Psychrometric Chart help with Air conditioning. 3

[jasno999]

This may sound like a silly question but I have totally forgotten how to use the Psyh- chart to follow a air conditioning system problem.

For example if I start with an outdoor condition of 95 degrees F and 80% Relative Humidity and I mix that air with return air that is at 80 degrees F and 50% relative humidity. The mix is 80% Return air and 20% supply air.

I then take that mixed air and run it thru a evaporator to cool it down to a disscharge temperature of 36 degrees F.

My question is how to properly do this but more importantly how to graph this on a Psychrometric Chart. I am confused as to how the lines would appear. I thought that in cooling you went horizontally from right to left on the chart and then followed the due pointline down to your disscharge temperature. As you did that you would lose water at the evaporator. Then when you got to the line associated with 36 degrees F at 100% realtive humidity you stoped going down and then moved horizontally to the right on the chart until you readed your room tempererature which in my case is 80 degrees F. But I don't knwo if that is the right way to do it or not.

I am confused right now and need soem explanation.

 

[jpankask]

There is an article on psychrometrics in the July 2002 edition of ASHRAE Journal. If you are an ASHRAE member or know one, I think you can still download it. Otherwise you might be able to get it at a university library or through inter-library loan. Alternatively, there is probably something in the ASHRAE Handbook of Fundamentals which you could get at a library or buy an older used copy off Ebay. Someone may be able to explain well with words but seeing it done on the chart would be much easier to understand.

 

[jasno999]

My problem is I need this information yesterday. LOL

I am working on an issue that I need to solve before the end of the day today and I need quick explanation.

 

[BronYrAur]

Your room condition is the return air. To find your mixed air, plot both the return and outside points. The mixed air will be 20% of the distance "up" the line. In other words, it will be 83 deg and roughly 60% RH.

When leaving the coil, you will move both to the right and up to get back to your room codition (return air), as you have both a sensible and latent load.

 

[Yorkman]

Jasno99,
Here are some rough numbers assuming sealevel 29.92" Hg. Your mixed air sould be about:
82 degrees Db; 71 degrees Wb; 58% Rh
35 degree supply air off the coil is pretty cool for standard comfort cooling maybe this is a special application.
At 35 degrees 95% Rh (few coils can produce 100% Rh)
11.0 Btu/Lbs dry air sensible heat
10.2 BTU/Lbs dry air latent heat
21.2 BTU/Lbs dry air Total heat
.518 sensible heat ratio
Moisture removal
.0093 Lbs water/Lbs dry air
Specific Volume @ 35 degrees
12.56 Cu.Ft./Lbs dry air

That should cover it there are some pretty good site on the net that go over the basics. Hope this helps

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[Yorkman]

I need to appologize I made an error in plotting the OSA the corrections are as follows
Mixed air: 83 degrees Db; 72.5 degrees Wb; 60% Rh
36 degree supply air off the coil is pretty cool for standard comfort cooling maybe this is a special application.
At 36 degrees 95% Rh (few coils can produce 100% Rh)
11.4 Btu/Lbs dry air sensible heat
11.3 BTU/Lbs dry air latent heat
22.7 BTU/Lbs dry air Total heat (this is load seen by the coil
.502 sensible heat ratio
Moisture removal
.0104 Lbs water/Lbs dry air
Specific Volume @ 36 degrees
12.56 Cu.Ft./Lbs dry air
I'm very sorry for the error hope that hasn't caused you too much trouble


I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[jasno999]

Ok I guess the question then is this:

How do I figure out what the required coolign capacity is for the system.

I know that I find a structural heat gain, electrical heat gain, human loading and maybe infiltration air if we have some. I can add those up and figure out what the heat gain in the room is. But how do then figure out the nessassary cooling required to maintain that room at say 80F?

Obviously I need to provide enough cooling to make up for the heat gain we see in the room. But I assume that I amso need to provide coolign to make up for the humidity in the outdoor air (or mixed air entering the evaporator)??? I am not sure how to figure out the total system loading.

 

[Yorkman]

Last night time was short to explain thouroughly.
This assumes your using an ASHRAE sealevel chart

1.Plot OSA point on chart at 95 degrees 90% Rh

2.Plot RA point on chart at 80 degrees 50% Rh

3.Connect the OSA point and the RA points for mixed air line.

4.Measure length of line 3.875" x 20% = .775"(there are other methods to find this and I can give a simple formula if you need one let me know.)

