Vacuum fan vs Air temperature

[ddsancho]

I have been studying the vacuum pressure capacity of a fan at different room temperatures (15 to 45 C deg).

As it was supposed, at higher room temperature I had less vacuum because the air has lower density.

The corrector factor that I use to link pressure and temperature is:
P1 / P2 = (T1+273) / (T2+273)

But this corrector factor seems to not work with my data. It is supposed that after the correction, all the pressures should be similiar, but then I have too much vacuum pressure at high temperatures or low vacuum pressure at low temperatures.
Maybe the air temperature is heated by bearings and winding when going through the fan?

Daniel

 

[m777182]

The fan performs adiabatic work on air. Secondly, due to high Re number within it, there is an unavoidable friction loss that is expressed through a temperature change. Probably those two effects spoils your calculation.
m777182

 

[fredt]

The density correction for temperature that you have used is OK for practical purposes.

Fan pressure rise is proportional to INLET density and in the case of an exhaust fan the density should also be corrected for pressure at the fan inlet.

There are a number of other factors that could be throwing out your calculations.

The main one is that fans have a varying pressure/volume characteristic. The simple density correction works only if the system resistance is unchanged.

 

[25362]

To ddsancho.

Atmospheric temperature may also affect atmospheric pressure readings. Mercurial column barometric pressure readings are to be corrected according to temperature, elevation and gravity.

The correction by gravity, usually negligible in engineering calculations, is a function of latitude and elevation.

Using as basis 62^oF, a reading of 30 in. Hg at sea level should be corrected at 90^oF, by deducting 0.17 in. Hg.

Question is in what manner could these corrections affect your findings.

At an elevation of, say, 500 ft, the next correction is a deduction of 500 X 0.1/100 = 0.5 in. Hg. At 35^o north latitude, the further correction due to gravity would be a reduction of 0.03 in.

The true atmospheric pressure would be:

30-0.17-0.5-0.03 = 29.3 in. Hg​

 

[ddsancho]

Finnally I could understand my problem.

In my test the fan was controled at constant voltage, but not at constant speed, so the the flow was not constant, and then it is not possible to apply the formula:

P1 / P2 = (T1+273) / (T2+273)

Thanks all for your dedicated time,
Daniel