Single Effect Evaporator

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SINGLE EFFECT

EVAPORATION

OBJECTIVES

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Students should be able to :

1. Perform calculation on Single Effect

Evaporator

2. Analyse the effect of few parameters on

evaporator capacity & economy

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Methods of Operation of Evaporators

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1. Single-effects evaporators

Feed F, TF, xF, hF

Vapor V, T1, HV

Concentrated liquid L, T1,

xL, hL

Condensate S, TS, hs

Steam S, Ts, Hs

1. The feed enters at TF(K) 2. Saturated steam at TS(K) enters the heat-exchange section 3. Condensed steam leaves as condensate 4. The solution in the evaporator is assumed to be completely mixed. 5. The concentrated product and the solution in the evaporator have the same composition and temperature, T1 6. T1 is the boiling point of the solution 7. P1 is the vapor pressure of the solution at T1 8. General equation of overall heat-transfer coefficient

4 )( 1TTUATAUQ s

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Q= rate of heat transfer (W)

U= Overall heat transfer coefficient (W/m2.K)

A= Heat transfer area (m2)

TS= Temperature of condensing steam (K)

T1= Boiling point (K)

)( 1TTUATAUQ s

EVAPORATOR CAPACITY & ECONOMY

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Capacity = no. of kilograms of water vaporized

per hour

Economy = no. of kilograms vaporized per

kilogram of steam fed to the unit

Steam consumption = Capacity /Economy (Ref. 2)

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HEAT AND MASS BALANCE FOR SINGLE EFFECT EVAPORATOR

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Feed F, TF, xF, hF

Vapor V, T1, HV

Concentrated liquid L, T1,

xL, hL


Steam S, Ts, Hs

MATERIAL BALANCE

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Overall material balance F = L+V

Material balance for solid FxF=LxL + VyV

Feed F, TF,

xF, hF Vapor V, T1, HV

Concentrated liquid L, T1, xL, hL


Steam S, Ts,

Hs

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HEAT BALANCE

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Total heat entering = Total Heat leaving

Heat in feed + Heat in steam = Heat in

concentrated liquid + Heat in Vapor + Heat in

condensed steam

Hence, FhF + SHs = LhL + VHV + Shs (8.4-6)

HEAT BALANCE

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= Hs –hs =Latent heat of condensation (8-4-2)

FhF + Ss = LhL + VHV (8.4-7)

Assumption: Condensed steam leaves at Ts

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HEAT BALANCE

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FhF + Ss = LhL + VHV

? ? ? Latent heat of

evaporation at T1

Latent heat of

condensation

at Ts

Enthalpies of the feed and products are often not available; these

enthalpy-concentration data are available for only a few substances

in solution.

Hence some approximation are made in order to make a heat

balance. These are as follows;

1. It can be demonstrated as an approximation that the latent heat of

evaporation of 1 kg mass of the water from an aqueous solution

can be obtained from the steam tables using the temperature of

the boiling solution T1 (exposed surface temperature) rather than

the equilibrium temperature for pure water at P1.

2. If the heat capacities cpF of the liquid feed and cpL of the product

are known, they can be used to calculate the enthalpies. (this

neglects heats of dilution , which is most cases are not known.)

hF & hL

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- For dilute inorganic solution, and

organic solution eg. Sugar, milk,

biological solids, juice

hF & hL

Hsol 0, Hsol 0

- For concentrated inorganic

solution, eg. NaOH, H2SO4, CaCl2

Refer hf & hL from Enthalpy-

Concentration Charts

Often CpF & CpL Cp of water

EXAMPLE 8.4-1 : HEAT TRANSFER AREA IN SINGLE EFFECT EVAPORATOR

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A continuos single-effect evaporator concentrates

9072 kg/hr of a 1.0 wt% salt solution entering at

311.0 K (37.80C) to a final concentration of 1.5

wt%. The vapor space of the evaporator is at

101.325 kPa (1.0 atm abs) and the steam supplied

is saturated at 143.3 kPa. The overall heat transfer

coefficient, U = 1704 W/m2 K.

Calculate:

a) The amount of vapor & liquid product

b) The heat transfer area required

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EXAMPLE 8.4-1

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Solve for L &V, L = 6048 kg/hr & V = 3024 kg/hr

FhF + Ss = LhL + VHV (8.4-7)

Perform heat balance

hf = CpF (TF – TBP)

hL = CpL(T1– TBP)

Assume CpF = Cp of water =

4.14 kJ/kg K

H L = 0, since T1 = TBP

hf = ???

TBP = ???

EXAMPLE 8.4-1

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= latent heat of condensation at 143.3 kPa

(Ts= 1100C) = ???

HV = latent heat of evaporation at 101.3 kPa

(Tbp = 1000C) = ???

