EVAPORATION

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CONTENT

1. Type of Evaporation equipment and Methods

2. Overall Heat Transfer Coefficient in Evaporators

3. Calculation Methods for Single Effect Evaporators

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3. Calculation Methods for Single Effect Evaporators

4. Calculation Methods for Multiple Effects Evaporators

5. Condenser for Evaporator

6. Evaporation using Vapor Recompression

Topic Outcomes� Define Evaporation process� Describe and determine the processing

factors that involved in evaporation process � identify the types of evaporators and � identify the types of evaporators and

distinguish the suitable evaporator for certain operations

� Describe other specific types of evaporators that are used in certain industries

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EVAPORATION� Heat is added to a solution to vaporize the solvent, which is usually water.

� Case of heat transfer to a boiling liquid.

� Vapor from a boiling liquid solution is removed and a more concentrated solution remains.

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solution remains.

� Refers to the removal of water from an aqueous solution.

� Example: concentration of aqueous solutions of sugar. In these cases the crystal is the desired product and the evaporated water is discarded.

Foaming or frothing

Pressure and temperature

Scale deposition

Materials of construction

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Processing Factors

Concentration in the liquid

solubility

Temperature sensitivityof materials

Scale deposition

PROCESSING FACTORS

(1) Concentrationdilute feed, viscosity µ↓, heat transfer coefficient, h↑concentrated solution/products, µ↑, and h ↓.

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(2) Solubilityconcentration ↑, solubility ↓ , crystal formed.solubility ↑ with temperature ↑.

(3) Temperature.heat sensitive material degrade at higher temperature & prolonged heating.

(4) Foaming/frothing.caustic solutions, food solutions, fatty acid solutions form

foam/froth during boiling.entrainment loss as foam accompany vapor.

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(5) Pressure and Temperaturepressure ↑, boiling point ↑.concentration ↑, boiling point↑.heat-sensitive material operate under vacuum.

(6) Material of constructionminimize corrosion.

TRAYGrowing PACKING

Growing foam leaves tower

Lean soln

Sweet gas Sweet gas

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LI

TRAYGrowing foam slows/ stops liquid from flowing down

LI

PACKING

Foam effects less severe, but can cause some channeling

Sour gasSour gas

Rich soln Rich soln

Lean soln

Forming Effects

Type of Evaporation Equipment

� Horizontal tube type� Vertical tube type� Long tube vertical type

Forced-circulation type

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� Forced-circulation type� Open kettle or pan� Open-pan solar evaporator� Falling-film-type evaporator� Agitated-film evaporator

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Wiped film evaporator

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Operating parameters Operating window

Feed rate kg/h 20 - 100.000

Evaporation rate kg/h up to 40.000

Heating temperature °°°°C up to 380 1)

Pressure on process side barg -1 to 30

Product viscosity at operating temperature mPas up to 70.000

Residence time min < 1 2)

Evaporation ratio (concentrate/feed) up to 1:50 3)

Through put % 20 - 100 3)

1) evaporators with electrical inductive heating up to 600 °°°°C2) in horizontal evaporators longer residence times possible3) depending on evaporator type

Effect of Processing Variables on Evaporator Operation

(1) TF

TF < Tbp, some of latent heat of steam will be used to heat up the cold

feed, only the rest of the latent heat of steam will be used to vaporize

the feed.

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Is the feed is under pressure & TF > Tbp, additional vaporization

obtained by flashing of feed.

(2) P1

desirable ∆T ↑ [Q = UA(TS – T1)]

A ↓ & cost ↓.

T1 depends on P1 will ↓ T1.

(3) PS

↑ PS will ↑ TS but high-pressure is costly.

optimum TS by overall economic balances.

(4) BPR

The concentration of the solution are high enough so that the

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The concentration of the solution are high enough so that the

cP and Tbp are quite different from water.

BPR can be predict from Duhring chart for each solution such

as NaOH and sugar solution.

(5) Enthalpy–concentration of solution.

for large heat of solution of the aqueous solution.

to get values for hF and hL.

