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.

 

[Yorkman]

Yes, there is a set of tables in the ASHRAE Fundamentals Handbook, in the Psychrometrics chapter 6 they are called the Thermodynamic Properties of Moist Air Tables. Then with the use of equation number 32 from that chapter and perhaps a little reading, this may make a little more sense. h = .240t + W(1061 + .444t)
Where: h = specific enthalpy BTU/Lbs of the moist air
t = the dry bulb temperature of the indegrees F
W = the humdity ratio of the moist air Lbs
Lbs water/lbs dry air
Using the table locate the W value and multiply it by the decimal %Rh to get the humidity ratio of the air at that condtion then enter that value into the equation. Solving for h will give you a very close representation of what the chart is providing.

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

 

[jasno999]

Ok. Well I do nto have that book. I wil llook over my Trane Air Conditioning Manual to see what that has in it. Otehrwise I may have to look at purchasing the ASHRE Book.

Do you know the exact name of the ASHRAE Fundamentals Handbook? What I mean by that is does it have a code number or some type of numberign associated wiht it?

 

[Yorkman]

These books are published by ASHRAE on a four year cycle. the name of the book is:
2005 ASHRAE Handbook Fundamentals. Go to Ebay and look under books ASHRAE, often they are there pretty cheap or amazon books may have one don't expect to pay any more than 30.00 to 60.00 dollars for an older edition 2001 will have the same tables as the 2005. You can go directly to

http://www.ashrae.org/

to purchace it new but they ask for quite a bit. You might do a seach on Google for a listing of the tables I've never tried that.

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

 

[Yorkman]

Jasno999, I just looked in the Trane Manual they do have a set a of tables there PG 400 Table 4-1 The ASHRAE Handbook is still a good idea but this might get you through for now. I would copy down the equation from the post as that came out of ASHRAE not the Trane book

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

 

[jasno999]

Ok here is antoehr good question for you.

How do you incorporate leakage?

Example with made up numbers:

-1 room
-Sensiable load 10,000BTUH
-Laten Load 5,000BTUH
-Total load 15,000BTUH (Includes structural, human, infiltration, solar and electrical)

-100% recirculation air system (no outdoor air to the evaporator)

-Supply fan is 1,000 CFM

-Assume that 200CFM air infiltrates the system.


Now how do you factor in leakage? My thought would be that since we have 200CFM or 20% infiltration that must mean we have aroudn the same amount of air leaking from the room to the outside. So we can say that leakage is 20%.

I would think we would calculate the leakage BTUH by this equation:

Leakage = Total Load + [Total Load X Leakage %]

or

15,000BTUH + [15,000BTUH X 20%] = 18,000BTUH

IS THIS CORRECT?


If so do I add the extra 3000BTUH that we found for leakage to the sensiable side of the load? I need to know that cause it will drastically effect my supply air temperature when I go to calculate that using the sensiable load and sensiable equation..

 

[jasno999]

Or maybe I am not supposed to factor leakage into the equation at all??? If not let me know...

 

[Yorkman]

If you know the sensible and latent loads in BTU/HR as stated and that included infiltration load then the 200 CFM that you mention had to be already factored in the way I see it. The 200 CFM had to be calculated using the amount of crack based on window and door design along with climate data and average wind velocity. It should have been used to find the BTU/Hr loads for the space. That was all part of the space load design.

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

 

[jasno999]

YEs I factored in the infiltration of outdoor air into the space.

However if you have a system that is usign a mixture of outdoor air and return air then you are basically adding air to the system and therfore pressurizing the space to some small degree. In doing that you will then be forcing some of the air out of the room.

So if you have 75% return air and 25% outdoor air then I would think that you are goign to have 25% of the air flow needign to escape the room via cracks in the doors/windows/wall/etc.

I would think even with pressurization you could still have infiltration alone with the leakage.

For example a strong wind on one side of the room could cause some infiltration air to come in. However since the space is being pressurized then on the other ide of the room you will have air leakign out to the outside. I woudl think that rate would be the infiltration rate plus the 25% of the flow that is from outdoor air.

