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Effectiveness to calculate the outlet temperatures of a heat exchanger

  • Thread starter PCal
  • Start date
31
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Problem Statement
The data below relates to a specific heat exchanger. A reliable colleague has looked up an effectiveness chart and says that the effectiveness in the given operating conditions is 0.82.
For hot fluid: mass flowrate qm= 0.7 kg s -1, specific heat capacity cp=1.8 kJ kg-1 K-1, inlet temp= 140°C . For cold fluid: mass flowrate qm= 0.6 kg s -1, specific heat capacity cp=4.2 kJ kg-1 K-1, inlet temp= 15°C . Area of heat transfer surface=14m^2, Overall heat transfer coefficient 360 W m-2 K-1.
Determine:
i. the two outlet temperatures.
ii. the heat transfer rate.
I've spent ages on this but can't seem to find an equation to find both outlet temperatures! Any help would be greatly appreciated.
Relevant Equations
ε=(T_HI-T_H2)/(T_HI-T_C1 )
Because the specific heat capacity is lower for the hot fluid:
ε=(T_hi-T_h2)/(T_hi-T_C1 )

0.82=(140-T_h2)/(140-15)

140-T_h2= 102.5

T_h2= 140-102.5=37.5°

I'm not sure if I'm actually on the right track here, if I am this is as far as I've gotten. I can't seem to then find an equation to calculate T_c2 other than the equation which my lessons tell me is only used when the specific heat capacity is lower for the cold fluid.
 

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This is where I'm currently at
241395
 

BvU

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In my lessons ε has been used to define effectiveness
My bad -- long time ago :sorry:

To make up: Got the same result as you did.
 
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241404
one more thing, how's this for the heat transfer rate please?
 
19,087
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Is this a countercurrent heat exchanger or a co-current heat exchanger? I assume that it is countercurrent, correct?
 
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Is this a countercurrent heat exchanger or a co-current heat exchanger? I assume that it is countercurrent, correct?
The question doesn't actually specify, I've assumed it's countercurrent/ cross flow
 
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Is this a countercurrent heat exchanger or a co-current heat exchanger? I assume that it is countercurrent, correct?
Part C of the question also refers to the correction factor which leads me to think it's cross flow
 

BvU

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one more thing, how's this for the heat transfer rate please?
In #1 you calculated 129 kW
##Q_{\rm max} ## = 157.2, so I find 218 kW hard to believe ...

The 218 comes from UA*LMTD which applies to a countercurrent heat exchanger. Here you have a less effective heat transfer configuration, so a higher LMTD and your expression isn't valid.
Must say the exercise composer doesn't make it easy for you. Good you did the UA*LMTD check -- I did it too and had to look more than twice :wink:


Is this a countercurrent heat exchanger or a co-current heat exchanger? I assume that it is countercurrent, correct?
Hi Chet! I was kind of struggling with that too. With ##\ C_{\rm min} =1.26,\ ## and ##\ UA =5.04##, NTU = 4. C_r=0.5 so that ##\ \varepsilon## for co-current would be 0.66 and counter 0.93 with the 0.82 in between.
 
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In #1 you calculated 129 kW
##Q_{\rm max} ## = 157.2, so I find 218 kW hard to believe ...

The 218 comes from UA*LMTD which applies to a countercurrent heat exchanger. Here you have a less effective heat transfer configuration, so a higher LMTD and your expression isn't valid.
Must say the exercise composer doesn't make it easy for you. Good you did the UA*LMTD check -- I did it too and had to look more than twice :wink:


Hi Chet! I was kind of struggling with that too. With ##\ C_{\rm min} =1.26,\ ## and ##\ UA =5.04##, NTU = 4. C_r=0.5 so that ##\ \varepsilon## for co-current would be 0.66 and counter 0.93 with the 0.82 in between.
Of course, I didn't click onto that! Does 129.5kW sound like a more believable heat transfer rate please? Apologies this modules taking a lot to get my head around!
 
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Yes. If I assume counterflow, I get outlet temperatures of 73 for the cold stream and 24.1 for the hot stream.
 

BvU

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Of course, I didn't click onto that! Does 129.5kW sound like a more believable heat transfer rate please? Apologies this modules taking a lot to get my head around!
The 129.15 follows from the given information, yes.
 
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Yes. If I assume counterflow, I get outlet temperatures of 73 for the cold stream and 24.1 for the hot stream.
Sorry I'm so confused, I got 66.25 for the cold stream outlet and 37.5 for the hot, this seems way off the mark!
 
19,087
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Sorry I'm so confused, I got 66.25 for the cold stream outlet and 37.5 for the hot, this seems way off the mark!
My calculation was for countercurrent flow. If you check your numbers, the heat transfer rate based on the log mean temperature difference does not match the heat transfer rate based on the flows and heat capacities (if the flow were assume to be countercurrent).
 
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Ohh that makes sense thank you, I'll look back over my lesson and calculations.
 
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My calculation was for countercurrent flow. If you check your numbers, the heat transfer rate based on the log mean temperature difference does not match the heat transfer rate based on the flows and heat capacities (if the flow were assume to be countercurrent).
Still completely baffled I'm afraid, it seems you've used 1-e^-ntu(1-c)/1-Ce^-ntu(1-c) to get effectiveness= 0.92 and used that in your equations. Going through the same equation I got the same answer, however my question states that the effectiveness is 0.82. Given that, my outputs of cold= 66.25 and hot=37.5 would be correct wouldn't they?
 
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Still completely baffled I'm afraid, it seems you've used 1-e^-ntu(1-c)/1-Ce^-ntu(1-c) to get effectiveness= 0.92 and used that in your equations. Going through the same equation I got the same answer, however my question states that the effectiveness is 0.82. Given that, my outputs of cold= 66.25 and hot=37.5 would be correct wouldn't they?
Yes. That would be correct. But it looks like their providing U and A in the problem statement is totally extraneous sine you never need to use it in solving this problem.

Incidentally, I didn't use the equation you referred to in solving the countercurrent results; I derived the results from scratch.
 
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Thank you, so using that I'd get a heat transfer rate of 129.15 kW? Given that, am I correct in thinking I'd then use ∆Tc=UA/Φ then F=∆Tc x LMTD to get the correction factor?
 
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Thank you, so using that I'd get a heat transfer rate of 129.15 kW? Given that, am I correct in thinking I'd then use ∆Tc=UA/Φ then F=∆Tc x LMTD to get the correction factor?
Sorry. I don't fully understand what you are saying. The 129.15 is correct for the problem as posed.
 
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Sorry. I don't fully understand what you are saying. The 129.15 is correct for the problem as posed.
Sorry, the next part of the question asks...
241477

Would the equations I posted above prove this wrong please? I worked out that the correction factor=1.68 but have no idea if this is correct.
 
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0.59! Why have I had to take the reciprocal though?
 
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Got it! I was doing F=∆Tc x LMTD as opposed to F=∆Tc/LMTD
Thank you for all your help!
 
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