Question

A 3 phase line of 4 km length delivers 4000 kW at a p.f of 0.8 lagging to a load the resistance and reactance per km of each conductor are 0.2 Ω and 0.5 Ω respectively if the voltage at the supply end is maintained at 11 kV. Calculate the received end voltage and efficiency of line.

Answer 1

To calculate the **received end voltage** and the **efficiency** of the transmission line, we need to follow these steps:

### 1. **Given data:**
- Load power (\( P_L \)) = 4000 kW = 4000 × 10³ W
- Power factor (p.f.) = 0.8 lagging
- Load voltage at the sending end (\( V_s \)) = 11 kV = 11 × 10³ V
- Line length (\( l \)) = 4 km
- Resistance per km (\( R \)) = 0.2 Ω/km
- Reactance per km (\( X \)) = 0.5 Ω/km

### 2. **Calculate total line impedance:**
For a 3-phase line, the total resistance and reactance are calculated by multiplying the per km values by the line length:

- Total resistance (\( R_{\text{total}} \)) = \( R \times l = 0.2 \times 4 = 0.8 \, \Omega \)
- Total reactance (\( X_{\text{total}} \)) = \( X \times l = 0.5 \times 4 = 2.0 \, \Omega \)

Thus, the total impedance (\( Z_{\text{total}} \)) of the line is:
\[
Z_{\text{total}} = R_{\text{total}} + jX_{\text{total}} = 0.8 + j2.0 \, \Omega
\]

### 3. **Calculate the load current:**

First, calculate the apparent power \( S_L \) delivered to the load:
\[
S_L = \frac{P_L}{\text{p.f.}} = \frac{4000 \, \text{kW}}{0.8} = 5000 \, \text{kVA}
\]

The load current for a 3-phase system is:
\[
I_L = \frac{S_L}{\sqrt{3} \times V_r} = \frac{5000 \times 10^3}{\sqrt{3} \times V_r}
\]
Where \( V_r \) is the received end voltage (unknown at this point). Since we don't yet have \( V_r \), we'll approximate it using the sending end voltage \( V_s \).

Let’s assume \( V_r \approx V_s = 11 \, \text{kV} \) for now to estimate \( I_L \):

\[
I_L = \frac{5000 \times 10^3}{\sqrt{3} \times 11 \times 10^3} = \frac{5000}{\sqrt{3} \times 11} = 262.43 \, \text{A}
\]

### 4. **Calculate voltage drop across the line:**

The voltage drop across the transmission line is:
\[
\Delta V = I_L \times Z_{\text{total}}
\]
Using the impedance (\( Z_{\text{total}} = 0.8 + j2.0 \)):

\[
\Delta V = 262.43 \times (0.8 + j2.0) = 262.43 \times 0.8 + j262.43 \times 2.0 = 209.94 + j524.86 \, \text{V}
\]

Now, calculate the magnitude of this voltage drop:
\[
|\Delta V| = \sqrt{209.94^2 + 524.86^2} = \sqrt{44067.1 + 275476.9} = \sqrt{319544} = 565.15 \, \text{V}
\]

### 5. **Calculate the received end voltage:**

The sending end voltage is 11 kV, and the voltage drop across the line is \( \Delta V \). The received end voltage \( V_r \) is the difference between the sending end voltage and the voltage drop:

For a 3-phase system, we must convert the line voltage to the phase voltage. The phase voltage for the sending end is:
\[
V_s^{\text{phase}} = \frac{V_s}{\sqrt{3}} = \frac{11 \times 10^3}{\sqrt{3}} = 6350.85 \, \text{V}
\]

Subtracting the voltage drop:
\[
V_r^{\text{phase}} = V_s^{\text{phase}} - |\Delta V| = 6350.85 - 565.15 = 5785.7 \, \text{V}
\]

Now, convert the received phase voltage back to the line voltage:
\[
V_r = V_r^{\text{phase}} \times \sqrt{3} = 5785.7 \times \sqrt{3} = 10020 \, \text{V} = 10.02 \, \text{kV}
\]

### 6. **Calculate the efficiency of the line:**

The efficiency \( \eta \) is given by:
\[
\eta = \frac{P_r}{P_s} \times 100\%
\]
Where \( P_s \) is the sending end power, and \( P_r = P_L = 4000 \, \text{kW} \).

