Construction of Circle Diagram of Induction Motor

Digital Electronics Electron Electronics & Electrical

Circle Diagram of Induction Motor

The circle diagram is a graphical representation of the performances of an induction motor. It is very useful to study the performance of an induction motor under all operating conditions. The construction of the circle diagram of an induction motor is based on the equivalent circuit of the motor which is shown in Figure-1.

By applying KCL in the equivalent circuit, we can write,

$$\mathrm{𝐼_1 = 𝐼_0 + 𝐼′_2 … (1)}$$

If the phase voltage V1 is taken along the vertical axis of OY as shown in Figure-2. Then, the no-load current (I₀) lags behind the voltage V₁ by an angle ϕ₀. Where, ϕ₀ is the no-load power factor angle and is of the order of 60°-80° because to produce the required flux per pole in the air gap of the machine, a large magnetising current is needed.

Now, at no-load, the slip s = 0, then

$$\mathrm{\frac{𝑅_2}{𝑠}= ∞ \:𝑖.𝑒.\: open\: circuited \:at \;no − load}$$

$$\mathrm{\therefore 𝐼_0 =\frac{𝑉_1}{𝑍_{𝑁𝐿}}… (2)}$$

Where,

$$\mathrm{𝑍_{𝑁𝐿} = (𝑅_0 \:||\: 𝑗𝑋_0) = No \:load \:impedance}$$

Here, all the rotational losses are taken into account by R0. Then, the no-load losses are given by,>

$$\mathrm{𝑃_0 = 𝑉_1𝐼_0 cos \varphi_0 … (3)}$$

The rotor current referred to the stator side is given by,

$$\mathrm{𝐼′_2 =\frac{𝑉_1}{𝑍′_{𝑒1}}=\frac{𝑉_1}{\sqrt{(𝑅_1 +\frac{𝑅′_2}{𝑠})^2+ (𝑋_1 + 𝑋′_2)^2}}… (4)}$$

This rotor current I′₂ lags behind the voltage V1 by the impedance angle ϕ.

Refer Figure-3,

$$\mathrm{sin\:\varphi =\frac{𝑋_1 + 𝑋′_2}{\sqrt{(𝑅_1 +\frac{𝑅′_2}{𝑠})^2+ (𝑋_1 + 𝑋′_2)^2}}… (5)}$$

Hence, by combining the eqns. (4) & (5), we get,

$$\mathrm{𝐼′_2 =\frac{𝑉_1}{𝑋_1 + 𝑋′_2}sin \varphi; … (6)}$$

Eqn. (6) is in the form of 𝑟 = 𝑎 sin θ, which represents a circle with diameter a. Therefore, the locus of current (𝐼′₂) is a circle whose diameter is

$$\mathrm{diameter =\frac{𝑉_1}{𝑋_1 + 𝑋′_2}}$$

It is shown in Figure-4.

The radius of the circle is,

$$\mathrm{O′C =\frac{𝑉_1}{2(𝑋_1 + 𝑋′_2)}}$$

From eqn. (1), it is clear that the stator current I1 is found by combining the results shown in Figures-2, 3, and 4. Then, the resulting figure obtained is shown in Figure-5 which is known as the circle diagram of the induction motor.