232 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

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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.

232 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.