Re: Heating air in a wind tunnel.


"John Larkin" <jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:jvsfm1t7dkms20qql0g6g1ml09tn8u99qh@xxxxxxxxxx
..
..
>
>
> This from my COOL.TXT cheat-*** file...
>
> ==========
>
> If moving air removes heat from a volume, the exhaust air becomes
> hotter than the intake air.
>
> The forced-air cooling equation is...
>
> P = 169 * Qa * (T2/T1 -1)
>
> P = power, watts
> Qa = air flow, CFM (cubic feet per minute)
> T2 = outlet temp, K (absolute temperature, Kelvins)
> T1 = inlet temp, K
>
> For 1 deg C rise at 1 CFM, we get...
>
> P = 169 * 1 * (301/300 - 1) = 0.5633 w
>
> So, near room temp, an air flow of 1 CFM will be heated 1.775 deg
> c/watt.
>
>
> Since 1 cubic foot = 28.31 liters, for Qm in LPM the equation becomes
>
> P = 5.69 * Qm * (T2/T1 -1)
>
> So one LPM has a heatsink capacity of 50.25 degC/watt
>
> ===============
>
>
> So, 250 lpm, 20 to 70 degrees C rise, takes
>
> P = 5.69 * 250 * ( (273+70)/(273+20) -1)
>
> = 260 watts,
>
> I think.
>

Thanks to John P, I found the following data (my figures are
approximations):

Density of air at 70C 1kg/m3
Constant Pressure specific heat 1000J/kg.K

CPSH = 1000W.s/kg.K
= 1W.s/l.K
= 0.0167W.min/l.K

So for 250l/min and 50C rise in temperature

250 * 50 * 0.0167 = 208.75W

Pretty close.

Thanks to all for your help.

--
John B



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