5.Measure along the MA line starting from the RA point .775" this is your MA temperature and humidity.Read 83 degrees 60% Rh

6.Plot the supply air point (SA) 36 degrees 100% Rh (depends on the efficiency of the cooling coil) I used 95%

7.Connect straight line from the MA point to the SA point

8.Form a right triangle using the SA point and the MA point,making a horizontal line fom the SA point and intersecting it with a vertical line from the MA point.

9. Locate enthalpy values at the SA point, the MA point, and at the 90 degree intersection of the horizotal and vertical lines of the triangle. These enthalpy lines run diagonally care must be used in accuratly locating.

10. At SA point read 13.4 BTU/Lbs, at MA point read
36 BTU/Lbs, at 90 degre intersection read 24.8 BTU/Lbs

11.The difference between 24.8 and 13.4 is sensible heat BTUs, the difference between 36 and 24.8 is latent heat BTUs, the total heat is the difference between 36 and 13.4 This is the load at the cooling coil, not the load in the room. The load in the room would be figured from the RA point.

12.To find the LBS of water removed extend horizotal lines across chart from the SA point and the MA point to the humidity ratio index at far right of the chart. For the SA point read .0042LBS+/- and the MA point read .0104+/- the difference is the LBS of water removed per Lbs of dry air .

Hope that wasn't too detailed point acrossed. Good luck

PS. ASHRAE sells a real nice psychrometris program for about 150.00 USD


I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[Yorkman]

Item 12 should read

12.To find the LBS of water removed extend horizotal lines across chart from the SA point and the MA point to the humidity ratio index at far right of the chart. For the SA point read .0042LBS+/- and the MA point read .0146+/- the difference is the LBS of water removed per Lbs of dry air. Aprx. .0104 Lbs of water /Lbs of dry air
Sorry way too many numbers for 6:00 a.m.


I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[Yorkman]

One last point, you need to determine the CFM based on the room load and required outside air. Once you know the CFM, you can use the formula
BTU/HR = 4.5 x CFM x [Δ]H
to find the total cooling coil load

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[jasno999]

Yorkman - Wow that was very helpfull. I have to now go use your explanation on a few of the issues I currently have.

Right nwo I am tryign to figure out if a quote I have for a project has the correct numbers in it. They gave me the room BTUH but I need to go back and see if that number is right and then I need to see what the real heat load of the evaporator is to see if they sized their condenser jsut right. I appreciate the help.

Now again where I am seeign an issue is you gave me the equation BTU/HR = 4.5 x CFM x ?H

But I think the real equation should read BTU/HR = 4.5 x CFM x ?T X 0.24 (I thought this was a delta Temperature not an enthalpy).

But what I was tryign to do was look at the sensiable heat and latent heat using equations to find the total heal load at the evaporator.

I was using the BTU/HR = 4.5 x CFM x ?T X 0.24 for my sensiable load and I was using the equation 0.7 X CFM X ?HR(grains/lb.)as my latent load equation. But I think this is nto the right way to do it for some reason. I was using the Mixed air properties and the Supply air properties to fill in the ?T and ?HR numbers.

 

[jasno999]

How about a situation where I am given the following:

I know the OA conditions, I know the internal Temperature requirment, I am given a capacity of the system, and I have some data for what the manufacturer says the internal humidity can be maintained at. I also knwo the fan flow of the system. Last but not least I knwo what the structural heat loads are, human healt loads are and electrical loads are. I am still running a 80% return air system.

How do I go about figering out what the supply temperature would be given that information? How would I go about determinng if the loads given to me by the manufacturer were good numbers and would be able to handle the conditions and give me the temperature requirment in the room?

 

[BronYrAur]

I haven't read everyone's reply, but here are my 2 cents:

The coil load will be the difference between the mixed air conditions and the coil leaving air conditions. In your example, the mixed air should be roughly 83 deg and 60%RH. The mixed air enthalpy (h) is around 36 but/lb, and you coil enthalpy is around 13 btu/lb.

Q (total)=(4.5)*(cfm)*(delta-h)

you can break up sensible and latent if you wish:

Q (sensible)=(1.08)*(cfm)*(delta-T), where T is temp diff
Q (latent)=(0.7)*(cfm)*(delta-HR), where HR is humidity ratio in grains/lb

Either way, if you know cfm, you can find Q.

