Running Torque of Three-Phase Induction Motor

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Torque of 3-Phase Induction Motor under Running Condition

Let the rotor circuit of 3-phase induction motor at standstill has per phase resistance R2, per phase reactance X2 and per phase induced EMF E2. If ‘s’ is the slip under running condition of the motor, then,

$$\mathrm{Rotor \:reactance/phase , 𝑋′_2 = 𝑠 𝑋_2}$$

$$\mathrm{Rotor \:EMF/phase , 𝐸′_2 = 𝑠 𝐸_2}$$

$$\mathrm{\therefore \:Rotor \:impedance/phase , 𝑍′_2 = \sqrt{𝑅_2^2 + (𝑠 𝑋_2)^2}}$$

$$\mathrm{Rotor\:impedance/phase ,𝐼′_2 =\frac{𝐸'_2}{𝑍′_2}=\frac{𝐸'_2}{\sqrt{𝑅_{2}^{2} + (𝑠 𝑋_2)^2}}… (1)}$$

$$\mathrm{Rotor\: power \:factor, cos \varphi′_2 =\frac{𝑅_2}{𝑍′_2}=\frac{𝑅_2}{\sqrt{𝑅_{2}^{2} + (𝑠 𝑋_2)^2}}… (2)}$$

Therefore,

$$\mathrm{Running\:torque, \tau_𝑟 \propto 𝐸′_2 𝐼′_2 cos \varphi′_2 … (3) +}$$

$$\mathrm{\because 𝐸′_2 \propto Magnetic\:flux (\varphi)}$$

$$\mathrm{\therefore \tau_𝑟 = 𝐾 \varphi 𝐼′_2 cos \varphi′_2}$$

$$\mathrm{⇒ \tau_𝑟 = 𝐾 \varphi ×\frac{𝐸'_2}{\sqrt{𝑅_{2}^{2} + (𝑠 𝑋_2)^2}}\times\frac{𝑅_2}{\sqrt{𝑅_{2}^{2} + (𝑠 𝑋_2)^2}}}$$

$$\mathrm{⇒ \tau_𝑟 =\frac{𝐾 \varphi 𝑠 𝐸_2 𝑅_2}{𝑅_2^2 + (𝑠 𝑋_2)^2}… (4)}$$

$$\mathrm{\because 𝐸_2 \propto \varphi}$$

$$\mathrm{\therefore \tau_𝑟 =\frac{𝐾 𝑠 𝐸_2^2 𝑅_2}{𝑅_2^2 + (𝑠 𝑋_2)^2}… (5)}$$

Eqn. (4) gives the value of running torque. It can be seen that

  • The running torque is directly proportional to slip, i.e., if slip increases, the torque will increase and vice-versa.
  • The running torque is directly proportional to square of supply voltage since (E2 ∝ V).

As the running torque of a 3-phase induction motor is given by,

$$\mathrm{\tau_𝑟 =\frac{𝐾 \varphi 𝑠 𝐸_2 𝑅_2}{𝑅_2^2 + (𝑠 𝑋_2)^2}}$$

Since the supply voltage (V) is constant, then stator flux and hence the EMF E2 will be constant.

$$\mathrm{\therefore \tau_𝑟 =\frac{𝐾_1 𝑠 𝑅_2}{𝑅_2^2 + (𝑠 𝑋_2)^2}… (6)}$$

Where, K1 = K ϕ E2 is a constant.

$$\mathrm{⇒ \tau_𝑟 =\frac{𝐾_1 𝑅_2}{\frac{𝑅_2^2}{𝑠}+ 𝑠𝑋_2^2}… (7)}$$

To be the maximum of the running torque, the denominator of eqn. (6) should be minimum. Hence, differentiating denominator of eqn. (6) with respect to slip 's' and equating it to zero, i.e.,

$$\mathrm{\frac{𝑑}{𝑑𝑠} (\frac{𝑅_2^2}{𝑠}+ 𝑠𝑋_2^2) = 0}$$

$$\mathrm{⇒ −\frac{𝑅_2^2}{𝑠^2}+ 𝑋_2^2 = 0}$$

$$\mathrm{⇒ 𝑅_2 = 𝑠 𝑋_2 … (8)}$$

Thus, for maximum torque under running conditions,

$$\mathrm{Rotor\:Resistance/Phase= Per\:unit\:slip \times standstill\:rotor\:reactance\:per\:phase}$$

Now,

$$\mathrm{\tau_𝑟 \propto\frac{𝑠 𝑅_2}{𝑅_2^2 + (𝑠 𝑋_2)^2}}$$

Putting R2 = s X2 in the above expression, the maximum torque under running condition is given by,

$$\mathrm{⇒ \tau_𝑟 \propto\frac{1}{2 𝑋_2}… (9)}$$

The slip corresponding to maximum torque is given by,

$$\mathrm{𝑠_𝑚 =\frac{𝑅_2}{𝑋_2}… (10)}$$

Numerical Example

A 6-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.03 Ω per phase and standstill reactance of 0.5 Ω per phase. Determine the slip corresponding to maximum torque and speed of the motor at which maximum torque is developed.

Solution

(1) Slip corresponding to maximum running torque is,

$$\mathrm{𝑠_𝑚 =\frac{𝑅_2}{𝑋_2}=\frac{0.03}{0.5}= 0.06 Ω}$$

(2) Speed of motor corresponding to maximum torque is,

$$\mathrm{Synchronous\:speed, 𝑁_𝑠 =\frac{120𝑓}{𝑃}=\frac{120 \times 50}{6}= 1000 RPM}$$

$$\mathrm{\therefore 𝑁_𝑟 = 𝑁_𝑠(1 − 𝑠_𝑚) = 1000 × (1 − 0.06) = 940 RPM}$$

raja
Published on 28-Aug-2021 14:12:16
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