Many times the use of a psychrometric chart can help us make an analysis of a HVAC systemsÆ performance, or help us predict how well a design will function under different operating conditions. This FAQÆs main focus is to provide some tips on how to use a psychrometric chart to calculate typical heating, cooling/dehumidification, and mixed air problems encountered in the field or at the early stages of design.
This exercise assumes the use of an ASHRAE sea level 29.92ö Hg chart.
Question 1:
I'm starting with an outdoor condition of 95 degrees F and 80% Relative Humidity and I'm mixing that air with return air that is at 80 degrees F and 50% relative humidity. If the mixed air is 80% return air and 20% outside air, how do I find the correct mixed air temperature and % Rh and graph this on the psychrometric chart properly?
To find the solutions for this question I would approach the problem like this:
1. Plot OSA point on chart at 95 degrees 80% Rh
2. Plot RA point on chart at 80 degrees 50% Rh
3. Connect the OSA point and the RA points to form the mixed air line.
4. Measure length of line 5.625" x 20% = 1.125ö(This measurement will depend on the size of your chart I'm using an 11" x 17" yours may be smaller)
5. Measure along the MA line starting from the RA point 1.125" this is your MA temperature and humidity. Read 83 degrees 60% Rh
6. Or use the following simple formula to find the mixed air Db temperature.
(R.A. degrees F x %) + (OSA degrees F x %) = Mixed Air degrees F, plot this on the MA line to find the corresponding %Rh
Question 2:
How do I figure out the required supply air temperature for the room? I know that I need to find the structural heat gains, electrical heat gain, occupant load and infiltration load. I can add the sensible heat and latent heat gains for the space and figure out what the total heat gain for the room is. But how do I figure out the necessary supply air temperature and % Rh to maintain that room at say 80F and 50% Rh?
To find this answer 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 .65 based on your rooms' sensible heat loads and latent heat loads: (sensible heat + latent heat = total heat) so that sensible heat divided by total heat = The Sensible Heat Ratio (SHR)
1. Draw a line from the center point on the base line of the protractor through the .65 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 (the RA point) and travels to the left at the .65 SHR slope and intersects the 95% curve on the chart, at 95%Rh (read 50 degrees Db).
4. The point of intersection will establish the required supply air temperature (the SA point) 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.
Question 3:
Now that I know what the required supply air temperature is how do figure out what my CFM requirements are?
Once you have established your supply air temperature, use the supply air temperature and the desired room air temperature to find the CFM, use the following formula:
1. BTU/Hr = 1.08 x CFM x [Δ]T with the known sensible heat load for BTU/HR, T[sub]1[/sub] as the space temperature and T[sub]2[/sub] as supply temperature, you can now solve for the CFM.
2. Another method would involve the total heat formula:
BTU/Hr = 4.5 x CFM x [Δ]H, this requires you to find the enthalpy of the return air and supply air from the psychrometric chart.
3. Connect a straight line from the RA point to the SA point.
4. Form a right triangle using the SA point and the RA points, making a horizontal line to the right from the SA point and intersecting it with a vertical line down from the RA point.
5. Locate enthalpy values at the SA point, the RA point, and at the 90 degree intersection of the horizontal and vertical lines of the triangle. These enthalpy lines run diagonally care must be used in accurately locating.
6. At SA point read 19.8 BTU/Lbs, at RA point read
31.4 BTU/Lbs, at 90 degree intersection read 27.6 BTU/Lbs
7. The difference between 27.6 and 19.8 is sensible heat BTU/Lbs, the difference between 31.4 and 27.6 is latent heat BTU/Lbs, and the total heat is the difference between 31.4 and 19.8 this is the load in the room BTU/Lbs.
8. Using the known total heat load BTU/Hr for the room, the return air enthalpy and supply air enthalpy find the[Δ]H, apply to: BTU/Hr = 4.5 x CFM x [Δ]H to find the CFM
Question 4.
In my original problem I was bringing in 20% OSA at 95 degrees and 80% Rh, to achieve a mixed air temperature of 83 degrees 60 %Rh does this affect the load in the room?
Simply stated NO! The mixed air temperature of 83 degrees 60 %Rh is higher than the desired space air temperature, it does put an additional load on the cooling coil but the room never sees this added load. So we must size the coil to handle this extra heat. HereÆs a method using the psychrometric chart to find the load on the coil.
1. Locate the MA point on the mixed air line and find the enthalpy corresponding to this point (read 36 BTU/Lbs)
2. Locate the SA point on the chart and find the corresponding enthalpy read (19.8 BTU/Lbs)
3. Using the difference of these two values and the CFM that we found in question 3 apply them to the Total Heat formula and solve for total BTU/Hr:
BTU/Hr = 4.5 x CFM x [Δ]H to find Total BTU/HR
4. Notice in this problem the total heat the coil sees is going to be more than the load of the room, the CFM will remain the same.
Question 5.
Is there a way to find the amount of moisture that the coil is removing from the air?
Yes there is, the psychrometric chart that weÆve been using lists these values as, Lbs of water/ Lbs of dry air. This is referred to as the humidity ratio W. Some charts list this as grains of moisture; note there are 7000 grains/lbs. HereÆs how to find the answer using our problem as an example.
1. Locate the MA point on the chart and extend a horizontal line to the right side of the chart and find the MA humidity ratio W (read .0145)
2. Locate the SA point on the chart and extend a horizontal line to the right side of the chart and find the SA humidity ratio W (read .0072)
3. Take the difference (.0073) and use this modified formula:
Lbs/Hr = 4.5 x CFM x [Δ]W to find the pounds of water per hour removed.
Let me finish off by saying, this site is my favorite source for technical information that I have found. I want to thank Quark for encouraging me to post this as a FAQ. I only hope I didn't come up too short. Please feel free to provide any added data that I may have over looked.