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feed, FTF , xF , hF.

vapor,V to condenserT1 , yV , HV

P1

T1

heat-exchangertubes

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steam, STS , HS

concentrated liquid, LT1 , xL , hL

condensate, STS , hS

T1

Simplified Diagram of single-effect evaporator

steam, T

feed, TF

vapor T1

(1)

T1

(2)

T2

(3)

T3

vapor T2 vapor T3to vacuum condenser

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steam, TS

concentratefrom firsteffect.

concentratefrom secondeffect.

concentratedproduct

condensate

Simplified diagram of forward-feed triple-effect evaporator

steam, TS

feed, TF

vapor T1

(1)

T1

(2)

T2

(3)

T3

vapor T2 vapor T3to vacuum condenser

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Simplified diagram of backward-feed triple-effect evaporator

concentratedproduct

condensate

� The feed (usually dilute) enters at TF and saturated steam at TSenters the heat-exchange section.

� Condensed leaves as condensate or drips.� The solution in the evaporator is assumed to be completely

mixed and have the same composition at T1.

SINGLE EFFECT EVAPORATORS

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mixed and have the same composition at T1.� The pressure is P1, which is the vapor pressure of the solution at

T1.� Wasteful of energy since the latent heat of the vapor leaving is

not used but is discarded.� Are often used when the required capacity of operation is

relatively small, but it will wasteful of steam cost.

CALCULATIONS a) vapor, V and liquid, L flowrates.b) heat transfer area, Ac) overall heat-transfer coefficient, U.

CALCULATION METHODS FOR SINGLE-EFFECT EVAPORAT

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c) overall heat-transfer coefficient, U.d) Fraction of solid content, xL.

(1) To calculate V & L and xL,- solve simultaneously total material balance &

solute/solid balance.F = L + V total material balanceF (xF) = L (xL) solute/solid balance

(2) To calculate A or U,- No boiling point rise and negligible heat of solution:

calculate hF, hL, Hv and λ.where, λ = (HS – hs)

h = cP(T – Tref)

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h = cP(T – Tref)where, Tref = T1 = (as datum)

cPF = heat capacity (dilute as water)HV = latent heat at T1

solve for S:F hF + S λ = L hL + V HV

solve for A and U:q = S λ = U A ∆T = UA (TS – T1)

(3) To get BPR and the heat of solution:

1) Calculate T1 = Tsat + BPR2) Get hF and hL from Figure 8.4-3.

3) Get λS & HV from steam tables for superheated vapor orH = H + 1.884 (BPR)

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HV = Hsat + 1.884 (BPR)4) Solve for S:

F hF + S λ = L hL + V HV

5) Solve for A and U:

q = S λ = U A ∆T = UA (TS – T1)

A continuous single-effect evaporator concentrates 9072 kg/h of a 1.0 wt % salt solution entering at 311.0 K (37.8 ºC) 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 coefficient U = 1704 W/m2 .K.

Example 8.4-1: Heat-Transfer Area in Single-Effect Evaporator

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Calculate the amounts of vapor and liquid product and the heat-transfer area required. Assumed that, since it its dilute, the solution has the same boiling point as water.

U = 1704 W/m2

P1 = 101.325 kPa

F = 9072 kg/hTF = 311 KxF = 0.01hF.

V = ?T1 , yV , HV

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U = 1704 W/m2

T1 A = ?S , TS , HS

PS = 143.3 kPa L = ?T1 , hL

xL = 0.015

S, TS , hS

Figure 8.4-1: Flow Diagram for Example 8.4-1

SolutionRefer to Fig. 8.4-1 for flow diagram for this solution.For the total balance,

F = L + V9072 = L + V

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For the balance on the solute alone,F xF = L xL

9072 (0.01) = L (0.015)L = 6048 kg/h of liquid

Substituting into total balance and solving,V = 3024 kg/h of vapor

Since we assumed the solution is dilute as water;cpF = 4.14 kJ/kg. K

From steam table, (A.2-9)At P1 = 101.325 kPa, T1 = 373.2 K (100 ºC).

HV = 2257 kJ/kg.

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HV = 2257 kJ/kg.At PS = 143.3 kPa, TS = 383.2 K (110 ºC).

λ = 2230 kJ/kg.

