SOLUTION FOR ACR 3933 2ND ASSIGNMENT - PSYCHROMETRICS

Question 1:
The air vapor mixture leaving the saturator will be saturated at 76 F. The relative humidity at saturation is 100%.

Question 2:
Sensible Heat Removed = 1.1*1 000cfm * (80-50) = 33 000 Btu/hr
Latent Heat removed = 0.69 * 1 000cfm * (77-54) = 15 870 Btu/hr
The factor 1.1 used in the sensible heat formula is a combination figure which converts cfm to lbs. per hour, and also takes into account the amount of heat required to raise 1 lb. of air 1 degree in temperature.
The factor 0.69 used in the latent heat formula is also a combination figure which converts cfm into lbs. per hour and also takes into account the amount of heat removed to condense moisture.

Question 3:
The mixture will be closest to the larger air quantity. In this case, it is the 80 DB condition. The total cfm in the mixture is 2 000 cfm. 500 cfm of outdoor air represents 500/2,000 or 1/4 of the mixture. The temperature difference between point 1 (80 DB) and point 2 (96 DB) is 96-80 or 16 degrees. One-quarter of 16 is 4 degrees. Therefore, the mixture is at 80+4 or 84 DB or point 3.
By locating 84F on line 1-2, we can determine the other properties by using the psychrometric chart.
a. DB = 84 F
b. WB = 69.2 F
c. DP = 62 F
d. Spec. hum. = 83.75 grains.

Remarks:
The temperature of the mixture can also be determined by calculation as follows:
tm = [(cfm1*t1)+(cfm2*t2)]/(cfm1 + cfm2) = [(1500*80)+(500*96)]/(1500+500) = 84 F