Why iron loss neglected in SC test; parameters

Why iron loss is neglected in the SC test

In the short-circuit test only a small voltage (about 5 to 10% of the rated value) is applied, just enough to circulate full-load current in the shorted secondary. The core flux is proportional to the applied voltage, so it is only about 5 to 10% of normal flux. Core loss varies approximately as the square of the flux, so it becomes about 0.25 to 1% of the rated core loss, which is negligible. The wattmeter therefore reads the copper loss alone. In this problem $30/400 = 7.5\%$ of rated voltage, so the core loss is about $0.075^2\times110 \approx 0.6$ W against 350 W.

Equivalent circuit parameters (referred to the primary)

Primary = HV side (400 V), secondary = LV side (200 V).

Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{110}{200\times1.5} = 0.3667 \\ I_w &= I_0\cos\phi_0 = 0.55\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.396\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 363.6\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 143.3\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{30}{18} = 1.667\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{350}{18^2} = 1.08\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.269\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the HV side

The turns ratio is $K = 400/200 = 2$, so $K^2 = 4$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_0' &= 363.6 \times 4 = 1455\ \Omega \\ X_0' &= 143.3 \times 4 = 573.3\ \Omega \end{aligned}$$

Equivalent circuit referred to the HV side

   Req=1.08 ohm   Xeq=1.27 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=1455] [X0=573.3]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 1.08\ \Omega$, $X_{eq} = 1.269\ \Omega$ ($Z_{eq} = 1.667\ \Omega$), shunt branch $R_0 = 1455\ \Omega$ in parallel with $X_0 = 573.3\ \Omega$.

The shunt branch carries $I_0$ referred to the primary $= 1.5/2 = 0.75$ A.

Answer (referred to primary): $R_{01} = 1.080\ \Omega$, $X_{01} = 1.269\ \Omega$, $R_0 = 1454.5\ \Omega$, $X_0 = 573.3\ \Omega$.