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47. A 15 hp motor has a total loss of 1,130 W when operating at full-load. What is the percent efficiency?

A. 89.5*
B. 96.5
C. 97.8
D. 88.5

η = \frac{P_{out}}{P_{out} + P_{loss}}

Answer :

To determine the percent efficiency of the motor, we use the efficiency formula provided:

[tex]\eta = \frac{P_{\text{out}}}{P_{\text{out}} + P_{\text{loss}}}[/tex]

Where:

  • [tex]P_{\text{out}}[/tex] is the output power of the motor in watts.
  • [tex]P_{\text{loss}}[/tex] is the power loss in watts.

First, we need to convert the motor's horsepower (hp) into watts because horsepower is a common unit used for engines and motors, while watts are a standard unit of power in the International System of Units (SI). The conversion factor is:

[tex]1 \text{ hp} = 746 \text{ W}[/tex]

Therefore, a 15 hp motor is equivalent to:

[tex]15 \text{ hp} \times 746 \text{ W/hp} = 11,190 \text{ W}[/tex]

Now, we'll plug the values into the formula:

[tex]\eta = \frac{11,190}{11,190 + 1,130}[/tex]

Calculate the denominator:

[tex]11,190 + 1,130 = 12,320[/tex]

Calculate the efficiency:

[tex]\eta = \frac{11,190}{12,320} \approx 0.908[/tex]

To convert this to a percentage, multiply by 100:

[tex]0.908 \times 100 = 90.8\%[/tex]

Given the options:

  • A. 89.5
  • B. 96.5
  • C. 97.8
  • D. 88.5

None of the options exactly match our computed result of 90.8%, but the closest option is A. 89.5%.

Therefore, based on the closest match, the chosen option is A. 89.5%.

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