g The pump inlet is located 1 m above an arbitrary datum. The pressure and velocity at the inlet are 100 kPa and 2 m/s, respectively. The pump exit is located 4 m above the same datum. The pressure and velocity are 500 kPa and 3 m/s, respectively. How much power is required to drive this pump assuming and efficiency of 75%

The power needed to drive this pump assuming and efficiency of 75% is 1874.0 watts (W)

Explanation:

Solution:

Given that:

Velocity = 2 m/s

Pressure = 100 kPa

The pump exit is = 4 m

Efficiency 75%

Thus,

We apply the method called the Bernoulli's equation between two reservoirs

p₁ / ps + v₁²/2g + z₁

=p₂/ps/v₂²/2g + z₂ + hL

The density of gasoline (pg) is = 680 kg m³

The gravity of acceleration is known to be = 9.81 m/s²

So,

100/680 * 9.81 + 2²/2 * 9.81 + 1 = 500 / 680 * 9.81 + 3²/2 * 9.81 + 4 + hL

16. 2 = 79.41 + hL

hl = 79.41 - 16.2

hL = 63.21 m

the unit weight of gasoline is (γ) = 680 * 9.81 = 6670.8 m/s

Now we find the efficiency

The efficiency (η) = The output power / Input power

Where hL = H

The input power = γ * Q * H/0.75

=6670.8 * 12/3600 * 63.21

=6670.8 * 0.333 * 63.21

=6670.8 * 0.2107

=1405.5/0.75

The input power = 1874.0 Watts