An electric motor is used to drive a harmonic vibrating screen. Due to extensive repairs, mass was added, and thus the natural frequency changed. The shaft drive speed of the rotating mass has to be decreased from the present 970 r/min to 910 r/min. The vibrating shaft is directly connected to the motor.

The power input to the 415 V, three-phase, six-pole, 50 Hz induction motor is 50 kW when running at 970 r/min. The stator losses are 2 kW, and the friction and windage losses are 1.5 kW. Calculate the following:

a) Rotor I²R loss

b) Gross torque in N·m

c) Power output of the motor

d) Rotor resistance per phase if the rotor phase current is 110 A

e) Resistance to be added to each phase to achieve the reduced speed if the motor torque and rotor current are to remain constant

a) Rotor IR loss: 46.5 kW. b) Gross torque: 458.37 N.m. c) Power output: 0 kW (unrealistic). d) Rotor resistance per phase: 1.571 Ω. e) Resistance to be added per phase: 0.079 Ω.

The rotor I'R loss and gross torque of an induction motor are calculated. The power output and rotor resistance per phase are found, as well as the resistance required to achieve a reduced speed.

Given:

- Motor speed before repairs = 970 rpm

- Motor speed after repairs = 910 rpm

- Power input to motor = 50 kW

- Stator losses = 2 kW

- Friction and windage losses = 1.5 kW

- Supply voltage = 415 V

- Number of poles = 6

- Frequency = 50 Hz

- Rotor phase current = 110 A

(a) To calculate the rotor I'R loss, we need to first find the total losses in the motor. The total losses are the sum of the stator losses, friction and windage losses, and rotor losses. We can find the rotor losses by subtracting the total losses from the power input:

Total losses = 2 kW + 1.5 kW = 3.5 kW

Rotor losses = 50 kW - 3.5 kW = 46.5 kW

The rotor I'R loss is given by:

I'R loss = rotor losses / (3 * rotor phase current^2)

Substituting the given values, we get:

I'R loss = 46.5 kW / (3 * (110 A)^2) = 0.122 ohms

Therefore, the rotor I'R loss is 0.122 ohms.

(b) To calculate the gross torque, we can use the formula:

P = 2πNT/60

where P is the power in watts, N is the motor speed in rpm, and T is the torque in N.m. Solving for T, we get:

T = (60P) / (2πN)

At 970 rpm, the gross torque is:

T1 = (60 * 50 kW) / (2π * 970 rpm) = 458.37 N.m (rounded to 3 decimal places)

At 910 rpm, the gross torque is:

T2 = (60 * P) / (2π * 910 rpm)

Since the rotor current and torque remain constant, the power output must also remain constant. Therefore, we can write:

P = T2 * 2π * 910 rpm / 60

Substituting the given values, we get:

50 kW - 3.5 kW = T2 * 2π * 910 rpm / 60

Solving for T2, we get:

T2 = 45.06 kW / (2π * 910 rpm / 60) = 1,440 N.m (rounded to the nearest integer)

Therefore, the gross torque is 458.37 N.m at 970 rpm and 1,440 N.m at 910 rpm.

(c) The power output of the motor is given by:

Pout = Pin - losses

Substituting the given values, we get:

Pout = 50 kW - 3.5 kW = 46.5 kW

Therefore, the power output of the motor is 46.5 kW.

(d) The rotor resistance per phase is given by:

R'R = I'R loss / rotor phase current^2

Substituting the given values, we get:

R'R = 0.122 ohms / (110 A)^2 = 0.001 ohms

Therefore, the rotor resistance per phase is 0.001 ohms.

(e) To achieve the reduced speed while keeping the torque and rotor current constant, we need to add resistance to the rotor. The additional resistance per phase is given by:

ΔR'R = (1 - N2/N1) * R'R

where N1 and N2 are the original and new speeds, respectively. Substituting the given values, we get:

ΔR'R = (1 - 910/970) * 0.001 ohms = 0.07934 ohms (rounded to 5 decimal places)

Therefore, the resistance to be added to each phase to achieve the reduced speed is 0.07934 ohms.

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