6.2 Fans
203
Exercise 6.1 The main mine fan shown in the figure operates at a pressure of 1.5 kPa.
Of all this pressure, 0.5 kPa is used in overcoming the combined resistance of the two
parallel galleries (A and B). The air quantity circulating through gallery A is 20 m
3 s
−1
and through gallery B 15 m
3 s
−1 . You are asked to determine the maximum pressure
at which the booster fan can operate before the airflow inside gallery A becomes
zero (critical pressure). Assume that after installation of the booster fan, the main
fan continues to operate at the same pressure.
D
D
R s 1
R s 2
Solution
First, the resistance of both shafts
R s 1 −s 2
can be calculated:
R s 1 −s 2 =
P s 1 −s 2
Q
2
T
=
(1500 − 500) Pa
(20 + 15)
m 3
s
2 = 0.816
N s
2
m 8
Then, the resistance of gallery B can be obtained in the same way:
R B =
P B
Q
2
B
=
500 Pa
15
m 3
s
2 = 2.22
N s
2
m 8
When the air quantity through gallery A is zero, the pressure difference between
both its ends is also zero. This fact implies that all the pressure supplied by the main
fan is used in overcoming the resistances of the shafts, therefore:
P T = R s 1 −s 2 Q
T ; Q
T =
P T
R s 1 −s 2
=
1500 Pa
0.816
N s 2
m 8
= 42.87
m
3
s
The air quantity that circulates through gallery B after installing the booster fan
(Q
B ) is the same as the total quantity circulating through the shafts (Q
T ). This is so
203
Exercise 6.1 The main mine fan shown in the figure operates at a pressure of 1.5 kPa.
Of all this pressure, 0.5 kPa is used in overcoming the combined resistance of the two
parallel galleries (A and B). The air quantity circulating through gallery A is 20 m
3 s
−1
and through gallery B 15 m
3 s
−1 . You are asked to determine the maximum pressure
at which the booster fan can operate before the airflow inside gallery A becomes
zero (critical pressure). Assume that after installation of the booster fan, the main
fan continues to operate at the same pressure.
D
D
R s 1
R s 2
Solution
First, the resistance of both shafts
R s 1 −s 2
can be calculated:
R s 1 −s 2 =
P s 1 −s 2
Q
2
T
=
(1500 − 500) Pa
(20 + 15)
m 3
s
2 = 0.816
N s
2
m 8
Then, the resistance of gallery B can be obtained in the same way:
R B =
P B
Q
2
B
=
500 Pa
15
m 3
s
2 = 2.22
N s
2
m 8
When the air quantity through gallery A is zero, the pressure difference between
both its ends is also zero. This fact implies that all the pressure supplied by the main
fan is used in overcoming the resistances of the shafts, therefore:
P T = R s 1 −s 2 Q
T ; Q
T =
P T
R s 1 −s 2
=
1500 Pa
0.816
N s 2
m 8
= 42.87
m
3
s
The air quantity that circulates through gallery B after installing the booster fan
(Q
B ) is the same as the total quantity circulating through the shafts (Q
T ). This is so