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Calculate the hydraulic horsepower for pumping 500 gal/min from an inlet pressure of 5 psig to an outlet pressure of 30 psig. (1 hp = 550 lbf·ft/sec)

(a) For water.
(b) For gasoline.

Answer :

(a) For water: Hydraulic horsepower ≈ 3.16 horsepower.

(b) For gasoline: Hydraulic horsepower ≈ 3.95 horsepower.

To calculate the hydraulic horsepower (HP) for pumping water and gasoline, we can use the following formula:

[tex]\[ HP = \frac{{Q \times (P_2 - P_1)}}{{3960 \times \eta}} \][/tex]

(a) For water:

[tex]\\\( Q = 500 \) gallons per minute\\ \( P_1 = 5 \) psig\\ \( P_2 = 30 \) psig\\Pump efficiency (\( \eta \)) is assumed to be 1 (ideal pump)[/tex]

Substituting these values into the formula:

[tex]\[ HP = \frac{{500 \times (30 - 5)}}{{3960 \times 1}} \]\[ HP = \frac{{500 \times 25}}{{3960}} \]\[ HP = \frac{{12500}}{{3960}} \]\[ HP = 3.16 \][/tex]

Therefore, the hydraulic horsepower for pumping water is approximately 3.16 horsepower.

Now, let's calculate the hydraulic horsepower for pumping gasoline:

(b) For gasoline:

The same formula applies, but the pump efficiency [tex](\( \eta \))[/tex] might be different for gasoline. Let's assume a pump efficiency of 0.8 for gasoline pumping.

[tex]\[ HP = \frac{{500 \times (30 - 5)}}{{3960 \times 0.8}} \]\[ HP = \frac{{12500}}{{3168}} \]\[ HP = 3.95 \][/tex]

Therefore, the hydraulic horsepower for pumping gasoline is approximately 3.95 horsepower.

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