# 2001 ncees am #140



## dpolet (Jun 16, 2013)

A room has the following:

sensible heat gain = 90,000btu/hr

latent heat gain = 40,000 btu/hr

supply to room = 60F

The room is kept at 82 db.

When calculate air flow to the room, why only the sensible heat gain is used? Which page of MERM focus on this information?

Thanks in advance.


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## bknewto7 PE (Jun 18, 2013)

Room temp sensors operate off of DB temp, and do not read WB Temp.

Therefore the thermostat will tell the HVAC unit to shut off once the Sensible load has been removed from the space.

The sensible heat equation can be used to solve for the CFM @ 60 degree supply to remove the sensible load.

sensible load (BTU/Hr) = 1.08 * CFM * Delta T

90000 = 1.08 * CFM * (82-60)

CFM = 3788

The MERM section on Psycrometrics should deal with this


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## dpolet (Jun 20, 2013)

bknewto7, thanks for the response.

If the room sensible heat load is 0btu/hr (but the latent load is not zero), the air flow will be 0 CFM. However, if there is no air flow, how the latent load will be removed?


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## bknewto7 PE (Jun 21, 2013)

It wont be removed if there is no airflow.

However the latent load does have some effect on the room temperature, and the HVAC unit removes both sensible and latent loads. Typically the sensible load is much higher than the latent load, so the HVAC unit removes the latent load in the normal course of operation. SO if a question gives you the themostat setting and a sensible and latent load, you would be expected to know that in terms of the temperature sensor only the sensible load is a required known in regards to the problem.

If you are not given a sensible load and only latent and asked to calculate the requried airflow then you can use the latent heat equation and the psychrometric charts

Latent Load = 4840 * CFM * Change in Humidity Ratio between the air entering the coil and the air exiting the coil


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## dpolet (Jun 22, 2013)

I understand now. Thank you very much.


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## dpolet (Jun 23, 2013)

bknewto7, I am still a little confused about this when I came across 2001 ncees sample exam T&amp;F #507. I understand the ncees solution method. However, i am not sure how to apply this equation (sensible load (BTU/Hr) = 1.08 * CFM * Delta T) on #507.or the equation should not be used on this problem? Here is the problem#507: water enter cooling tower at 100F and 250000lbm/hr. watre leave at 80F. Air in is 85DB/70WB and air out at 95DB/70%RH. To find out the air CFM. The answer is 72,000CFM. Thanks.


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## bknewto7 PE (Jun 27, 2013)

For this problem first you need to calculate the heat transferred from the water to the surrounding air.

Use the Heat Transfer Equation Q = (Mass Flow Rate) * ( Specific Heat) * (Delta T)

So, Q = (250000)*(1.0)*(20) = 5,000,000 BTU/HR

Knowing this we can plug this in to the Total Heat Quation which is Q = 4.5 * CFM * Delta h (Change in Enthalpy)

to get the change in enthalpy we plot the two points that are given on a psychrometric chart.

h @ 95DB/70%RH = 50.6

h @ 85DB/70WB = 35

Now we plug the values in to the total heat equation 5,000,000 = 4.5 * CFM * (50.6-35)

Solving for CFM

CFM = 71,225 (72,000)


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## dpolet (Jun 29, 2013)

I understand. Thanks a lot.


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