- #1

Naraneer

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## Homework Statement

A double effect forward feed evaporator is to concentrate a 12.5% solution of a certain chemical compound "YZ". The feed is to be at and is to be 120 F of such an amount as to contain 6250 "YZ" per hour. Saturated steam is available at 250 F (29.7 psi ). The accepted value of the overall coefficient for the first effect is ( 400 Btu/hr ft2 F ). Condensate may be assumed to leave at the saturation temperature of the steam. The evaporator is provided with a barometric counter-current condenser fed with 357500 Ib/hr cooling water entering at 85 F and leaving at 136 F

i) Calculate the values of the concentration of the "YZ" solution leaving the 1st and 2nd effects, respectively. ii) Determine the temperature of the 2nd effect.

iii) What heating surface must be used if all effects are to have the same area?

iv) Compute the value of the overall coefficient of the 2nd effect.

## Homework Equations

[/B]

Cp steam=0.48 Btu/Ib F Latent Heat of steam at 250 F=945.5 Btu/Ib

Feed Enthalpy= 70 Btu/Ib

Evaporation per Ib of Steam = 1.45

Effect I II

T 230 T

t sat 216 141

Lf (latent heat) 967 1011.32

h (enthalpy of liquid) 192 175

## The Attempt at a Solution

(945.5)*m(steam) + 5*10^4(70)=(5*10^4-m(vapor 1)(192)+m(vapor 1)Hv1

Hv1=0.48(230-216)+967 + Enthalpy of Saturated liquid at T=216 F

Effect 2

(0.48(230-216)+967) *m(vapor 1)+(5*10^4-m(vapor 1)(192)=(5*10^-4-m(vapor 1)-m(vapor 2))(175)+m(vapor 2)*Hv2

Hv2=0.48 (T-141)+1013.2 + Enthalpy of Saturated liquid at T=141 F

How can I determine the values of Enthalpy of saturated liquid at 216 F and 141 F?

Secondly, How can I perform an Energy balance over the condenser?