Answer:
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