If you are having trouble finding the mixed air condition, here is an easy way. Plot both on the psyc chart and draw a straight line connecting them. Then use simple math the determine the mixed air temp:

Mixed T = (80 deg)*(80%)+ (95 deg)*(20%) = 83 deg

Go to 83 deg on the line you drew and find the corresponding RH and other properties. This is your mixed air point.

 

[jasno999]

BronYrAur - I have been readign the posts but my last question was a little different because I was searching for the disscharge temperature off the coil with a different set of given conditions. It is basically a different question from the first one I asked.

I think that Yorkman did a great job explaining that first example and showing me how to look at things. YOu sort of restated it and showed me the breakdown for the sensiable verses the latent. When I have done thee calculations before with the breakdown I have notieced that my total for sensiable and laten typically does nto equal my total for enthalpy alone. They are close but not the same. Maybe my numbers are jsut off rome not being perfect when readign the psyh chart. I can double check tomorrow.

EIther way I still appreciate you help but and looking for answeres to a sort of different question.

 

[Yorkman]

Hey Jasno999,
Just got home from work, lets look at what you have.
BTU/Hr = 4.5 x CFM x [Δ]H is a total heat formula for air at sealevel. It does not accept [Δ]T's in this form, but can be adjusted the this one in my opinion is easier. You can use it to find required CFM,or total BTU/Hr depending on the data you have, provided you know the enthalpy values of your supply and return/space air.
BTU/Hr = 1.08 x CFM x (T[sub]1[/sub]-T[sub]2[/sub]) is used for sensible heat problems generally.
To find the answers you are looking for requires the use of the sensible heat ratio. The sensible heat ratio is a ratio of sensible heat to total heat based on your space load. The protractor on the upper left side of the psychrometric chart lets you plot the ratio or slope on the chart. Look closely at the protractor, the numbers on the inside radius are the ratios of sensible heat. Suppose your sensible heat ratio for your problem was .60 based on your rooms' sensible and latent heat load

1.Draw a line from the center point on the base line of the protractor through the .6 on the radius of the protractor. This establishes your slope.

2.Now plot your desired Db and %RH for the space on the chart, 80 degrees Db 50%RH.

3.Transfer the sensible heat line or slope (remember the protractor?) so that it starts at your desired room temperature and travels to the left and intersects the 90% curve on the chart at 90%Rh (read 45 degrees Db).

4.The point of intersection will establish the required supply air temperature to maintain the desired space temperature.

5.You could select any desired supply air temperature located on this line in theory and adjust your CFM, but most A/C equipment delivers air at 90% Rh or higher so you are some what limited, so much for theories.

6.Once you have established your supply air temperature, use the enthalpy values for the supply air and desired space condition and enter them into the :
BTU/Hr = 4.5 x CFM x [Δ]H formula along with your total room load in BTU/Hr to find the required CFM.

If that don't make your head hurt I don't know what will!!
I might suggest a good book on the subject by The Trane Company "Trane Air Conditioning Manual" excellent resource for this type of work as well as just all around good A/C therory. As I posted before ASHRAE has a very good and affordable "Psychrometric Analysis" program.
This was fun hoped it helped, that is about the limit of my skills without spending a couple of nights of exciting reading %-) Good luck!!

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[Yorkman]

After reading your post with the formula
"But I think the real equation should read :
BTU/HR = 4.5 x CFM x [Δ]T X 0.24 (I thought this was a delta Temperature not an enthalpy)".

Is correct, but:This might help in understanding the theory.

BTU/HR = 4.5 x CFM x [Δ]T X 0.24 will work, but it is only for sensible heat applications)
Note the breakdown:
.075 (density of air) x 60 min./Hr = 4.5
4.5 x .24(specific heat of air, BTU to change 1 Lbs of air 1 degree F) = 1.08 (Lbs/Cu.Ft./Hr/degree change) ouch!

When using: BTU/Hr = 4.5 x CFM x ?H keeps the units in Lbs/Cu.Ft./Hr while allowing us to use Enthalpy BTU/Lb. A value not limited to only sensible heat changes it can also represents latent heat changes or both.. Hope that makes sense.

