# I need help with a NCEES question



## cityeng (Oct 27, 2010)

Problem :

Water at 100 degrees F enters a cooling tower at the rate of 250,000 lb/hr. water leaves at 80 degrees F

Atmospheric conditions:

85 F db

70 F Wb

29.92 in. Hg

Air leaves the cooling tower at :

95F db

70% relative humidity

It is asking to solve for the volume flow rate(cfm) of atmospheric air required to cool the water

The solution guide gets the following equation

250,000(68) +Mair*h1 = (250,000 – Mevap)(48) +Mair*h2

Could you please tell me where the 68 and 48(numbers in red) come from? I can get all the enthalpies and humidity ratios from the Psychometric charts but cannot figure out where these numbers come from. I would really appreciate your help

This is number 507 out of the 2001 mechanical PE NCEES sample problems.


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## mizzoueng (Oct 27, 2010)

Enthalpy of the saturated liquid for the 100 deg water is 68.03

48 is likely the enthaply of the makeup at 80 deg f, assuming thats the average temperature for the water.

These came from the steam tables, not the psych chart.

See the MERM, page 38-12, or your local ASHRAE Fundamentals book. I am not sure the equations in the MERM are all that accurate, I am not doing HVAC.


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## MadDawg (Oct 27, 2010)

I spent quite a while looking at this last night, and I hope I am correct. I looked in the steam tables at 29.92 in Hg = 14.7 psia and found that the saturation temperature is 212 and the enthalpy of a saturated liquid (hf) is ~180 btu/lbm. Then I calculated how many degrees subcooled it is (212-100= 112 and 212-80=132) and found the enthalpy of the subcooled liquid as h=hf-cp(T). That gives me 180-112=68 and 180-132=48.

Can't wait to forget all of this Friday night...


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## Rafa (Oct 27, 2010)

MadDawg said:


> I spent quite a while looking at this last night, and I hope I am correct. I looked in the steam tables at 29.92 in Hg = 14.7 psia and found that the saturation temperature is 212 and the enthalpy of a saturated liquid (hf) is ~180 btu/lbm. Then I calculated how many degrees subcooled it is (212-100= 112 and 212-80=132) and found the enthalpy of the subcooled liquid as h=hf-cp(T). That gives me 180-112=68 and 180-132=48.
> Can't wait to forget all of this Friday night...


In the solution they are considering the make up water, but how do you solve for make up water if you have to calculate the air flow first and then make up water can be calculated. I just took the make-up water out of the equation and came up with 70281 CFM of air, then calucalted the make up water: 3855 lmb/hr with is only a 1.5% of the water flow into the tower. Hope I have good approach to the problem. Thanks.


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## axiom (Oct 28, 2010)

The above seems like a long way home, the enthalpies direct from steam tables are:

[email protected]=68

[email protected] =48

You can approximate the enthalpy of any subcooled liquid using [email protected] temp.

I can e-mail you my solution if you want it.


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## MadDawg (Oct 28, 2010)

axiom said:


> The above seems like a long way home, the enthalpies direct from steam tables are:
> [email protected]=68
> 
> [email protected] =48
> ...



I agree...when I pulled out my thermo book to look at the steam tables to write my earlier response, I saw that the [email protected] temp was the same, but I hesitated to write that since I wasn't sure that's a correct way to go about it...


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## cityeng (Oct 28, 2010)

Thanks so much to all of for all your help, this board forum is awasome. I will be glad when this is over tommorrow. God! I hope I pass.

These nthalpies do work for this equation, I also found an equation in an old thremo book that I had handwritten in:

Mwater*delta T water = Mair*(delta h - Tambient air*change in humidity ratio). This equation does getthe right answer but not sure if it is correct or nor do I have any other examples to check it. The last mechanical T&amp;F had several cooling tower problems and I was trying to get some problem experience in this area. Does any know if this equation will work for all?


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## jamiecta (Aug 22, 2011)

cityeng said:


> Thanks so much to all of for all your help, this board forum is awasome. I will be glad when this is over tommorrow. God! I hope I pass.These nthalpies do work for this equation, I also found an equation in an old thremo book that I had handwritten in:
> 
> Mwater*delta T water = Mair*(delta h - Tambient air*change in humidity ratio). This equation does getthe right answer but not sure if it is correct or nor do I have any other examples to check it. The last mechanical T&amp;F had several cooling tower problems and I was trying to get some problem experience in this area. Does any know if this equation will work for all?


I know this topic is old...but maybe it will help some people now as well:

I do not believe that *Mwater*delta T water = Mair*(delta h - Tambient air*change in humidity ratio)* is correct. Just a simple look at the units should prove to you that this is not correct. I believe what you meant to say is more along the lines of:

mdot_w*(delta h water) = mdot_air*(delta h - h2w*(delta W))

If you want to know where this comes from, it is a solved/rearranged equation that can be found a number of places, including ASHRAE. It is based on a simple energy balance.

Look at the original formula you posted from their solution:

250,000(68) +Mair*h1 = (250,000 – Mevap)(48) +Mair*h2 -&gt; mdot_w*h1w + mdot_a*h1a = (mdot_w-mdot_evap)(h2w) + mdot_a*h2a (*Equation (1)*)

I noticed someone asked: _"In the solution they are considering the make up water, but how do you solve for make up water if you have to calculate the air flow first and then make up water can be calculated."_

For this you need to understand more than just plug and chug. You need to utilize two equations. The 2nd equation is a basic fundamental psych chart equation that states mdot_water = mdot_air*(delta W) where:

mwater is the amount of water added (say lb/hr)

mair is the amount of dry air flowing (lb/hr)

and W is the amount of moisture (lbw/lba) at the inlet (W1) and outlet (W2) conditions of the air

mdot_evap is the same as mwater here. The amount of water that evaporates from the water stream is = to the amount of moisture added to the air stream (in this basic exam type problem). So, now let's plug mdot_air*(delta W) in for mdot_evap in* Equation (1)*

This gives us:

mdot_w*h1w + mdot_a*h1a = (mdot_w-mdot_air*(delta W))(h2w) + mdot_a*h2a

You can now rearrange this equation to get:

*mdot_w*(delta h water) = mdot_air*(delta h - h2w*(delta W))* - *Correct simplified equation for this problem to save steps on the test*

In this case, you have everything given to you and you only need to solve for mdot_air and then check units for what they want (lb/min? lb/hr? ft³/hr?).

ha and Ws comes from psych charts.

hw comes from steam table at saturated liquid for the given water temperatures.

I hope this helps anyone in the future with this problem. I felt like the previous answers to this problem were a little misleading and incorrect. Someone please correct me if I am wrong.


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