Selection Procedure - Carrier 42BH Series Product Data

Selection procedure (cont)
Component Static Resistance
External Static Pressure
Total Static Pressure
B. Refer to the Fan Performance table on page 22
and determine that a size 20 unit will deliver 1800
cfm at 1.0 total static pressure with a
tor and mid-range fan rpm of 840.
NOTE: The rpm's shown in the Fan Performance
table are for reference only. The appropriate drives
will be selected at the factory based on specified
cfm and total static pressure.
Example 2 — Determining unit capacity for
specified water flow (gpm)
I Determine job requirements.
Given:
Maximum Chilled Water Flow Rate . . . . . . 15.0 Gpm
Entering Water Temperature. . . . . . . . . . . . . . . . . 42 F
Air Delivery . . . . . . . . . . . . . . . . . . . . . . . . . . 1800 Cfm
Entering Air Temperature. . . . . . . . 70 F db, 65 F wb
External Static Pressure. . . . . . . . . . . . . . . 0.30 in. wg
I Make initial selection of 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 the air delivery correction factors us-
ing actual unit air delivery.
Enter the 42BH Cooling Capacity Correction Factors
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 sen-
sible heat (SH) factor of 0.93.
I Determine water temperature rise.
Arbitrarily select a total cooling capacity and apply
the correction factor to obtain a beginning rating curve
capacity.
Rating Curve Capacity =
Enter the cooling capacity curve for a size 20 unit with
standard 4-row coil at 68.0 MBtuh and project verti-
cally to the specified entering wet bulb (65 F). Project
from this point horizontally to the right to the speci-
fied entering water temperature (42 F) and then verti-
cally to read water temperature rise of 11.0 degrees F.
6
0.68
+ 0.30
0.98
3 ⁄ hp mo-
4
Selected Total Capacity
TH Factor
64.0 MBtuh
=
0.94
= 68.0 MBtuh
I Calculate actual water flow for the selected to-
tal capacity by using the following formula:
Actual Total Cooling
Actual Gpm =
0.5 Temp Rise
64.0
=
0.5 x 11.0 (From Step IV)
= 11.6
Since 11.6 is less than the specified gpm, another
higher total cooling capacity must be selected and Steps
IV and V repeated until the resulting actual gpm matches
the specified gpm.
I Determine rating curve sensible capacity and the
actual sensible capacity.
Once the specified gpm is matched and the total cool-
ing capacity is known, the actual sensible capacity may
be determined using Step IV of Example 1. The water
pressure drop may be read from the water pressure
drop curve.
Example 3: Heating selection
I Determine job requirements. (Unit previously se-
lected for cooling.)
Given:
Heating Load . . . . . . . . . . . . . . . . . . . . . . . 54.0 MBtuh
Entering Water Temperature . . . . . . . . . . . . . . . 110 F
Entering Air Temperature . . . . . . . . . . . . . . . . 70 F db
Hot Water Flow Rate (same as cooling) . . . 14.6 gpm
I Determine actual heating capacity.
Enter Heating Capacity curve (page 23) for 42BH size
20 unit with a 4-row coil at 14.6 gpm and find 168.0
MBtuh rated heating capacity.
Refer to Correction Factors tables on pages 7 and 8.
Actual Heating Capacity = Rated Capacity
This slightly exceeds required heating capacity of 54.0
MBtuh.
IMPORTANT: Check to ensure leaving air tempera-
ture does not exceed 104 F.
MBtuh x 1000
Leaving =
Air Temp
1.08 x cfm
56.7 x 1000
=
1.08 x 1800
= 99 F.
Leaving air temperature is within the limit. (Consult
factory for special motors if the temperature exceeds
the limit.)
x SH Factor (from
Cooling Capacity
Correction Factors)
x Hot Water Capacity
Correction Factor
= 168.0 x 0.93 x 0.363
= 56.7
+ Entering
Air Temp
+ 70