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.