Question

Define voltage regulation of alternator. A 400V, 10 KVA, 3 phase star connected alternator has resistance per phase of 1.0 ohm. Open circuit voltage per phase of 90V is obtained for field current of 1.0 A. For the same field current, short circuit current per phase is 15A. Calculate: i) Synchronous impedance, ii) synchronous reactance, iii) open circuit voltage per phase and iv) Regulation while supplying a load current of 15A at 0.8 power factor lag.

Answer 1

**Voltage Regulation of Alternator:**

Voltage regulation of an alternator refers to the change in terminal voltage when the alternator shifts from no-load to full-load conditions while the field current remains constant. It's a measure of the alternator's ability to maintain a constant output voltage under varying load conditions.

The formula for voltage regulation is:

\[
\text{Voltage Regulation} = \frac{E_{oc} - V_{FL}}{V_{FL}} \times 100\%
\]

where \( E_{oc} \) is the no-load (open-circuit) voltage, and \( V_{FL} \) is the full-load (loaded) voltage.

**Given Data:**

- Rated voltage, \( V_L \) = 400 V (line-to-line)
- Rated power, \( S \) = 10 KVA
- Star connected alternator
- Resistance per phase, \( R \) = 1.0 Ω
- Open circuit voltage per phase, \( E_{oc} \) = 90 V
- Short circuit current per phase, \( I_{sc} \) = 15 A
- Load current per phase, \( I_{FL} \) = 15 A
- Power factor (p.f.) = 0.8 lagging

Let's calculate each part:

### i) Synchronous Impedance (\( Z_s \)):

Synchronous impedance is calculated from the short-circuit test data.

The formula for synchronous impedance is:

\[
Z_s = \frac{V_{ph}}{I_{sc}}
\]

Where \( V_{ph} \) is the open-circuit voltage per phase. However, the open-circuit voltage per phase given here is not the same as the rated line-to-line voltage divided by √3.

In a star-connected alternator, the phase voltage \( V_{ph} \) is:

\[
V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400 \text{ V}}{\sqrt{3}} \approx 230.94 \text{ V}
\]

However, we have \( E_{oc} = 90 \text{ V} \) for the open circuit voltage per phase. We'll use this in our synchronous impedance calculation.

\[
Z_s = \frac{E_{oc}}{I_{sc}} = \frac{90 \text{ V}}{15 \text{ A}} = 6 \text{ Ω}
\]

### ii) Synchronous Reactance (\( X_s \)):

Synchronous reactance can be found from the synchronous impedance and the resistance per phase.

The formula is:

\[
X_s = \sqrt{Z_s^2 - R^2}
\]

Substituting the values:

\[
X_s = \sqrt{6^2 - 1^2} = \sqrt{36 - 1} = \sqrt{35} \approx 5.92 \text{ Ω}
\]

### iii) Open Circuit Voltage per Phase (\( E_{oc} \)):

The open-circuit voltage per phase is given directly as \( E_{oc} = 90 \text{ V} \).

### iv) Voltage Regulation:

To find the voltage regulation, we need to find the full-load terminal voltage \( V_{FL} \) first.

The formula for the full-load voltage is:

\[
V_{FL} = E_{oc} - I_{FL} (R \cos \phi + X_s \sin \phi)
\]

where \( \phi \) is the load angle whose cosine is the power factor.

Here:

\[
\cos \phi = 0.8 \text{ and } \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - 0.64} = 0.6
\]

So:

\[
V_{FL} = 90 - 15 \left[1.0 \times 0.8 + 5.92 \times 0.6\right]
\]

Calculate the term inside the brackets:

\[
1.0 \times 0.8 + 5.92 \times 0.6 = 0.8 + 3.552 = 4.352
\]

Thus:

\[
V_{FL} = 90 - 15 \times 4.352 = 90 - 65.28 = 24.72 \text{ V}
\]

Now, the voltage regulation is:

\[
\text{Voltage Regulation} = \frac{E_{oc} - V_{FL}}{V_{FL}} \times 100\%
\]

\[
\text{Voltage Regulation} = \frac{90 - 24.72}{24.72} \times 100\% \approx \frac{65.28}{24.72} \times 100\% \approx 264.9\%
\]

**Summary:**

1. **Synchronous Impedance (\( Z_s \))**: 6 Ω
2. **Synchronous Reactance (\( X_s \))**: 5.92 Ω
3. **Open Circuit Voltage per Phase (\( E_{oc} \))**: 90 V
4. **Voltage Regulation**: Approximately 264.9%

Answer 2

To address the problem regarding the voltage regulation of an alternator, we'll need to calculate several parameters based on the given data. Here's a step-by-step breakdown:

### Given Data

1. **Alternator Rating**: 400V, 10 kVA, 3-phase, star-connected.
2. **Resistance per Phase**: \( R = 1.0 \, \Omega \).
3. **Open Circuit Voltage per Phase**: \( V_{oc} = 90V \) for a field current of \( I_f = 1.0A \).
4. **Short Circuit Current per Phase**: \( I_{sc} = 15A \) for the same field current.
5. **Load Current**: \( I_L = 15A \) at a power factor (pf) of 0.8 lagging.

