# NCEES 2011 Practice Exam HVAC & Refrigeration Problem 509



## GregoryPE (Oct 14, 2016)

I am hoping you guys can help me. Question 509 on the 2011 NCEES Practice HVAC &amp; Refrigeration Exam seems simple, but I can't quite figure it out. Basically you have a window facing north, east and west, with Cooling Load Temperature Difference (CLTD) at 26, 55 &amp; 72 respectively, and Solar Heat Gain Factor (SHGF) at 47, 215 &amp; 215 respectively. Indoor temp is 75F db and outdoor temp is 95F db. U_window = 1.1 Btu/hr-ft^2-F. A_window = 40SF and A_total = 120SF. We are looking for total heat gain in Btu/hr through the windows.

I am using the MERM 13th Edition. I am using Equation 43.9 as the governing equation, which states that qdot = U*A*(CLTD_corrected) + A(SC)(SCL).

Assuming that SC=1, and SCL~SHGF, I thought the solution should be qdot = 1.1*40*(26+55+72) + (40*47+40*215+40*215) = 6732 + 19080 = 25812 Btu/hr.

It turns out I am right on the 2nd half of the equation, but wrong on the 1st half. Instead of U*A*(CLTD_corrected), they are using U*A*(T_indoor - T_outdoor). = 40*3*1.1*(95-75) + 19080 = 21270 Btu/hr.

Can someone please explain to me why they are using deltaT in lieu of CLTD? These NCEES solutions don't explain things very well!

Thank you


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## GregoryPE (Oct 14, 2016)

My original post should read (T_outdoor - T_indoor), as this gives the positive result.


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## Habib (Oct 15, 2016)

Attached is the solution, hope that will help.

View attachment 8713


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## GregoryPE (Oct 18, 2016)

Habib, could you expand on your answer? That didn't do much for me.


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## Habib (Oct 18, 2016)

My solution above was not posted completely due to size limit I guess and only showed heat load through roof which might have confused you more( sorry about that). 

I was trying to post a solution with heat load through roof, walls and windows. Anyways let's just talk about windows:

The CLTD is only given for walls and roof. If it is given for windows then your approach would be right to calculate heat load through windows by using the following equation:

Q=Qrad +Qcond

Q=(S)(SHGF)(A)+UA(CLTD)

but in this question CLTD is not given for windows and just like for walls it will depend on glass material and properties etc. We can't assume CLTD for walls to be same as for windows.

In a situation like this, it is often a practice to use delta T between outdoor and indoor to calculate conduction load through windows.

hope that clears the confusion.

.


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## GregoryPE (Oct 19, 2016)

Thanks so much for your help. Not just on this question but on the others too!


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