6.4 Calculation by the Simplified Expression of the Regulator Area
237
6.4 Calculation by the Simplified Expression
of the Regulator Area
In order to solve the problems of regulator area selection, the expression previously
deduced for the equivalent orifice (Eq. 6.23) (Murgue 1873) is often used:
A =
1.2
√ P
Q
(6.23)
Different coefficients are applicable depending on the geometry involved. The
above expression with coefficient 1.2 is the most popular, although it is more
accurately adapted to box-type regulators.
21
Exercise 6.9 Determine the approximate area of a sliding panel regulator which is
capable of letting a flow rate of 12 m
3 s
−1 pass when the static pressures windward
and leeward of it are 530 Pa and 340 Pa, respectively.
Solution
Applying Eq. 6.24, we have:
A =
1.2
√
Q =
1.2
√
(530 − 340) Pa
· 12
m
3
s
= 1.04 m
2
It should be noted that shock losses have been neglected in this expression, leading
to the consequent error.
Exercise 6.10 The mining district in the figure is traversed by an airflow rate of
32 m
3 s
−1 when the pressure difference between the intake and the return airways
is 250 Pa. This airflow rate is to be reduced to 27 m
3 s
−1 by means of a regulator
installed in the return airway. Determine the area of the regulator.
32
250 Pa
21 Sliding panel ventilation door.
237
6.4 Calculation by the Simplified Expression
of the Regulator Area
In order to solve the problems of regulator area selection, the expression previously
deduced for the equivalent orifice (Eq. 6.23) (Murgue 1873) is often used:
A =
1.2
√ P
Q
(6.23)
Different coefficients are applicable depending on the geometry involved. The
above expression with coefficient 1.2 is the most popular, although it is more
accurately adapted to box-type regulators.
21
Exercise 6.9 Determine the approximate area of a sliding panel regulator which is
capable of letting a flow rate of 12 m
3 s
−1 pass when the static pressures windward
and leeward of it are 530 Pa and 340 Pa, respectively.
Solution
Applying Eq. 6.24, we have:
A =
1.2
√
Q =
1.2
√
(530 − 340) Pa
· 12
m
3
s
= 1.04 m
2
It should be noted that shock losses have been neglected in this expression, leading
to the consequent error.
Exercise 6.10 The mining district in the figure is traversed by an airflow rate of
32 m
3 s
−1 when the pressure difference between the intake and the return airways
is 250 Pa. This airflow rate is to be reduced to 27 m
3 s
−1 by means of a regulator
installed in the return airway. Determine the area of the regulator.
32
250 Pa
21 Sliding panel ventilation door.