Thank you for visiting 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. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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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Rewritten by : Jeany
