Calculation of Fan work by VdP and the adiabatic process where PV^(gamma) = Constant; which is OK?

[flsanders]

Dear forum

I have been trying to calculate the work done by a fan on a gas flow in two ways, but I get two different values of power.

The first method is the W=V * dP
(I guess this is assuming isothermal compression). I then use the power calculated as a measure of the internal energy increase and calculate the outlet temperature this way.


The second method I used is adiabatic compression. Seems on for an insulated fan.
I followed

https://en.wikipedia.org/wiki/Adiabatic_process

but did not arrive to the same fan work.

My question is when to use the first method and when to use the second.

What about for a compressor, I guess the same model shall be used for this?

 

[georgeverghese]

Also include the kinetic energy loss / gain between inlet and outlet nozzles for fans.

 

[flsanders]

Yes, this I know. But for now I am confused on a more general level about why the adiabatic compression model is not suitable for my fan calculation.

 

[25362]

Get hold of Fan Engineering, an Engineer's Handbook on Fans and Their applications.
Edited by Robert Jorgensen
Published by BUFFALO FORGE COMPANY, Buffalo, New York

 

[Tunalover]

I have Fan Engineering and it dates back to the 1930's! Is it still published?

Tunalover
Electro-Mechanical Product Development
UMD 1984
UCF 1993

 

[25362]

My "old" copy is dated 1983.

 

[chicopee]

Are you talking about a real fan or a blower?

 

[flsanders]

I was talking about a fan; axial or radial with a dP of max 20kPa. But in principle I want to understand why the work I calculate wit the adiabatic model is less than for the W=VdP calculation.

 

[rense]

it is difficult to provide comments without knowing what you are doing,
show us your calc's in both cases (for example for air) and someone may help.

 

[chicopee]

Under adiabatic conditions, the entropy of the system is considered constant but under isothermal conditions the entropy of the system is not constant. It would make sense that under adiabatic condition the efficiency value should be greater than under isothermal condition. What's puzzling to me is your reason not to use the efficiency calculations based either on static or total pressure which is the trend in the industry.

 

[georgeverghese]

If developed dp is small, then also include isothermal compression eff for dW=VdP formula. If dp is higher, use the appropriate polytropic or isentropic eff also for the adiabatic compression formula.
In the case of the isothermal approximation, integrate between the limits of P1 and P2 to get the total work of compression - here V = nRT/P, so dW = nRT dP/P, then integrate between limits and divide the result by isothermal compression eff.

 

[flsanders]

Thanks Chicopee and georgeverghese

I think I am a bit further now in my understanding. When I was comparing the
1. adiabatic model
2. and the isothermal model

I was making one crucial mistake (I think); I used the same efficiency coefficient. But in fact for the isothermal model I should use the static fan efficiency (with static head) and for the adiabatic model I should use the isentropic efficiency (which is not known to me because as Chicopee says industry standard is to provide the efficiency as static or total)

The numericla difference in static efficiency and isentropic effieciency should explain why I get different values of power of the two calculations. So if I could get the isentropic efficiency from the vendor it would be of lower magnitude than the static???

 

[lilliput1]

The formula used to estimate FAN motor HP = (cfm X in w.g. pressure)/(6362 x Efficiency)
Efficiency as ist approximation = 0.65

After the fan total pressure is estimate is refined we select the fan using manufacturer (like Twin City) computer program. We yry to pick the most efficient fan that would fit.