# NCEES 2001 Prob 140



## slickjohannes (Oct 22, 2015)

can someone please explain the steps involved in dolving this problem?


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## Ramnares P.E. (Oct 22, 2015)

Would be helpful if you posted the problem...


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## P-E (Oct 22, 2015)

...so we could help dolve it.


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## slickjohannes (Oct 22, 2015)

I knew someone would say that.

I'd rather not post copyrighted material. If someone has got the book and can dolve the problem, great- but if not then no worries.


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## Ramnares P.E. (Oct 23, 2015)

No one is asking you to post the book. Post a picture of the problem or simply type it out entirely.


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## P-E (Oct 23, 2015)

You could copy it wrong.


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## Ramnares P.E. (Oct 23, 2015)

power-engineer said:


> You could copy it wrong.




But then I'd dolve it wrong :dunno:


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## matt267 PE (Oct 23, 2015)

dtop being sick heads and help the guy out.


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## P-E (Oct 23, 2015)

matt267 said:


> dtop being sick heads and help the guy out.


Shouldn't you be studying?


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## matt267 PE (Oct 23, 2015)

Nope. The April 2016 exam is months away.


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## Ramnares P.E. (Oct 23, 2015)

That's the spirit Matt


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## Ramnares P.E. (Oct 23, 2015)

matt267 said:


> dtop being sick heads and help the guy out.




Here's the step by step solution:

1. Given dry bulb conditions, we're dealing with Sensible heat.

2. Sensible flow heat equations, in English units: Qs = 1.08 * q * delta T

Qs = sensible heat flow (Btu/hr)

q = air flow (cfm)

delta T = temperature difference, °F

Given Qs, delta T in the problem, solve for q.

q = Qs / (1.08 * delta T)

Qs = 90, 000 Btu/hr

delta T = 82 - 60 = 22°F

q = 90,000/(1.08*22) = 3787.9 cfm

Nearest answer is B - 3750 cfm


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## matt267 PE (Oct 23, 2015)

Yuck, I hate thermo.


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## Ramnares P.E. (Oct 23, 2015)

More of an HVAC give me points question than thermo I'd think.


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## P-E (Oct 23, 2015)

Ramnares P.E. said:


> More of an HVAC give me points question than thermo I'd think.


Yes.


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## matt267 PE (Oct 23, 2015)

power-engineer said:


> Ramnares P.E. said:
> 
> 
> > More of an HVAC give me points question than thermo I'd think.
> ...


looks like thermo to me.

I'd rather deal with pounds of primary sludge being removed.


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## P-E (Oct 23, 2015)

matt267 said:


> power-engineer said:
> 
> 
> > Ramnares P.E. said:
> ...


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## matt267 PE (Oct 23, 2015)

Ah yes, Uncle Eddy.


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## slickjohannes (Oct 23, 2015)

Dhanks guys.


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## Audi Driver P.E. (Oct 24, 2015)

Prior to my exam prep, I hadn't had much exposure to HAVAC, so when I came across this problem, I tried for a long time to figure out why the answer given in the book used 1.09, when MERM uses 1.08. Little things like this bother me, so I use them as a learning opportunity. If you are unfamiliar with HVAC calcs as well (as it seems from your question), I think it's important for an engineer to know how that factor is derived. The MERM points out that some references use 1.09 and others 1.1 but does not explain how that factor is calculated. So here you go:

It's a ratio of the specific heat to specific volume of air multiplied by the minute to hour conversion factor required since the heat load is given in units per hour (i.e.: (cp/specific volume)*60 min/hr).

Results depend upon precision used in the calculation, and upon the conditions of the air (temp and moisture content).

In the end, whether you use 1.09 or 1.08 is pretty inconsequential because it doesn't really affect your answer enough to matter for the exam.


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## P-E (Oct 24, 2015)

Audi_driver said:


> Prior to my exam prep, I hadn't had much exposure to HAVAC, so when I came across this problem, I tried for a long time to figure out why the answer given in the book used 1.09, when MERM uses 1.08. Little things like this bother me, so I use them as a learning opportunity. If you are unfamiliar with HVAC calcs as well (as it seems from your question), I think it's important for an engineer to know how that factor is derived. The MERM points out that some references use 1.09 and others 1.1 but does not explain how that factor is calculated. So here you go:
> 
> It's a ratio of the specific heat to specific volume of air multiplied by the minute to hour conversion factor required since the heat load is given in units per hour (i.e.: (cp/specific volume)*60 min/hr).
> 
> ...


Yup you have to get by the minutiae and go with the rule of thumb sometimes. Another one is using 2.31 as the conversion from psi to feet. Most of the time you'll get close enough. Sometimes time management is more important on these exams. If you have a shortcut, use it.


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## JHW 3d (Oct 28, 2015)

The 1.08 is just a conversion factor for the units.

The governing equation is just sensible heat transfer on air:

q_dot = m_dot * cp * Delta T (MERM Eq 40.18, as well as many other places)


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## Audi Driver P.E. (Nov 12, 2015)

JHW3 said:


> The 1.08 is just a conversion factor for the units.
> 
> The governing equation is just sensible heat transfer on air:
> 
> q_dot = m_dot * cp * Delta T (MERM Eq 40.18, as well as many other places)


As I noted, it's more than just a conversion factor for the units, it is a ratio of the specific heat to the specific volume which has dependencies upon the air/moisture conditions.


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