Question

A 3.5 hp pump delivers 1140 lbf of ethylene glycol at 20°C in 12 seconds, against a head of 17 ft. Calculate the efficiency of the pump.

Solution

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The Efficiency of the Pump is the ratio of power performed by the pump to the total power it produces. It can be solved by dividing the power output by the power input which is 3.5\ hp.

Power output can be solved by the formula Poutput=γQHP_{output}=\gamma \cdot Q \cdot H where γ\gamma is the unit weight of water, Q is the discharge and H is the head of the pump.

Covert the density ρ=1117kgm3\rho = 1117 \frac{kg}{m^3} based on Table A.3 of Appendix A for ethylene glycol into lbft3\frac{lb}{ft^3} to get the unit weight.

γ=1117kgm3(2.205lb1kg)(13m33.283ft3)\gamma=1117 \frac{kg}{m^3}\left(\frac{2.205lb}{1kg}\right)\left(\frac{1^3m^3}{3.28^3ft^3}\right)

γ=69.8lbft3\gamma=69.8 \frac{lb}{ft^3}

Then, solve for the discharge by dividing the weight by time and unit weight.

Q=WtγQ=\frac{W}{t \cdot \gamma}

Q=1140lbf12s69.8lbft3Q=\frac{1140 lbf}{12s \cdot 69.8 \frac{lb}{ft^3}}

Q=1.36ft3sQ=1.36 \frac{ft^3}{s}

After solving the discharge, the power output can now be solved, by substituting the γ\gamma, Q and H.

Poutput=γQHP_{output}=\gamma \cdot Q \cdot H

Poutput=69.8lbft31.36ft3s17ftP_{output}=69.8 \frac{lb}{ft^3} \cdot 1.36 \frac{ft^3}{s} \cdot 17ft

Poutput=1613.78lbftsP_{output}=1613.78 \frac{lb-ft}{s}

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