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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.

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hamzaahmeds

Answer:

Lost Mechanical Power = 7.7565 KW

Head Loss = 26.35 m

Explanation:

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

[tex]P_{useful} = \rho ghV[/tex]

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,

[tex]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[/tex]

Now, the lost mechanical power will be:

[tex]Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\[/tex]

Lost Mechanical Power = 7.7565 KW

Now, for the head loss:

[tex]Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = \frac{Lost\ Mechanical\ Power}{\rho gV} \\\\Head\ Loss = \frac{7756.5\ W}{(1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)} \\[/tex]

Head Loss = 26.35 m

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