The enthalpy of the feed can be calculated from,hF = cpF (TF – T1)hF = 4.14 (311.0 – 373.2)

= -257.508 kJ/kg.

Substituting into heat balance equation;F hF + S λ = L hL + V HV

with hL = 0, since it is at datum of 373.2 K.9072 (-257.508) + S (2230) = 6048 (0) + 3024 (2257)

S = 4108 kg steam /h

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S = 4108 kg steam /hThe heat q transferred through the heating surface area, A is

q = S (λ)q = 4108 (2230) (1000 / 3600) = 2 544 000 W

Solving for capacity single-effect evaporator equation;q = U A ∆T = U A (TS – T1)

2 544 000 = 1704 A (383.2 – 373.2)Solving, A = 149.3 m2.

Example 8.4-3: Evaporation of an NaOH Solution.

An evaporator is used to concentrate 4536 kg/h of a 20 % solution of NaOH in water entering at 60 ºC to a product of 50 % solid. The pressure of the saturated steam used is 172.4 kPa and the pressure in the vapor space of the evaporator is 11.7 kPa. The overall heat-transfer coefficient is 1560 W/m2.K.

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is 1560 W/m .K.

Calculate the steam used, the steam economy in (kg vaporized/kg steam)used, and the heating surface area in m2.

U = 1560 W/m2

P1 = 11.7 kPa

F = 4536 kg/hTF = 60 ºCxF = 0.2hF.

S = ?

V, T1 , yV , HV

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U = 1560 W/m2

T1 A = ?S = ?TS , HS

PS = 172.4 kPaL, T1 , hL

xL = 0.5

S, TS , hS

Figure 8.4-4: Flow Diagram for Example 8.4-3

Solution

Refer to Fig. 8.4-4, for flow diagram for this solution.For the total balance,

F = 4536 = L + VFor the balance on the solute alone,

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For the balance on the solute alone,F xF = L xL

4536 (0.2) = L (0.5)L = 1814 kg/h of liquid

Substituting into total balance and solving,V = 2722 kg/h of vapor

To determine T1 = Tsat + BPR of the 50 % concentrate product, first we obtain Tsat of pure water from steam table. At 11.7 kPa, Tsat = 48.9 ºC.

From Duhring chart (Fig. 8.4-2), for a Tsat = 48.9 ºC and 50 % NaOH , the boiling point of the solution is T1 = 89.5 ºC. hence,

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NaOH , the boiling point of the solution is T1 = 89.5 ºC. hence, BPR = T1 - Tsat = 89.5-48.9 = 40.6 ºC

From the enthalpy-concentration chart (Fig.8.4-3), forTF = 60 ºC and xF = 0.2 get hF = 214 kJ/kg.T1 = 89.5 ºC and xL = 0.5 get hL = 505 kJ/kg.

For saturated steam at 172.4 kPa, from steam table, we getTS = 115.6 ºC and λ = 2214 kJ/kg.

To get HV for superheated vapor, first we obtain the enthalpy at Tsat = 48.9 ºC and P1 = 11.7 kPa, get Hsat = 2590 kJ/kg. Then using heat capacity of 1.884 kJ/kg.K for superheated steam. So H = H + c BPR

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1 satcapacity of 1.884 kJ/kg.K for superheated steam. So HV = Hsat + cP BPR

= 2590 + 1.884 (40.6) = 2667 kJ/kg.

Substituting into heat balance equation and solving for S,F hF + S λ = L hL + V HV

4535 (214) + S (2214) = 1814 (505) + 2722 (2667)S = 3255 kg steam /h.

The heat q transferred through the heating surface area, A isq = S (λ)q = 3255 (2214) (1000 / 3600) = 2 002 000 W

Solving for capacity single-effect evaporator equation;

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Solving for capacity single-effect evaporator equation;q = U A ∆T = U A (TS – T1)

2 002 000 = 1560 A (115.6 – 89.5)

Solving, A = 49.2 m2.Steam economy = 2722/3255

= 0.836

EVAPORATION

feed, TF

vapor T1

(1)

T

(2)

T

(3)

T

vapor T2 vapor T3to vacuum condenser

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steam, TS

concentratefrom firsteffect.

