Fan pumping power requirement to overcome heat exchanger pressure drop

[MLA65]

Given an air to water heat exchanger with the following characteristics:

Inlet temperature of Hot air: 750F
Outlet temperature of hot air: 400F
Flow: 10,000 SCFM
Pressure drop: 1" of Water

My question is: How to calculate the pumping power requirement for a fan that will help me to circulate the air through the heat exchanger?

I have found the folowing equation:
Power[HP] = Flow[CFM]*Pressure_Drop["W]/(6356*efficiency)

1- I'm not sure if this is the right equation that I should use. Could someone comment?
2- The flow in CFM at the inlet will be different than the one I will see in the outlet, because the density is different. So how should I calculate the required fan pumping power? Which value of CFM should I consider, inlet or outlet?

 

[desertfox]

Hi MLA65

Is this for a factory?

 

[MLA65]

Hi Desertfox

Yes, it's for a factory.

 

[TheTick]

Measure the pressure drop at two or three different flow rates, then you can derive the resistance factor to use.

 

[GregLocock]

In proper units your equation should be

Power=volumetricflowrate*pressuredrop/efficiency

so your equation looks OK but I haven't checked the scale factor.

It is the flowrate at the fan that is important.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[MintJulep]

The best way is to look at the fan curves for the fan that is doing the work.

Your equation is ok, but you need the fan data for the efficiency.

What is the density at the fan?

 

[MLA65]

The density at the fan is 0.746Kg/m^3.

The fan will be located after the heat exchanger, where the temparature is about 400F.

If I use the flow before the Heat Exchanger I get a higher power requirement for the fan than what I would get if I apply the same formula but with the flow after the heat exchanger. Is it ok to use the flow I would have after the heat exchanger, where the fan will be located?

 

[desertfox]

Hi MLA65

Not really my area of expertise but if the fan is after the heat exchanger then what pushes the air through it?

desertfox

 

[zekeman]

"If I use the flow before the Heat Exchanger I get a higher power requirement for the fan than what I would get if I apply the same formula but with the flow after the heat exchanger. Is it ok to use the flow I would have after the heat exchanger, where the fan will be located?"
Probably better to locate where air is cooler but the power should be independent of the position and BTW how did you get the pressure drop of 1 inch or is that a design goal? Also is this a static or dynamic pressure drop?
I also have a problem with that formula since it is generally used by ventilation people assume adiabatic incompressible flow ( neither of which you have), but your huge range of temperatures makes me wonder about its validity.
My take using the energy equation where the flow equation involves pressure, velocity, work and heat transfer,Q is quite a problem . I think you will have to start with an assumed flow and develop the Q rejected during the flow, and then get the work done.
V1^2/2g+H1 + Q +W=V2^2/2g+H2
V velocity
H enthalpy
Q heat added (in your case rejected Negative number)
W work done (by fan)

 

[GregLocock]

His fan is not the length of the temperature drop

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[MintJulep]

Your trying to calculate the pumping power of the fan. What density do you thing you should use?

 

[MLA65]

In reply to some of the comments, please note that fan will be installed after the heat exchanger, where the temperature will be around 400F.
The pressure drop of 1" is the total of the drop produced by the heat exchanger (this number is given by the heat exchanger supplier) and it also includes the drop produced by the duck that is before the heat exchanger.

 

[zekeman]

"His fan is not the length of the temperature drop"

But less than a fortnight.

 

[ARenko]

desertfox,
The fan is pulling the air through the heat exchanger.

MLA65,
There is no ductwork after the fan? Is the fan sized yet? Look at the fan curve - you need fan speed and corresponding torque curve or you can't size the motor. If the fan isn't sized have the fan supplier (e.g. multi-wing) help you size the motor.

You have to use conditions at the plane of the fan.

See below link for equations and more.

http://www.hudsonproducts.com/products/tuflite/flowfans.pdf

 

[zekeman]

I should have added, to complete the solution, you have to also use the momentum equation:
dp=f*rho*V^2/2Dg+rho/g*V*dV for each width of duct
where
f= friction factor (obtained from Reynolds number)
The parameters p,V,rho, T are related by
rho*A*V= constant
p/rho=R*T
taking the log derivative
drho/rho)+dV/V=0
dp/p-drho/rho-dT/T=0
since there is heat transfer from the fluid we also get
dT=-dQ/cp
Starting from x=0 we have p1,T1,V1 and from these equations going along the length of the duct you get to pf,Tf,Vf
Now you can invoke the energy equation I previously gave;
V1^2/2g+H1 + Q +W=V2^2/2g+H2
and get W which is the work for 1 lb of fluid. Multiplied by the weight flow rate you get your fan power.