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Single Effect Evaporator Calculation

  • Thread starter Starlord01
  • Start date
  • #1
Hi All.
Please can someone 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.
 
Last edited by a moderator:

Answers and Replies

  • #2
Below I am unsure of.
Guessing I am wrong though but would require some explanation as to where I am going wrong please:

Calculate

Required Heat Surface Area:

Using Q = U A ΔT

Q = Heat Load
U = Overall heat transfer coefficient 3000W m-2 K-1
A = Heat Transfer Area
ΔT = temp difference between steam and solution 109.3 - 90 = 19.3 degrees C


A = Q
U ΔT

Q = m x hfg
where m = mass of steam condensed 1kg s-1
hfg = latent heat of vaporisation 2232kJ Kg-1

Q = 2232000W

so

A = 2232000
3000 x 19.3

A = 38.55m2


Something does not seem right here?
 
  • #3
THIS HOMEWORK WAS POSTED IN ANOTHER FORUM, SO NO TEMPLATE
It goes on to request evaporator economy

Economy = Vapour product
steam supplied

=0.2934
1

= 0.2934

Again, appears suspect?


Other parts require to calculate heat required to raise the feed to boiling temp - I have attempted this but don't have the workings to hand

And available heat for evaporation and hence evaporation rate expected (I think this would be far easier if my previous sections were correct)


Please help asap.
I have atleast tried and am not asking for the actual answers but a clear demonstration and explanation to clarify where I am wrong

Thanks in advance
 
  • #4
19,918
4,095
This is a homework problem, correct?
 
  • #5
Yes it is
 
  • #6
19,918
4,095
Hi All.
Please can someone 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.
I'm wondering why you just didn't write Product Flow Rate = (1.1)(0.08)/0.3=0.2933

What was their point in providing a boiling point elevation of 30 C?
 
Last edited by a moderator:
  • #7
19,918
4,095
Below I am unsure of.
Guessing I am wrong though but would require some explanation as to where I am going wrong please:

Calculate

Required Heat Surface Area:

Using Q = U A ΔT

Q = Heat Load
U = Overall heat transfer coefficient 3000W m-2 K-1
A = Heat Transfer Area
ΔT = temp difference between steam and solution 109.3 - 90 = 19.3 degrees C
I see where you got the 109.3 C, but where did you get the 90 C? The problem statement indicates that the separator pressure is 0.1 bar. Isn't that the same thing as the evaporator pressure.
 
  • #8
Thanks for the input.
Boiling point elevation occurs as the solution concentrates.
When the solution approaches 30% NaOH, the boiling point rises by 30degrees C requiring an evaporator temperature of 90degrees C.
Delta T is the difference between steam temp and evaporator solution temp.

Or this was my understanding?
 
  • #9
I'm wondering why you just didn't write Product Flow Rate = (1.1)(0.08)/0.3=0.2933
My route just seemed more logical/easier in my own mind - personal preference I guess?
 
  • #10
19,918
4,095
Thanks for the input.
Boiling point elevation occurs as the solution concentrates.
When the solution approaches 30% NaOH, the boiling point rises by 30degrees C requiring an evaporator temperature of 90degrees C.
Delta T is the difference between steam temp and evaporator solution temp.

Or this was my understanding?
It isn't clear to me how this particular evaporator is operated. Can you supply a diagram? Is the vapor from the evaporator heated to provide the steam?
 
  • #11
19,918
4,095
The boiling point elevation BPE is equal to the concentrated liquid temperature in the evaporator minus the vapor temperature. At 0.1 bar pressure for the vapor, the vapor temperature is 45.8 C. So the concentrated liquid temperature is 75.8 C, not 90 C.
 

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