
239 posts
fan aerolic power calculation
kindly tell me the derivation of fan power consumption . In many books it is written but how the formula has been arrieved at & why does a factor 102 is written in the denominator? please send the complete derivation with logic & units used & why? at rajmadhuri1972@yahoo.co.in raj
Know the answer to this question? Join the community and register for a free guest account to post a reply.

138 posts
Re: fan aerolic power calculation
The power absorbed from an ideal fan is simply:
P[W] = dp[Pa] * q[m³/s]
where P is the power, dp is the pressure increase, q the volume flow.
There is no "102" factor to be seen.
For a real fan, you have to take the losses into account.
The losses cannot be calculated, you need to measure them or to ask the supplier.
Usually the supplier can provide you the fan curves: P(q) and dp(q) for a given rotation speed. You can use these to predict the power and the pressure also for other speeds using the "fan laws" for extrapolation. These extrapolation laws do not contain any "102" factor either.
I have never seem any such "102" anywhere.
Must be related to specific units.

239 posts
Re: fan aerolic power calculation
the formula for the calculation of fan eff. is written below fan mech. eff. = m3/s * pr. diff. (mmwg)*100/(102* fan shaft power) refer BEE book for fans & blowers please comment on this formula . what about fan shaft power & aerolic power

79 posts
Re: fan aerolic power calculation
Fan shaft power (watt) = Pressure(Pa) X Volume(m3/s)
This formula is valid and applicable to all fans and blowers.
The formula becomes [{pressure(mmwg) X Volume (m3/s)} /102] is derived by changing the pressure from Pa to mmwg and power from watt to Kwatt
1 mmwg = 9.81 Pa
1Kwatt = 1000 watt
so, the formula becomes [{pressure(mmwg) X Volume (m3/s)} X 9.81/1000]
which is = [{pressure(mmwg) X Volume (m3/s)} /102]
I think this satisfy you.