Product Overview

Page ___________ of ______________
Form C-HU - Page 5

To obtain Pressure Loss Feet of water for other GPM see the graphic data on page 6.
Example: Unit Reznor Model 23/33
Entering Water Temperature (EWT) 160°F
Entering Air Temperature (EAT) 40°F
Water Temperature Drop (WTD) 10°F
The heating output of any particular installation is a function of many different factors. It is very seldom that any
installation will exactly match the conditions described in the tables on the previous page. For those installa-
tions, correction factors must be used to determine heating output and other values.
Below is an example of conditions different from those given in TABLE A and B on the previous page. Fol-
lowing are procedures for determining heating output and other values at conditions other than “cataloged”
conditions.
I. In TABLE B nd the Heating Capacity for “catalog” conditions with High Speed Fan Setting 24,000 BTUH
II. Determine Heating Capacity for EWT at 160°F and EAT of 40°F
24,000 BUTH x 0.846 = 20,304 BTUH
Find the correction factor in TABLE C that satises the conditions listed. In this instance, it is 0.846.
Multiply original BTUH output by the correction factor.
III. Determine Heating Capacity for WTD of 10°F
20,304 BTUH x 1.15 = 23,350 BTUH
Find the correction factor in TABLE D that satises the conditions listed. In this instance it is 1.15.
Multiply BTUH output by the correction factor.
IV. Determine Gallons per Minute (GPM) at 200°F EWT, 60°EAT, but with WTD of 10°F
2.42 GPM
Find the GPM from TABLE B for “catalog” conditions with High speed Fan Setting
In TABLE D nd the GPM Correction Factor for WTD of 10°F. In this case it is 2.30. Multiply original
GPM by the correction factor.
2.42 GPM x 2.30 = 5.57 GPM
Note: This formula applies only to units with 200°F EWT and 60°F EAT. For all other applications, use
the formula shown (right):
GPM = BTUH ÷ (500 x WTD)
Determine GPM for installation described in step III above at 10°F 23350 ÷ (500 x 10°F WTD) = 4.67 GPM
V. Determine Water Pressure Drop (WPD) in Feet of Water at 10°F WTD
4.67 GPM
Find the GPM from step IV above
On the Heat Exchanger Resistance Chart on page 6, nd the WPD at 4.67 GPM on the left side axis.
Follow it until it meets the line for Model WS23/33. From that point, follow the line down to the bottom
axis to determine the WPD at 176°F mean water temperature.
0.14 FT H
2
O (as marked)
Determine the Correction Factor (K).
(160°F + 150°F) ÷ 2 = 155°F
The above example started with an EWT of 160°F and WTD of 10°F. That would result in water
temperature at 150°F as it leaves the heater. Find the mean (average water temperature).
Find the Correction Factor (K) for the value nearest 155°F. At 158°F the Correction Factor (K) is 1.05.
Multiply 1.05 by the WPD found on the chart 0.14 FT H
2
O.
0.14 FT H
2
O x 1.05 = 0.147 FT H
2
O
VI. Determine Heating Capacity for water ow rate of 3.03 GPM
24,000 BUTH
Determine the Heating Capacity from TABLE B for “catalog” conditions with High Speed Fan Setting
Divide actual ow rate in GPM by cataloged ow rate found in TABLE B. 3.03 GPM ÷ 2.42 GPM = 125%
In TABLE E nd the MBH Correction Factor for a ow rate of 125%. In this case it is 1.04. Multiply
original MBH by the correction factor.
24,000 BTUH x 1.04 = 24,960 BTUH
VII. Determine Leaving Air Temperature (LAT) using the formula shown (right): LAT = EAT+ BTUH ÷ (CFM x 1.085)
In TABLE B nd the Air Volume (cfm) for “cataloged” model and apply it conditions described in Step
III above.
40°F + (23,350 BTUH ÷ (500 cfm x 1.085)) = 83°F
ENGINEERING DATA (cont’d)
HOT WATER CAPACITIES, CALCULATIONS AND CORRECTION FACTORS