Induction motor equivalent circuit from two-wattmeter tests

In the two-wattmeter method the total 3-phase power is the algebraic sum of the readings. The no-load test gives the shunt branch, the blocked-rotor test the series branch. Star connection: $V_{ph} = V_L/\sqrt{3}$, $I_{ph} = I_L$.

No-load test

$$\begin{aligned} P_0 &= W_1 + W_2 = 5000 + (-3200) = 1800\ \text{W} \\ V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ \cos\phi_0 &= \frac{P_0}{\sqrt{3}V_L I_0} = \frac{1800}{\sqrt{3}\times 400 \times 20} = 0.1299, \quad \sin\phi_0 = 0.9915 \\ I_w &= I_0\cos\phi_0 = 20 \times 0.1299 = 2.598\ \text{A} \\ I_\mu &= I_0\sin\phi_0 = 20 \times 0.9915 = 19.83\ \text{A} \\ R_0 &= \frac{V_{ph}}{I_w} = \frac{230.94}{2.598} = 88.89\ \Omega \\ X_0 &= \frac{V_{ph}}{I_\mu} = \frac{230.94}{19.83} = 11.65\ \Omega \end{aligned}$$

(The negative reading $W_2$ confirms pf below 0.5, as expected at no load.)

Blocked-rotor test

$$\begin{aligned} P_{sc} &= W_1 + W_2 = 2300 + 750 = 3050\ \text{W} \\ Z_{eq} &= \frac{V_{sc}/\sqrt{3}}{I_{sc}} = \frac{28.87}{60} = 0.4811\ \Omega \\ R_{eq} &= \frac{P_{sc}}{3I_{sc}^2} = \frac{3050}{3 \times 60^2} = 0.2824\ \Omega \\ X_{eq} &= \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{0.4811^2 - 0.2824^2} = 0.3895\ \Omega \end{aligned}$$

Stator resistance is not given, so the usual split is assumed:

$$\begin{aligned} R_1 &= R_2' = \frac{R_{eq}}{2} = 0.1412\ \Omega \\ X_1 &= X_2' = \frac{X_{eq}}{2} = 0.1948\ \Omega \end{aligned}$$

Note: the power factors were found from $P/(\sqrt{3}VI)$ using the measured V, I and total power; the ratio formula $\tan\phi = \sqrt{3}(W_1 - W_2)/(W_1 + W_2)$ gives different values ($\cos\phi_0 \approx 0.126$ at no load and $\cos\phi_{sc} \approx 0.75$ in the blocked-rotor test, against 0.130 and 0.587 here) because the given readings are not consistent with the measured V and I.

 I1  0.1412  j0.1948      j0.1948  0.1412
 o--[====]--[~~~~]--+---+--[~~~~]--[====]--+
                    |   |                  |
 V=230.94 V    R0=88.89 jX0=11.65   0.1412(1-s)/s
                    |   |                  |
 o------------------+---+------------------+

Answer (per phase, referred to stator): $R_0 = 88.89\ \Omega$, $X_0 = 11.65\ \Omega$, $R_1 = R_2' = 0.1412\ \Omega$, $X_1 = X_2' = 0.1948\ \Omega$.