A 4-pole, 60-Hz, 690-V, delta-connected, three-phase induction motor develops 20 HP at full-load slip of 4%. 1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied. 2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

1. i. 20 Nm ii. 4.85 HP

2. 16.5 %

Explanation:

1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied.

i. Torque

Since slip is constant at 4 %,torque, T ∝ V² where V = voltage

Now, T₂/T₁ = V₂²/V₁² where T₁ = torque at 690 V = P/2πN where P = power = 20 HP = 20 × 746 W = 14920 W, N = rotor speed = N'(1 - s) where s = slip = 4% = 0.04 and N' = synchronous speed = 120f/p where f = frequency = 60 Hz and p = number of poles = 4.

So, N' = 120 × 60/4 = 30 × 60 = 1800 rpm

So, N = N'(1 - s) = 1800 rpm(1 - 0.04) = 1800 rpm(0.96) = 1728 rpm = 1728/60 = 28.8 rps

So, T = P/2πN = 14920 W/(2π × 28.8rps) = 14920 W/180.96 = 82.45 Nm

T₂ = torque at 340 V, V₁ = 690 V and V₂ = 340 V

So, T₂/T₁ = V₂²/V₁²

T₂ = (V₂²/V₁²)T₁

T₂ = (V₂/V₁)²T₁

T₂ = (340 V/690 V)²82.45 Nm

T₂ = (0.4928)²82.45 Nm

T₂ = (0.2428)82.45 Nm

T₂ = 20.02 Nm

T₂ ≅ 20 Nm

ii. Power

P = 2πT₂N'

= 2π × 20 Nm × 28.8 rps

= 1152π W

= 3619.11 W

converting to HP

= 3619.11 W/746 W

= 4.85 HP

2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

Since torque T ∝ sV² where s = slip and V = voltage,

T₂/T₁ = s₂V₂²/s₁V₁²

where T₁ = torque at slip, s₁ = 4% and voltage V₁ = 690 V and T₂ = torque at slip, s₂ = unknown and voltage V₂ = 340 V

If the torque is the same, T₁ = T₂ ⇒ T₂T₁ = 1

So,

T₂/T₁ = s₂V₂²/s₁V₁²

1 = s₂V₂²/s₁V₁²

s₂V₂² = s₁V₁²

s₂ = s₁V₁²/V₂²

s₂ = s₁(V₁/V₂)²

substituting the values of the variables into the equation, we have

s₂ = s₁(V₁/V₂)²

s₂ = 4%(690/340)²

s₂ = 4%(2.0294)²

s₂ = 4%(4.119)

s₂ = 16.47 %

s₂ ≅ 16.5 %

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explanation: