Induction Motor Torque: Solved Problems

Demonstrative Video

PROBLEM 01

Problem-1

Problem Statement

A 150-KW, 3000-V, 50-Hz, 6-pole star-connected induction motor has a star connected slip-ring rotor with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.1 Ω/phase and its per phase leakage reactance is 3.61 mH. The stator impedance may be neglected. On rated voltage with short-circuited slip rings, determine

starting current
starting torque

Solution

\[\begin{aligned} I_{st} & =\dfrac{V}{\sqrt{\left(R_{2}^{\prime}\right)^{2}+\left(X_{2}^{\prime}\right)^{2}}}=\dfrac{3000/\sqrt{3}}{\sqrt{1.3^{2}+14.7^{2}}}=117.4\mathrm{A}\\ N_{s} & =\dfrac{120f}{P}=\dfrac{120\times50}{6}=1000\mathrm{rpm}=\left(50/3\right)\mathrm{rps}\\ T_{st} & =\dfrac{3}{2\pi N_{s}}\cdot\dfrac{V^{2}R_{2}^{\prime}}{\left(R_{2}^{\prime}\right)^{2}+\left(X_{2}^{\prime}\right)^{2}} \\ & =\dfrac{3}{2\pi\left(50/3\right)}\times\dfrac{\left(3000/\sqrt{3}\right)^{2}\times1.3}{1.3^{2}+14.7^{2}}\\ & =513\mathrm{N-m} \end{aligned}\]

Solution-1

\[\begin{aligned} K & =\dfrac{1}{3.6}\\ R_{2}^{\prime} & =R_{2}/K^{2}=3.6^{2}\times0.1=1.3~\Omega\\ X_{2} & =2\pi\times50\times3.61\times10^{-3}=1.13~\Omega\\ X_{2}^{\prime} & =3.6^{2}\times1.13=14.7~\Omega \end{aligned}\]

PROBLEM 02

Problem-2

Problem Statement

Calculate the torque exerted by an 8-pole, 50-Hz, 3-phase induction motor operating with a 4% slip which develops a maximum torque of 150 kg-m at a speed of 660 rpm. The resistance per phase of the rotor is 0.5 \(\Omega\)

Solution

\[\begin{aligned} T_{max} & =k\Phi E_{2}\dfrac{s_{b}}{2R_{2}}=k\Phi E_{2}\dfrac{0.12}{2\times0.5}=0.12k\Phi E_{2}\\ T & =k\Phi E_{2}\dfrac{sR_{2}}{R_{2}^{2}+\left(sX_{2}\right)^{2}}=k\Phi E_{2}\dfrac{0.04\times0.5}{0.5^{2}+\left(0.04\times4.167\right)^{2}}\\ & =\dfrac{0.02k\Phi E_{2}}{0.2778}\\ \dfrac{T}{T_{max}} & =\dfrac{T}{150}=\dfrac{0.02}{0.2778\times0.12}\\ \Rightarrow T & =90\mathrm{kg-m} \end{aligned}\]

Solution-2

\[\begin{aligned} N_{s} & =120\times50/8=750\mathrm{rpm}\\ N\text{at maxm. torque} & =660\mathrm{rpm}\\ \text{Corresponding slip}~s_{b} & =\dfrac{750-660}{750}=0.12\\ \text{For maxm. torque} & \Rightarrow R_{2}=s_{b}X_{2}\\ & \rightarrow X_{2}=\dfrac{R_{2}}{s_{b}}=0.5/0.12=4.167\Omega \end{aligned}\]

PROBLEM 03

Problem-3

Problem Statement

A 3-phase, 50-Hz, 8-pole, induction motor has full-load slip of 2%. The rotor resistance and standstill rotor reactance per phase are 0.001 \(\Omega\) and 0.005 \(\Omega\), respectively. Find the ratio of the maximum to full load torque and the speed at which the maximum torque occurs

Solution

Solution-3

\[\begin{aligned} N_{s} & =120\times50/8=750\mathrm{rpm}\\ s_{b} & =\dfrac{R_{2}}{X_{2}}=\dfrac{0.001}{0.005}=0.2\\ N & =750\times\left(1-0.2\right)=600\mathrm{rpm}\\ \dfrac{T_{FL}}{T_{max}} & =\dfrac{2s_{b}s_{FL}}{s_{b}^{2}+s_{FL}^{2}}=\dfrac{2\times0.2\times0.02}{0.20^{2}+0.02^{2}}\\ & =0.1980 \end{aligned}\]

PROBLEM 04

Problem-4

Problem Statement

For a 3-phase slip ring induction motor, the maximum torque is 2.5 times the full-load torque and the starting torque is 1.5 times the full-load torque. Determine the percentage reduction in rotor circuit resistance to get a full-load slip of 3%. Neglect the stator impedance

Solution

\[\begin{aligned} \dfrac{T_{fl}}{T_{st}} & =\dfrac{2as}{a^{2}+s^{2}}\\ \Rightarrow\dfrac{2}{2.5} & =\dfrac{2a\times0.03}{a^{2}+0.03^{2}}\\ \Rightarrow a^{2}-0.15a+0.03^{2} & =0\\ \Rightarrow a & =0.1437=\dfrac{R_{2}^{\prime}}{X_{2}}\\ R_{2}^{\prime} & =0.1437X_{2} \end{aligned}\]
When F.L slip = 0.03
\[=\dfrac{\left(X_{2}/3\right)-0.1437X_{2}}{\left(X_{2}/3\right)}\times100=56.8\%\]
% reduction in rotor resistance is

Solution-4

\[\begin{aligned} T_{max} & =2.5T_{fl}\\ T_{st} & =1.5T_{fl}\\ \dfrac{T_{st}}{T_{max}} & =\dfrac{1.5}{2.5}=\dfrac{3}{5}=\dfrac{2s_{b}}{1+s_{b}^{2}}\\ & \Rightarrow s_{b}=1/3\\ & \Rightarrow s_{b}=R_{2}/X_{2}=1/3\\ & \Rightarrow R_{2}=X_{2}/3 \end{aligned}\]