Optimizing Counter Flow Heat Exchangers: Calculating Outlet Temperatures

[Mitch1]

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

 

[Chestermiller]

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

 

[Mitch1]

Hi Chet
Thanks for your reply
Would this be 0.05x(85-16)=3.45?
Is this the heat capacity rate?
Thanks

 

[Chestermiller]

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 WC_p(T_out-T_in).

Chet

 

[Mitch1]

You left out the heat capacity from your expression to calculate the heat load. The heat load is WC_p(T_out-T_in).

Chet

Ok so it simply 3.45x4.182=14.4279
Thanks

 

[Mitch1]

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

 

[Chestermiller]

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

 

[Mitch1]

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

 

[Mitch1]

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

 

[Chestermiller]

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

 

[blitzman]

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

 

[Chestermiller]

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

 

[Mitch1]

I believe it should be 420-305 which is 115 degrees