evaporator

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LT # 2

EVAPORATION

SIGNATURE

Lagamayo, Linear Larie B. ______________

Tirona, Lei An ______________

Laurenciana, Toni Rose ______________

PROBLEM 1 (EVAPORATOR)

A triple – effect forward – 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 O_{C = 1.78x + 6.22x}2

(BPR O_{F = 3.2x + 11.2x}2_{), where x is wt fraction of sugar in solution (K}₁_{). Saturated}

Steam at 205.5 kPa (29.8 psia) [121.1 OC (250 OF) saturation temperature] is being used.

The pressure in the vapor space of the third effect is 13.4 kPa (1.94 psia). The feed rate

is 22,680 kg/h (50,000 lbm/h) at 26.7 OC (80 OF). The heat capacity of the liquid

solutions is (K1) Cp = 4.19 – 2.35x kJ/kg K (1.0 – 0.56x Btu/lbm OF). The heat of the

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 or 550, 350, and 200 Btu/h

ft2_{. If each effect has the same surface area, calculate the area, the steam rate used, and}

the steam economy. Given:

F = 22,680 kg/h

BPR O_{C = 1.78x + 6.22x}2_{S = 205.5 kPa, 121.1}O_{C C}_p_{= 4.19 – 2.35x (K}₁₎

U1 = 3,123 W/m2 K U2 = 1987 W/m2 K U3 = 1136 W/m2 K

P3 = 13.4 kPa (3rd effect evaporator)

Required: Area, Steam Rate, and Steam Economy Solution:

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Overall Balance F = L3 + (V1 + V2 + V3) 22,680 = L3 + (V1 + V2 + V3) - - - > Equation 1 Solid Balance FxF = xsolidsL3 + (V1 + V2 +V3)(0) (0.1)(22,680) = 0.50L3 L3 = 4536 kg/h

Amount of Liquid Vaporized

18,144 = V1 + V2 + V3

Assume the each of the evaporators vaporize the same amount of liquid 3V1 = 18,144

V1 = 6,048 kg/h

Material Balances at Evaporators

Evaporator 1: F = V1 + L1 - - - -> L1 = 16,632 kg/h

Evaporator 2: L1 = V2 + L2 - - - > L2 = 10,584 kg/h

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Solid Balances at Evaporators Evaporator 1: FxF = L1x1 (0.1)(22,680) = 16,632x1 - - - > x1 = 0.136 Evaporator 2: L1x1= L2x2 (0.136)(16,632) = 10,584x2 - - - > x2 = 0.214 Evaporator 3: L2x2 = L3x3 (0.214)(10,584) = 4,536x3 - - - > x3 = 0.500 BPR Calculation BPR1 = 1.78x1 + 6.22 (1.78)(0.136) + 6.22(0.136)2 - - - > BPR1 = 0.36 OC BPR2 = 1.78x2 + 6.22 (1.78)(0.214) + 6.22(0.214)2 - - - > BPR2 = 0.65 OC BPR3 = 1.78x3 + 6.22 (1.78)(0.500) + 6.22(0.500)2 - - - > BPR3 = 2.45 OC ( ) ( ) ( )

Calculation of Temperature Drop on the Evaporators

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Assumption

*_{Since cold feed has entered effect 1, more heat should be added.} _{should be}

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Estimated Values of Temperature Drops

Boiling Point Calculation of Solution on the Evaporators

Heat Capacity Calculation of the Liquid on the Evaporators

F: Cp = 4.19 – 2.35xF - - - > 4.19 – 2.35(0.1) = 3.955 kJ/kg K

L1: Cp = 4.19 – 2.35x1 - - - > 4.19 – 2.35(0.136) = 3.869 kJ/kg K

L2: Cp = 4.19 – 2.35x2 - - - > 4.19 – 2.35(0.214) = 3.684 kJ/kg K

L3: Cp = 4.19 – 2.35x3 - - - - -> 4.19 – 2.35(0.500) = 3.015 kJ/kg K

Enthalpy Calculation of the Vapor Streams

( ) ( ) ( )

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Latent Heat Calculation

Energy Balance Calculation on the Evaporators

Evaporator 1: ( ) ( )

