A vertical cylindrical container contains 5590 gallons of gasoline and is 1.49 m

Question

A vertical cylindrical container contains 5590 gallons of gasoline and is 1.49 m in radius. Due to evaporation within the tank, the pressure on the top of the fluid is 2.5 times normal atmospheric pressure.  The density of gasoline is 737 kg/m3.

What is the pressure at the bottom of the tank? If a nozzle at the bottom of the tank is used to let fuel out, with what velocity would the gasoline exit the hole? If a nozzle at the bottom of the tank is used to let fuel out, what volume flow rate will result if the diameter of the hole is 1.2 cm? At what initial rate will the height of the fuel level in the tank drop as fuel is removed?

Explanation / Answer

A vertical cylinder radius=1.49 m

Volume of gasoline=5590 gallons=21.16 m3

density=  737 kg/m3.

weight of gasoline=Volume*density*9.8=21.16*737*9.8= 152830 N

area of cylindr=pi*r2=3.14*1.49^2= 6.9711 m2

So pressure at bottom due to weight =152830/ 6.9711=21923 N/m2

top pressure given 2.5*atmospheric pressure=2.5*100,000 N/m2=250000   ( 1atm =100,000N/m2)

so total pressure at the bottom=250000 +21923=271923 N/m2

ii) Height of the cyllinder H= Volume/area=21.16 / 6.9711 = 3.0354 m and p top pressure=250000 N/m2

this is excess pressure at top so

velocity will be= v = Cv (2 (g H + p / ))1/2 =0.97*sqrt(2*(9.8*3.03+(250000/737)))= 26.3479 m/s

iii) If a nozzle at the bottom of the tank is used to let fuel out, what volume flow rate will result if the diameter of the hole is 1.2 cm?

flow rate will be

Q = Cd A (2 (g H + p / ))1/2 where A area of nozzle

   =0.94*3.14*0.012^2*sqrt(2*(9.8*3.03+(250000/737)))= 0.0115 m3/s

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