A vertical cylindrical container contains 5840 gallons of gasoline and is 1.53 m

Question

A vertical cylindrical container contains 5840 gallons of gasoline and is 1.53 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.

B) If a nozzle at the bottom of the tank is used to let fuel out, with what velocity would the gasoline exit the hole?

C)

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?

Give your answer in gallons/minute.

D)

At what initial rate will the height of the fuel level in the tank drop as fuel is removed?

Give your answer in mm/minute.

Please show work for each part and thanks in advance !

Explanation / Answer

5840 gallons = 22.10 m^3

height of the tank = 22.10/ pi (1.53)^2 = 3.0 m

Apply Bernouilis equation

p+ 1/2 * rho v1^2 + eho gh = Pa + 1/2 rho v2^2

2.5 Pa + 0 + rho gh = pa + 1/2 * rho v2^2

(a)

v2 = sqrt 2( 1.5Pa + rho gh/ rho)

= sqrt 2 ( 1.5(1.013 * 10^5 + 737(9.8) ( 3)/737)

=21.7m/s

(b)

from the volume flow rate equation

Q = A v2 = V2 =21.7 * pi ( 0.012/2)^2 =0.002452968 m^3/s = 38.8 gal/ min

(d)

38.8 gallons / min = dV/dt = d(pi R^2 h ) /dt

pi R^2 dh / dt = 38.8 gallons / min   

1 gallon = 0.003785 m^3

38.8 gallon = 0.146874 m^3

pi R^2 dh / dt = 0.146874 m^3 / min

pi x 1.53^2 dh/dt = 0.146874 m^3 / min

dh/dt =0.146874/(pi*1.53^2)

= 0.01998 m / min

= 0.01998 x 1000

= 19.98 mm/min

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