Calculation of heat created during compression

Click For Summary
The discussion revolves around calculating the heat generated during steam compression in a milk evaporator. The steam enters at 60°C and 200 mbar, and is compressed to 250 mbar, during which it heats up to 100°C. Participants emphasize that the heat generated is primarily due to compression, not external heat addition, and that isentropic assumptions may not be appropriate. The need for accurate enthalpy values from steam tables is highlighted to assess the efficiency of the process. Understanding these calculations is crucial for maintaining the quality of the evaporated milk by avoiding protein denaturation.
Gosia
Messages
2
Reaction score
0

Homework Statement


Hello, I have a task to make a heat balance of the evaporator. To do this, I was told to carry a few smaller energy balances in different parts of the evaporator. One of the things I had to calculate was the heat generated during steam compression.

-steam is entering 12818,18 kg/h the fan at temp 60 C at pressure 200 mbar (saturated conditions)
-steam durninng compression is being heated to 100 C, we want to achive 250mbar
-after that steam is being cooled by sprayed water at 65 C

I have to calculate the work of the pump and the heat that is being generated.

Homework Equations


dU=Q-W

The Attempt at a Solution


[/B]
I was trying to approach it as it is isotropic compression going from 200mbar 60C to 250C 65
by simply divieding enthalpies from steam tables and omiting whole aspect of tempreture increse. But my supervisor said that this is actually hapaning in plants, that durning compression this much amount of heat is produced due to friction and comression and it must be calculated in the project. I coudnt find any similar exercises or exampels.Is this data sufficient? I feel like I am missing something. Please help!
 
Last edited by a moderator:
Physics news on Phys.org
Hello Gosia, :welcome:

Gosia said:
I was trying to approach it as it is isotropic compression going from 200mbar 60C to 250C 65
by simply divieding enthalpies from steam tables and omiting whole aspect of tempreture increse.
Hard to follow. PLease describe the steps you mentioned separately:
Gosia said:
-steam is entering 12818,18 kg/h the fan at temp 60 C at pressure 200 mbar (saturated conditions)
SO you have ##h_i## and ##s_i## there? namely : ?
-steam durninng compression
Assume isentropic (not isotropic -- that's something else): again: ##h_{is}## there? namely : ? So ##T_{is}## ?

In reality ##h## ends up higher than ##h_{is}## because there is an efficiency ##\eta_{is} = {h_i - h_{actual} \over h_i - h_{is} }##
again: ##h_{actual}## there? namely : ? So ##T_{actual}## ?
is being heated to 100 C, we want to achive 250mbar
heated as in 'heat is added' or is it heated because of the compresssion ?
(In the latter case your ##T_{actual}## deterimines ##\eta_{is}## )

-after that steam is being cooled by sprayed water at 65 C
Strange. Why would someone do such a destructive deed ?

[edit] sheet 20 here
 
BvU said:
Hello Gosia,

Hard to follow. PLease describe the steps you mentioned separately:

That is not necessary because this approach was totally wrong.
BvU said:
SO you have ##h_i## and ##s_i## there? namely : ?
Yes, evrything from steam tables
BvU said:
Assume isentropic (not isotropic -- that's something else): again: ##h_{is}## there? namely : ? So ##T_{is}## ?
Isentropic is not a good assumption.

BvU said:
In reality ##h## ends up higher than ##h_{is}## because there is an efficiency ##\eta_{is} = {h_i - h_{actual} \over h_i - h_{is} }##
again: ##h_{actual}## there? namely : ? So ##T_{actual}## ?
heated as in 'heat is added' or is it heated because of the compresssion ?
(In the latter case your ##T_{actual}## deterimines ##\eta_{is}## )

Heat is beeing generated ONLY becaouse of compression. NO additional heat is beeing given by surroundings.
BvU said:
Strange. Why would someone do such a destructive deed ?
This is a milk evaporator, in order to have no denaturation of proteins we have to maintain tempereture below 100 C
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
9K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
9K