Answer:
a) 149,270 m³
b) 4 hours 9 minutes
Explanation:
You want to know the volume of air in a 4 km tunnel 8 m high and 5 m wide with a semicircular cross section at the top. And you want to know the replacement time for that air if it is exchanged at 10 m³ per second.
Area
The area of the semicircular top of the tunnel is ...
A = π/8d²
A = π/8·(5 m)² = 25π/8 m²
The area of the rectangular bottom of the tunnel is ...
A = LW
A = (5 m)(8 -2.5 m) = 27.5 m²
So the total cross sectional area of the tunnels is ...
(25π/8 +27.5) m² ≈ 37.317477 m²
a) Volume
The volume of the tunnel is ...
V = Bh
V = (37.317477 m²)(4000 m) ≈ 149269.9 m³
The volume of the air in the tunnel is about 149269.9 m³.
b) Exchange time
At 10 m³ per second, the air will be replaced after ...
(149269.9 m³)/(10 m³/s) = 14926.99 s ≈ 4 hours 9 minutes
The given pump takes 4 hours 9 minutes to replace the air in the entire tunnel.
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Additional comment
This tunnel would not meet any standard of safety.
A typical building is supposed to have about 5 air exchanges per hour. That's about 21 times the rate given here. Some tunnel structures need to have the air exchanged once per minute. (Here, that would mean the wind in the tunnel is 149 mph.)
The breathing requirements of the tunnel occupants need to be considered, as well as removal of air pollutants.
There are 3600 seconds in 1 hour.
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