T1 T2 T3

concentratefrom secondeffect.

concentratedproduct

condensate

Simplified diagram of forward-feed triple-effect evaporator

EVAPORATION

� Forward-feed multiple/triple-effect evaporators;

- the fresh feed is added to the first effect and flows to the next in the same direction as the vapor flow.

- operated when the feed hot or when the final concentrated

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- operated when the feed hot or when the final concentrated product might be damaged at high temperature.

- at steady-state operation, the flowrates and the rate of evaporation in each effect are constant.

- the latent heat from first effect can be recovered and reuse. The steam economy ↑, and reduce steam cost.- the Tbp ↓ from effect to effect, cause P1 ↓.

EVAPORATION

Calculation Methods for Multiple-effect Evaporators.

� Objective to calculate;- temperature drops and the heat capacity of

evaporator.- the area of heating surface and amount of vapor leaving

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- the area of heating surface and amount of vapor leaving the last effect.

� Assumption made in operation;- no boiling point rise.- no heat of solution.- neglecting the sensible heat necessary to heat

the feed to the boiling point.

EVAPORATION

� Heat balances for multiple/triple-effect evaporator.- the amount of heat transferred in the first effect is

approximately same with amount of heat in the second effect,

q = U1 A1 ∆T1 = U2 A2 ∆T2 = U3 A3 ∆T3

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q = U1 A1 ∆T1 = U2 A2 ∆T2 = U3 A3 ∆T3

- usually in commercial practice the areas in all effects are equal,

q/A = U1 ∆T1 = U2 ∆T2 = U3 ∆T3

- to calculate the temperature drops in evaporator,Σ ∆T = ∆T1 + ∆T2 + ∆T3 = TS – T3

- hence we know that Σ ∆T are approximately inversely proportional to the values of U,

321

11 111

1

UUU

UTT

++Σ∆=∆

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- similar eq. can be written for ∆T2 & ∆T3

- if we assumed that the value of U is thesame in each effect, the capacity equation,

q = U A (∆T1 + ∆T2 + ∆T3 ) = UA Σ ∆T

EVAPORATION

steam, T

feed, TF

vapor T1

(1)

T1

(2)

T2

(3)

T3

vapor T2 vapor T3to vacuum condenser

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Simplified diagram of backward-feed triple-effect evaporator

steam, TST1 T2 T3

concentratedproduct

condensate

EVAPORATION

� Backward-feed multiple/triple-effect evaporators;- fresh feed enters the last and coldest effect and continues on until the concentrated product

leaves the first effect.- advantageous when the fresh feed is cold or

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- advantageous when the fresh feed is cold orwhen concentrated product is highly viscous.

- working a liquid pump since the flow is from lowto high pressure.

- the high temperature in the first effect reduce the viscosity and give reasonable heat-transfer coefficient.

Step-by-step Calculation Method for Triple-effect Evaporator (Forward Feed)

For the given x3 and P3 and BPR3

From an overall material balance, determine VT = V1 + V2 + V3

(1st trial – assumption)Calculate the amount of concentrated solutions & their concentrations in each effect using material

balances.

Find BPR & ∆T in each effect & Σ∆T.

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Find BPR & ∆T in each effect & Σ∆T.If the feed is very cold, the portions may be modified appropriately, calculate the boiling point in

each effect.

Calculate the amount vaporized and concentrated liquid in each effect through energy & material balances.

If the amounts differ significantly from the assumed values in step 2, step 2, and 4 must be repeated with the amounts just calculated.

Using heat transfer equations for each effect, calculate the surface required for each effect

If the surfaces calculated are not equal, revise the ∆TS . Repeat step 4 onward until the areas are distributed satisfactorily.

A triple-effect forward-feed evaporator is being used to evaporate a sugar solution containing 10 wt% solids to a concentrated solution of 50 %. The boiling-point rise of the solutions (independent of pressure) can be estimated from (BPR ºC = 1.78x + 6.22 x2 ), where x is wt fraction of sugar in solution. Saturated steam at 205.5 kPa and 121.1ºC saturation temperature is being used. The pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 22

EXAMPLE. 8.5-1 : Evaporation of Sugar Solution in a Triple- Effect Evaporator

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pressure in the vapor space of the third effect is 13.4 kPa. The feed rate is 22 680 kg/h at 26.7 ºC. the heat capacity of the liquid solutions is cP = 4.19 – 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, U2 = 1987, and U3 = 1136 W/m2.K. If each effect has the same surface area, calculate the area, the steam rate used, and the steam economy.

