Relative humidity and humidity ratio question. 2

[jasno999]

I am trying to set up a program on excel to look at air conditioning loads and what not. In doing this I would liek to use equations and data fro mcharts as much as possiable so that I do nto need to reference the psychrometric chart - I want to keep the model I am building dynamic and easy to use.

So my quetion is this. I have a portion of mixed air that is entering my evaporator.

My split is 80% return air and 20% outdoor air.

My outdoor air is 103F and 100%RH

My return air is 80F and 48%RH

I know how to find the mixed air temperature:

Tma = (103 X .2)+(80 X .8)= 84.6F

I also know that I can use the humidity as measured in grains/lb to find humidity:

H(gr/lb)= (75gr/lb X .8) + (325gr/lb X .2) = 125gr/lb

This all makes sense cause if I draw a line on the psyh chart between the points and then use the 80% mark away from the outdoor point I get the point that corresponds with what I jsut found.


HERE IS THE REAL QUESTION:

Is there any way to look at the humidity measured in grains/lb and use an equation and maybe a chart of values to convert it to a Relative Humidity Ratio measured in % ?

 

[DUMechEng]

Try this....

%RH=H*P/(4354*Psat+H*Psat)

Where:
Psat=3226*10^(k*(1-1165.67/T))
k=4.39553-3.469*(T/1000)+3.072*(T/1000)^2-0.883*(T/1000)^3
T=Dry Bulb Temp (deg R)
H=Humidity Ratio (gr/lb)
P=Atmospheric Pressure (psi)

if you wanted to i guess you could substitute k and Psat into the top equation to get a %RH equation based only on T, H, and atmospheric pressure.

 

[Yorkman]

The humidity ratio at saturation for a given Db temperature is proportional to the %Rh for that given Db temperature. For example at 70 degrees Db at saturation is .0158 Lbs/Lbs air when the ratio is dropped to .008 for that same Db temperature
%Rh = .008 [÷].0158 = 50.6% Or are you looking for some thing a bit more involved.













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

 

[MintJulep]

Formulas for moist air properties are well documented. ASHRAE fundamentals for starters, although Google will probably turn up both the forumulas and several user functions for excel.

 

[jasno999]

Here is what I did and the problem I needed to solve.

I was lookign at mixed air. I knew that I could use the equation:

Mixed Humidity (Grains/lb) = [Room Humidity(grains/lb) X % Recirculation air] + [Outdoor Humidity(grains/lb) X (1- % Recirculation air)]

That gave me my mixed air humidity in grains per pound.

I also could use the same equation to find my mixed air temperature.

If I knew those then what I needed to determine was my % Relative Humidity using the charts.

So what I figured out was that the humidity ration (lb/lb) was proportional to the grains per pound.

What I determined (and this may not be exact but it gets the job done) was that if you multiply the humidity ratio time 7000 then you get the grains/lb.

So for example at 80F and at saturation:

Humidity ratio = 0.0223400 (lb/lb)

0.0223400 X 7000 = 156.38 grains/lb

therfore if I have a mixed air condition at 80F and I calculate my humidity to be 112 grains/lb then I can say:

112/156.38 = 0.716 or 72%RH



IS IT OK TO DO IT THIS WAY?

 

[Yorkman]

Yes that will work. Is the conversion to grains nessecary for your problem? The same result will be achieved if you stayed with the hummidity ratios throughout and applied the math the same way. Just wondering

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

 

[jasno999]

I thought it woudl too but when I did some math it did not work out for me. It may be due to the fac that I ahve been pumping out the calculations allw week and my mind went numb.

But let me run thru a problem to prove thsi to myself.

Say I have:

80% recirculation system

Outdoor air Temp = 102F
Outdoor Air Humidity = 95%RH

Return air temp = 80F
Return air Humidity = 45%RH

Mixed temp = (102*.20)+(80*.80)= 84.4F

Mixed Humidity %RH = (95*.20)+(45*.80)= 55%RH

If I had gone to grains per pound

Gr.lb = gr/lb @ saturation X humidity ratio

Outdoor air Grains per pound = 303.597 X .95 = 288.417gr/lb

Return air Grains per pound = 156.38 X .45 = 70.371rg/lb

Using same formula- (288.417*.20)+ (70.371*.80)= 113.98gr/lb

Looking at the chart or using table data this tells me that at 84.4F and 113.98gr/lb the humidity ration will be 62.8%RH

So after doign this math I have come to determine that:

1) I am doign the calculations wrong somehow

OR

2) YOu can't use %RH in that calculation to determine the mixed air humidity. One ways says 55% and the other way says 62% - Coem to think of it you can't use %s because they are just that- they do nto tell you the full story about the total amount of humidity actually in the air - they only give the %.


Let me knwo if I am incorrect here.

 

[DUMechEng]

yes, that method will work. you just need to setup a lookup tables for saturation humidity ratio instead of using the formulas i gave you above.

you are correct, you cannot use %RH in the mixed air equation, however what yorkman is saying is that you dont need the gr/lb to lb/lb conversion. it just ads an extra step.

the way you did your example in your last post responding to yorkman is exactly correct. you didnt convert back and forth between gr/lb and lb/lb. you were doing the conversion correct however, 7000 grains = 1 pound.

also, i think you have the theory correct but you mixed up the terms humidity ratio and relative humidity in some instances. humidity ratio is the gr/lb or lb/lb number that you are talking about (both the same number just different units). its a measure of mass of moisture per mass of dry air. relative humidity is technically the ratio of partial pressure of water vapor to the partial pressure of water vapor at saturation. however, assuming an ideal gas you can use the ratio of humidity ratio to humidity ratio at saturation.

