Electrical Machines · Chapter 65

No-Load and Blocked-Rotor Tests

Part 4 · Induction Motors — measuring the equivalent circuit with two simple experiments.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  1. Explain why the no-load test isolates the shunt (magnetising) branch and the rotational losses.

  2. Separate stator copper loss, core loss and friction–windage loss from no-load readings.

  3. Explain why the blocked-rotor test isolates the series branch, and why it is done at reduced voltage.

  4. Compute every parameter of the approximate equivalent circuit from the two tests.

  5. Predict the starting current and starting torque at rated voltage from blocked-rotor data.

Section 65-1

Why Two Tests Are Enough

Chapter 64 gave the per-phase equivalent circuit: a series branch \(R_1 + jX_1\), a shunt branch \(R_c \parallel jX_m\), and the rotor branch \(R_2'/s + jX_2'\). Exactly as for the transformer in Part 3, we need two experiments that each make one part of the circuit dominate.

No-load test ≈ open-circuit test

At no load the slip is tiny, so \(R_2'/s\) is enormous and the rotor branch is effectively open. Current flows only through \(R_1 + jX_1\) and the shunt branch.

Blocked-rotor test ≈ short-circuit test

With the rotor held, \(s = 1\), so \(R_2'/s = R_2'\) is small and the rotor branch short-circuits the shunt branch. The series impedance dominates.

One important difference from the transformer. The induction motor has an air gap, so its magnetising current is large — typically 25–40 % of rated current, against 2–5 % for a transformer. The shunt branch therefore cannot be ignored as casually, and the stator voltage drop at no load is not negligible.
Section 65-2

The DC Resistance Test

The stator resistance is measured first by applying a small DC voltage across two line terminals. For a star-connected stator the current passes through two phases in series:

\[R_{1,\text{dc}} = \frac{V_\text{dc}}{2\,I_\text{dc}} \qquad \text{(star)}, \qquad R_{1,\text{dc}} = \frac{3}{2}\,\frac{V_\text{dc}}{I_\text{dc}} \qquad \text{(delta)}\]

The DC value is usually multiplied by 1.1–1.3 to allow for skin effect and for the higher operating temperature. Since no rotor current flows with DC excitation, this test tells us nothing about \(R_2'\).

Section 65-3

The No-Load Test

The motor runs uncoupled at rated voltage and frequency. Two wattmeters (or a power analyser) record the total input \(P_0\), together with the line voltage \(V_0\) and line current \(I_0\).

  • The input supplies only losses: stator copper loss \(3I_0^2R_1\), core loss \(P_\text{core}\), and friction and windage \(P_\text{fw}\). Rotor copper loss is negligible because \(s \approx 0\).

  • The rotational loss is \(P_\text{rot} = P_0 - 3I_0^2R_1 = P_\text{core} + P_\text{fw}\).

  • To split \(P_\text{rot}\), the test is repeated at decreasing voltage and \(P_\text{rot}\) is plotted against \(V^2\). The straight line extrapolated to \(V = 0\) gives \(P_\text{fw}\), which does not depend on voltage.

⚙
No-load relations (per phase)
The no-load reactance is almost \(X_1 + X_m\)
\[\cos\phi_0 = \frac{P_0}{\sqrt{3}\,V_0 I_0}, \qquad Z_\text{NL} = \frac{V_0/\sqrt{3}}{I_0}, \qquad X_\text{NL} \approx X_1 + X_m\]

Because \(\cos\phi_0\) is very low (about 0.1–0.25), \(X_\text{NL} \approx Z_\text{NL}\) to a good approximation.

Separating friction & windage from core loss V² (applied voltage squared) P_rot P_fw (intercept) core loss ∝ V²
Figure 65.1 — Rotational loss plotted against \(V^2\): the intercept at zero voltage is the friction and windage loss.
Section 65-4

The Blocked-Rotor Test

The rotor is locked and a reduced voltage is applied — just enough to circulate rated current, typically 10–25 % of rated voltage. Readings: \(V_\text{BR}\), \(I_\text{BR}\), \(P_\text{BR}\).

  • Core loss is negligible because the voltage, and hence the flux, is low.

  • The magnetising branch is bypassed by the low-impedance rotor branch, so the input is almost entirely copper loss.

  • IEEE Std 112 recommends performing the test at 25 % of rated frequency, because a running motor sees rotor frequency \(sf\), only a few hertz. Testing at 50 Hz overstates \(R_2'\) through skin effect, especially in deep-bar and double-cage rotors (Chapter 70).

\[Z_\text{BR} = \frac{V_\text{BR}/\sqrt{3}}{I_\text{BR}}, \qquad R_\text{eq} = \frac{P_\text{BR}}{3 I_\text{BR}^2} = R_1 + R_2', \qquad X_\text{eq} = \sqrt{Z_\text{BR}^2 - R_\text{eq}^2} = X_1 + X_2'\]
Table 65.1 — Empirical split of \(X_\text{eq}\) (IEEE Std 112 / NEMA design classes).
Rotor design\(X_1\)\(X_2'\)
Wound rotor; NEMA Design A and D0.5 \(X_\text{eq}\)0.5 \(X_\text{eq}\)
NEMA Design B0.4 \(X_\text{eq}\)0.6 \(X_\text{eq}\)
NEMA Design C0.3 \(X_\text{eq}\)0.7 \(X_\text{eq}\)
Section 65-5

Assembling the Equivalent Circuit

  1. DC test → \(R_1\).

  2. Blocked-rotor test → \(R_\text{eq}\) and \(X_\text{eq}\); then \(R_2' = R_\text{eq} - R_1\), and split \(X_\text{eq}\) using Table 65.1.