EVAPORATION

S = ?(2)

T3

T1 T2F = 22680 xF = 0.1TF = 26.7 ºC

V1 = 22,680 – L1

(1) (3)

V2 = L1 – L2 V3 = L2 - 4536

P3 = 13.7 kPa

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S = ?TS1 = 121.1 ºCPS1 = 205.5 kPa

T1 , L1 , x1

TS1 TS3 TS2

T2 , L2 , x2

T3

L3 = 4536x3 = 0.5

Fig. 8.5-1: Flow diagram for example 8.5-1

Solution,The process flow diagram is given in Fig. 8.5-1..

Step 1,From steam table, at P3 = 13.4 kPa, get Tsat = 51.67 ºC. Using the BPR equation for third effect with xL = 0.5,

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BPR3 = 1.78 (0.5) + 6.22 (0.52) =2.45 ºC.T3 = 51.67 + 2.45 = 54.12 ºC. (BPR = T – Ts)

Step 2, Making an overall and a solids balance.F = 22 680 = L3 + (V1 + V2 + V3)FxF = 22 680 (0.1) = L3 (0.5) + (V1 + V2 + V3) (0)L3 = 4536 kg/hTotal vaporized = (V1 + V2 + V3) = 18 144 kg/h

Assuming equal amount vaporized in each effect,V1 = V2 = V3 = 18 144 / 3 = 6048 kg/h

Making a total material balance on effects 1, 2, and 3, solvingF = 22 680 = V1 + L1 = 6048 + L1, L1 = 16 632 kg/h.

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F = 22 680 = V1 + L1 = 6048 + L1, L1 = 16 632 kg/h.L1 = 16 632 = V2 + L2 = 6048 + L2, L2 = 10 584 kg/h.L2 = 10 584 = V3 + L3 = 6048 + L3, L3 = 4536 kg/h.

Making a solids balance on each effect, and solving for x,22 680 (0.1) = L1 x1 = 16 632 (x1), x1 = 0.136

16 632 (0.136) = L2 x2 = 10 584 (x2), x2 = 0.21410 584 (0.214) = L3 x3 = 4536 (x3), x3 = 0.5 (check)

Step 3, The BPR in each effect is calculated as follows:BPR1 = 1.78x1 + 6.22x1

2 = 1.78(0.136) + 6.22(0.136)2

= 0.36ºC.BPR2 = 1.78(0.214) + 6.22(0.214)2 =0.65ºC.BPR3 = 1.78(0.5) + 6.22(0.5)2 =2.45ºC. then,Σ∆T available = T – T (sat) – (BPR + BPR + BPR )

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Σ∆T available = TS1 – T3 (sat) – (BPR1 + BPR2 + BPR3 )= 121.1 – 51.67 – (0.36+0.65+2.45) = 65.97ºC.

Using Eq.(8.5-6) for ∆T1 , ∆T2 , and ∆T3

∆T1 = 12.40 ºC ∆T2 = 19.50 ºC ∆T3 = 34.07 ºC

321

11 111

1

UUU

UTT

++Σ∆=∆

( ))11361()19871()31231(

)31231(97.65

++=

However, since a cold feed enters effect number 1, this effect requires more heat. Increasing ∆T1 and lowering ∆T2 and ∆T3 proportionately as a first estimate, so∆T1 = 15.56ºC ∆T2 = 18.34 ºC ∆T3 = 32.07 ºCThe above data ∆T1, ∆T2 and ∆T3 are getting from iteration-s

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To calculate the actual boiling point of the solution in each effect,T1 = TS1 - ∆T1 = 121.1 – 15.56 = 105.54 ºC. T2 = T1 - BPR1 - ∆T2 = 105.54 – 0.36 – 18.34 = 86.84 ºC. TS2 = T1 –BPR1 = 105.54 – 0.36 = 105.18 ºC. T3 = T2 - BPR2 - ∆T3= 86.84 – 0.65 – 32.07 = 54.12 ºC.TS3 = T2 –BPR2 = 86.84 – 0.65 = 86.19 ºC.

