CLTD/CLF ....whattt?

[NHEngineer037]

Question 118 of the 2016 NCEES practice exam references a CLTD/CLF way for calculating a simple cooling load through glass.  The solution refers to ASHRAE Fundamentals.  I have 2017 and 2013 and both don't discuss the method or provide formulas.  

I don't get it.

Has anyone figured out what this procedure is?  (I have been trying to figure it out to make sense of the provided solution with Google and my other books without luck.)

Thank you,

Ryan

 

[MikeGlass1969]

I will just provide a link to a site that you can download the information from.

https://hvac.okstate.edu/cltdsclclf-cooling-load-calculation-method 

 

[Slay the P.E.]

When I was an instructor for the P.P.I course, I noticed this problem in the NCEES practice exam under the 2008 specs. Back then it was problem #509 in the HVAC PM portion. It was so bizarre that they were evalua ting an antiquated method I decided to write them to see if they replied. Here's the letter I wrote:

 

[glenn90]

So does NCEES want you to use the CLTD or just DT across the wall / window? Problem 118 in the 2016 NCEES HVAC+R Practice exam gives CLTD information across a window and asks to calculate the heat gain but does not use the CLTD values in their solution. Just uses DT across the window (20F). Why?

 

[Slay the P.E.]

So does NCEES want you to use the CLTD or just DT across the wall / window? Problem 118 in the 2016 NCEES HVAC+R Practice exam gives CLTD information across a window and asks to calculate the heat gain but does not use the CLTD values in their solution. Just uses DT across the window (20F). Why?

Looks like one would use actual delta T.  From Chapter 15 of 2009 ASHRAE Fundamentals we have this:

​

So it looks like in the NCEES solution they used this equation. What is not at all clear is why they set the incident total irradiance, Et = 1 Btu/(h x ft^2). A value of 1 is not typical for this parameter. 

Looks like another letter to NCEES is in order here. This problem needs more explanation.