Question

A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height of 5 feet and a radius of 2 feet. Find the work required to pump the water to a height of 1 feet above the top of the tank

Answer 1

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

Answer 2

The work required to pump water to 1 ft above the top of the tank is 26,958.46 ft.lbf.

The given parameters:

• Density of the tank, $\rho$ = 40 lb/ft³
• Height of the tank, h = 5 ft
• Radius of the tank, r = 2 ft

The volume of the right circular cone tank is calculated as follows;

$V = \pi r^2 \frac{h}{3} \\\\ V = \pi (2)^2 \times \frac{5}{3} \\\\ V = 20.95 \ ft^3$

The work required to pump water to 1 ft above the top of the tank is calculated as follows;

$W = Fd\\\\ W = (\rho Vg )d\\\\ W = (40 \ lb/ft^3\times 20.95\ ft^3 \times 32.17 \ ft/s^2) \times (1 \ ft)\\\\ W = 26,958.46 \ ft.lbf$

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