Solved Problems on Single-Phase Induction Motor Testing - No-Load and Blocked-Rotor

Problem-1

A 220 V, single-phase induction motor gave the following results:

\[\begin{aligned} \text{Blocked-rotor test} & : 120~\mathrm{V},~9.6~\mathrm{A},~460~\mathrm{W} \\ \text{No-load test} & : 220~\mathrm{V},~4.6~\mathrm{A},~125~\mathrm{W} \end{aligned}\]

The stator winding resistance is \(1.5~\Omega\), and during the blocked-rotor test the starting winding is open. Determine the equivalent circuit parameters. Also, find the friction and windage losses.

Solution

  • Blocked-rotor test:

\[\begin{aligned} V_{\mathrm{sc}} & =120 \mathrm{~V}, ~\mathrm{I}_{\mathrm{sc}}=9.6 \mathrm{~A}, ~P_{\mathrm{sc}}=460 \mathrm{~W} \\ Z_e & =\frac{V_{\mathrm{sc}}}{I_{\mathrm{sc}}}=\frac{120}{9.6}=12.5 ~\Omega \\ R_e & =\frac{P_{\mathrm{sc}}}{I_{s c}^2}=\frac{460}{(9.6)^2}=4.99~ ~\Omega \\ \mathrm{X}_e & =\sqrt{Z_e^2-R_e^2}=\sqrt{(12.5)^2-(4.99)^2} =11.46 ~\Omega \\ X_{1 m} & =X_2^{\prime}=\frac{1}{2} \mathrm{X}_e=\frac{1}{2} \times 11.46=5.73~ ~\Omega \\ R_{1 m} & =1.5 ~\Omega \\ R_e & =R_{1 m}+R_2^{\prime} \\ R_2^{\prime} & =R_e-R_{1 m}=4.99-1.5=3.49~ ~\Omega \end{aligned}\]
  • \[V_0=220 \mathrm{~V}, ~I_0=4.6 \mathrm{~A}, ~ P_0=125 \mathrm{~W}\]
    No-load test

  • No-load power factor

    \[\cos \phi_0=\frac{P_0}{V_0 I_0}=\frac{125}{220 \times 4.6}=0.1235\]
    \[\begin{aligned} & \therefore \quad \sin \phi_0=0.9923 \\ & \\ & \quad \mathrm{Z}_0=\frac{V_0}{I_0}=\frac{220}{4.6}=47.83 ~\Omega \\ & \therefore \quad \mathrm{X}_0=\mathrm{Z}_0 \sin \phi_0=47.83 \times 0.9923=47.46 ~\Omega \end{aligned}\]

  • Core, friction and windage losses

\[\begin{aligned} & =\text { power input to motor at no load - no-load copper loss } \\ & =P_0-I_0^2\left(R_{1 m}+\frac{R_2^{\prime}}{4}\right) \\ & =125-(4.6)^2\left(1.5+\frac{3.49}{4}\right)=74.8 \mathrm{~W} \end{aligned}\]

Problem-2

A \(115~\mathrm{V}\), \(60~\mathrm{Hz}\), single-phase, split-phase induction motor is tested to yield the following data.

Voltage (V) Current (A) Power (W)
With auxiliary winding open
\(\quad\) No-load test 115 3.2 55.17
\(\quad\) Blocked-rotor test 25 3.72 86.23
With main winding open
\(\quad\) Blocked-rotor test 121 1.21 145.35

The main-winding resistance is \(2.5 ~\Omega\) and the auxiliary-winding resistance is \(100 ~\Omega\). Determine the equivalent circuit parameters of the motor.

Solution

  • From the blocked-rotor test on the main winding with the auxiliary winding open

    \[\begin{aligned} Z_{bm}& = \frac{25}{3.72} = 6.72 ~\Omega \\ R_{bm}& = \frac{86.23}{3.72^{2}} = 6.23 ~\Omega \\ X_{bm}& = \sqrt{6.72^{2} - 6.23^{2}} = 2.52 ~\Omega \\ X_{1}&= X_{2} = 0.5 \times 2.52 = 1.26 ~\Omega \\ R_{2}& = 6.23 - 2.5 = 3.73 ~\Omega \end{aligned}\]

  • From the no-load test on the main winding with auxiliary winding open

    \[\begin{aligned} &Z_{nL} = \frac{115}{3.2} = 35.94 ~\Omega \\ &R_{nL} = \frac{55.17}{3.2^{2}} = 5.39 ~\Omega \\ &X_{nL} = \sqrt{35.94^{2} - 5.39^{2}} = 35.53 ~\Omega \\ &X_{m} = 2 \times 35.53 - 0.75 \times 2.52 = 69.17 ~\Omega \\ &P_{r}=55.17-3.2^{2}(2.5+0.25\times3.73)\approx20 \mathrm{W} \end{aligned}\]

  • From the blocked-rotor test on the auxiliary winding with the main winding open

    \[\begin{aligned} &Z_{ba} = \frac{121}{1.2} = 100.83 ~\Omega\\&R_{ba} = \frac{145.35}{1.2^{2}} = 100.94 ~\Omega\\&R_{2a} = 100.94 - 100 = 0.94 ~\Omega\\ &a = \sqrt{\frac{0.94}{3.73}} = 0.5 \end{aligned}\]

Problem-3

A \(230 \mathrm{~V}, 50 \mathrm{~Hz}\), 4-pole single-phase split phase induction motor is tested to yield the following results (with auxiliary winding open):

\[\begin{aligned} \text{No load test} &: 230 \mathrm{~V}, 6.4 \mathrm{~A}, 220 \mathrm{~W} \\ \text{Block rotor test} &: 82.5 \mathrm{~V}, 9.3 \mathrm{~A}, 500 \mathrm{~W} \end{aligned}\]

The main winding resistance is \(2.5 ~\Omega\), then determine the rotor resistance referred to stator ?

Solution

Impedance under block rotor condition,

\[\mathrm{Z}_b=\mathrm{V}_b / \mathrm{l}_b=82.5 / 9.3=8.87 ~\Omega\]

Resistance under block rotor condition,

\[\mathrm{R}_b=\mathrm{W}_b /(\mathrm{l}_b)^ 2=500 /(9.3) 2=5.78 ~\Omega\]

Rotor resistance referred to stator,

\[\begin{aligned} & \mathrm{R}^{'}_ 2=\mathrm{R}_b-\mathrm{R}_1 \\ & =5.78-2.5=3.28 ~\Omega \end{aligned}\]