Heat Removsl By Ventilation 1

[NikT]

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

 

[Wareagl487]

Nikt -

Yep, you're complicating it - temperature is in degrees Fahrenheit.

Andy W.

 

[MintJulep]

Or Rankine. It's a delta, so it don't matter.

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.

 

[NikT]

To Wareagl487 & MintJulep;

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!"

 

[IRstuff]

Why would it be different? Delta of 1ºF is thermodynamically equivalent to delta of 1ºR

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.wikipedia.org/wiki/Newton's_law_of_cooling#Newton.27s_law_of_cooling

TTFN

FAQ731-376

 

[DUMechEng]

q = mass flow (lb/hr) x specific heat (btu/lb/deg R) x delta T (R)

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

 

[texanman]

to DUMechEng, I gave you a star for your explanation.
thanks,

 

[NikT]

Thanks to all, especially DUMechEng. Yor answer, which appears to be close to my solution made sense. I guess I danced around the conversion so much I lost my way. Thanks again, NikT

 

[RossABQ]

PLEASE note -- that 1.08 value is only valid at or near sea level -- in the Rockies we are using 0.89 at 5500 ft, somewhere around 0.75 for 7200 ft.

Which is exactly why it's important to know how that "constant" was derived....

 

[marcoh]

In metric units:

Q = ? . V . Cp . ?T

Where:
Q = heat transfer (Watts)
? = density (kg/m³)
V = volume flow rate (m³/sec)
Cp = Specific heat (J/kg/°C)
?T = temperature difference (°C)

Typically for air at sea level:
? . Cp = 1.213

therefore sensible heat transfer is
Q = 1.213 . V . ?T

 

[NikT]

Thanks to all who responded. I will try the suggested equations.

NikT