Question
Determine the overall efficiency of a motor that delivers 2 hp to a load if it draws 7.5 amperes when connected to a 240-volt supply. ______________________________________________________________________________________________________________________________________________________________
Step 1
Since 1 hp is equal to 746 watts, we calculate: \[ 2 \, \text{hp} \times 746 \, \text{W/hp} = 1492 \, \text{W} \] Show more…
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A hoist motor supplied by a $240-\mathrm{V}$ source requires $12.0 \mathrm{~A}$ to lift an $800-\mathrm{kg}$ load at a rate of $9.00 \mathrm{~m} / \mathrm{min}$. Determine the power input to the motor and the power output, both in horsepower, and the overall efficiency of the system. $$ \begin{array}{l} \text { Power input }=I V=(12.0 \mathrm{~A})(240 \mathrm{~V})=2880 \mathrm{~W}=(2.88 \mathrm{~kW})(1.34 \mathrm{hp} / \mathrm{kW})=3.86 \mathrm{hp} \\ \text { Power output }=F v=(800 \times 9.81 \mathrm{~N})\left(\frac{9.00 \mathrm{~m}}{\min }\right)\left(\frac{1.00 \mathrm{~min}}{60.0 \mathrm{~s}}\right)\left(\frac{1.00 \mathrm{hp}}{746 \mathrm{~J} / \mathrm{s}}\right)=1.58 \mathrm{hp} \\ \text { Efficiency }=\frac{1.58 \text { hpoutput }}{3.86 \mathrm{hp} \text { input }}=0.408=40.8 \% \end{array} $$
A hoist motor supplied by a 240-V source requires $12.0$ A to lift an 800 -kg load at a rate of $9.00 \mathrm{~m} / \mathrm{min}$. Determine the power input to the motor and the power output, both in horsepower, and the overall efficiency of the system.
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