The sending end power \( P_s \) can be approximated as:
\[
P_s = \sqrt{3} \times V_s \times I_L \times \text{p.f.}
\]
Substitute the values:
\[
P_s = \sqrt{3} \times 11 \times 262.43 \times 0.8 = 4009.75 \, \text{kW}
\]

Finally, the efficiency is:
\[
\eta = \frac{4000}{4009.75} \times 100\% = 99.76\%
\]

### **Final Results:**
- **Received end voltage**: \( V_r \approx 10.02 \, \text{kV} \)
- **Efficiency of the line**: \( \eta \approx 99.76\% \)

Answer 2

To solve this problem, we need to calculate the received end voltage and the efficiency of the 3-phase transmission line. We'll proceed through the following steps:

### Given Data:
- Length of the line \( L = 4 \) km
- Power delivered \( P = 4000 \) kW
- Power factor \( \text{pf} = 0.8 \) (lagging)
- Resistance per km \( R = 0.2 \, \Omega/\text{km} \)
- Reactance per km \( X = 0.5 \, \Omega/\text{km} \)
- Voltage at the supply end \( V_s = 11 \text{ kV} \) (line-to-line)

### Step 1: Calculate the Total Line Resistance and Reactance

1. **Total Line Resistance \( R_{total} \):**
   \[
   R_{total} = R \times L = 0.2 \, \Omega/\text{km} \times 4 \text{ km} = 0.8 \, \Omega
   \]

2. **Total Line Reactance \( X_{total} \):**
   \[
   X_{total} = X \times L = 0.5 \, \Omega/\text{km} \times 4 \text{ km} = 2.0 \, \Omega
   \]

### Step 2: Calculate the Load Current

1. **Calculate the load power in kVA:**
   \[
   S = \frac{P}{\text{pf}} = \frac{4000 \text{ kW}}{0.8} = 5000 \text{ kVA}
   \]

2. **Calculate the load current (line-to-line voltage):**

   First, find the line-to-neutral voltage \( V_{LN} \) at the supply end:
   \[
   V_{LN} = \frac{V_s}{\sqrt{3}} = \frac{11 \text{ kV}}{\sqrt{3}} \approx 6.35 \text{ kV}
   \]

   Then, calculate the load current \( I_L \):
   \[
   I_L = \frac{S}{\sqrt{3} \times V_{LN}} = \frac{5000 \text{ kVA}}{\sqrt{3} \times 6.35 \text{ kV}} \approx \frac{5000 \text{ kVA}}{11 \text{ kV}} \approx 454.5 \text{ A}
   \]

### Step 3: Calculate the Voltage Drop Across the Line

1. **Voltage drop in each phase:**
   \[
   V_{drop} = I_L \times (R_{total} + jX_{total})
   \]

   Substitute the values:
   \[
   V_{drop} = 454.5 \text{ A} \times (0.8 \, \Omega + j2.0 \, \Omega)
   \]

2. **Calculate the magnitude of the voltage drop:**
   \[
   |V_{drop}| = 454.5 \times \sqrt{0.8^2 + 2.0^2} = 454.5 \times \sqrt{0.64 + 4.0} = 454.5 \times \sqrt{4.64} \approx 454.5 \times 2.15 \approx 977.2 \text{ V}
   \]

### Step 4: Calculate the Received End Voltage

1. **Received end voltage \( V_R \):**
   \[
   V_R = V_s - V_{drop}
   \]

   Convert \( V_{drop} \) to line-to-line voltage:
   \[
   V_{drop} \text{ (line-to-line)} = |V_{drop}| \times \sqrt{3} = 977.2 \text{ V} \times \sqrt{3} \approx 1690 \text{ V}
   \]

   Finally, calculate \( V_R \):
   \[
   V_R = 11 \text{ kV} - 1.69 \text{ kV} = 9.31 \text{ kV}
   \]

### Step 5: Calculate the Efficiency of the Line

1. **Calculate the input power \( P_{in} \):**
   \[
   P_{in} = \sqrt{3} \times V_s \times I_L \times \text{pf}
   \]

   Substitute values:
   \[
   P_{in} = \sqrt{3} \times 11 \text{ kV} \times 454.5 \text{ A} \times 0.8 \approx \sqrt{3} \times 11 \times 454.5 \times 0.8 \approx 11 \times 454.5 \times 1.38 \approx 6926 \text{ kW}
   \]

2. **Calculate the efficiency:**
   \[
   \text{Efficiency} = \frac{P_{out}}{P_{in}} \times 100\% = \frac{4000 \text{ kW}}{6926 \text{ kW}} \times 100\% \approx 57.8\%
   \]

### Summary

- **Received End Voltage**: Approximately \( 9.31 \text{ kV} \)
- **Efficiency of the Line**: Approximately \( 57.8\% \)