Heat Removsl By Ventilation
Heat Removsl By Ventilation
(OP)
In Thread 403-17260, Willard3 gave the formula CFM X 1.085 X Delta T = BTU/ hr. Can willard3 (or anyone else) please show step by step how this equation was derived? Is delta T in degrees-F, or C, or R(Rankine)? Surely, the units of delta T would have an effect on the magnitude of the CFM calculation. I can see that one would use the specific heat of air and the inverse of air density to get lbs/ft. cubed to find heat flow. But I'm a bit stymied because I believe specific heat is expressed in some absolute temp scale like the Rankine or Kelvin scale, and I believe air density is in pounds mass, not pounds weight, and I haven't been able to reconcile all the dimensions and units. I might be complicating this whole business unnecessarily, but I don't think so.
Thank you,
NikT
Thank you,
NikT





RE: Heat Removsl By Ventilation
Yep, you're complicating it - temperature is in degrees Fahrenheit.
Andy W.
RE: Heat Removsl By Ventilation
The 1.085 factor has all of the things needed to make the units work out.
If you are working with air substantially away from standard conditions then you'll need a bit different factor to account for the density change.
RE: Heat Removsl By Ventilation
Thanks for your thoughts, but I really want to know how the units are manipulated to arrive at the 1.085 constant. In other words, what's in the black box that makes 1.085. While, I agree that temperature being a delta should not be a problem, isn't the magnitude of a delta of one degree Fahrenheit a lot different than a delta of one degree Rankine in terms of heat energy?
I sorry if I'm am muddying up the water, but like the bank robber staring up at Dirty Harry's 44 Magnum wondering if there were 5 shots or 6, "I gots to know!"
RE: Heat Removsl By Ventilation
The assumption is that there is a factor that has units of BTU/hr/ºF. A larger delta T results in a larger heat flow. That's the basis of Newton's Law of cooling: http://en.
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RE: Heat Removsl By Ventilation
in order to convert from mass flow to cfm you need to multiply by density and by 60 mins/hr.
so....
q = volumetric flow (cubic feet/min) x 60 mins/hr x 0.075 lb/cubic foot x 0.24 btu/lb/R x delta T (R)
or...
q = 1.08 x CFM x Delta T
if you use the actual density of the air at your conditions the factor will change slightly
RE: Heat Removsl By Ventilation
thanks,
RE: Heat Removsl By Ventilation
RE: Heat Removsl By Ventilation
Which is exactly why it's important to know how that "constant" was derived....
RE: Heat Removsl By Ventilation
Q = ρ . V . Cp . ΔT
Where:
Q = heat transfer (Watts)
ρ = density (kg/m³)
V = volume flow rate (m³/sec)
Cp = Specific heat (J/kg/°C)
ΔT = temperature difference (°C)
Typically for air at sea level:
ρ . Cp = 1.213
therefore sensible heat transfer is
Q = 1.213 . V . ΔT
RE: Heat Removsl By Ventilation
NikT