The temperatures in the three effects are as follows:Effect 1 Effect 2 Effect 3 CondenserTS1 = 121.1ºC TS2 = 105.18 TS3 = 86.19 TS4 = 51.67T1 = 105.54 T2 = 86.84 T3 = 54.12

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Step 4,The heat capacity of the liquid in each effect is calculated from the equation cP = 4.19 – 2.35x.

F: cPF = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.KL1: cP1 = 4.19 – 2.35 (0.136) = 3.869 kJ/kg.KL2: cP2 = 4.19 – 2.35 (0.214) = 3.684 kJ/kg.KL3: cP3 = 4.19 – 2.35 (0.5) = 3.015 kJ/kg.K

The values of the enthalpy H of the various vapor streams relative to water at 0 ºC as a datum are obtained from the steam table as follows:

Effect 1:H1 = HS2 + 1.884 BPR1 = 2684 + 1.884(0.36) 2685 kJ/kg.

λ = H – h = 2708 – 508 = 2200 kJ/kg.

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λS1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.

Effect 2:H2 = HS3 + 1.884 BPR2= 2654 + 1.884(0.65) = 2655 kJ/kg.

λS2 = H1 – hS2 = 2685 – 441 = 2244 kJ/kg.

Effect 3:H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.λS3 = H2 – hS3 = 2655– 361 = 2294 kJ/kg.

Write the heat balance on each effect. Use 0ºC as a datum.FcPF (TF –0) + SλS1 = L1cP1 (T1 –0) + V1H1 ,, ………(1)

22680(3.955)(26.7-0)+2200S = 3.869L1(105.54-0)+(22680-L1)2685

L1cP1 (T1 –0) + V1λS2 = L2cP2 (T2 –0) + V2H2 ………(2)3.869L (105.54-0)+(22680-L )2244=3.684L (86.84-0)+(L -L )2655

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3.869L1(105.54-0)+(22680-L1)2244=3.684L2(86.84-0)+(L1-L2)2655

L2cP2 (T2 –0) + V2λS3 = L3cP3 (T3 –0) + V3H3 ………(3) 3.68L2(86.84-0)+(L1-L2)2294=4536(3.015)(54.1-0)+(L2-4536)2600

Solving (2) and (3) simultaneously for L1&L2 and substituting into(1)L1 = 17078 kg/h L2 = 11068 kg/h L3 = 4536 kg/h

S = 8936kg/h V1 = 5602kg/h V2 = 6010kg/hV3 = 6532kg/h

EVAPORATION

Step 5, Solving for the values of q in each effect and area,

( ) WxxSq S6

11 10460.5100022003600

8936 =

== λ

( ) WxxVq S6

212 10492.3100022443600

5602 =

== λ

( )6010

51

( ) WxxVq S6

323 10830.3100022943600

6010 =

== λ

( )2

6

11

11 4.112

65.153123

10460.5m

x

TU

qA ==

∆= ( )

26

22

22 8.95

34.181987

10492.3m

x

TU

qA ==

∆=

( )2

6

33

33 1.105

07.321136

10830.3m

x

TU

qA ==

∆= 2321 4.104

3

)(m

AAAAm =++=

EVAPORATION

Am = 104.4 m2, the areas differ from the average value by less than 10 % and a second trial is really not necessary. However, a second trial will be made starting with step 6 to indicate the calculation methods used.

52

Step 6,Making a new solids balance by using the new L1 = 17078, L2 = 11068, and L3 = 4536, and solving for x,

22 680 (0.1) = L1 x1 = 17 078 (x1), x1 = 0.13317 078 (0.130) = L2 x2 = 11 068 (x2), x2 = 0.20511 068 (0.205) = L3 x3 = 4536 (x3), x3 = 0.5 (check)

EVAPORATION

Step 7. The new BPR in each effect is then,BPR1 = 1.78(0.133) + 6.22(0.13)2 =0.35ºC.BPR2 = 1.78(0.205) + 6.22(0.205)2 =0.63ºC.BPR3 = 1.78(0.5) + 6.22(0.5)2 =2.45ºC. then,Σ∆T available = 121.1 – 51.67 – (0.35+0.63+2.45) = 66.0 ºC.

