Calculation of heat created during compression

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SUMMARY

The discussion focuses on calculating the heat generated during the compression of steam in a milk evaporator system. The steam enters at a flow rate of 12818.18 kg/h, at a temperature of 60°C and a pressure of 200 mbar, and is compressed to 250 mbar while being heated to 100°C. The participants emphasize the importance of using enthalpy values from steam tables and correcting for efficiency during the isentropic process, noting that heat generation occurs solely due to compression without additional heat from the surroundings.

PREREQUISITES
  • Understanding of steam tables and thermodynamic properties
  • Knowledge of isentropic vs. isotropic processes
  • Familiarity with energy balance equations, specifically dU=Q-W
  • Basic principles of heat transfer in evaporators
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  • Study the calculation of enthalpy changes using steam tables
  • Learn about the efficiency of isentropic processes in thermodynamics
  • Research the principles of heat generation during compression in fluid systems
  • Explore the design and operation of milk evaporators to understand temperature control
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Students and professionals in chemical engineering, thermodynamics, and food processing, particularly those involved in the design and optimization of evaporator systems.

Gosia
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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!
 
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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
 

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