Fin/Fan heat exchanger mass and energy balance

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The discussion centers on calculating the mass and energy balance for a fin/fan heat exchanger, specifically addressing discrepancies in heat rates between water and air. The user is struggling to apply fin efficiency to determine heat dissipation through the fins, with given mass flow rates and temperature differences for both fluids. Concerns are raised about the accuracy of the air flow rate obtained from an anemometer, suggesting potential misinterpretation of the readings. The importance of fin efficiency in this context is questioned, as it typically pertains more to heat exchanger design than to mass and energy balance calculations. Clarification on the anemometer's measurement units is sought to ensure accurate data for the project.
HethensEnd25
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I am currently working on a project which requires a mass and energy balance on a fin/ fan heat exchanger. The problem I am running into is finding how to apply the fin efficiency to find out the amount of hea tbeing dissipated through the fins. The hot fluid is water and the cooling fluid is air.

Question: How do you find the amount of heat going through the fin/fan?

Problem :

Given:
Base area of fan A=.62 m2
OD of tubes= 15.8mm
ID=unknown
number of tubes = picture included.
Cross Flow one pass fluids unmixed

Water Data:
mass flow rate: 0.898 lb/s
Inlet Temp: 127F
Outlet Temp: 100F
Density @ (avg of temps) = 61.986lb/ft3
Cp= 1Btu/lb*F

Air Data:
Mass flow rate : 310.7 lb/s
Inlet Temp:79.1F
Outlet Temp: 88.95F
Density @ (avg temp): 0.08 lb/ft3
Cp=0.25 Btu/lb*F

Values of density and Cp taken from Unit Operations of Chemical Engineering Seventh Edition.Equations used:

q=mCpΔT

where;
q=heat rate
m=mass flow rate
Cp= specific heat
ΔT=temperature difference between inlet and outlet

q=U0AΔTL
q=heat rate
U= overal heat coefficient
A=area
ΔTL= log mean temperature difference.

Results:

Looking at water

q=24.24 Btu/s

Looking at air
q=765.09 Btu/s

Upon observation the two heat rates clearly do not equal one another.

Thus my problem. How do I find out the true amount of heat passing through the fins? Do I use the NTU method and work backwards for efficiency?

Best Regards,

D

Attached find a picture of the exchanger base with tubes
 

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250,000 CFM through that tiny little coil? Not likely. Take another look at how you got your air flow rate.

And why are you looking at fin efficiency? I'm curious because that is normally of interest only to the person designing the heat exchanger, not somebody doing a mass and energy balance.
 
jrmichler said:
250,000 CFM through that tiny little coil? Not likely. Take another look at how you got your air flow rate.

And why are you looking at fin efficiency? I'm curious because that is normally of interest only to the person designing the heat exchanger, not somebody doing a mass and energy balance.

This is being done as a project in class. The air flow rate was acquired by use of an anemometer. All the units were in GPM the bulk fluid properties were looked up. A typical anemometer reading from our experiment showed the inlet and outlet flow ranging from 1200-1600 GPM where the anemometer reads that as “1600x100”. That was a topic of discussion in class as the instructor wasn’t sure if the device was already multiplying the figures by 100 to get the readings. An example being “ is it 1600x100 to get 160,000? Or was it 16x100 to get the reading 1600”. If you could clarify that issue it would be tremendous. I plan on going back into the lab to find the model of the anemometer.

I personally am curious about the fin efficiency because I stumbled upon it in a book I had and figured it would contribute something to the overall heat coefficient.
 
HethensEnd25 said:
A typical anemometer reading from our experiment showed the inlet and outlet flow ranging from 1200-1600 GPM
Are you certain about units in GPM? An anemometer measures air velocity.

Edit: 310.7 pounds/sec of air mass flow works out to 3884 cubic foot/second (233025 CFM) at 0.08 pounds per cubic foot. Heat exchanger air area (not figuring for the reduction of open area by the fins and piping) is 0.62 m^2 (6.67 sq.ft). This yields an air velocity of 34936 FPM; an air speed of 397 MPH doesn't seem likely.
 
Last edited:
Had my central air system checked when it sortta wasn't working. I guess I hadn't replaced the filter. Guy suggested I might want to get a UV filter accessory. He said it would "kill bugs and particulates". I know UV can kill the former, not sure how he thinks it's gonna murder the latter. Now I'm finding out there's more than one type of UV filter: one for the air flow and one for the coil. He was suggesting we might get one for the air flow, but now we'll have to change the bulb...

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