Question

At full load, a commercially available 200 hp, threephase induction motor operates at an efficiency of \(96 \%\) and a power factor of 0.92 lag. The motor is supplied from a three-phase outlet with a line-voltage rating of \(208 \mathrm{V}\) a) What is the magnitude of the line current drawn from the \(208 \mathrm{V}\) outlet? \((1 \mathrm{hp}=746 \mathrm{W} .)\) b) Calculate the reactive power supplied to the motor.

Short answer

a) The line current drawn is approximately 469.72 A. b) The reactive power supplied is approximately 75,236.86 VAR.

Step-by-step solution

Convert Horsepower to Watts

To determine the electrical conditions, we first need to convert the motor power rating from horsepower to watts. Given that \[1 \text{ hp} = 746 \text{ W}, \]we multiply:\[200 \text{ hp} \times 746 \text{ W/hp} = 149,200 \text{ W}.\]

Calculate Input Power

Next, we account for the efficiency of the motor to determine the input power. Since the motor operates at 96% efficiency, the input power required is:\[\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} = \frac{149,200\,\text{W}}{0.96} = 155,416.67\,\text{W}.\]

Calculate Apparent Power

The apparent power \( S \) (in volt-amperes, VA) in a three-phase system can be calculated using the formula:\[ S = \frac{P}{\text{Power Factor}} = \frac{155,416.67\,\text{W}}{0.92} \approx 168,934.42\,\text{VA}.\]

Calculate Line Current

Using the formula for apparent power in a three-phase system, \( S = \sqrt{3} \cdot V \cdot I \), we can solve for the line current \( I \):\[ I = \frac{S}{\sqrt{3} \cdot V} = \frac{168,934.42\,\text{VA}}{\sqrt{3} \cdot 208\,\text{V}} \approx 469.72\,\text{A}.\]

Calculate Reactive Power

Finally, calculate the reactive power \( Q \) using the relationship \[ Q = \sqrt{S^2 - P^2}. \]First, calculate output power \( P \):\[ P = 155,416.67\,\text{W}. \]Then:\[ Q = \sqrt{168,934.42^2 - 155,416.67^2} \approx 75,236.86\,\text{VAR}.\]

Key concepts

Motor Efficiency

When discussing motor efficiency, we're referring to how well a motor converts electrical power into mechanical power. In simple terms, it's a measure of how much input energy is effectively used to perform work without being wasted. Efficiency is expressed as a percentage; the higher the percentage, the more efficient the motor.
For the three-phase induction motor in the given exercise, the efficiency is 96%. This means that 96% of the electrical energy supplied is converted into mechanical energy, while only 4% is lost to inefficiencies such as heat or friction.
This high efficiency is typical of well-designed electric motors and is crucial for cost savings in applications where the motor runs continuously. To find the input power required by a motor, you can use the formula:

• \( \text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} \)

Understanding motor efficiency helps in selecting the right motor for the job, ensuring environmental sustainability, and optimizing cost savings over the motor's operational life.

Power Factor

Power factor is an essential concept when working with alternating current (AC) systems like three-phase induction motors. It measures how effectively electrical power is converted into useful work output, a number ranging from 0 to 1. When the power factor is close to 1, more input power is effectively used to do work. When it's much less than 1, more power is wasted.
In our exercise, the motor has a power factor of 0.92 lag. "Lag" indicates that the current lags the voltage—a common scenario in inductive loads such as motors. Improving power factor can make your electrical system more efficient and reduce electricity costs.
The power factor is calculated using the formula:

• \( \text{Power Factor} = \frac{\text{True Power (P)}}{\text{Apparent Power (S)}} \)

In practical terms, a higher power factor signals that your motor uses electricity more effectively, lowering net costs and reducing energy consumption.

Reactive Power

Reactive power is the portion of electricity that establishes and sustains the electric and magnetic fields of a motor or other inductive loads. Unlike active power, which performs actual work, reactive power oscillates between the source and load, essentially building up and collapsing fields without doing useful work.
In the context of a three-phase induction motor, reactive power is measured in volt-amperes reactive (VAR). It's calculated using the formula:

• \( Q = \sqrt{S^2 - P^2} \)

Here, \(S\) is the apparent power, and \(P\) is the true power. In our example, the reactive power calculates to approximately 75,236.86 VARs, indicating the amount of power oscillating back and forth.
Understanding reactive power is crucial for managing power flows in electrical grids as it impacts voltage levels. While reactive power doesn't do work, it's vital for voltage regulation and maintaining grid stability.

Line Current Calculation

Line current is the current flowing through one line of a three-phase system. Calculating it helps in selecting the appropriate wiring and protective devices for the motor operation. In our example, this three-phase induction motor is supplied by a 208 V outlet, and we need to find the magnitude of the line current.
To calculate the line current for a three-phase system, we use the formula:

• \( I = \frac{S}{\sqrt{3} \cdot V} \)

where \(S\) is the apparent power and \(V\) is the line voltage. Substituting the known values gives a line current \(I\) of approximately 469.72 A.
Calculating line current is essential to ensure the motor is properly wired and protected, contributing to the overall safety and efficiency of the electrical system.