A 2-kW pump is used to pump kerosene ( rho = 0. 820 kg/L) from a tank on the ground to a tank at a higher elevation. Both tanks are open to the atmosphere, and the elevation difference between the free surfaces of the tanks is 30 m. The maximum volume flow rate of kerosene is

In fluid dynamics, the maximum volume flow rate (Q) is the volume or amount of fluid flowing via a required cross-sectional area per unit time.

In fluid mechanics, using the following relation, we can determine the maximum volume flow rate of kerosene.

Power = mass flow rate(m) × specific work(w) --- (1)
Specific work = acceleration due to gravity (g) × head (h) ---- (2)
Mass flow rate (m) = density (ρ) × volume flow rate (Q) --- (3)

By combining the three equations together, we have:

The power gained through the fluid pump to be:

P = ρ × Q × g × h

Making Q the subject, we have:

[tex]\mathbf{Q = \dfrac{P}{\rho \times g \times h}}[/tex]

where:

P = 2 kW = 2000 W
ρ = 0.820 kg/L
g = 9.8 m/s
h = 30 m

[tex]\mathbf{Q = \dfrac{2000 \ W }{820 \ kg/m^3 \times 9.8 m/s \times 30 \ m}}[/tex]

Q = 0.008296 m³/s

Q ≅ 8.3 L/s

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