Answer:
Required diameter of hose pipe = 0.2864 mm
Solution:
From the continuity eqn, the fluid flow rate is given by:
Av = [tex]\frac{V}{t}[/tex]
where
A = cross-sectional area = [tex]\pi r^{2}[/tex]
r = hose pipe radius
v = velocity of gas
Also, [tex]v = \frac{displacement, d}{time, t}[/tex]
Using:
1 gallon = 3.854 l
1 mile = 1609.34 m
[tex]1 m^{3} = 1000 l[/tex]
Therefore,
[tex]A\frac{d}{t} = \frac{V}{t}[/tex]
[tex]\pi r^{2} = \frac{V}{d}[/tex]
[tex]\pi r^{2} = \frac{(1 gal).(\frac{3.7854 l}{gal}).(\frac{10^{- 3} m^{3}}{l})}{37 miles(\frac{1609.34 m}{miles})}[/tex]
[tex]6.357\times 10^{- 8} = \pi r^{2}[/tex]
[tex]r^{2} = 2.024\times 10^{- 8}[/tex]
[tex]r = 1.423\times 10^{- 4} m = 0.1423 mm[/tex]
The diameter of the hose pipe = 2r = [tex]2\times 1.423\times 10^{- 4}[/tex]
The diameter of the hose pipe = [tex]2.846\times 10^{- 4} m = 0.2846 mm[/tex]