heat exchanger exergy analysis

[zdej]

i am having a problem with heat exchanger (tube and fin) exergy analysis...

the problem which i am having looks like this:

in one of the cases air (mass flow of air: m,a=0,71 kg/s) is entering the heat exchanger with temperature T,aIN=21C and relative humidity RH,aIN=57%, air is cooled to the state where it has temperature T,aOUT=19C and RH,aOUT=64%. Water entering the heat exchanger has T,wIN=7C and on the outlet T,wOUT=14,91C, mass flow of water is 0,043 kg/s.

for exergy calculation:
air inlet entalpy and entropy h,aIN=43,55 kJ/kg and s,aIN=0,1552 kJ/kgK. Air inlet conditions are the same as dead end state conditions. Air outlet enthalpy and entropy are h,aOUT=41,50 kJ/kg and s,aOUT=0,1487 kJ/kgK.
water enthalpy and entropy: h,wIN=29,52 kJ/kg; s,wIN=0,1064 kJ/kgK; h,wOUT=62,68 kJ/kgK;s,wOUT=0,2231 kJ/kgK. Water enthalpy and entropy at dead end state (21C): h,w0=88,18 kJ/kg and s,w0=0,3107 kJ/kgK.

calculated exergies, air: ex,aIN=0 kj/kg;(ex,aIN*m,a): Ex,aIN=0 kW, ex,aOUT=0,1330 kJ/kg, Ex,aOUT=0,0944 kW
calculated exergies, water: ex,wIN=1,4348 kJ/kg, Ex,wIN=0,0627, ex,wOUT=0,2675 kJ/kg, Ex,wOUT=0,0117 kW

if i calculate second law efficiency ( outlet exergies/inlet exergies), i get efficiency: 1,69 !?

i do not know where could be a mistake... enthalpy and entropy values seems to be correct... any ideas?

Thanks a lot!

 

[chicopee]

Where did you guys ever get the term "exergy"?

 

[zdej]

and what is wrong with it?

 

[chicopee]

I've taken thermo and heat transfer courses and this term never came up and that may be the reason for the lack of meaningful responses.

 

[MisterE2]

Yeah, I kindof just stopped using exergy as a grad student in thermal fluid sciences, I have to say though, why do you care so much about the relative humidity, it doesn't change all that much in the energy transfer analysis and my guess would be that's what might be part of the error. But, there's another problem, apparently your water is heating the air and the air is doing exactly the same to the water by your problem definition.

 

[DRWeig]

zdej,

Are these values that you have measured?

Best to you,

Goober Dave

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[zdej]

MisterE2, could you explain this: 'But, there's another problem, apparently your water is heating the air and the air is doing exactly the same to the water by your problem definition.'? Thanks.

DRWeig, they are not measured. I calculated them, I compared my results with some heat exchanger calculation software, and they are quite similar...

 

[MisterE2]

ah, i c, misread due to the format, the air heats the cooled water, you should really put Ta and Tw inlet and outlet on different lines directly above, its just hard to read now.

Anyway, wouldn't it be easier to just calculate the loss from the system sum(enthalpy outlet)/sum(enthalpy inlet) which would be the system power loss %, you didn't specify whether the exit values were desired or not.