Question

Problem 05.086 - Water pumped from a lower reservoir to a higher reservoir Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 21 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. Take the density of water to be 1000 kg/m3. The irreversible head loss of the system is 26.35 m. The lost mechanical power in this process is kW.

Answer 1

Lost Mechanical Power = 7.7565 KW

Head Loss = 26.35 m

Step-by-step explanation:

First, we will find the useful mechanical power used to transport water to the higher reservoir:

`P_(useful) = \rho ghV`

where,

P_useful = Useful mechanical Power = ?

ρ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height = 45 m

V = Volume flow rate = 0.03 m³/s

Therefore,

`P_(useful) = (1000\ kg/m^3)(9.81\ m/s^2)(45\ m)(0.03\ m^3/s)\\P_(useful) = 13243.5\ W = 13.2435\ KW`

Now, the lost mechanical power will be:

`Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\`

Lost Mechanical Power = 7.7565 KW

Now, for the head loss:

`Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = (Lost\ Mechanical\ Power)/(\rho gV) \\\\Head\ Loss = (7756.5\ W)/((1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)) \\`

Head Loss = 26.35 m