The following measurements were taken on a 5 HP, four-pole, 208 V, three-phase, 60 Hz, star-connected induction motor.

No load
(60 Hz) Locked rotor
(15 Hz) DC test

Voltage (V) 208 35 20

Current (A) 4 12 25

Three-phase power (W) 250 450

Determine the full per-phase equivalent circuit parameters of the induction motor.

Question

The following measurements were taken on a 5 HP, four-pole, 208 V, three-phase, 60 Hz, star-connected induction motor.

	No load (60 Hz)	Locked rotor (15 Hz)	DC test
Voltage (V)	208	35	20
Current (A)	4	12	25
Three-phase power (W)	250	450

Determine the full per-phase equivalent circuit parameters of the induction motor.

Accepted Answer

(a) DC test. For this star-connected motor, the measured stator resistance is the value for two phases connected in series. Thus

[latex]R_1=\frac{1}{2}\left(\frac{20}{25} ight)=\mathbf{0 . 4} \boldsymbol{\Omega}[/latex]

(b) No-load test. On no-load the power delivered is due to stator copper loss, iron loss and friction and windage loss. Since no separate data are taken to obtain the friction and windage loss from a graph, they will be incorporated in the core loss for the purposes of this calculation.

[latex]P_{ ext {no-load }}=250 \mathrm{~W}[/latex]

[latex] ext { Stator loss }=3 imes 4^2 imes 0.4=19.2 \mathrm{~W}[/latex]

Iron loss + windage and friction [latex]=230.8 \mathrm{~W}[/latex]

This is a three-phase quantity, therefore

Real power dissipated in [latex]R_{\mathrm{c}}=76.9 \mathrm{~W} / \mathrm{ph}[/latex]

From Fig. 36.15(b), the series equivalent circuit, [latex]76.9 \mathrm{~W}[/latex] is dissipated in [latex]R_{\mathrm{c}}^{\prime}[/latex]. The current through [latex]R_{\mathrm{c}}^{\prime}[/latex] is [latex]4 \mathrm{~A}[/latex].

Thus

[latex]\quad R_{\mathrm{c}}^{\prime}=\frac{76.9}{4^2}=4.8 \Omega \quad[/latex] and [latex]\quad R_1+R_{\mathrm{c}}^{\prime}=5.2 \Omega[/latex]

The no-load impedance per phase is obtained from

[latex]Z_{\mathrm{NL}}=\frac{208 \mathrm{~V}}{\sqrt{3} imes 4 \mathrm{~A}}=30 \Omega[/latex]

Since [latex]\quad Z_{\mathrm{NL}}^2=\left(R_1+R_{\mathrm{c}}^{\prime} ight)^2+\left(X_1+X_{\mathrm{m}}^{\prime} ight)^2 \quad[/latex] (see Fig. 36.15(b)

Hence [latex]X_1+X_{\mathrm{m}}^{\prime}=\sqrt{ }\left(30^2-5.2^2 ight)=29.5 \Omega[/latex] (at [latex]60 \mathrm{~Hz}[/latex] )

(c) Locked rotor test. Refer to Fig. [latex]36.17[/latex] and equation [36.16].

Real power is dissipated in the stator and rotor resistance, hence

[latex]\begin{aligned}\left(R_1+R_2^{\prime} ight) &=\frac{450}{3(12)^2}=1.04 \Omega \R_2^{\prime} &=1.04-0.4=0.64 \Omega\end{aligned}[/latex]

The locked-rotor impedance per phase [latex](15 \mathrm{~Hz})[/latex] is obtained from

Since [latex]\begin{aligned}Z_{\mathrm{LR}} &=\frac{35 \mathrm{~V}}{\sqrt{3} imes 12 \mathrm{~A}}=1.68 \Omega \Z_{\mathrm{LR}}^2 &=\left(R_1+R_2^{\prime} ight)^2+\left(X_1+X_2^{\prime} ight)^2 \quad( ext { see Fig. 36.17) }\X_1+X_2^{\prime} &\left.=\sqrt{ }\left(1.68^2-1.04^2 ight)=1.32 \Omega \quad ext { (at } 15 \mathrm{~Hz} ight)\end{aligned}[/latex]

Reactance at [latex]60 \mathrm{~Hz}=\frac{60}{15} imes 1.32=5.28 \Omega[/latex]

Since it can be assumed that [latex]X_1=X_2^{\prime}[/latex], each reactance is [latex]2.64 \Omega[/latex].

From the no-load test:

[latex]X_1+X_{\mathrm{m}}^{\prime}=29.5 \Omega[/latex]

So [latex]\quad X_{\mathrm{m}}^{\prime}=29.5-2.64=26.9 \Omega[/latex]

To replace [latex]X_{\mathrm{m}}^{\prime}[/latex] and [latex]R_{\mathrm{c}}^{\prime}[/latex] by [latex]X_{\mathrm{m}}[/latex] and [latex]R_{\mathrm{c}}[/latex] we calculate the admittance of [latex]X_{\mathrm{m}}^{\prime}[/latex] in series with [latex]R_{\mathrm{c}}^{\prime}[/latex]; the real part of this will correspond to [latex]1 / R_{\mathrm{c}}[/latex] and the imaginary part corresponds to [latex]1 / X_{\mathrm{m}}[/latex].

[latex]\begin{aligned}\mathbf{Y} &=\frac{1}{R_{\mathrm{c}}^{\prime}+\mathrm{j} X_{\mathrm{m}}^{\prime}}=\frac{1}{4.8+\mathrm{j} 26.9} \&=\frac{4.8-\mathrm{j} 26.9}{4.8^2+26.9^2} \&=(6.42-\mathrm{j} 36) 10^3 \Omega^1\end{aligned}[/latex]

Hence [latex]\begin{aligned}&R_{\mathrm{c}}=\frac{1}{6.42 imes 10^3}=155 \Omega \&X_{\mathrm{m}}=\frac{1}{36 imes 10^3}=27.76 \Omega\end{aligned}[/latex]

The full per-phase equivalent circuit of the star-connected motor can now be drawn as shown in Fig. 36.18.

Question 36.5: The following measurements were taken on a 5 HP, four-pole, ...

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