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[quark]

Jasno,

Your latent heat formula is also approximately correct. The latent heat is due to moisture gain and if you want to express the latent heat as a multiple of cfm and moisture gain,

Q = k x cfm x gr/lb

If you expand the equation for dimensional stability,

Q (btu/hr) = k*cu.ft/min*(60min/hr)*(gr/lb)*(7000/7000)*(1000btu/lb)* (0.075lb/cu.ft)

upon simplification and rearranging the terms,
Q (btu/hr) = k*60*0.075*1000/7000 (btu/hr)
or k = 0.64 (which is close to your constant)

I would like to explain one part of Yorkman's excellent post. The SHR line cutting at 90% curve is an indication of coil bypass factor (BPF). When you pass the mixed air through the cooling coil some portion of it gets bypassed (with better coil fin arrangement, a safe bet can be about 5%). So, 5% is untreated mixed air and 95% is treated mixed air. This point can be marked on the SHR line (drawn from the room condition), in the same way as explained for mixed condition.

Actually coil ADP is identified after you do this exercise. I am puzzled as to how you could get it well in advance.

Sometimes, if the latent heat portion is high, the SHR line may not intersect the saturation line and runs almost parallel to it. In this case, you have to assume a coil ADP (below room entry point)and include reheat into the system. The entry point is calculated by assuming some DB temperature difference (say, 15 to 20F) from room entry point and room condition.

Yorkman,

Excellent work (and energy - in physiological sense). For quite sometime, I have been thinking of writing a FAQ in HVAC forum about this but couldn't do it due to laziness. We all will be greatful if you can you convert this excellent post into a FAQ.

Best regards,

 

[Yorkman]

Quark,
Thanks for the personel comments and the added explaination on SHR, by the way what is ADP? :~/ I' not familiar with the acronim please expand. I've been chided by some for using a psychrometric chart for analysis rather the a packaged program. While I think the programs are very valuable and fast, I still enjoy the act of plotting out the points and crunching the numbers, although I fear the process may be going the way of the slide rule. :-( ;-)
One of the reasons I enjoy this site so much is the information gained is always more than what one individauls' experience could provide. I've never posted a FAQ and the request means alot. I'm leaving for a long weekend but when I get back I will do my best to assemble some of the more coherent thoughts, into a quick primer on using the psychrometric chart. I may touch base with you on a few points, Thanks

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[jasno999]

Ok this is what I did. (NEW EXAMPLE)

I looked at my structural losses (at the given outdoor conditions of say 105 & 60%RH) thru the walls & ceiling of the room. I then added that to my electrical heat gains in the room. I then added my sensiable heat gains from people in the room who would be working with a room temperature of 80F (I used charts to figure that out with the proper equation). And finally I added a 10% leakage loss factor on to the entire equation.

I took all of those values and added them together. For this new example those values came out to be 10,000 BTUH. That should be the total heat gain the room sees so we need to make sure the cooling system is able to cool the air enough to equal this amoutn of heat gain.


I was also told that my room conditions would be maintained at 80F and 50%RH. Knowing that I figured I could use the sensicabel equation to find out what the temperature of the air was enetering the room.

SO Qs = Flow X 60 X 0.24 X Delta T

Qs = 10,000 BTUH
Flow = 37.5 lb/min
Indoor air temp = 80F

Therfore:

Tsa = 80F - (10,000/(37.5 X 60 * 0.24))

Tsa = 61.5F

IS THAT THE RIGHT WAY TO DO THAT? OR DO I NEED SOME TYPE OF LATENT HEAT IN THIS EQUATION FOR THE ROOM? I figured that since moisture is nto being removed between the supply air and the room that latent did nto factor in here.

I figured this was all sensiable and that the latent factored in where moisture was being removed and that woudl be at the evaporator where you would need to look at bout the sensiabel and the latent loads.

I could be wrong but that is the question. Did I figure out the supply air temperature Tsa correctly?

If I had the supply air temperature I sould then assuem a 95% RH at the supply air out of the evaporator and then used the mixed air conditions prior to the evaporator along wit hthe supply air conditions to determine the heat rejection required at the evaporator.

 

[Yorkman]

The latent load of the people and from infiltration must be factored in as load in the room. Take the 10,000 BTU/HR sensible plus the latent BTU/Hr from infiltration and humans added together to get total heat. Find the SHR based on these totals. Plot desired room Db and Wb temps draw SHR line to 95% curve on psychrometric chart this will be required supply air temp. Then use Qs = Flow X 60 X 0.24 X Delta T where you know Qs; Supply air and room air to solve for CFM, you shouldn't choose the CFM you need to calculate it
Next use : BTU/Hr = 4.5 x CFM x ?H where you know the required CFM that you just found, the enthalpy of the mixed air, and the supply air enthalpy to find the coil capacity need for the system load. Thats how I would approach it.

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI