OC and SC tests for efficiency and regulation

The open-circuit (OC) and short-circuit (SC) tests find the equivalent-circuit parameters, iron loss and copper loss of a transformer without actually loading it, so very little energy is used.

Open-circuit (no-load) test

  AC   +--(A)--+--[W]--+--------+      +------o
 supply|       |       |        |  LV  |  HV  open
 (var.)|      (V)      |   ||   | wdg  | wdg
       +-------+-------+--------+      +------o

• Done on the LV side with HV open, at rated voltage and frequency (using an autotransformer/variac).
• Readings: $V_0$, $I_0$, $P_0$.
• Since $I_0$ is only 2-5% of rated current, copper loss is negligible, so $P_0$ = iron loss $P_i$.
• From readings: $\cos\phi_0 = \frac{P_0}{V_0I_0}$, $I_w = I_0\cos\phi_0$, $I_\mu = I_0\sin\phi_0$, $R_0 = \frac{V_0}{I_w}$, $X_0 = \frac{V_0}{I_\mu}$.

Short-circuit test

  AC   +--(A)--+--[W]--+--------+      +------+
 supply|       |       |        |  HV  |  LV  | short
 (low) |      (V)      |   ||   | wdg  | wdg  | (thick
       +-------+-------+--------+      +------+  link)

• Done on the HV side with LV shorted. Voltage is raised slowly from zero until rated current flows; this needs only 5-10% of rated voltage.
• Readings: $V_{sc}$, $I_{sc}$, $P_{sc}$.
• Flux is very small at this low voltage, so iron loss is negligible and $P_{sc}$ = full-load copper loss.
• From readings: $Z_{01} = \frac{V_{sc}}{I_{sc}}$, $R_{01} = \frac{P_{sc}}{I_{sc}^2}$, $X_{01} = \sqrt{Z_{01}^2 - R_{01}^2}$ (referred to HV).

Finding efficiency

At a fraction $x$ of full load and power factor $\cos\phi$:

$$\eta = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2P_{cu,FL}}\times 100$$

where $P_i$ is from the OC test and $P_{cu,FL}$ from the SC test (scaled by $(I_{rated}/I_{sc})^2$ if the test was not at rated current). Maximum efficiency occurs at $x = \sqrt{P_i/P_{cu,FL}}$.

Finding voltage regulation

Using $R_{01}$ and $X_{01}$ (or $R_{02}$, $X_{02}$) from the SC test:

$$\%\text{Reg} = \frac{I(R\cos\phi \pm X\sin\phi)}{V}\times 100$$

(+ for lagging, − for leading pf). Equivalently, $\%\text{Reg} = \%R\cos\phi \pm \%X\sin\phi$ where $\%R = \frac{P_{sc}}{S}\times100$ and $\%Z = \frac{V_{sc}}{V_{rated}}\times100$ when the test is at rated current.

Example: if a 10 kVA transformer has $P_i = 100$ W and $P_{cu,FL} = 200$ W, full-load efficiency at upf = $\frac{10000}{10300} = 97.1\%$.

Advantages: low power consumption, no need for a real load, and both losses found separately so efficiency at any load and pf can be predicted.