180
5 Main Ventilation
R =
P n
Q 2
n
=
P n
250
m 3
min
2
Equalizing:
P n
250
m 3
min
2 =
6000 Pa + P n
800
m 3
min
2
Therefore:
P n = 6000 Pa
250
m
3
min
2
800
m 3
min
2 −
250
m 3
min
2 = 649.35 Pa
Operating the main fan at two different speeds
In this case, the resistance is the same for both speeds. Therefore, for the main fan
drift we have:
R =
P v 1 + P n
Q
2
1
=
P v 2 + P n
Q
2
2
where the subscripts 1 and 2 represent the two rotation speeds to which the motor
fan is subjected.
So, in this case, the P n will be (Eq. 5.13):
P n =
P v 1 Q
2
2 − P v 2 Q
2
1
Q
2
1 − Q
2
2
(5.13)
Exercise 5.6 The main fan in a mine moves 1000 m
3 min
−1 of air operating at a
pressure of 8000 Pa. When its rotational speed is reduced, the flow drops to 700 m
3
min
−1 and the pressure developed is 3500 Pa. Determine the NVP of this mine.
Solution
For the first rotation velocity, we have:
R =
P v 1 + P n
Q
2
1
=
8000 Pa + P n
1000
m 3
min
2
For the second velocity:
5 Main Ventilation
R =
P n
Q 2
n
=
P n
250
m 3
min
2
Equalizing:
P n
250
m 3
min
2 =
6000 Pa + P n
800
m 3
min
2
Therefore:
P n = 6000 Pa
250
m
3
min
2
800
m 3
min
2 −
250
m 3
min
2 = 649.35 Pa
Operating the main fan at two different speeds
In this case, the resistance is the same for both speeds. Therefore, for the main fan
drift we have:
R =
P v 1 + P n
Q
2
1
=
P v 2 + P n
Q
2
2
where the subscripts 1 and 2 represent the two rotation speeds to which the motor
fan is subjected.
So, in this case, the P n will be (Eq. 5.13):
P n =
P v 1 Q
2
2 − P v 2 Q
2
1
Q
2
1 − Q
2
2
(5.13)
Exercise 5.6 The main fan in a mine moves 1000 m
3 min
−1 of air operating at a
pressure of 8000 Pa. When its rotational speed is reduced, the flow drops to 700 m
3
min
−1 and the pressure developed is 3500 Pa. Determine the NVP of this mine.
Solution
For the first rotation velocity, we have:
R =
P v 1 + P n
Q
2
1
=
8000 Pa + P n
1000
m 3
min
2
For the second velocity: