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HVAC - Mixture Question


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#1 Krakosky

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Posted 10 April 2012 - 12:11 PM

I was working an HVAC mixture problem and got it wrong. Very discouraging bc I was sure I knew how to do these types of problems. I'll list the problem and my approach below.

"An air handler that is equipped with a cooling coil receives 2,000 cu. ft/min of outdoor air that is maintained at 71 deg F and 50% relative humidity. The air is mixed with 4,000 cu. ft/min of outdoor air at 95 deg F dry bulb having a humidity ratio of 0.012 lb of moisture per lb of dry air prior to entering the cooling coil. The enthalpy of the mixture entering the cooling coil (Btu/lb dry air) is most nearly?

I found the enthalpy from the psych chart for the 2 conditions listed and then tried to use Eq. 38.21. Is this equation incorrect to use to find the change in enthalpy? I thought it could be used to find all the properties of a mixture as long as the mass or volumetric flow rates are known. Thanks for any input.

#2 MetsFan

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Posted 10 April 2012 - 02:48 PM

No, you can't use that equation directly to find the enthalpy. It's only for the relationship between mass flow and temperature or volumetric flow and temperature. I do feel like that equation makes it confusing. Here's what I use to find the mixed air of two air streams when temperature and CFM are given:

T1 x (CFM1/CFMtotal) + T2 x (CFM2/CFMtotal)

In your problem, you first need to find the point of the mixed air. Here's the steps I would take:

1. Find the two given points on the psych chart.
2. Draw a line between the two points. The mixed air temperature will fall somewhere along this line.
3. Find the dry bulb temperature of the mixed air:
Using the equation above, 71 x (2000/6000) + 95 x (4000/6000) = 23.67 + 63.33 = 87.0 degree F dry bulb.
4. Find the point on the line you drew in step 2 that corresponds to 87 deg. F dry bulb. That point represents your mixed air.
From there, you can find any mixed air property. In your problem it would give me an enthalpy of roughly 32.8 BTU/lb.

#3 ksprayberry

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Posted 10 April 2012 - 03:01 PM

I'd go at it like this, I'm sure there are other quicker ways...Maybe there are a couple ways actually.

1st way
Plot and draw a line between the two state points
Then lever rule (weighted average) your OA & RA ... (2000*71 + 4000*95) /6000 = 87
Having your line drawn between the two state points, you could project up from the bottom the 87 and see where it intercepts the line between the two, then project that to the saturation line to get your enthalpy.

2nd way.
You have you mixed db temp same as above. You have one of your humidity ratios, get the second by projecting the point (71/50%) it to the right. You can use these two humidity ratio and the same lever rule / weighted average to get (2000*.012 + 4000*.0081) /6000 = .0094

Now that you have your humidity ratio, you could really use the formula hmix = Hdry air + W*Hg.... H dry air from the table in ASHRAE at 87 and Hg is from the steam table at 87 degrees saturated vapor. Should look something like this. ..20.905+.0094*1099.11 ~ 31.2 a little on the light side, but straight out of the ashrae book
Also, one other formula for arriving at the enthalpy of the air is (.24*T) + W(1061+.444t)

I've attached a cheat sheet that might help as well. Check it for yourself thought. I don't want to be responsible for causing someone to miss a problem.

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#4 Krakosky

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Posted 10 April 2012 - 03:04 PM

Yep the answer is h = 33 Btu/lbm dry air. The solution used the same method as you did but I understand your explanation a lot better. Thanks!

#5 ksprayberry

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Posted 10 April 2012 - 03:06 PM

The cheat sheet allows you to use your wetbulb temperature and get your enthalpy directly from it instead of using the psychrometric chart. Maybe a bit quicker. It doesn't really help you in this case, but could come in handy.

#6 Krakosky

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Posted 10 April 2012 - 03:21 PM

Thanks for the cheat sheet. How do you use the table on the first page?

