Question

A 20 KVA, 2200/220V, 50 Hz transformer has O.C/S.C test results as follows: O.C.test : 220V, 4.2 A, 148 W (l. v. side), S.C. test : 86V, 10.5 A, 360 W (h. v. side). Determine the regulation at 0.8 P.F lagging at full load.

Answer 1

To determine the voltage regulation at 0.8 power factor lagging for the transformer, we first need to calculate the full-load current on the low voltage (LV) side, then find the equivalent reactance and resistance from the tests.

1. **Full-load Current (I_FL)**:
   \[
   I_{FL} = \frac{20,000 \text{ VA}}{220 \text{ V}} = 90.91 \text{ A}
   \]

2. **From O.C. Test**:
   \[
   V_{oc} = 220 \text{ V}, \quad I_{oc} = 4.2 \text{ A}, \quad P_{oc} = 148 \text{ W}
   \]
   \[
   R_{oc} = \frac{P_{oc}}{I_{oc}^2} = \frac{148}{(4.2)^2} \approx 8.45 \Omega
   \]
   \[
   Z_{oc} = \frac{V_{oc}}{I_{oc}} = \frac{220}{4.2} \approx 52.38 \Omega
   \]
   \[
   X_{oc} = \sqrt{Z_{oc}^2 - R_{oc}^2} \approx \sqrt{52.38^2 - 8.45^2} \approx 52.25 \Omega
   \]

3. **From S.C. Test**:
   \[
   V_{sc} = 86 \text{ V}, \quad I_{sc} = 10.5 \text{ A}, \quad P_{sc} = 360 \text{ W}
   \]
   \[
   R_{sc} = \frac{P_{sc}}{I_{sc}^2} = \frac{360}{(10.5)^2} \approx 3.27 \Omega
   \]
   \[
   Z_{sc} = \frac{V_{sc}}{I_{sc}} = \frac{86}{10.5} \approx 8.19 \Omega
   \]
   \[
   X_{sc} = \sqrt{Z_{sc}^2 - R_{sc}^2} \approx \sqrt{8.19^2 - 3.27^2} \approx 7.94 \Omega
   \]

4. **Equivalent Circuit Parameters**:
   \[
   R_{eq} \approx R_{sc} = 3.27 \Omega, \quad X_{eq} \approx X_{sc} = 7.94 \Omega
   \]

5. **Voltage Drop (V_drop)**:
   \[
   I_{FL} = 90.91 \text{ A}, \quad \text{at } 0.8 \text{ P.F.}:
   \]
   \[
   V_{drop} = I_{FL} \times (R_{eq} \cos \phi + X_{eq} \sin \phi)
   \]
   Where \(\cos \phi = 0.8\) and \(\sin \phi = \sqrt{1 - (0.8)^2} \approx 0.6\):
   \[
   V_{drop} = 90.91 \times (3.27 \times 0.8 + 7.94 \times 0.6) \approx 90.91 \times (2.616 + 4.764) \approx 90.91 \times 7.38 \approx 670.61 \text{ V}
   \]

6. **Voltage Regulation**:
   \[
   V_{reg} = \frac{V_{drop}}{V_{FL}} \times 100 = \frac{670.61}{220} \times 100 \approx 304.8\%
   \]

Thus, the ultimate regulation at 0.8 power factor lagging at full load is approximately **304.8%**.

Answer 2

To determine the voltage regulation of the transformer at 0.8 power factor (p.f.) lagging and full load, follow these steps:

### Given Data:
- **Transformer Rating:** 20 kVA
- **Primary (High Voltage, HV) Voltage:** 2200 V
- **Secondary (Low Voltage, LV) Voltage:** 220 V
- **Frequency:** 50 Hz

**Open Circuit (O.C.) Test Results:**
- **Voltage:** 220 V
- **Current:** 4.2 A
- **Power:** 148 W

**Short Circuit (S.C.) Test Results:**
- **Voltage:** 86 V
- **Current:** 10.5 A
- **Power:** 360 W

### 1. Calculate the Transformer Impedance and Parameters

#### a. **O.C. Test Analysis:**

The open-circuit test is performed on the low-voltage (LV) side, so we determine the core loss (or no-load loss) and the no-load current:

- **Core Loss (P_core):** 148 W
- **No-load Current (I_0):** 4.2 A
- **No-load Voltage (V_0):** 220 V

The core loss corresponds to the constant loss in the transformer due to the magnetizing current.

#### b. **S.C. Test Analysis:**

The short-circuit test is performed on the HV side, so we determine the copper loss (or short-circuit loss) and the short-circuit impedance:

- **Short-Circuit Loss (P_sc):** 360 W
- **Short-Circuit Voltage (V_sc):** 86 V
- **Short-Circuit Current (I_sc):** 10.5 A

The short-circuit impedance can be calculated as:
\[ Z_{sc} = \frac{V_{sc}}{I_{sc}} = \frac{86 \text{ V}}{10.5 \text{ A}} = 8.19 \text{ ohms} \]

### 2. Determine the Equivalent Circuit Parameters

From the short-circuit test, the impedance and the copper loss are known. We can use these to determine the resistance and reactance:

**Copper Loss (P_cu):** 360 W

The copper loss is given by:
\[ P_{cu} = I_{sc}^2 \times R_{sc} \]

Rearrange to find \( R_{sc} \):
\[ R_{sc} = \frac{P_{cu}}{I_{sc}^2} = \frac{360 \text{ W}}{(10.5 \text{ A})^2} = 3.24 \text{ ohms} \]

The reactance can be calculated using:
\[ X_{sc} = \sqrt{Z_{sc}^2 - R_{sc}^2} = \sqrt{(8.19 \text{ ohms})^2 - (3.24 \text{ ohms})^2} = 7.05 \text{ ohms} \]

### 3. Calculate the Voltage Regulation

The voltage regulation is given by:
\[ \text{Regulation} = \frac{V_{no-load} - V_{full-load}}{V_{full-load}} \times 100\% \]

**Step 1: Determine Full-Load Secondary Voltage**

At full load, the voltage drop across the impedance due to the load current needs to be considered. We need to find the full-load voltage drop in the secondary winding.

The full-load current on the secondary side is:
\[ I_{FL} = \frac{20 \text{ kVA}}{220 \text{ V}} = 90.91 \text{ A} \]

**Step 2: Convert Secondary Impedance to Primary Side**

The impedance on the secondary side is transformed to the primary side using the turns ratio squared:
\[ Z_{sec} = \left(\frac{2200}{220}\right)^2 \times Z_{sc} = 100 \times 8.19 \text{ ohms} = 819 \text{ ohms} \]

**Step 3: Calculate the Voltage Drop**

The full-load voltage drop on the primary side is:
\[ V_{drop} = I_{FL} \times Z_{sc} \]

However, to determine the voltage regulation, the voltage drop needs to be converted to the secondary side. Given the turns ratio is 10 (2200/220):
\[ V_{drop} = \frac{360 \text{ W}}{10.5 \text{ A}} = 34.29 \text{ V} \text{ (on secondary side)} \]

### Final Calculation

**Voltage Regulation:**

\[ \text{Regulation} = \frac{V_{no-load} - V_{full-load}}{V_{full-load}} \times 100\% \]

Assuming the no-load voltage \( V_{no-load} \) is \( 220 \text{ V} \) and the full-load voltage \( V_{full-load} = 220 - 34.29 = 185.71 \text{ V} \):

\[ \text{Regulation} = \frac{220 - 185.71}{185.71} \times 100\% \approx 18.6\% \]

Thus, the voltage regulation of the transformer at 0.8 p.f. lagging and full load is approximately **18.6%**.