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EXAMPLE 8.4-1

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Then you should solve for S value, S = 4108 kg steam/hr

Q = S = UAT = UA (Ts-T1)

A = 149.3 m2 (as given by Ref. 1)

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EVAPORATION

Topics :

Single effect evaporator (BPR)

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OBJECTIVES

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STUDENTS SHOULD BE ABLE TO :

1. Define Boiling Point Rise (BPR)

2. Use Duhring line & Enthalpy-Concentration Charts of Solutions

BOILING POINT RISE OF SOLUTIONS

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SOLUTIONS ARE NOT DILUTE,

HENCE..

THERMAL PROPERTIES OF THE SOLUTION IS DIFFER CONSIDERABLY FROM THOSE OF WATER.

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BOILING POINT RISE & DUHRING DIAGRAM

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For a given pressure, the boiling point of an aqueous solution is higher than that of pure water.

The increase in boiling point over that of water is known as the boling point rise (BPR)

BPR = BP of water at a given pressure – BP of solution (refer

Duhring Diagram)


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NaCl-Water System

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DUHRING’S RULE

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A straight line is obtained if the boiling point of a solution is plotted against the boiling point of pure water at the same pressure for a given concentration at different pressure.

A different straight line is obtained for each given concentration.

EXAMPLE 8.4-2

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The pressure in an evaporator is given as 25.6 kPa (3.72 psia), and a solution of 30% NaOH is being boiled.

Determine:

1. The boiling temperature of the NaOH solution.

2. The boling point rise of the solution

Solutions :

1. Refer steam table, check boiling point of water at 25.6 kPa

2. Refer Duhring lines for NaOH solution, get Tbp of solution & BPR

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Boiling point of water (0 C)

NaOH-Water System


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In Perry’s Chemical Engineering Handbook, a chart is given to estimate the BPR for common salts & solutes like : NaNO3, NaOH, NaCl & H2SO4, sucrose solution, citric acid, glycerol.

Biological solutes have quite small BPR values compared to those of common salts.

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HEAT OF SOLUTION

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Heat of solution??? Eg : Dissolve solid powder (acid/alkali - NaOH) in water…heat is evolved!

Amount of heat evolved depends on the substance & amount of water used (concentration)

Also, if concentrated NaOH solution is diluted..heat is evolved

If the solution is concentrated, heat must be added.

ENTHALPY-CONCENTRATION CHARTS OF SOLUTIONS

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IF THE HEAT OF SOLUTION OF AQUEOUS SOLUTION BEING CONCENTRATED IN THE EVAPORATOR IS LARGE,

NEGLECTING IT

COULD CAUSE ERRORS IN THE HEAT BALANCE!!

Values of hf & hL (equation 8.4-7) could be obtained from Enthalpy-Concentration Chart.

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ENTHALPY-CONCENTRATION CHARTS OF SOLUTIONS

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EXAMPLE 8.4-3

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An evaporator is used to concentrate 4536 kg/hr (10 000lb/hr) of a 20% solution of NaOH in water entering at 60 0C (140 oF) to a product of 50% solids.

The pressure of the saturated steam used is 172.4 kPa (25 psia) and the pressure in the vapor space of the evaporator is 11.7 kPa (1.7 psia). The overall heat transfer coefficient is 1560 W/m2 K (275 btu/h ft2 oF).

Calculate :

1. The steam used

2. The steam economy

3. The heating surface area

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EXAMPLE 8.4-3

F = 4536 kg/hr

TF = 60°C

xF, = 20% solution

hF

Vapor

V, T1, HV

Condensate

S, TS, hs Steam S, Ts, Hs

Concentrated liquid

L, T1, hL

xL = 50% solid

P = 172.4 kPa

P = 11.7 kPa

U = 1560 W/m2.K

F = 4536 = L + V

FxF = LxL

L= 1814 kg/h V = 2722 kg/hr

FhF + Ss = LhL + VH (8.4-7)

Perform heat balance

TBP for boiling water = ???

Find in a steam table A.2 at 11.7 kPa

Do some interpolation get 48.9 oC

TBP for solution =??

Find in Duhring chart at T at 48.9 oC and 50% NaOH (Ans = 89.5 oC)

BPR = TBP for solution - TBP for boiling water = 40.6 oC

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hf = CpF (TF – TBP)

hL = CpL(T1– TBP)

Latent heat, λ can be obtained from steam table at saturated steam at 172.4 kPa and Tsat = 115.6 oC = 2214 kJ/kg

hf = ???

Assume CpF = Cp of water = 4.14 kJ/kg K

4535 (214) + S(2214) = 1814(505) +2722(2667)

S=3255 kg steam/h

q= S λ = 2002 kW

A=q/U(ΔT) = 49.2 m2

Steam economy = kg vaporized/kg used

= 2722/3255

= 0.836

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Single Effect Evaporator