Answer:
A. 50.93 m/s
B. 201 kW
Explanation:
The pump moves a volume of 0.1 m³ of water every second. The diameter of the nozzle is 5 cm, which translates to 0.05 m, or a radius of 2.5 cm, equivalent to 0.025 m. Thus, the cross-sectional area of the nozzle is calculated as A = πr²
π×0.025²
= 0.001963 m²
Consequently, for 0.1 m³ to pass through this nozzle in 1 second, the velocity must be
0.1 / 0.001963
= 50.93 m/s
The pump is designed to provide energy to water to lift it 4 m above sea level to the nozzle and to accelerate it from rest to 50.93 m/s through the nozzle. Assuming no energy losses, the dynamic energy of the water equals the potential energy it would possess if it remained stationary at a height h.
P.E = k.E
mgh = 0.5mv²
h = v² / 2g
h = 50.93² / (2 × 9.81)
h = 132.21 m
Thus, to facilitate the acceleration of water, a head of 132.21 m is required. The static head is 4 m since the nozzle is raised 4 m above sea level, with an additional heads loss of 3 m due to friction along the system. Therefore, the total head the pump needs to deliver is
132.21 + 4 + 3
= 139.21 m
Using the relationship density = mass / volume implies that mass = density × volume.
The pump moves water at a rate of 0.1 m³ per second. Therefore, in one second, the mass of the water pumped is
0.1 × 1030 = 103 kg
The overall energy transferred from the pump to the water corresponds to
Potential energy, which is expressed as = mgh
= 103 × 9.81 × 139.21
= 140,662.0 Nm/s
= 140,662.0 Watts
However, since the pump operates at 70% efficiency, we find
140,662 = 0.7P where P is the power rendered to the pump
Thus, P = 140,662 / 0.7
= 200,946 Watts
= 201 kW