Balance evaporador 2012

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Module 9 – Steady state simulation 1

First Example: A Single Effect Evaporator

(to be done in Excel)

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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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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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• 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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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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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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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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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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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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Duhring lines (sodium chloride)

http://www.nzifst.org.nz/unitoperations/evaporation4.htm


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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

First Example Tier 2.xls

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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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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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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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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)


• 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)

http://www.nzifst.org.nz/unitoperations/appendix8.htm


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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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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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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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Find the Economy

= ṁV/ṁS

6802.5 kg/hEconomy = = 0.788

8626.5 kg/h

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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

Balance evaporador 2012

Technology

Transcript of Balance evaporador 2012