Signed For Easy Mounting. One-In. Selection Procedure - Carrier 42BH Series Product Data

Selection procedure
Example 1: Selection for specified total
cooling
I Determine job requirements:
Given:
Total Cooling Load . . . . . . . . . . . . . . . . . . 67.0 MBtuh
Sensible Cooling Load . . . . . . . . . . . . . . . 41.0 MBtuh
Air Delivery . . . . . . . . . . . . . . . . . . . . . . . . . . 1800 Cfm
Entering Air Temperature . . . . 75.0 F db, 64.5 F wb
Entering Water Temperature. . . . . . . . . . . . . . . . . 43 F
External Static Pressure. . . . . . . . . . . . . . . 0.30 in. wg
Maximum Chilled Water Flow Rate . . . . . . 15.0 Gpm
I Select unit based on required air delivery.
From Fan Performance table on page 22, select a unit
that delivers required airflow. Model 42BH size 20 de-
livers 1800 cfm.
I Determine water temperature rise.
This step is used to determine the actual water tem-
perature rise required to provide the specified cooling
capacity.
(This can be done to meet either specified sensible or
specified total capacity. For this example, we will de-
termine water temperature rise required to meet the
specified total capacity after it has been corrected to
the rating curve capacity. The resulting rating curve
sensible capacity will be determined in Step IV and then
corrected to actual capacity. If the sensible capacity
were the critical requirement, the reverse procedure
would be followed.)
A. Determine the air delivery correction factors using
actual unit air delivery (1800 cfm).
Enter the 42BH Cooling Capacity Correction Fac-
tors table, page 7, at 1800 cfm and find that a
42BH size 20 unit has a total heat (TH) factor of
0.94 and a sensible heat (SH) factor of 0.93.
B. Correct the specified total capacity to the required
rating curve capacity.
Rating Curve Capacity =
C. Enter the cooling capacity curve for a size 20 unit
with standard 4-row coil at 71.3 MBtuh and project
vertically to the specified entering wet bulb (65 F).
Project from this point horizontally to the right to
the specified entering water temperature (43 F) and
then vertically to read water temperature rise of 9.2
degrees F.
I Determine the rating curve sensible capacity and
the actual sensible capacity..
A. Enter the cooling curve at the intersection of total
capacity (71.3
bulb temperature (65 F). Read horizontally to the
left to the specified dry bulb (75 F). Read down to
the rating curve sensible capacity of 46.0 MBtuh.
Specified Total Cap.
TH Factor
67.0 MBtuh
=
0.94
= 71.3 MBtuh
MBtuh) and entering
B. Correct the rating curve capacity to actual
capacity:
Actual Sensible Cooling
= SH x Rating Curve Sensible Cooling
= 0.93 (From Step III A) x 46.0 MBtuh
= 42.8 MBtuh
I Determine water flow and pressure drop.
A.
Actual Gpm =
B. Pressure drop is obtained from the curve on page
18 for the 4-row cooling coil and is 10 ft of
water.
I Find Leaving Air Dry Bulb (LDB) and Leaving
Wet Bulb (LWB) Temperatures.
A.
LDB = EDB −
= 75
= 53.0 F
B. Leaving wet bulb is computed by referring to the
Enthalpy at Saturation table on page 21. First de-
termine that the enthalpy at saturation for the en-
tering (h
ent
of dry air.
h = h
lvg ent
= 29.69 −
= 21.33
Again referring to the table, the temperature cor-
responding to 21.33, by interpolation, is 51.8 F,
which is the leaving wet bulb.
I Verify that the selected unit will provide the re-
quired external static pressure of 0.30 in. wg
with the selected coil, options, and ductwork.
A. Determine total static pressure. For an application
with ductwork, the total static pressure is equal to
the sum of the external static pressure, duct grilles,
and component static pressures.
Assume for this example that the selected 42BH
size 20 unit with 4-row cooling coil has a double
deflection grille and throwaway filter. Enter the
42BH Component Static Resistance table (page 21)
at unit size 20, 1800 cfm, and find the static resis-
tance (in. wg) for the following components:
Cabinet
wet 4-Row Coil
Throwaway Filter
Double Deflection Grille
Actual Total Cooling
.5 x Temp Rise
67.0
=
.5 x 9.2 (From Step III C)
= 14.6
Actual Sensible MBtuh
Actual Cfm x .00108
42.8 (Step IV B)
−
1800 x .00108
) wet bulb of 64.5 is 29.69 Btu per lb
Actual Total Capacity
−
.00445 x Actual Cfm
67.0
.00445 x 1800
0.12
0.37
0.12
+ 0.07
0.68
5