Torque Equation of Three-phase Induction Motor

Torque Equation of Three-phase Induction Motor

Let us define some parameters to find the torque produce by the induction motor.
\tau_d = Developed torque
n_s = Synchronous speed
n_r = Rotor speed
P_g = Power transfer from stator to rotor = air gap power
P_m = Total mechanical power developed by rotor

(1)   \begin{equation*} P_g = \omega_s \tau_d = 2 \pi n_s \tau_d \end{equation*}

(2)   \begin{equation*} P_m = \omega_r \tau_d = 2 \pi n_r \tau_d \end{equation*}

Developed Torque (\tau_d)

The torque generated by the conversion of electrical to mechanical power is known as the developed torque. This torque is also known as the electromagnetic torque. The developed torque define as ratio of the mechanical power developed by rotor to the mechanical angular velocity of the rotor.


(3)   \begin{equation*} \begin{align} & \tau_d = \frac {P_m}{\omega_r} \\ & P_m = (1-s) P_g \\ & \omega_r = (1-s) \omega_s \\ \end{align} \end{equation*}

So, developed torque

(4)   \begin{equation*} \tau_d=\frac{(1-s) P_g}{(1-s) \omega_s}=\frac{P_g}{\omega_s} \end{equation*}

This torque is differs from the torque actually available at the terminals by the friction and windage torque in the induction motor. As shown in equation (4), the developed torque depends on the air gap power P_g and synchronous speed \omega_s. The synchronous speed is independent of loading condition and it is constant. Hence, we get torque \tau_d if air gap power P_g is known.

Torque equation

Electrical power generate in rotor

(5)   \begin{equation*} \begin{align} &= 3\:E_2_s\: I_2_s\: cos\phi_2_s \\ &= 3\:E_2_s\:\frac{E_2_s}{Z_2_s}\:\frac{R_2}{Z_2_s} \: =\: \frac{3\:E_2_s^2\:R_2}{Z_2_s^2} \\ &= \frac{3\:s^2\:E_2_0^2\:R_2}{R_2^2 + s^2\:X_2_0^2} \end{align} \end{equation*}

The power generates in rotor is same as rotor copper loss.


Rotor copper loss = Electrical power generate in rotor

Input power to rotor = 2\pi n_s \tau_d

slip \times rotor input = rotor copper loss

(6)   \begin{equation*} \begin{align}  s \times 2 \pi n_s \tau_d &= \frac{3\:s^2\:E_2_0 ^2\:R_2}{R_2^2 + s^2\:X_2_0^2} \\ \tau_d &= \frac{3\:E_2_0^2}{2\:\pi\:n_s} \frac{s\:R_2}{R_2^2 + s^2\:X_2_0^2} \\ So, \tau_d &=\frac{k\:s\:E_2_0^2\:R_2}{R_2^2 + s^2\:X_2_0^2} \\ where \: k &= \frac{3}{2\pi n_s}= \frac{3}{\omega_s}= constant \end{align} \end{equation*}

Starting Torque

At starting condition, rotor speed is zero. Therefore, slip is equal to 1. So, by putting, s=1 in above equation, we can obtain starting torque.

(7)   \begin{equation*} \tau_s_t = \frac{3\:E_2_0^2\:R_2}{2\:\pi\:n_s(R_2^2 + \:X_2_0^2)} \end{equation*}

Torque at synchronous speed

At synchronous speed, slip is equal to zero. So, from equation (6) \tau_d=0.

Condition for maximum torque

From equation (6), torque in running condition is given as,

(8)   \begin{equation*} \tau_d = \frac{k\:s\:R_2\:E_2_0^2}{R_2^2 + s^2\:X_2_0^2} \end{equation*}

For the given supply voltage V_1, E_2_0 remains constant, if the impedance of stator winding assume as negligible.

Hence, k\:E_2_0=k_1= constant

(9)   \begin{equation*} \begin{align} \tau_d &= \frac{k_1\:s\:R_2}{R_2^2 + s^2\:X_2_0^2} \\ \tau_d &= \frac{k_1 R_2}{\frac{R_2^2}{s}+s\:X_2_0^2} \\ \end{equation*}

(10)   \begin{equation*}  \tau_d &=\frac{k_1\:R_2}{(\frac{R_2^2}{\sqrt s} - X_2_0 \sqrt s)^2+2\:R_2\:X_2_0} \end{align} \end{equation*}

For maximum value of developed torque \tau_d develop when the right hand side of equation (10) is a maximum which is possible when

(11)   \begin{equation*} \frac{R_2}{\sqrt s} - X_2_0 \sqrt s = 0 \end{equation*}

(12)   \begin{equation*}  R_2 = s\:X_2_0 \end{equation*}

(13)   \begin{equation*} R_2 = X_s \end{equation*}

So, The maximum torque is

(14)   \begin{equation*} \begin{align} \tau_d_m_a_x &= \frac{k\:s\:R_2\:E_2_0^2}{R_2^2+R_2^2} \\ &= \frac{k\:s\:E_2_0^2}{2\:R_2} \\ &= \frac{k\:s\:E_2_0^2}{2\:s\:X_2_0} \end{equation*}

(15)   \begin{equation*}  \tau_d_m_a_x = \frac{k\:E_2_0^2}{2\:X_2_0} \end{equation*}

So, from equation (15), we can conclude two things,

  1. \tau_d_m_a_x is independent of the rotor resistance.
  2. \tau_d_m_a_x is inversely proportional to the standstill reactance of rotor.

Related article:

Working of Three-phase Induction Motor

Advantages and Disadvantages of Induction Motor

Induction Motor: Construction and Classification

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