|
Cooling Coil
Calculations Actual Air vs. Standard Air CFM |
The
question begins this way:
“The
performance of your unit is not correct.”
Why?
“Because when I calculate the coil load from the stated
conditions I do not get
the capacity shown”
Answer:
When calculating the
Total Capacity do not use Qt = 4.5 * cfm * (h1 – h2)
Because
4.5 is derived for standard air as follows:
ma = cfm *
Density * 60 where the density of standard air = .075 lba/ft³
At 100 DB and 78 WB, the
W =
.015601 lbw/lba
h = ha +
Whg = cpa*T + W*(1061 + .444*T) Btu/lba
h1
=.24*100 + .015601*(1061 + .444*100) = 41.25 Btu/lba
At 57.50 DB and 57.30
WB, the W =
.0099622 lbw/lba
h2
= .24*57.50 + .0099622*(1061 + .444*57.50) = 24.62 Btu/lba
If you use
this equation then you will get the following:
Qt = 4.5 * 15000 * (41.25 – 24.62) = 1,122,525 Btu/hr, or a perceived error of 8.3
At 100 DB and 78 WB, the
W =
.015601 lbw/lba, Density = .06914 ft³/lba
When calculating the sensible
capacity do not use Qs = 1.1 CFM *(T1-T2):
The
above shows a sea level calculation.
The
same equations apply for altitude however the true density must
include DB, WB, and PB = altitude PB. PB altitude is calculated from
the
following equation: PB
= 14.696 * (1 - ALTITUDE*6.8753E-6)^5.2559