Honeywell AUTOMATIC CONTROL SI Edition, Engineering Manual

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ENGINEERING MANUAL OF AUTOMATIC CONTROL
VALVE SELECTION AND SIZING
442
QUANTITY OF STEAM
To find the quantity of steam (Q) in kilograms per hour use
one of the following formulas:
1. When heat output is known:
Simplifying:
STEAM VALVE PRESSURE DROP
Proportional Applications
When specified, use that pressure drop (
∆
P) across the valve.
When not specified:
1. Calculate the pressure drop (
∆
P) across the valve for good
modulating control:
∆
P = 80% x (P
1
– P
2
)
NOTE: For a zone valve in a system using radiator orifices
use:
∆
P = (50 to 75)% x (P
1
– P
2
)
Where
P
1
= Pressure in supply main in kPa absolute.
P
2
= Pressure in return in kPa absolute.
2. Determine the critical pressure drop:
∆
P
critical = 50% x P
1
a
Where:
P
1
a = Pressure in supply main in kPa (absolute
pressure)
Use the smaller value
∆
P or
∆
Pcritical when calculating Kv.
Two-Position Applications
Use line sized valves whenever possible. If the valve size
must be reduced, use:
∆
P = 20% x (P
1
-P
2
)
Where
P
1
= Pressure in supply main in kPa absolute.
P
2
= Pressure in return in kPa absolute.
Where:
Heat output = Heat output in kJ/kg.
2325 kJ/kg = The approximate heat of vaporization of
steam.
2. For sizing steam coil valves:
Where:
m 3 /s = Volume of air from the fan in cubic meters
per second.
∆
Ta = Temperature difference of air entering and
leaving the coil.
4330 = A scaling constant.
2325 kJ/kg = The approximate heat of vaporization of
steam.
3. For sizing steam to hot water converter valves:
Q = L/s x DTw x 6.462
Where:
L/s = Water flow through converter in liters per
second.
∆
Tw = Temperature difference of water entering
and leaving the converter.
6.462 = A scaling constant.
4. When sizing steam jet humidifier valves:
Where:
W1 = Humidity ratio entering humidifier, grams of
moisture per kilogram of dry air.
W2 = Humidity ratio leaving humidifier, grams of
moisture per kilogram of dry air.
1.2045 kg/m
3
= The specific weight of air at standard
conditions of temperature (20
°
C) and
atmospheric pressure (101.325 kPa).
m 3 /s = Volume of air from the fan in cubic meters
per second.
3600 s/h = A conversion factor from second to hours.
Q =
Heat Output
2325 kJ/kg
Q =
m3/s x
∆
Ta x 4330
2325 kJ/kg
Q =
( )
W1 – W 2 g moisture
kg air
•
1.2045 kg air
m3
•
m3
s
•
3600s
h
Q = 4.3362
( )
W1 – W 2 kg moisture
hr