### Calculations

#### i) Synchronous Impedance

**Synchronous Impedance** (\(Z_s\)) is calculated from the open circuit voltage and short circuit current.

1. **Find the Synchronous Reactance** (\(X_s\)):

   Given the short circuit current (\(I_{sc}\)) and the resistance per phase (\(R\)), we can find \(X_s\) using the formula:

   \[
   I_{sc} = \frac{V_{oc}}{Z_s}
   \]

   Rearranging, we get:

   \[
   Z_s = \frac{V_{oc}}{I_{sc}}
   \]

   \[
   Z_s = \frac{90V}{15A} = 6 \, \Omega
   \]

2. **Calculate the Synchronous Reactance** (\(X_s\)):

   Using the resistance (\(R\)) and synchronous impedance (\(Z_s\)):

   \[
   Z_s^2 = R^2 + X_s^2
   \]

   \[
   X_s^2 = Z_s^2 - R^2
   \]

   \[
   X_s^2 = 6^2 - 1^2 = 36 - 1 = 35
   \]

   \[
   X_s = \sqrt{35} \approx 5.92 \, \Omega
   \]

3. **Synchronous Impedance** (\(Z_s\)):

   \(Z_s = 6 \, \Omega\)

#### ii) Synchronous Reactance

The synchronous reactance \(X_s\) has been calculated as:

   \[
   X_s \approx 5.92 \, \Omega
   \]

#### iii) Open Circuit Voltage Per Phase

The open circuit voltage per phase \(V_{oc}\) is already given as 90V for the field current of 1A. So:

   \[
   V_{oc} = 90V
   \]

#### iv) Regulation While Supplying a Load Current of 15A at 0.8 Power Factor Lag

1. **Calculate the Terminal Voltage** (\(V_t\)):

   The load is supplied at a power factor of 0.8 lagging. The voltage regulation is given by:

   \[
   \text{Voltage Regulation} = \frac{E - V_t}{V_t} \times 100\%
   \]

   Where \(E\) is the open circuit voltage and \(V_t\) is the terminal voltage under load.

2. **Determine the Current Component**:

   \[
   I_L = 15A \text{ at } \text{pf} = 0.8
   \]

   The load current can be resolved into real and reactive components:

   \[
   I_{L\text{(real)}} = I_L \times \text{pf} = 15A \times 0.8 = 12A
   \]

   \[
   I_{L\text{(reactive)}} = I_L \times \sqrt{1 - \text{pf}^2} = 15A \times \sqrt{1 - 0.8^2} \approx 15A \times 0.6 = 9A
   \]

   The load impedance per phase is:

   \[
   Z_{load} = \frac{V_t}{I_L}
   \]

   The voltage drop due to reactance and resistance is:

   \[
   V_{drop} = I_L \times (R + jX_s)
   \]

   \[
   V_{drop} = I_L \times (1 \Omega + j5.92 \Omega)
   \]

   The magnitude of the voltage drop is:

   \[
   |V_{drop}| = \sqrt{(I_L \times R)^2 + (I_L \times X_s)^2}
   \]

   \[
   |V_{drop}| = \sqrt{(15 \times 1)^2 + (15 \times 5.92)^2} = \sqrt{225 + 527.04} \approx \sqrt{752.04} \approx 27.4V
   \]

   Finally, the terminal voltage \(V_t\) is:

   \[
   V_t = E - |V_{drop}|
   \]

   \[
   V_t = 90V - 27.4V = 62.6V
   \]

3. **Calculate the Voltage Regulation**:

   \[
   \text{Voltage Regulation} = \frac{E - V_t}{V_t} \times 100\%
   \]

   \[
   \text{Voltage Regulation} = \frac{90V - 62.6V}{62.6V} \times 100\% \approx 44.1\%
   \]

### Summary

- **Synchronous Impedance**: \(6 \, \Omega\)
- **Synchronous Reactance**: \(\approx 5.92 \, \Omega\)
- **Open Circuit Voltage Per Phase**: \(90V\)
- **Voltage Regulation**: \(\approx 44.1\%\)