DO you see what I am saying?

I woudl think that since you are losing some of the coolign load to air escaping out of the room that you would have to somehow factor that into the overall heat loading.

How would you do that? I don't see how that would be part of the infiltration load cause that load consists of a sensiable load that has no association with flow right? and the latent portion that is multiplied times the amount of return air being sent back to the evaporator.

I guess I am confused...

 

[jasno999]

In my application we are not sure what the infiltration rate is goign to be. SO we have to make some assumptions. At the very least I am buildign a program that will allow us to change the infiltration amount and see how it affects the overall load on the evaporator.

Yo uare correct that the infiltration rate wil laffect loading because the more you have of it the more hot air gets in and the more work the evaporator has to do.

However you are also goign to have air that leaks out. First off if you bring in 25% outdoor air (for example) and send that along with 75% of the return air thru the evaporator and into the room - then you will return 75% of it but the otehr 25% must escape fro mthe room due to pressuriziation of the space and leakage thru the cracks and spaces.

At the same time if you bring in say 50CFM of infiltration air then you will need to get rid of antoehr 50CFM of room air on top of the 25% that is already escaping.

Now I don't think we have incorporated this leakage of air from the room to the outside itno any of our equations. I see this air that escapes as a loss in my mind. If you are losing soem portion of conditionid air then that shoudl be a heat loss right???

Infiltration as I see it is a heat gain to the system. Leakage is a cooling loss. It does seem like using both is sort of a double hit to the system but I would think that you need to include it in the math.

Any help here would be appreciated cause I think that thsi is the last step that I am stuck on. after I have this figured out I can finish my model and I can really look at all the systems I have in great detail.

Thanks

 

[Yorkman]

If the Hvac system is pressurizing the space then you will not have any heat gains due to infiltration, correct. So that the issue of sensible heat and latent heat gains affecting the space load goes away. But in order to obtain that positve pressure to stop this infiltration to the space, we had to bring in outside air, ventilation air through the unit this is seen as a direct load to the unit. The load is still there for the coil, the only difference is we get to treat it before the occupant sees it. As to not knowing what the leak rate is going to be, yes I suppose thats true, but based on the type of structure, windows ,doors,and quality of construction there is data out there that will allow you to make an educated guess as to the possible amount. Here goes my plug for the ASHRAE Fundamentals Manual again.

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

 

[jasno999]

I was able to download the ASHRAE manual via my work computer and software my company has set up. So I hae that in PDF form and I have been readign thru it. Still I am left with some confusion.


I understand what you are saying. The load of the outdoor air is seen by the evaporator coil not the room. That is the reason why the evaporator coil heat rejection load is higher than the room heat rejection load required.

I guess what I see is that the room has heat gains from different things (electrical/strustural/solar/human/maybe infiltration)- But I woudl think that thru leakage it would also have a heat loss. Do you have to add that heat loss into the room load calculation somehow?

And I think as odd as it sounds in my application we may possiably have infiltration loads (infiltration air) coming into the space eeven though the space is goign to be somewhat pressurized. I can't get itto all the details but the space I am dealign with is not goign to be well sealed and it could see some high wind flow rates on oen side. SO one end coudl see infiltration air and the other end could see leakage of the infiltration air along with the pressurized air due to the intake of outdoor air to the evaporator.

So do we factor leakage (from inside to outside) into the room load calculation at all????

How exactly?


All the leakage data and calculations I see the the Trane manual and in the ASHRAE manual talk abotu infiltration and the leakage of air into the space. I need to knwo about leakage out of the space to the outside and hwo to factor that in.

Again maybe my head is screwed up fro mal lthe calculations and I don't need to do that- but I would think that I would. I would think in the end that the room air that is escaping from the room to the outside should be taken as a coolign loss and should somehow be factor into the room heat loading.

 

[jasno999]

Like on page 34 of the trane manual (bottom of the first column and top of the 2nd column).

