Which air density to use?

[GregoryPE]

Can someone please help me out here as I feel like I see contradicting answers. Let's say you have to heat 21,000 CFM from 0F to 50F. When using qdot = vdot*rho*cp*deltaT, do I use the density of 0F air (.086 lbm/ft^3) or 50F air (.077 lbm/ft^3)? The difference is substantial.

Thank you for your help!

 

[Habib]

I assume out door is zero degree and indoor is 50 F. Now if air flow is given for indoor condition, use density at 50F or adjust air flow with outdoor conditions with density at 0 F.

 

[GregoryPE]

I'm sorry, but what do you mean by "adjust air flow with outdoor conditions..."? Thank you for responding.

 

[Habib]

First of all thanks for asking such great questions, really appreciate that as it helps everyone to understand the basics.

As stated previously, the best thing is to use the condition which is given ( 21,000 cfm at 50 degree F and use density at 50 degree as well).

However, if you are interested in using density at out door condition ( 0 degree F in this case) then adjust the air flow for out door condition. 

Adjusted air flow will be calculated as below:

21000x.077(density at 50 degree F)=adjusted air flow x .086 ( density at 0 degree F)

This will result in adjusted air flow of 18,802 cfm.

 

[MikeGlass1969]

Heating is just sensible heat.  you should use 1.08*cfm*deltaT...   You should know this equation frontwards and backwards.

 

[Habib]

That equation uses .075 as density and .24 as Cp ( 60x.075x.24=1.08) and will probably work for temp ranges which have density and Cp value close to mentioned above. With low temp ranges or very high humidity, you will get very different answers by using this equation.

 

[GregoryPE]

MikeGlass1969...Using 1.08 you need to understand the underlying assumptions. You can't always use it.

Habib...I guess I just assumed that it would be more accurate to use an average density over the two temperatures, but apparently not!

 

[JHW 3d]

Can someone please help me out here as I feel like I see contradicting answers. Let's say you have to heat 21,000 CFM from 0F to 50F. When using qdot = vdot*rho*cp*deltaT, do I use the density of 0F air (.086 lbm/ft^3) or 50F air (.077 lbm/ft^3)? The difference is substantial.Thank you for your help!

If you take the average of 0.077 and 0.086 that is within 5% of either one. The heat transfer is linear with air density (in this case) so the answer would also be within 5% -- is 5% that "substantial"?

Is there a practice problem you answered incorrectly or is this just a hypothetical?