Question

a 3560rpm, three-phase, 60hz, 460v, 100hp induction motor is going to be controlled using a variable- frequency drive. a. approximately how much torque would the motor provide at its rated operating conditions. b. when operating at 25hz what would you expect the output torque, speed, and power to be assuming that the variable-frequency drive was properly configured for the motor. c. when operating at 45hz what would you expect the output torque, speed, and power to be assuming that the variable-frequency drive was properly configured for the motor. d. when operating at 85hz what would you expect the output torque, speed, and power to be assuming that the variable-frequency drive was properly configured for the motor

Answer 1

a. Rated Operating Conditions:

- Torque
`(\(T\))`: Approximately
`\(314 \, \text{Nm}\)`

b. Operating at 25 Hz:

- Torque
`(\(T\))`:: Approximately
`\(54.98 \, \text{Nm}\)`

- Speed
`(\(n\))`: Approximately
`\(2.62 \, \text{rad/s}\)`

- Power
`(\(P\))`: Approximately
`\(144.13 \, \text{W}\)`

c. Operating at 45 Hz:

- Torque
`(\(T\))`:: Approximately
`\(142.12 \, \text{Nm}\)`

- Speed
`(\(n\))`: Approximately
`\(4.71 \, \text{rad/s}\)`

- Power
`(\(P\))`: Approximately
`\(669.09 \, \text{W}\)`

d. Operating at 85 Hz:

- Torque
`(\(T\))`:: Approximately
`\(422.82 \, \text{Nm}\)`

- Speed
`(\(n\))`: Approximately
`\(8.88 \, \text{rad/s}\)`

- Power
`(\(P\))`: Approximately
`\(3754.2 \, \text{W}\)`

Let's go through the calculations for each part of the question.

a. Rated Operating Conditions:

At its rated operating conditions (3560 rpm), the motor provides mechanical power (P) of:

`\[ P = 100 \, hp * 746 \, \text{W/hp} = 74600 \, \text{W} \]`

The speed in radians per second (\(n\)) is calculated as:

`\[ n = (3560 \, \text{rpm}) * \left((2 \pi)/(60)\right) \approx 372.75 \, \text{rad/s} \]`

Now, the torque
`(\(T\))` can be calculated using the formula:

`\[ T = (P)/(2 \pi n) = (74600)/(2 \pi * 372.75) \approx 314 \, \text{Nm} \]`

b. Operating at 25 Hz:

The output speed
`(\(n\))` at 25 Hz is:

`\[ n = 25 \, \text{Hz} * \left((2 \pi)/(60)\right) \approx 2.62 \, \text{rad/s} \]`

The output torque
`(\(T\))` at 25 Hz is proportional to the square of the frequency:

`\[ T = \left((25)/(60)\right)^2 * 314 \, \text{Nm} \approx 54.98 \, \text{Nm} \]`

The output power
`(\(P\))` can be calculated as:

`\[ P = T * n = 54.98 \, \text{Nm} * 2.62 \, \text{rad/s} \approx 144.13 \, \text{W} \]`

c. Operating at 45 Hz:

The output speed
`(\(n\))` at 45 Hz is:

`\[ n = 45 \, \text{Hz} * \left((2 \pi)/(60)\right) \approx 4.71 \, \text{rad/s} \]`

The output torque
`(\(T\))` at 45 Hz is proportional to the square of the frequency:

`\[ T = \left((45)/(60)\right)^2 * 314 \, \text{Nm} \approx 142.12 \, \text{Nm} \]`

The output power
`(\(P\))` can be calculated as:

`\[ P = T * n = 142.12 \, \text{Nm} * 4.71 \, \text{rad/s} \approx 669.09 \, \text{W} \]`

d. Operating at 85 Hz:

The output speed
`(\(n\))` at 85 Hz is:

`\[ n = 85 \, \text{Hz} * \left((2 \pi)/(60)\right) \approx 8.88 \, \text{rad/s} \]`

The output torque
`(\(T\))` at 85 Hz is proportional to the square of the frequency:

`\[ T = \left((85)/(60)\right)^2 * 314 \, \text{Nm} \approx 422.82 \, \text{Nm} \]`

The output power
`(\(P\))`can be calculated as:

`\[ P = T * n = 422.82 \, \text{Nm} * 8.88 \, \text{rad/s} \approx 3754.2 \, \text{W} \]`

Answer 2

Answer: At its rated operating conditions, the motor would provide 100 hp * 746 W/hp = 74600 W of mechanical power. To calculate the torque, we can use the formula:

Explanation:

T = P / (2 * pi * n)

Where T is the torque in Nm, P is the power in watts, and n is the speed in radians per second. At 3560 rpm, the speed in radians per second is:

n = (3560 rpm) * (2 * pi / 60) = 372.75 rad/s

Therefore, the torque at rated operating conditions would be:

T = 74600 / (2 * pi * 372.75) = 314 Nm

b. When operating at 25 Hz, the output speed would be:

n = 25 Hz * (2 * pi / 60) = 2.62 rad/s

To calculate the output torque, we can use the same formula as before, but we need to take into account that the motor is now operating at a different frequency. Assuming that the variable-frequency drive is properly configured for the motor, the voltage and current supplied to the motor should be adjusted to maintain a constant flux level. This means that the torque will be proportional to the square of the frequency. Therefore, the output torque at 25 Hz would be:

T = (25/60)^2 * 314 Nm = 54.98 Nm

The output power can be calculated as:

P = T * n = 54.98 Nm * 2.62 rad/s = 144.13 W

c. When operating at 45 Hz, the output speed would be:

n = 45 Hz * (2 * pi / 60) = 4.71 rad/s

Using the same formula as before, the output torque at 45 Hz would be:

T = (45/60)^2 * 314 Nm = 142.12 Nm

The output power can be calculated as:

P = T * n = 142.12 Nm * 4.71 rad/s = 669.09 W

d. When operating at 85 Hz, the output speed would be:

n = 85 Hz * (2 * pi / 60) = 8.88 rad/s

Using the same formula as before, the output torque at 85 Hz would be:

T = (85/60)^2 * 314 Nm = 422.82 Nm

The output power can be calculated as:

P = T * n = 422.82 Nm * 8.88 rad/s = 3754.2 W

SPJ11