It says:

"When outdoor air is taken itno a central air handling unit... pressure in the room can eaisly build up high enough to offset the sind pressures. This stops infiltration; actually, air is leakign out.."

Ok in my case I think I will still have some infiltration along with a pressurized system that needs to push air out of the room (or leak air out). However there is not statment after this one to describe how that leaky air is supposed to be taken into account....

 

[jasno999]

On top of those questions here is another one.

IS it possiable to have a higher heat load for the room than for the evaporator in a given problem.

Right now I have a situation where my total room load is like 22,000BTUH where 20,000 is sensiable and 2,000 is latent.

So I use the sensiabel portion to find the disscharge air temperature. After finidng that I use the discharge air conditions compared to the mixed air conditions enertign the evaporator to find the total heat lost for the evaporator using the Qt=lb/hr X (delta Enthalpy) and I am gettign a evaparator load of 19,000BTUs which is less that the original load of the room.


IS that possiable?

 

[Yorkman]

#1 First understand that if you are presurizing the room the only way you can do it is to bring in outside air, this outside air is part of the mixed air you are seeing at the coil so the energy to heat or cool this has been accounted for in your earlier calculations. Go to the HVAC forum and check out the FAQ there, question 4.
#2 If your situation is unique in that even with resonable building pressurizaton you are still going to get some infiltration due to unusual crack conditions and high wind velocities you will have to find a way to calculate that amount. I'am not sure of how to do that but if you post in the HVAC forum somebody with more exposure to this maybe able to help.
#3 Maybe it was done or maybe you overlooked it but when or how did you find the CFM. After you found your supply air temperature, using the SHR. You would use the supply air temperature and the room air temperature to find the CFM.: BTU/Hr (sensible)[÷] (1.08 x [Δ]T) = CFM Then use the CFM and the supply air enthalpy and mixed air enthalpy to find the total load of the coil.
#4 If your mixed air condition has an enthalpy value less than the space condition but greater than the supply air enthalpy then you could have a load at the coil less than the load of the room. Isn't that what the economizer cycle is all about? Or it is possible you ommitted finding the CFM using the sensible BTU/Hr [&Delta];T


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

 

[jasno999]

Well as we had discussed earlier looking at the design conditions and the real world are two different things.

Basically I had my CFM set for me by the system suppliers. THe CFM they used is close to what I found for my worst case condition. But I had the CFM already ste upo and I used it to find out what the supply air temperatures had to be for the system being proposed.

I also did look at the system based on purly design and found my CFMs the way you wanted me to. However those CFM were al ldifferent for each outdoor air condition. My fan is only goign to have one speed and since I was given the fan info I used it to solve for SA temp and then used that data to solve for evaporator Q.

But in doing that I am not using the SHR.

What I did was this:

FOund my total load and my breakdown of sensiable to latent load.

then I used: Temp Supply Air = Temp Room air - [BTU/Hr (sensible) ÷ (lb/min X 60 X 0.24)]

Which is Tsa = Troom - [Qs/(flow X Constant)]


So using that I found my supply air temperature and made the assumption that it was coming off the coil at 95%RH.

I then wrote an equation to solve for my mixed air conditions (temperature & Humidity):

Temp mixed air = [Room temp X % recirc]+[Outdoor Temp X (1-% recirc)]

And I then used the humidity (grains/lb) to calculate my humidity values:

Humidity Mixed (Gr/lb) = [Room Humidity(gr/lb) X % recirc]+[Outdoor Humidity(gr/lb) X (1-% recirc)]

Knowing the mixed air humidity in grains per pound would then allow me to use the charts from ASHRAE to determine what the Relative Humidity was for each point. (Basically at the mixed air temp you can find the grains/lb at saturation and use that to devide the mixed air humidity you just found to get a %.

Now I had the %RH and could use that in the equation you gave me earlier to find the enthalpy:

h = .240t + W(1061 + .444t)

When I had the enthalpy of the mixed air I would then use it along with teh enthalpy of the supply air in this equation:

Q(evaporator) = Flow(lb/hr) X Delta h(btu/lb)

And that is how I found my evaporator load.