#7 MetsFan

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Posted 10 April 2012 - 03:37 PM

Thanks for the chart! Another one for my binder :)

Tenths of degrees marks off 35.1, 35.2, or 35.3. So if you're looking for the enthalpy of 60.5 deg. F wb, the answer is 26.8 BTU/lb

#8 MizzouMatt

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Posted 10 April 2012 - 04:04 PM

Thanks. I will add it to my binder too.

#9 Krakosky

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Posted 10 April 2012 - 04:49 PM

It's going to suck if there is a mixture question on the exam bc we can't write on the psych charts we bring. So I'll have to attempt to use the tiny one they provide.

#10 ksprayberry

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Posted 10 April 2012 - 05:10 PM

Tell me about it! I bought a pair of +2.5 reading glasses last night in hopes of reading one better.

#11 Outlaw44

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Posted 10 April 2012 - 05:34 PM

I'm not HVAC, so take this with a grain of salt, but I always use the method described above by drawing a line b/t the two points being mixed and then calculate the DB temp from the mixing equation. Then you can find that point on your line and find any other variable tied to it.

Now, I seem to remember Mr. Lindeburg stating in the MERM that the Tmix equation that Krak references can be used for anything that changes linearly with temperature. I'm not saying that enthalpy is one of those items, but there are a few he mentions in the paragraph just above or below the equation. Likewise, there are a few things he states it can't be used for. And I don't remember what those are! Enthalpy must be one.

But, thinking in thermo terms, Cp*deltaT is equal to enthalpy, so if Cp doesn't change for a fluid over various temps, then deltaT would be linear to enthalpy, correct? Isn't Cp for air almost always 0.24 Btu/lbm-degF? Does humidity ratio or water vapor in the air screw up being able to use the Tmix equation with enthalpy?

Either way, it's too close to test time for me to switch my process and to be honest, I don't think it matters that I understand why enthalpy can't be used in the equation, so please feel free to completely ignore. :) Just curious.

#12 Krakosky

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Posted 10 April 2012 - 05:43 PM

I feel the same way. All of a sudden it feels like there are 100 things I could use a little more clarity on. Probably psyching myself out more than anything. In the problem I referred to they used that equation to find T_mix and w_mix.

#13 Krakosky

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Posted 10 April 2012 - 05:57 PM

http://www.eng.buffa...539/W12Ch03.pdf

Look at page 27. Now I'm confused. They use the mixture equation to find h_mix.

#14 MizzouMatt

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Posted 10 April 2012 - 06:05 PM

Well Tdb, Twb. Enthalpy, Humidity ratio, Dew point all vary linearly so that makes sense.
my method for these problems is to:
1) find both points on a psych chart
2) find the ratio of type of air i am looking for / total air
3)measure the distance between the points in cm
4) multiply this distance by the calculated ratio.
5) measure that distance on the chart from the reference point
6) get desired info about that point

#15 ksprayberry

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Posted 10 April 2012 - 06:19 PM

Outlaw44 you're correct about the CpT = enthalpy, but its' the enthalpy for dry air. It's relevant for all ranges covered in the psychrometric chart, once you get beyond that, there is a slight chage to the Cp say around what 400-800 degrees, it changes to .244 or something like that, but you have to add to it the enthalpy of the water vapor in the air mixture as well, that is where the (1061* +.444T)*W figure comes from.

At 87 degrees, Cp*T = about 20.6 + enthalpy of the vapor. Your exactly right on the dry air side. I think the CpT is mean for a "Ideal gas". Not sure. I'd let someone else chime in to be certain I'm right.
Your also correct about the linear thing the lever rule (weighted average) can be used for all things linear, but not things like relative humidity or wetbulb don't use it for those. I've seen people make the mistake of figuring wetbulb temps this way. It doesn't work quote right.

