What is the maximum horsepower developed if a 1/2 horsepower engine operates at 90% efficiency?

What is the maximum horsepower developed if a 1/2 horsepower engine operates at 90% efficiency?

• A. 0.45 hp
• B. 0.5 hp
• C. 0.6 hp
• D. 0.65 hp

Answer: (C)

To determine the maximum horsepower developed by a 1/2 horsepower engine operating at 90% efficiency, it is essential to understand the relationship between horsepower and efficiency. Horsepower is a measure of the engine's output power, while efficiency indicates how much of the input power is effectively converted into useful work. In this case, the engine has a nominal rating of 1/2 horsepower, which is equivalent to 0.5 horsepower. When considering the efficiency of the engine, we multiply the rated horsepower by the efficiency percentage, which is expressed as a decimal. In this scenario, the calculation would be: 0.5 horsepower (rated) x 0.90 (efficiency) = 0.45 horsepower. Thus, when operating at 90% efficiency, the engine actually develops 0.45 horsepower of useful output power. The maximum horsepower still remains at the engine's rated capacity of 0.5 horsepower, since this is the maximum output under ideal conditions without considering losses. Therefore, the available maximum horsepower at that level of efficiency would not exceed 0.5 horsepower. As a result, the correct answer reflects that the effective output, given the operating conditions, is indeed the original horsepower rating of 0.

To determine the maximum horsepower developed by a 1/2 horsepower engine operating at 90% efficiency, it is essential to understand the relationship between horsepower and efficiency.

Horsepower is a measure of the engine's output power, while efficiency indicates how much of the input power is effectively converted into useful work. In this case, the engine has a nominal rating of 1/2 horsepower, which is equivalent to 0.5 horsepower. When considering the efficiency of the engine, we multiply the rated horsepower by the efficiency percentage, which is expressed as a decimal.

In this scenario, the calculation would be:

0.5 horsepower (rated) x 0.90 (efficiency) = 0.45 horsepower.

Thus, when operating at 90% efficiency, the engine actually develops 0.45 horsepower of useful output power.

The maximum horsepower still remains at the engine's rated capacity of 0.5 horsepower, since this is the maximum output under ideal conditions without considering losses. Therefore, the available maximum horsepower at that level of efficiency would not exceed 0.5 horsepower.

As a result, the correct answer reflects that the effective output, given the operating conditions, is indeed the original horsepower rating of 0.