁsλ = ṁLhL + ṁVHV
q = ṁs(Hs-hs) = ṁsλ
ṁsλ – ideal heat
transferred in evaporator
Module 9 – Steady state simulation 6

7. NAMP PIECE
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.
Module 9 – Steady state simulation 7

8. NAMP PIECE
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.
Module 9 – Steady state simulation 8

9. NAMP PIECE
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
Module 9 – Steady state simulation 9

10. NAMP PIECE
Duhring lines (sodium chloride)
http://www.nzifst.org.nz/unitoperations/evaporation4.htm
Module 9 – Steady state simulation 10

11. NAMP PIECE
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
Module 9 – Steady state simulation 11

12. NAMP PIECE
1. Draw Diagram and Label Streams
9070 kg/h feed, Vapor V Tv, 0% solids,
0oC, 5% solids, Hv, ṁv
hF
U = 1400
Feed F W/m2 oC
P= 100
mm Hg
Ts, Hs, 1.37 atm
gauge, ṁs q=? Condensate S
Ts, hs, ṁs
Steam S A=?
Liquor L TL, 20% solids,
hL, ṁL
Module 9 – Steady state simulation 12

13. NAMP PIECE
2. Perform Mass Balances
ṁF = 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
Module 9 – Steady state simulation 13

14. NAMP PIECE
3. Perform Heat Balances to find the Economy
The economy is defined as the mass of water
evaporated per mass of steam supplied.
ṁFhF + ṁSHS = ṁLhL + ṁVHV+ ṁShS
ṁFhF + ṁSλ = ṁLhL + ṁVHV
q = ṁS(HS- hS) = ṁSλ
Module 9 – Steady state simulation 14

15. NAMP PIECE
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)
Module 9 – Steady state simulation 15

16. NAMP PIECE
Finding the enthalpy of the feed
1. Find the heat capacity of the liquid feed yNaCl=0.05
feed is 5% sodium chloride, 95% water ywater=0.95
Cp,water=4.18 kJ/kgoC
Cp,mix = x iCpi
all mixture Cp,NaCl=0.85 kJ/kgoC
components
(Cp)F = 0.05*0.85 + 0.95*4.18 = 4.01 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
hF = Cp,F (TF - Tref )
hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kg
Module 9 – Steady state simulation 16

17. NAMP PIECE
Finding the enthalpy of the liquor
yNaCl=0.20
1. Find the heat capacity of the liquor
ywater=0.80
feed is 20% sodium chloride, 80% water
Cp,water=4.18 kJ/kgoC
Cp,mix = x iCpi
all mixture
Cp,NaCl=0.85 kJ/kgoC
components
Cp,L = 0.20*0.85 + 0.80*4.18 = 3.51 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
hL = Cp,L (TL - Tref )
hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kg
Module 9 – Steady state simulation 17

18. NAMP PIECE
Heat Balances
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)
ṁs=8626.5 kg/h
q = 8626.5 kg/h*2182 kJ/kg = 1.88x107 kJ/h =
5228621 W = 5.23 MW
Module 9 – Steady state simulation 18

19. NAMP PIECE
Find the Economy
= ṁV/ṁS
6802.5 kg/h
Economy = = 0.788
8626.5 kg/h
Module 9 – Steady state simulation 19

20. NAMP PIECE
4. Calculate Required Heating Surface
Condensing temperature of steam (1.37 atm gauge =
126.1oC
q = UAΔT
A = q/UΔT
5228621 W
A= = 98.02 m2
W
1400 2 o (126.1 - 88) o C
m C
Module 9 – Steady state simulation 20