Give me a quick check and let me know if you see somehtign wrong in that process.

I understand that the outdoor air is taken into accoutn at the evaporator so I do nto need it in the room loading. Was hard for me to grasp that but I get it now.

 

[Yorkman]

The formulas amd the method all seem to look good to me. Do the numbers work out when plotted on the psychrometric chart, using the vendors CFM?

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

 

[jasno999]

When you ask if the numbers look good when plotted out what exactly do you mean?

I mean I tried to plt them out but I don't knwo if they look good or not. I had a SHR line that went thru the due point line? Is that ok?

Tell me what sort of things I need to look for if I plot the points and tell me how you would plot your points out.

 

[Yorkman]

I don't mean to be so vague, I just don't have enough data to know what results your getting. We've thrown a lot of examples around for heat loads and CFM, but I'm not sure what was a senairio and what was a valid number?
Why not try using the formulas and psychrometric charts on a system that is all ready up and running? It will give you some good practice and allow you to verify your calculations against a functioning system.



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

 

[jasno999]

That is good advice. However I do nto have a system up and runnign tht I could use as a model.

I know that I have been vague but I can't tell you all the details of what I am working on. That is why I was throwing out all the different examples.

I have a feeling that something is off or wrong because my results are not matching up with what I have received from some known AC providers.

At the same tiem maybe my numbers are correct and I need to ask some questions. I am pretty confident I used al lof the proper equations in the proper places. But when I get my final Evaporator loads I am seeing some different values than what I am being told.

So I am goig nto review my calculations again today to see what I can come up with. Then I wil ltry to get anotehr set of eyes on it where I work. After that hopefully we can coem to a conclusion one way or the other.

My concearn is that there is a lower temperature condition which high humidity that is turning out to be the most critical point requiring the most heat rejection at the evaporator. It makes sense cause the latent load at the evaporator for this condition is high. However the evaporator load for this condition turns out to be about 10,000BTU higher than for the other conditions that I have. And that is higher than what any of the systems being offered can do. So it comes down to one condition that has me scratching my head and then looking at that along with the offered systems to figure out what is goign on.


I will post more questions if something hits me while I am reviewing my numbers.

 

[jasno999]

Ok I have come to a conclusion. Somehting is definitly wrong with my math.

I did all the the stuff you have told me to do. I found my disscharge temperatures using the sensiable load required for the room. Then I used the supply air temperature and mixed air temperature and humidity to get the evaporator Q.

However I have that one ccondition that is a high temperature and a high humidity. Say 102F and 100%RH -

When I do the math I get a disscharge temperature of around 64F.

However I was told the room conditins could be maintained at 80F and 50% RH. When you go and plot this on the psyh chart you get a line that slopes downward from left to right. Meaning somehow between the supply air and the room air we are losing humidity???? That is not possiable cause we should be gaining humidity here.

I am confused.

What I am finding is that outdoor air humidity effects the evaporator load the mixed air enthalpy and the latent side of the room Q.

Room Humidity effects the evaporator load the mixed air enthalpy and the latent side of the room Q.

Since these do not affect the sensiable load at all and the sensiable load is the only number being used to find the supply air temperature using Qs=Flow(lb/hr)*0.24*deltaT That means the supply air temperature and humidity is never affected by the laten loads (inside or outside the room). Therefor the evaporator is the only item that is affected by leten loading. IT makes sense until you realise that if you have supply air coming into the room and it is drying out (losing humidity) as it gets heated up to the room temperature of 80F that makes little to no sense and it tells me that somehting is wrong in the equation.

If anyhitng we are addign moisture to the room via human presperation and infiltration air. I realize that the evaporator is supposed ot be cooling the air and gettign rid of that moisture but I am confused as to how I lose moisture as my supply air gets heated back up. I would think in order to do that you would need some sort of dehumidification device which we do nto have on this project.