8.18 Sizing the Auxiliary Fan
359
(d) So the shaft power of the fan:
Pw shaft =
FTP Q v
η T
=
459.10 Pa · 10
m
3
s
0.7
= 6558.57 W = 6.6 kW
Exercise 8.16 The construction, by drilling and blasting, of a main gallery is planned.
This gallery will have a 30 m
2 cross section, 17 m perimeter and a final length of
200 m with K 1.2 = 0.015 N s
2 m
−4 . The following machinery will be used during
the works:
• 1 × 150 kW loader,
• 1 backhoe loader of 100 kW,
• 1 jumbo of 120 kW,
• 1 telescopic boom lift of 60 kW,
• 3 dumpers of 90 kW each,
• 1 gunite robot of 50 kW, and
• 1 concrete mixer truck of 100 kW.
The maximum number of people working on the face at the same is estimated
to be 5, and an optimum air velocity in the gallery has been specified as 1 m s
−1 .
The ducts used in the ventilation system will be made of glass wool (K 1.2 = 0.0028
Ns
2 m
−4 ) with joints every 50 m. Each joint creates pressure losses equivalent to
those produced by 2 m of duct length. The system extends 5 m into the existing main
gallery by means of a 90º elbow (as in the case of the Exercise 8.15). The fan set up
is shown in the figure below:
Silencer
Silencer
Tests performed by the manufacturer did not include shock losses in the bellshaped inlet. The fan has a diameter of 1.27 m with a total efficiency of 83% (obtained
for the conditions outlined in this exercise). The duct is 1.3 m in diameter.
Determine:
(a) Necessary airflow rate to supply adequate fresh air for operators to breathe.
(b) Minimum airflow rate to return the gases through the gallery at the specified
minimum speed.
(c) Airflow rate required to dilute the engine exhaust gases assuming a combination
of 1 loader, 1 backhoe and 3 dumpers working at the same time.
(d) Airflow rate necessary to dilute gases generated by the explosives used in
blasting.
(e) The total airflow required at the gallery face.
(f) Airflow rate to be provided by the fan (estimate 25% leak rate at the joints).
(g) Air velocity at the duct outlet.
(h) Dynamic pressure at the face.
359
(d) So the shaft power of the fan:
Pw shaft =
FTP Q v
η T
=
459.10 Pa · 10
m
3
s
0.7
= 6558.57 W = 6.6 kW
Exercise 8.16 The construction, by drilling and blasting, of a main gallery is planned.
This gallery will have a 30 m
2 cross section, 17 m perimeter and a final length of
200 m with K 1.2 = 0.015 N s
2 m
−4 . The following machinery will be used during
the works:
• 1 × 150 kW loader,
• 1 backhoe loader of 100 kW,
• 1 jumbo of 120 kW,
• 1 telescopic boom lift of 60 kW,
• 3 dumpers of 90 kW each,
• 1 gunite robot of 50 kW, and
• 1 concrete mixer truck of 100 kW.
The maximum number of people working on the face at the same is estimated
to be 5, and an optimum air velocity in the gallery has been specified as 1 m s
−1 .
The ducts used in the ventilation system will be made of glass wool (K 1.2 = 0.0028
Ns
2 m
−4 ) with joints every 50 m. Each joint creates pressure losses equivalent to
those produced by 2 m of duct length. The system extends 5 m into the existing main
gallery by means of a 90º elbow (as in the case of the Exercise 8.15). The fan set up
is shown in the figure below:
Silencer
Silencer
Tests performed by the manufacturer did not include shock losses in the bellshaped inlet. The fan has a diameter of 1.27 m with a total efficiency of 83% (obtained
for the conditions outlined in this exercise). The duct is 1.3 m in diameter.
Determine:
(a) Necessary airflow rate to supply adequate fresh air for operators to breathe.
(b) Minimum airflow rate to return the gases through the gallery at the specified
minimum speed.
(c) Airflow rate required to dilute the engine exhaust gases assuming a combination
of 1 loader, 1 backhoe and 3 dumpers working at the same time.
(d) Airflow rate necessary to dilute gases generated by the explosives used in
blasting.
(e) The total airflow required at the gallery face.
(f) Airflow rate to be provided by the fan (estimate 25% leak rate at the joints).
(g) Air velocity at the duct outlet.
(h) Dynamic pressure at the face.