 

[jasno999]

Ok thank you.

 

[Jabba007]

Wow... where is it 103F and 100% RH?

Not thta I'm doubting that, but I just want to know where NOT to ever go!

Jabba

 

[jasno999]

YOrkman you said:

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.


Now after lookign at that in more detail I have realized that the equation only works if you use the charto find the W value. YOu had told me previously to use %RH as the W value and it woudl get me close. It does get your close but it is not exact and can really scew the numbers on you. I was hoping to use the table data to build a program that would calculate my enthalpys and my gr/lb for different temperature and humidity conditions but it appears that this can nto be done.

YOu have to use the chart or deal with numbers that are close but not exact.

 

[KiwiMace]

The ideal gas law doesn't save you from the difference between the ratios of partial pressure (pw/pws) and the ratios of absolute humidity (W/Ws). To paraphrase ASHRAE fundamentals Chap 6: Both start at zero and finish at 1 for completely dry and completely saturated, but the values in the middle differ and significantly at warmer temperatures.

I had to deal with this issue in a computer modelling exercise some years ago. At 20degC or 68degF, the error is about 0.6% at 50%. I was dealing with temps around freezing so i neglected it in the end.

If you know temp and RH, the process is as follows:
1. pws = f(T) using hyland and wexler empirical eqn#6.
2. pw = pws * RH
3. W can be determined from pw (I only know the metric eqn)
4. Same eqn for Ws from pws (eqns#22,23 in chap 6)
5. degree of saturation = W/Ws

This can be reversed, iterating the pw from W.

ASHRAE have a new book specifically on psychrometrics which is a bit easier to follow.

 

[jasno999]

I have no idea what you are saying.

 

[Yorkman]

Jasno, re-read the post I said; 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 did not say this YOu had told me previously to use %RH as the W value and it woudl get me close. It does get your close but it is not exact and can really scew the numbers on you. I was hoping to use the table data to build a program that would calculate my enthalpys and my gr/lb for different temperature and humidity conditions but it appears that this can nto be done. Further more I said it was a close representaion of what the (psychrometric) chart is providing. I have know idea what degree of accuracy you where shooting for but the method and formula does reflect what the sea level psychrometric chart shows and that was all I promised. I've tried several numbers in the formula and compaired them and see very little if any variation them.
So maybe you need to recheck your method?


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

 

[jasno999]

YOrkman. I am not trying to be a jerk I am just tryign to understand this. I appreciate all of your help but that is exactly what you said. I copied and pasted that from a post you had on the other topic were were discussing 3 weeks ago. Here I will copy your post again this is exactly what you said:

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



Now can you hlep me in understanding this again. When I look at say a temperature of 112F and 40%RH my enthalpy values and my grains per lb are close but not exact when using those equations. If I look at the PSY chart it shows the numbers different form what the equations say.

 

[insult2injury]

h = h_air + w*h_water.vapor

so approximately: h = Cp_air*T_air + w*h_sat.water

h_sat.water can be found through an equation of state or interpolation table for best accuracy. Across a small range of temperature, the vapor curve is fairly linear, so:

h_sat.water = 1061.5 + 0.435*T for T in degrees F and h_sat,water in BTU/lbm

This produces minimal error (+/- 0.2 abs) from about 15 to 120 degrees F. I hope that you are not exceeding 120 degrees.

 

[Yorkman]

Jasno,
It is my turn to apologize for getting short with you, and harder still to admit that through my ignorance I have mislead you unintentionally. The few temperatures and humidity values that I tried using in the formula that I provided did work reasonably well. Unfortunatly the ranges that you are asking for increased the degree of error.
What you need to do is first look at the CinciMace post, he/she is correct and it sounds like she/he has done a similiar calculation with success. It's probably a lot more involved than you'd like.
It looks to me like you will need to work with the Hyland and Wexler empirical equations, specifically equation 6 for temperatures between 32[°]F and 392[°]F. And then use equations 22 and 23 to find the actual humidity ratio for the desired temperature based on the partial pressure of dry air and partial pressure of water vapor, (a little bit of Daltons Law of partial pressures). Then armed with the correct W (humidity ratio) I think you can use the original formula that I gave you. All of this comes from The ASHRAE Fundamentals Handbook Chapter 6
I would like to guide you further but I have exceeded my comfort zone by about 20[°]. I truely am sorry for any difficulty I've caused you.


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

 

[Yorkman]

Insult,
You posted while I was typing my retraction (man I hate that)! I had offered that same equation not realizing, the error that occured as temperatures and %Rh became higher. Thanks to CinciMace for pointing me in the right direction. I'am going to have to dust of the text book and explore this topic a bit further.

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

 

[insult2injury]

If that's the equation he was using, he should only be seeing errors of a fractions of a percent, which is more accurate than most psychrometric evaluations.

I2I

 

[Yorkman]

This is the formula from The ASHRAE Fundamentals Handbook, like I said there is some errors when I used higher temps in the equation. As you approach 100[°]F + the enthalpy values go a bit astray,like you said. At temperatures in the "comfort zone" the accuracy is pretty close.



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

Where:
h = specific enthalpy BTU/Lbs of the moist air
t = dry bulb temperature of the indegrees F
W = humdity ratio of the moist air Lbs water/lbs dry air


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