Honeywell AUTOMATIC CONTROL - Quantity of Water

ENGINEERING MANUAL OF AUTOMATIC CONTROL
VALVE SELECTION AND SIZING
442
REPRINTED BY PERMISSION FROM ASHRAE HANDBOOK—
1996 HVAC SYSTEMS AND EQUIPMENT
Fig. 14. Pressure Drop Correction for
Ethylene Glycol Solutions.
Table 3. Water Flow Formula Table.
0
0.8
1.0
1.2
1.4
1.6
40 80 120 160
PRESSURE DROP CORRECTION FACTOR
TEMPERATURE, °F
ETHYLENE
GLYCOL
SOLUTION
WATER
M12226
10%
20%
30%
40%
50% BY MASS
0
0.8
1.0
1.2
1.4
1.6
40 80 120 160
PRESSURE DROP CORRECTION FACTOR
TEMPERATURE, °F
PROPYLENE
GLYCOL
SOLUTION
WATER
M12227
10%
20%
30%
40%
50% BY MASS
REPRINTED BY PERMISSION FROM ASHRAE HANDBOOK—
1996 HVAC SYSTEMS AND EQUIPMENT
Fig. 15. Pressure Drop Correction for
Propylene Glycol Solutions.
QUANTITY OF WATER
To find the quantity of water (Q) in gallons per minute use
one of the following formulas:
1. When Btu/hr is known:
Where:
Btu/hr = Heat output.
K=Va lue from Table 3; based on temperature of
water entering the coil. The value is in
pounds per gallon x 60 minutes per hour.
TD
w
=Temperature difference of water entering
and leaving the coil.
Q =
Btu/hr
K x TD
w
2. For hot water coil valves:
Where:
cfm = Airflow through the coil.
1.08 = A scaling constant. See Note.
TDa = Temperature difference of air entering and
leaving the coil.
K=Va lue from Table 3; based on temperature of
water entering the coil (pounds per gallon x
60 minutes per hour).
TDw = Temperature difference of water entering
and leaving the coil.
NOTE: The scaling constant 1.08 is derived as
follo ws:
Q =
cfm x 1.08 x T
D
a
K x TD
w
Where:
1 lb air
13.35 ft
3
=
The specific volume of air at standard
conditions of temperature and atmospheric
pressure.
Simplifying the equation:
To find the scaling constant for air conditions other
than standard, divide 14.40 Btu by specific volume of
air at those conditions.
3. For fan system chilled water coil valves:
Where:
cfm = Airflow through the coil.
Btu/lb = Heat per pound of dry air removed. Includes
both sensible and latent heat.
113 = A scaling constant.
TDw = Temperature difference of water entering
and leaving the coil.
1.08 =
0.24 BTU
lb air ˚F
x
60 min
1 hr
x
1 lb ai
r
13.35 ft
3
1.08 =
14.40 Btu mi
n
˚F hr 13.35 f
3
Q =
cfm x Btu/l
b
113 x TD
w
Water
Temp F K
Water
Temp F K
40
60
80
100
120
150
180
502
500
498
496
495
490
487
200
225
250
275
300
350
400
484
483
479
478
473
470
465