Hi all

Please help

I have a question regarding calculating heat load and required surface area for a single effect evaporator.

Details:

Solution Feed rate 1.1kg s-1

Solution feed concentration 8%, concentrated to 30%

Feed solution at 60 degrees Celsius with Specific Heat Capacity of 4.2kJ kg-1 K-1

Boiling point elevation of 30 degrees Celsius

Heating steam is provided at 1.4bar and separator pressure is 0.1 bar

Condensate collected at rate of 1kg s-1 (assumed to be at condensing temp)

Overall heat coefficient 3000W m-2 K-1

Calculate:

Vapour flowrate:

w c₁ = (w - W) c₂

w = solution feed rate kg s-1 1.1

c₁ = feed concentration kg solute per kg solution 0.08

W = Vapour flow rate kg s-1

c₂ = product concentration kg solute per kg solution 0.30

1.1 x 0.08 = (1.1 - W) 0.30

W = 1.1 - (1.1 x 0.8)

0.3

W = 0.8066 kg s-1

Product flow rate:

w - W = Product flow rate

1.1 - 0.8066 = 0.2934 kg s-1

Product flow rate = 0.2934 kg s-1

These I am confident with.