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Σ∆T available = 121.1 – 51.67 – (0.35+0.63+2.45) = 66.0 ºC.

The new ∆T are obtained using Eq.(8.5-11),

( )C

A

ATT

m

°==∆=∆ 77.164.104

4.11256.1511'1

( )C

A

ATT

m

°==∆=∆ 86.164.104

8.9534.1822'2

( )C

A

ATT

m

°==∆=∆ 34.324.104

1.10507.3233'3

CT °=++=Σ∆ 97.6534.3286.1677.16

These ∆T’ values are readjusted so that ∆T 1`= 16.77,

∆T 2`= 16.87, ∆T 3` = 32.36, and Σ∆T = 66.0 ºC. Tocalculate the actual boiling point of the solution in eacheffect,

(1) T1 = TS1 + ∆T 1` = 121.1 – 16.77 = 104.33ºC(2) T2 = T1 – BPR1 - ∆T 2` = 104.33 – 0.35 – 16.87 = 87.11 ºC

TS2 = T1 – BPR1 = 104.33 – 0.35 = 103.98ºC(3) T3 = T2 – BPR2 - ∆T 3` = 87.11 – 0.63 – 32.36 = 54.12 ºC

54

(3) T3 = T2 – BPR2 - ∆T 3` = 87.11 – 0.63 – 32.36 = 54.12 ºCTS3 = T2 – BPR2 = 87.11 – 0.63 = 86.48 ºC.

Step 8;Following step 4 to get cP = 4.19 – 2.35x,

F: cPF = 4.19 – 2.35 (0.1) = 3.955 kJ/kg.KL1: cP1 = 4.19 – 2.35 (0.133) = 3.877 kJ/kg.KL2: cP2 = 4.19 – 2.35 (0.205) = 3.705 kJ/kg.KL3: cP3 = 4.19 – 2.35 (0.5) = 3.015 kJ/kg.K

Then the new values of the enthalpy are,(1) H1 = HS2 + 1.884 BPR1 = 2682 + 1.884(0.35) = 2683 kJ/kg.

λS1 = HS1 – hS1 = 2708 – 508 = 2200 kJ/kg.(2) H2 = HS3 + 1.884 BPR2 = 2654 + 1.884(0.63) = 2655 kJ/kg.

λS2 = H1 – hS2 = 2683 – 440 = 2243 kJ/kg.(3) H3 = HS4 + 1.884 BPR3 = 2595 + 1.884(2.45) = 2600 kJ/kg.

λS3 = H2 – hS3 = 2655– 362 = 2293 kJ/kg.

55

Writing a heat balance on each effect,and solving,

(1) 22680(3.955)(26.7-0)+2200S = 3.877L1(104.33-0)+(22680-L1)2683(2) 3.877L1(104.33-0)+(22680-L1)2243=3.708L2(87.11-0)+(L1-L2)2655(3) 3.708L2(87.11-0)+(L1-L2)2293=4536(3.015)(54.1-0)+(L2-4536)2600

L1 = 17005 kg/h L2 = 10952 L3 = 4536 S = 8960V1 = 5675 V2 = 6053 V3 = 6416

EVAPORATIONSolving for q and A in each effect,

( ) WxxSq S6

11 10476.5100022003600

8960 =

== λ

( ) WxxVq S6

212 10539.3100022433600

5675 =

== λ

( ) WxxVq S6

323 10855.3100022933600

6053 =

== λ

56

( ) WxxVq S 323 10855.3100022933600

=

== λ

( )2

6

'11

11 6.104

77.163123

10476.5m

x

TU

qA ==

∆=

( )2

6

'22

22 6.105

87.161987

10539.3m

x

TU

qA ==

∆=

( )2

6

'33

33 9.104

36.321136

10855.3m

x

TU

qA ==

∆=

EVAPORATION

The average area Am = 105.0 m2 to use in each effect.

57

steam economy = 025.28960

641660535675321 =++=++S

VVV