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A 78-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s through a constant-diameter pipe. The free surface of the pool is 32 ft above that of the lake. Determine the irreversible head loss of the piping system, in ft, and the mechanical power used to overcome it. Take the density of water to be 62.4 lbm/ft3.

Respuesta :

Answer:

irreversible head loss is 38.51 ft

mechanical power 6.55 hp

Explanation:

Given data:

Pump Power 12 hp = 8948.39 watt = 6600 lbs ft/sec

Q = 1.5 ft^3/s

Pump actually delivers [tex]P'  = \eta P = 0.78 \times 8948.39 = 6979.74 watt[/tex]

Power that water gains[tex] = mgh = \rho \phi gh = r \phi h[/tex]

[tex]P = 62.4 \times 1.5 \times 32 = 2995.2 lbs ft/sec[/tex]

hence Power lost = 6600 - 2995.2  3604.8 lbs ft/sec = 6.55 hp

head loss[tex] hl = \frac{3604.8}{r \phi} = \frac{3604.8}{62.4 \times 1.5}[/tex]

hl = 38.51 ft

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