Thanks
Kelly

#16 ksprayberry

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Posted 10 April 2012 - 06:42 PM

Not that I'm trying to advertise or anything but this company offers a trial version of their psychrometric program that has a mixing option. It's small and straight forward for figuring state points and mixes. It might be good to play around with and compare or confirm your calculations made by hand. Make up a couple of mixes and see what you have.

http://www.linric.com/downloads.htm

Seriously, here are three examples right now that I'm working on at work.
Given (Not much) 3 direct replacement McQuay Multizone units approximately 15 tons each.
Design conditions - 93db/73wb OA - 75db/62.5wb RA
MZ-1 4000 cfm total - 3,615 cfm RA / 385 cfm OA
MZ-2 4,600 cfm total - 4,125 cfm ra / 475 cfm OA
MZ-3 4,200 cfm total - 2,025 crm RA / 2,175 cfm OA

What is the mixed air condition to the DX coil disreguarding any heat gains from the blower motor?

#17 Krakosky

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Posted 10 April 2012 - 06:44 PM

So is that why this problem can't be solved using the mixture equation? Because you have to add the w*hvapor term to the calculated hair? I'd rather use the other method if it's going to work every time.

#18 ksprayberry

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Posted 10 April 2012 - 06:57 PM

Getting it off the psychrometric chart is more straightforward than any other method. How accurate the results are depend more on you accurately projecting to the saturation / Enthalpy line properly. I would go with the plot and go method on the psychrometric rather than try to calculate it long hand.

Psychrometric way.
1 plot the two state points
2 draw the line between.
3 lever rule the dry bulb temperature (weighted average)
4 project db temp from the bottom of the chart up to where it intersects the line between the two points. (The mixed condition will always lie on that line)
4. Project up to the enthalpy scale.
Easy peasy lemon squeesy.

I will post a couple of my cheat sheets that may help with the long hand method too.

#19 MetsFan

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Posted 10 April 2012 - 07:01 PM

I'm an HVAC engineer and this spreadsheet is what I've used when dealing with any Psych problems:

http://dl.dropbox.co...blem - Copy.xls

I don't know if it will help anybody at this point, but figured I'd give an alternative to the linric download.

ksprayberry, I'm not sure about the ideal gas law, but it sounds like something I read during my studies. I'll see if I can find it.

krakosky, have you tried to figuring out the enthalpy with the mixture equation using mass instead of mass flow? The equation you pointed out had mass flow instead of volumetric flow, so maybe you have to use the specific volume of the air and calculate the mass flow to get the enthalpy? I could be completely wrong though.
You may be over thinking this too. If you stick with using the mixture equation for dry bulb and then using the psych chart for everything else, you should be good.

#20 Krakosky

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Posted 10 April 2012 - 07:05 PM

The method you listed is the one I plan to use. Now I hope there is one of these problems on the exam after our day long discussion.

#21 ksprayberry

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Posted 10 April 2012 - 07:17 PM

At the end of the day, use what feels good to you. When I'm doing it the long method (ha + Whg), I used the attached tables to help me. It has both h of the dry air and Hg of the water vapor in the mix right in the same table. All you're missing is the W humidity ratio...By the way, IT IS NOT Wg on the table don't make that mistake.!
KS

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#22 Outlaw44

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Posted 11 April 2012 - 12:11 AM

Thanks Kelly!

Man, I went away to a meeting this afternoon and responses started flying! I think we're all just a tad anxious, eh? :)

#23 ksprayberry

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Posted 11 April 2012 - 12:20 AM

Yeah, tell me about it. It's a touch hard to focus this week. It's time to be done with this, this time.

#24 r_mojo1

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Posted 11 April 2012 - 01:02 PM

My brain is saturaded with information at this point. I have noticed that lately my ears cannot stop ringing. It is a constant noise. I need to take the test and relax.

#25 MetsFan

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Posted 11 April 2012 - 02:15 PM

I've been seeing "engineering" everywhere. I look at a boat floating in the ocean: What is the buoyant force? I see an expansion tank: I wonder what the axial forces are? Look at a structural beam: What would the moment of inertia be for that?

I think the most important thing is to breathe and realize we have studied and should be well prepared for this exam. If the problem looks too complicated, you're probably looking at it the wrong way. It should only take you a couple of equations to solve it.

Now I hope I can use my own advice and not panic :)




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