Evaporator 2: ( ) ( ) Evaporator 3: ( ) ( )

Substitution and Calculation of Corresponding Values

22,680(3.955)(26.7 – 0) + S(2,200) = L1(3.869)(105.54 – 0) + (22,680 – L1)(2,685)

L1(3.869)(105.54 – 0) + (22,680 – L1)(2,244) = L2(3.684)(86.84 – 0) + (L1 – L2)(2,655)

L2(3.684)(86.84 – 0) + (L1 – L2)(2,294) = 4,536(3.015)(54.12 – 0) + (L2 – 4,536)(2,600)

Obtaining the Values

L1 = 17,078 kg/h L2 = 11,068 kg/h L3 = 4,536 kg/h S = 8,936 kg/h

V1 = 5,602 kg/h V2 = 6,010 kg/h V3 = 6,532 kg/h

Heat Transfer Rate Calculation

Evaporator 1: ( ) ( )( ) Evaporator 2: ( ) ( )( ) Evaporator 3: ( ) ( )( )

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Area Calculation Evaporator 1: _{( )( )} Evaporator 2: _{( )( )} Evaporator 3: _{( )( )} * Trial 2

New Solid Balances at Evaporators

Evaporator 1: FxF = L1x1 (0.1)(22,680) = 17,078x1 - - - > x1 = 0.133 Evaporator 2: L1x1= L2x2 (0.133)(17,078) = 11,068x2 - - - > x2 = 0.205 Evaporator 3: L2x2 = L3x3 (0.205)(11,068) = 4,536x3 - - - > x3 = 0.500 New BPR Calculation BPR1 = 1.78x1 + 6.22 (1.78)(0.133) + 6.22(0.133)2 - - - > BPR1 = 0.35 OC BPR2 = 1.78x2 + 6.22 (1.78)(0.205) + 6.22(0.205)2 - - - > BPR2 = 0.63 OC BPR3 = 1.78x3 + 6.22 (1.78)(0.500) + 6.22(0.500)2 - - - > BPR3 = 2.45 OC ( ) ( ) ( ) ( )( ) ( )( )

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( )( )

New Boiling Point Calculation of Solution on the Evaporators

New Heat Capacity Calculation of the Liquid on the Evaporators

F: Cp = 4.19 – 2.35xF - - - > 4.19 – 2.35(0.1) = 3.955 kJ/kg K

L1: Cp = 4.19 – 2.35x1 - - - > 4.19 – 2.35(0.133) = 3.877 kJ/kg K

L2: Cp = 4.19 – 2.35x2 - - - > 4.19 – 2.35(0.205) = 3.708 kJ/kg K

L3: Cp = 4.19 – 2.35x3 - - - - -> 4.19 – 2.35(0.500) = 3.015 kJ/kg K

New Enthalpy Calculation of the Vapor Streams

( ) ( ) ( )

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New Latent Heat Calculation

New Energy Balance Calculation on the Evaporators

Evaporator 1: ( ) ( )

Evaporator 2: ( ) ( ) Evaporator 3: ( ) ( )

New Substitution and Calculation of Corresponding Values

22,680(3.955)(26.7 – 0) + S(2,200) = L1(3.877)(104.33 – 0) + (22,680 – L1)(2,683)

L1(3.877)(104.33 – 0) + (22,680 – L1)(2,243) = L2(3.708)(87.11 – 0) + (L1 – L2)(2,655)

L2(3.708)(87.11 – 0) + (L1 – L2)(2,293) = 4,536(3.015)(54.12 – 0) + (L2 – 4,536)(2,600)

Obtaining the New Values

L1 = 17,005 kg/h L2 = 10,952 kg/h L3 = 4,536 kg/h S = 8,960 kg/h

V1 = 5,675 kg/h V2 = 6,053 kg/h V3 = 6,416 kg/h

New Heat Transfer Rate Calculation

Evaporator 1: ( ) ( )( ) Evaporator 2: ( ) ( )( ) Evaporator 3: ( ) ( )( )

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New Area Calculation

Evaporator 1: _{( )( )}

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Steam Economy Calculation