Complete and Approximate Equivalent Circuits of Induction Motor

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Complete Equivalent Circuit of Induction Motor

In order to obtain the complete per-phase equivalent circuit of an induction motor, it is required to refer the rotor part to the stator side frequency and voltage level.

The transformation of the rotor circuit of the induction motor can be done by the means of effective turns ratio of the induction motor.

The figure shows the complete equivalent circuit of the induction motor.

Let suffixes "s" and "r" be used for stator and rotor quantities respectively.

Then,

$$\mathrm{π‘Ž_{𝑒𝑓𝑓}= Effective\:turns\: ratio \:of \:induction \:motor}$$

$$\mathrm{𝑅′_{π‘Ÿ} = Resistance \:of \:the \:rotor\: winding\: per\: phase\: referred \:to \:stator}$$

$$\mathrm{𝑋′_{π‘Ÿ0} = Standstill \:rotor \:reactance \:per \:phase\: referred\: to \:stator}$$

Therefore, the rotor EMF referred to stator side is given by,

$$\mathrm{\frac{𝐸_π‘Ÿ}{𝑁_{π‘’π‘Ÿ}}=\frac{𝐸′_π‘Ÿ}{𝑁_{𝑒𝑠}}… (1)}$$

$$\mathrm{⇒𝐸′_π‘Ÿ =\frac{𝑁_{𝑒𝑠}}{𝑁_{π‘’π‘Ÿ}}𝐸_π‘Ÿ = π‘Ž_{𝑒𝑓𝑓} . 𝐸_π‘Ÿ = 𝐸_𝑠 … (2)}$$

Similarly, the rotor current referred to stator side is,

$$\mathrm{𝐼′_π‘Ÿ =\frac{𝐼_π‘Ÿ}{π‘Ž_{𝑒𝑓𝑓}}… (3)}$$

Rotor impedance referred to stator side is given by,

$$\mathrm{𝑍′_{π‘Ÿ0} = π‘Ž_{𝑒𝑓𝑓}^{2}(\frac{π‘…π‘Ÿ}{𝑠}+ 𝑗𝑋_{π‘Ÿ0}) … (4)}$$

Where, s is the fractional slip of the rotor.

The standstill rotor reactance referred to stator side is given by,

$$\mathrm{𝑋′_{π‘Ÿ0} = π‘Ž_{𝑒𝑓𝑓}^{2} 𝑋_{π‘Ÿ0} … (5)}$$

Approximate Equivalent Circuit of Induction Motor

The circuit shown in the figure below is known as the approximate equivalent circuit per phase of the induction motor. The approximate equivalent circuit of the induction motor is obtained by shifting the shunt branches R0 and X0 in the equivalent circuit. In the approximate equivalent circuit, the only component that depends upon the slip (s) is the resistance which is representing the developed mechanical power by the rotor. All other quantities are constant and the reactances correspond to those quantities at the fixed stator frequency (fs).

This approximate equivalent circuit is used as the standard for all performance calculation of an induction motor.

By referring the approximate equivalent circuit of the induction motor, the following equations can be written down for one phase at a slip s.

The impedance beyond the terminals A and B is given by

,

$$\mathrm{𝑍_{𝐴𝐡} = (𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠}) + 𝑗(𝑋_𝑠 + 𝑋′_π‘Ÿ) … (6)}$$

$$\mathrm{𝐼′_π‘Ÿ =\frac{𝑉_𝑠}{𝑍_{𝐴𝐡}}=\frac{𝑉_𝑠}{(𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠}) + 𝑗(𝑋_𝑠 + 𝑋′_π‘Ÿ)}\:… (7)}$$

$$\mathrm{\therefore \:Magnitude\: of \:𝐼′_π‘Ÿ = |𝐼′_π‘Ÿ| =\frac{𝑉_𝑠}{\sqrt{(𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠})^2 + (𝑋_𝑠 + 𝑋′_π‘Ÿ)^2}}\:… (8)}$$

Therefore,

$$\mathrm{𝐼′_π‘Ÿ = |𝐼′_π‘Ÿ|∠−\varphi_{r} = 𝐼′_π‘Ÿ cos \varphi_{r} − 𝑗𝐼′_π‘Ÿ sin\varphi_{r} … (9)}$$

Where,

$$\mathrm{\varphi_{r} = tan^{−1} (\frac{𝑋_𝑠 + 𝑋′_π‘Ÿ}{𝑅_𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠}}) … (10)}$$

Then, the power factor from the approximate equivalent circuit is,

$$\mathrm{cos\varphi_{r} =\frac{(𝑅_𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠})}{|𝑍_{𝐴𝐡}|}… (11)}$$

The no-load current is given by,

$$\mathrm{𝐼_0 = 𝐼_𝑀 + 𝐼_π‘š}$$

$$\mathrm{⇒ 𝐼_0 =\frac{𝑉_𝑠}{𝑅_0}+\frac{𝑉_𝑠}{𝑗𝑋_0}… (12)}$$

Hence, the total stator current is given by the phasor sum of rotor current referred to stator and no-load current, i.e.,

$$\mathrm{𝐼_𝑠 = 𝐼′_π‘Ÿ + 𝐼_0 … (13)}$$

$$\mathrm{Total\: core \:losses,\: 𝑃_𝑐 = 𝑃_β„Ž + 𝑃_𝑒 = 3𝑉_𝑠𝐼_0 cos \varphi_0 … (14)}$$

The input power to the stator is given by,

$$\mathrm{𝑃_{𝑖𝑛𝑝𝑒𝑑} = 3𝑉_𝑠𝐼_𝑠 cos \varphi_s = 3𝑉_𝑠𝐼'_π‘Ÿ cos \varphi_{r} + 𝑃_𝐢= 3𝐼′_{π‘Ÿ}^{2}(𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠}) + 𝑃_𝐢 … (15)}$$

The air-gap power per phase in the induction motor is given by,

$$\mathrm{𝑃_𝑔 = 𝑉_𝑠𝐼′_π‘Ÿ cos \varphi_{r} = 𝐼′_{π‘Ÿ}^{2} \times (\frac{𝑅′_π‘Ÿ}{𝑠})}$$

$$\mathrm{⇒ 𝑃_𝑔 =\frac{𝑉_𝑠^2}{(𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠})^2+ (𝑋_𝑠 + 𝑋′_π‘Ÿ)^2}\times (\frac{𝑅′_π‘Ÿ}{𝑠}) … (16)}$$

Therefore, the torque developed by the motor is given by,

$$\mathrm{\tau_𝑑 =\frac{𝑃_𝑔}{\omega_S}= \frac{𝑉_𝑠^2}{\omega_S[(𝑅𝑠 +\frac{𝑅′_π‘Ÿ}{𝑠})^2+ (𝑋_𝑠 + 𝑋′_π‘Ÿ)^2}\times (\frac{𝑅′_π‘Ÿ}{𝑠})… (17)}$$

raja
Published on 24-Aug-2021 07:35:21
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