Honeywell AUTOMATIC CONTROL SI Edition, Инструкция по эксплуатации

Страница: 452 / 512

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