Question

Two 11KV, 3-phase, 3000KVA generators having reactance of 15% operate in parallel. The generators supply power to a transmission line through a 6000KVA transformer of ratio 11/22KV and having leakage reactance of 5%. Calculate fault current and fault KVA on H.T. side of the transformer.

Answer 1

To calculate the fault current and fault KVA on the High Tension (H.T.) side of the transformer for the given system, we'll follow these steps:

1. **Calculate the equivalent reactance of the generators:**
   - Each generator has a reactance of 15%, which is expressed as \( X_{\text{gen}} = 0.15 \) per unit (pu).

2. **Determine the total equivalent reactance of the generators in parallel:**
   - Since the generators are operating in parallel, their combined reactance is determined by:
     \[
     \frac{1}{X_{\text{eq}}} = \frac{1}{X_{\text{gen1}}} + \frac{1}{X_{\text{gen2}}}
     \]
     For identical generators:
     \[
     X_{\text{eq}} = \frac{X_{\text{gen}}}{\text{Number of Generators}} = \frac{0.15}{2} = 0.075 \text{ pu}
     \]

3. **Calculate the reactance of the transformer:**
   - The transformer has a leakage reactance of 5% on its own rated voltage, which is \( X_{\text{tr}} = 0.05 \) pu.

4. **Convert the transformer's reactance to the generator side (Low Tension side to High Tension side):**
   - The transformer has a ratio of 11 kV/22 kV. The reactance needs to be converted to the H.T. side:
     \[
     X_{\text{tr\_HT}} = X_{\text{tr}} \times \left(\frac{V_{\text{LT}}}{V_{\text{HT}}}\right)^2
     \]
     Where \( V_{\text{LT}} \) is 11 kV, and \( V_{\text{HT}} \) is 22 kV.
     \[
     X_{\text{tr\_HT}} = 0.05 \times \left(\frac{11}{22}\right)^2 = 0.05 \times \left(\frac{1}{2}\right)^2 = 0.05 \times 0.25 = 0.0125 \text{ pu}
     \]

5. **Determine the total equivalent reactance on the H.T. side:**
   - The total reactance is the sum of the generator's reactance and the transformed reactance of the transformer:
     \[
     X_{\text{total}} = X_{\text{eq}} + X_{\text{tr\_HT}} = 0.075 + 0.0125 = 0.0875 \text{ pu}
     \]

6. **Calculate the fault current on the H.T. side:**
   - The fault current can be found using the formula:
     \[
     I_{\text{fault}} = \frac{1}{X_{\text{total}}}
     \]
     So,
     \[
     I_{\text{fault}} = \frac{1}{0.0875} \approx 11.43 \text{ pu}
     \]
     Since the base current is related to the base KVA and base voltage, the actual fault current in amperes will depend on these values. However, in per unit (pu) terms, it is 11.43.

7. **Calculate the fault KVA on the H.T. side:**
   - Fault KVA can be found by:
     \[
     \text{Fault KVA} = \text{Fault Current (pu)} \times \text{Base KVA}
     \]
     The base KVA on the H.T. side is the transformer's rating, which is 6000 KVA.

     Thus,
     \[
     \text{Fault KVA} = 11.43 \times 6000 = 68580 \text{ KVA}
     \]

### Summary:
- **Fault Current (pu):** 11.43 pu
- **Fault KVA on the H.T. side:** 68,580 KVA

This calculation assumes that the system is balanced and that the fault occurs at the point where these calculations are relevant. The actual fault current in amperes can be derived from the base values and the per unit fault current.

Answer 2

To calculate the fault current and fault KVA on the High Tension (HT) side of the transformer, follow these steps:

### 1. **Determine the Fault Level on the Generator Side:**

**Generator Data:**
- **Voltage (V_g):** 11 kV (line-to-line)
- **Power (S_g):** 3000 kVA
- **Reactance (X_g):** 15% (or 0.15 per unit)

**Generators in Parallel:**
There are two generators operating in parallel, so their total reactance in parallel will be:

\[ X_{g,\text{total}} = \frac{X_g}{n} \]

where \( n \) is the number of generators.

For two generators:

\[ X_{g,\text{total}} = \frac{0.15 \times (11 \text{ kV})^2}{2 \times 3000 \text{ kVA}} \]
\[ X_{g,\text{total}} = \frac{0.15 \times 121}{6000} \]
\[ X_{g,\text{total}} = \frac{18.15}{6000} \]
\[ X_{g,\text{total}} = 0.003025 \text{ per unit} \]

### 2. **Calculate the Transformer Reactance:**

**Transformer Data:**
- **Voltage Ratio:** 11/22 kV
- **Rated Power (S_t):** 6000 kVA
- **Leakage Reactance (X_t):** 5% (or 0.05 per unit)

To find the reactance on the high tension (HT) side, convert it to the HT side. The transformer's reactance in per unit on the HT side can be calculated using the following formula:

\[ X_{t, \text{HT}} = X_t \times \left(\frac{V_{HT}}{V_{LT}}\right)^2 \]

where \( V_{HT} \) is the high voltage side (22 kV) and \( V_{LT} \) is the low voltage side (11 kV).

\[ X_{t, \text{HT}} = 0.05 \times \left(\frac{22}{11}\right)^2 \]
\[ X_{t, \text{HT}} = 0.05 \times 4 \]
\[ X_{t, \text{HT}} = 0.20 \text{ per unit} \]

### 3. **Calculate the Total Reactance on the HT Side:**

The total reactance on the HT side is the sum of the generators' reactance and the transformer's reactance:

\[ X_{total} = X_{g, \text{HT}} + X_{t, \text{HT}} \]

Since \( X_{g, \text{HT}} \) is already in per unit on the HT side, we add it directly:

\[ X_{total} = 0.003025 + 0.20 \]
\[ X_{total} = 0.203025 \text{ per unit} \]

### 4. **Calculate the Fault Current:**

The fault current in per unit can be calculated as:

\[ I_{fault} = \frac{1}{X_{total}} \]

So,

\[ I_{fault} = \frac{1}{0.203025} \]
\[ I_{fault} \approx 4.93 \text{ per unit} \]

To find the fault current in kA, we convert it to the actual fault current using the base current:

Base current on the HT side:

\[ I_{base, \text{HT}} = \frac{S_{base}}{\sqrt{3} \times V_{HT}} \]
\[ I_{base, \text{HT}} = \frac{6000 \text{ kVA}}{\sqrt{3} \times 22 \text{ kV}} \]
\[ I_{base, \text{HT}} \approx 157.5 \text{ A} \]

Thus,

\[ I_{fault} = 4.93 \times 157.5 \]
\[ I_{fault} \approx 776.5 \text{ A} \]

### 5. **Calculate the Fault KVA:**

The fault KVA on the HT side is:

\[ S_{fault} = I_{fault} \times V_{HT} / 1000 \]
\[ S_{fault} = 776.5 \text{ A} \times 22 \text{ kV} / 1000 \]
\[ S_{fault} \approx 17.1 \text{ MVA} \]

### Summary:

- **Fault Current (HT side):** Approximately 776.5 A
- **Fault KVA (HT side):** Approximately 17.1 MVA