Optimizing Counter Flow Heat Exchangers: Calculating Outlet Temperatures

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Discussion Overview

The discussion revolves around calculating the outlet temperatures of exhaust gases in a counter flow heat exchanger, specifically focusing on the heat transfer between process water and exhaust gases. Participants explore various methods and equations related to heat load and temperature changes, while addressing uncertainties in the calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a homework problem involving the calculation of outlet temperatures for a heat exchanger, providing specific parameters such as flow rates and temperature changes.
  • Another participant suggests determining the heat load from the process water's flow rate and temperature change.
  • There is a discussion about the correct calculation of heat load, with one participant initially omitting the heat capacity in their expression.
  • Participants discuss the implications of the heat exchanger being a double pipe type and in counter flow, questioning how this affects the equations used.
  • One participant expresses uncertainty about how to proceed without knowing the heat exchanger area, considering the effectiveness method for calculating outlet temperatures.
  • A participant shares their calculations for the heat load and outlet temperature, arriving at a specific value for the outlet temperature based on their approach.
  • Another participant challenges the calculated outlet temperature, suggesting a different value based on their interpretation of the equations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct outlet temperature, with differing calculations presented. There is ongoing uncertainty regarding the appropriate method to use given the lack of area information and the implications of the heat exchanger configuration.

Contextual Notes

Participants highlight limitations in their calculations, such as missing assumptions about the heat exchanger area and the dependence on specific heat capacities. The discussion reflects various approaches to the problem without resolving the mathematical steps or assumptions involved.

Mitch1
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Homework Statement



Process water with a specific heat capacity of 4.182 kJ kg–1 K–1 flows at a rate of 0.050 kg s–1 through a heat exchanger where its temperature is increased from 16°C to 85°C. Heat is supplied by exhaust gases (mean specific heat capacity 1.075 kJ kg–1 K–1) which enter the heat exchanger at a temperature of 420°C. If the mass flowrate of the exhaust gases is 0.044 kg s–1, determine their outlet temperature

Homework Equations



∆T=(TH2 –TC1)–(TH1 –TC2) ln⎛TH2 –TC1⎞

⎜ ⎟ ⎝TH1 –TC2⎠

The Attempt at a Solution



I understand the basis on how to calculate the Temperature difference and heat loss along with area however I am unsure on how to calculate the outlet hot temperature Th2 and how to rearrange this equation to find the outlet temp
Any guidance or equations that may be of use would be much appreciated
 
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What is the rate of heat flow from the hot gas to the process water? This is the "heat load" of the heat exchanger. You can determine this directly from the flow rate and temperature change of the process water.

Chet
 
Hi Chet
Thanks for your reply
Would this be 0.05x(85-16)=3.45?
Is this the heat capacity rate?
Thanks
 
Mitch1 said:
Hi Chet
Thanks for your reply
Would this be 0.05x(85-16)=3.45?
Is this the heat capacity rate?
Thanks
You left out the heat capacity from your expression to calculate the heat load. The heat load is WCp(Tout-Tin).

Chet
 
Chestermiller said:
You left out the heat capacity from your expression to calculate the heat load. The heat load is WCp(Tout-Tin).

Chet
Ok so it simply 3.45x4.182=14.4279
Thanks
 
Chet, please note the heat exchanger is a double pipe type and the fluids are in counter flow - not sure if this has an impact on which equations that were going to be used
Thanks in advance
 
Mitch1 said:
Chet, please note the heat exchanger is a double pipe type and the fluids are in counter flow - not sure if this has an impact on which equations that were going to be used
Thanks in advance
You can figure that out yourself if you can articulate what is happening physically. Please give it a try.

Chet
 
Chestermiller said:
You can figure that out yourself if you can articulate what is happening physically. Please give it a try.

Chet
Okay thanks for your help I will give it a go
 
Hi Chet
Thanks for the help with the heat exchanger question
I would be grateful if you pointed me in the right direction on which method to use as I do not have the area, i could easily work the temp outlet hot out using the effectiveness method although I do not have the area. I am rather stuck and do not know which way to look
Again thanks
 
  • #10
Mitch1 said:
Hi Chet
Thanks for the help with the heat exchanger question
I would be grateful if you pointed me in the right direction on which method to use as I do not have the area, i could easily work the temp outlet hot out using the effectiveness method although I do not have the area. I am rather stuck and do not know which way to look
Again thanks
You would have to do that kind of analysis if you didn't know the inlet and outlet temperatures for one of the streams. But, in this problem, you do. So you can get the heat load, and you can get the outlet temperature of the other stream. If you were designing a heat exchanger, you would not be finished here. To design the heat exchanger, you would have to find the heat transfer area to make good on this heat load. You could use the method you alluded to in order to do this.

Chet
 
  • #11
Hi
this is my attempt

Φ=qmc cpC (TC1-TC2) = qmH cpH (TH1-TH2)
Therefore:
Φ = qmc cpC (TC1-TC2)
Φ = 0.050 x 4.184 x (85-16)
Φ = 0.209 x 69
Φ = 14.4W
Now:
14.4 =0.044 x 1.075 x (410 - TH2)
14.4/0.044 x 1.075 = (410 - TH2)
= 410 – 304
TH2 = 106 degrees
 
  • #12
blitzman said:
Hi
this is my attempt

Φ=qmc cpC (TC1-TC2) = qmH cpH (TH1-TH2)
Therefore:
Φ = qmc cpC (TC1-TC2)
Φ = 0.050 x 4.184 x (85-16)
Φ = 0.209 x 69
Φ = 14.4W
Now:
14.4 =0.044 x 1.075 x (410 - TH2)
14.4/0.044 x 1.075 = (410 - TH2)
= 410 – 304
TH2 = 106 degrees
Looks good.

Chet
 
  • #13
I believe it should be 420-305 which is 115 degrees
 

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