Synchronous Motor Gross Torque Calculation

The question asks us to calculate the gross torque developed by a three-phase salient pole synchronous motor, given its power output and speed. We are asked to neglect losses.

We are given the following information:

• Total power developed (P) = 100 kW
• Speed (N) = 1000 RPM

We need to find the gross torque (T) developed by the motor.

The relationship between power (P), torque (T), and angular speed ($\omega$) is given by the formula:

$$P = T \omega$$

Here, P is in Watts, T is in Newton-meters (N-m), and $\omega$ is in radians per second (rad/s).

First, we need to convert the given power from kW to Watts:

$$P = 100 \text{ kW} = 100 \times 1000 \text{ W} = 100,000 \text{ W}$$

Next, we need to convert the given speed from RPM (revolutions per minute) to rad/s. The formula for converting speed from RPM (N) to angular speed ($\omega$) is:

$$\omega = \frac{2 \pi N}{60}$$

Substitute the given speed (N = 1000 RPM) into the formula:

$$\omega = \frac{2 \pi \times 1000}{60} = \frac{2000 \pi}{60} = \frac{100 \pi}{3} \text{ rad/s}$$

Now we can rearrange the power formula ($P = T \omega$) to solve for torque (T):

$$T = \frac{P}{\omega}$$

Substitute the values of P (in Watts) and $\omega$ (in rad/s) into this formula:

$$T = \frac{100,000}{\frac{100 \pi}{3}} \text{ N-m}$$

$$T = \frac{100,000 \times 3}{100 \pi} \text{ N-m}$$

$$T = \frac{1000 \times 3}{\pi} \text{ N-m}$$

$$T = \frac{3000}{\pi} \text{ N-m}$$

Using the approximate value of $\pi \approx 3.14159$, we calculate the torque:

$$T \approx \frac{3000}{3.14159} \approx 954.93 \text{ N-m}$$

Rounding this value to the nearest whole number or considering the given options, the calculated gross torque is approximately 955 N-m.