Calculate Apparent Power Of A 5 HP Electric Motor With 0.8 Efficiency And 0.7 Power Factor

Introduction

In the realm of electrical engineering, understanding the relationship between power, efficiency, and power factor is crucial for designing and operating electrical systems effectively. This article delves into calculating the apparent power of a 5 HP electric motor, considering its efficiency and power factor. This is a very important aspect for those who work with electrical installations and also for those who are preparing for technical tests in the electrical field.

Understanding the Problem

To determine the apparent power, we need to consider the motor's output power, efficiency, and power factor. The output power is given as 5 HP (horsepower), the efficiency is 0.8, and the power factor is 0.7. Our goal is to calculate the apparent power in VA (volt-amperes).

Converting Horsepower to Watts

First, we need to convert the output power from horsepower (HP) to watts (W), as the calculations for electrical power are typically done in watts. The conversion factor is 1 HP = 746 W. Therefore:

$$Output Power (W) = 5 HP \times 746 W/HP = 3730 W$$

This conversion gives us the actual mechanical power delivered by the motor, which we will use as a crucial parameter in further calculations.

Accounting for Efficiency

Efficiency is the ratio of output power to input power. An efficiency of 0.8 means that 80% of the input electrical power is converted into mechanical output power, while the remaining 20% is lost as heat due to factors like friction and electrical resistance. We can represent this relationship as:

$$Efficiency = \frac{Output Power}{Input Power}$$

To find the input power, we rearrange the formula:

$$Input Power (W) = \frac{Output Power}{Efficiency} = \frac{3730 W}{0.8} = 4662.5 W$$

This input power represents the real power that the motor draws from the electrical source. Real power, measured in watts, is the power that performs actual work.

Incorporating the Power Factor

The power factor is the ratio of real power (W) to apparent power (VA). It indicates how effectively the electrical power is being used. A power factor of 1 means that all the power supplied is used for doing work, while a power factor less than 1 indicates that some power is wasted due to the phase difference between voltage and current in AC circuits. The formula for power factor is:

$$Power Factor = \frac{Real Power}{Apparent Power}$$

In our case, the power factor is 0.7. We can rearrange the formula to solve for apparent power:

$$Apparent Power (VA) = \frac{Real Power}{Power Factor} = \frac{4662.5 W}{0.7}$$

Calculating Apparent Power

Now, we can calculate the apparent power:

$$Apparent Power (VA) = \frac{4662.5 W}{0.7} ≈ 6660.7 VA$$

Thus, the apparent power of the motor is approximately 6660.7 VA. This value represents the total power that the motor appears to draw from the electrical source, including both the real power used for work and the reactive power that circulates in the circuit but does not perform work.

Understanding Apparent Power

Apparent power, measured in volt-amperes (VA), is the product of the voltage and current in an AC circuit. It is the vector sum of real power (W) and reactive power (VAR). Reactive power is the power that oscillates between the source and the load without doing any useful work. It is caused by inductive and capacitive elements in the circuit.

The concept of apparent power is crucial in electrical system design and analysis. Electrical equipment, such as transformers and generators, are rated in VA or kVA (kilo-volt-amperes) because they must be able to handle the total current, regardless of the power factor. Ignoring the apparent power can lead to overloading equipment and potentially causing damage or failure. Furthermore, understanding how real and reactive power combine to form apparent power provides key insights into the efficiency and performance of AC electrical systems.

Practical Implications

The calculation of apparent power is not just a theoretical exercise; it has significant practical implications. For instance, when selecting a generator or a transformer for a motor, it is essential to consider the apparent power rather than just the real power. If the apparent power is higher than the rating of the equipment, it can lead to overloading and potential damage. Furthermore, utilities often charge consumers based on apparent power consumption, especially for large industrial loads. Therefore, optimizing the power factor is crucial for reducing energy costs and improving the efficiency of electrical systems.

Conclusion

In summary, the apparent power of the 5 HP electric motor, with an efficiency of 0.8 and a power factor of 0.7, is approximately 6660.7 VA. This calculation involved converting horsepower to watts, accounting for efficiency to find the real power, and then using the power factor to determine the apparent power. This value is essential for selecting appropriate electrical equipment and understanding the overall power consumption of the motor. Understanding these concepts is crucial for electrical engineers and technicians to ensure the efficient and safe operation of electrical systems.

Final Answer

The apparent power of the motor is approximately 6660.7 VA, which corresponds to option (E).