  3. No-load test → \(X_m \approx X_\text{NL} - X_1\); the core-loss resistance follows from \(R_c = 3E_0^2/P_\text{core}\) if it is to be retained.

Order matters. \(X_1\) is needed to extract \(X_m\), so the blocked-rotor test must be reduced before the no-load test.
1 Worked Example 65.1 — Parameters of a 400 V Motor

Problem. A 400 V, 50 Hz, 4-pole, star-connected induction motor (NEMA Design A) gave the following results. DC test: \(R_1 = 0.5~\Omega\) per phase (corrected). No-load: 400 V, 8 A, 1200 W. Blocked rotor: 100 V, 25 A, 1500 W. Find the rotational loss, the no-load power factor and the equivalent-circuit parameters.

No-load test.

\[P_{\text{Cu},0} = 3(8)^2(0.5) = 96~\text{W}, \qquad P_\text{rot} = 1200 - 96 = 1104~\text{W}\]
\[\cos\phi_0 = \frac{1200}{\sqrt{3}(400)(8)} = 0.217, \qquad Z_\text{NL} = \frac{230.9}{8} = 28.87~\Omega\]

Blocked-rotor test.

\[Z_\text{BR} = \frac{100/\sqrt{3}}{25} = 2.309~\Omega, \qquad R_\text{eq} = \frac{1500}{3(25)^2} = 0.800~\Omega\]
\[X_\text{eq} = \sqrt{2.309^2 - 0.800^2} = 2.166~\Omega \;\Rightarrow\; X_1 = X_2' = 1.083~\Omega\]
\[R_2' = 0.800 - 0.5 = 0.300~\Omega, \qquad X_m \approx 28.87 - 1.08 = 27.78~\Omega\]

The no-load power factor of about 0.2 is typical: almost all of the no-load current magnetises the air gap.

2 Worked Example 65.2 — Starting Current and Torque

Problem. For the motor of Example 65.1, estimate the starting current and starting torque at rated voltage.

The blocked-rotor impedance is linear, so current scales with voltage:

\[I_\text{st} = 25 \times \frac{400}{100} = 100~\text{A}\]

At standstill the air-gap power equals the rotor copper loss, and \(\omega_s = 2\pi(1500)/60 = 157.1\) rad/s:

\[T_\text{st} = \frac{3 I_\text{st}^2 R_2'}{\omega_s} = \frac{3(100)^2(0.3)}{157.1} = 57.3~\text{N·m}\]

This neglects the magnetising current drawn through \(X_m\); at standstill it is small compared with 100 A, so the estimate is good.

Section 65-6

Summary and Common Mistakes

Key Points
  • No-load test (rated V, no load): gives rotational loss and \(X_1 + X_m\).

  • Blocked-rotor test (reduced V, rated I): gives \(R_1 + R_2'\) and \(X_1 + X_2'\).

  • DC test: gives \(R_1\), which separates \(R_2'\).

  • Starting current scales linearly, and starting torque as the square, of the applied voltage.

Common Mistakes
  • Using line voltage with per-phase current when computing impedance of a star-connected motor.

  • Forgetting to subtract stator copper loss from the no-load input.

  • Dividing the DC reading by one phase instead of two for a star connection.

Section 65-7

Chapter Review

Practice Problems
  1. P65.1 A 415 V, star-connected motor draws 10 A and 1500 W at no load. Its stator resistance is 0.6 Ω per phase. Find the rotational loss and no-load power factor.

    Show answer
    \[P_\text{rot} = 1500 - 3(10)^2(0.6) = 1320~\text{W}, \qquad \cos\phi_0 = \frac{1500}{\sqrt{3}(415)(10)} = 0.209\]
  2. P65.2 In a blocked-rotor test a star-connected motor took 80 V, 20 A, 1200 W. With \(R_1 = 0.4~\Omega\), find \(R_2'\) and \(X_1 = X_2'\).

    Show answer
    \[Z = \frac{46.19}{20} = 2.309~\Omega,\; R_\text{eq} = \frac{1200}{1200} = 1.0~\Omega,\; X_\text{eq} = 2.082~\Omega\]

    Hence \(R_2' = 0.6~\Omega\) and \(X_1 = X_2' = 1.041~\Omega\).

Multiple-Choice Questions
  1. MCQ 1. The blocked-rotor test is performed at:
    (a) rated voltage   (b) reduced voltage, rated current   (c) rated voltage, reduced current   (d) zero voltage

    Show answer
    (b) — rated voltage with the rotor locked would draw five to eight times rated current.
  2. MCQ 2. Friction and windage loss is found from the no-load test by:
    (a) the DC test   (b) extrapolating \(P_\text{rot}\) versus \(V^2\) to zero voltage   (c) the blocked-rotor test   (d) measuring slip

    Show answer
    (b) — core loss vanishes at zero voltage; mechanical loss does not.
Conceptual Questions
  1. Why is the no-load current of an induction motor much larger, as a fraction of rated current, than that of a transformer?

  2. Why does testing at reduced frequency give a more accurate \(R_2'\) for deep-bar rotors?

  3. Why must the blocked-rotor results be used before \(X_m\) can be found?

Looking Ahead

With measured parameters in hand, Chapter 66 draws the circle diagram — a graphical construction, built from the same two tests, that reads off current, power factor, torque and efficiency at any load.