# engineering physics question?

A vertical cylindrical container contains 5640 gallon of gasoline and is 1.48 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/m^3.

1.What is the pressure at the bottom of the tank?

p= N/m^2

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

v =m/s

3.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?

VFR = gallons/minute.

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

delta h/delta t = mm/minute.

### 1 Answer

- JohnLv 77 years agoFavorite Answer
1 Gallon [Fluid, US] = 0.00378541178 Cubic Meters

1 Gallon [UK] = 0.004546 Cubic Meters - taking US gallons

5640 US Fluid Gallons = 21.349 m^3 this weighs 15734.213 kg and creates a pressure of

15734 * 9.81 / area of base. And area = 1.48 ^2 * pi() = 6.881 m^2

P = 22431 Newtons / m^2

since 1982 the IUPAC has recommended that the standard for atmospheric pressure should be harmonized to 100,000 Pa = N/m^2

To this we add the pressure in the top of the tank = 2.5 * 100,000 =250,000

so total P = 22431 + 250000 = 272431 Pa

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