Answer:
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 %
