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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.
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Let's break down the problem step by step and calculate the received end voltage and the efficiency of the line.

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

### Step 1: Calculate the Current at the Sending End
The active power delivered to the load is given by the formula:
\[
P = \sqrt{3} V_{\text{line}} I_{\text{line}} \cos(\theta)
\]
Where:
- \( V_{\text{line}} = 11 \, \text{kV} \) (line-to-line voltage)
- \( \cos(\theta) = 0.8 \) (the power factor)
- \( I_{\text{line}} \) is the current at the sending end.

First, convert the power into watts:
\[
P = 4000 \, \text{kW} = 4000 \times 10^3 \, \text{W}
\]

Now, solving for \( I_{\text{line}} \):
\[
4000 \times 10^3 = \sqrt{3} \times 11 \times 10^3 \times I_{\text{line}} \times 0.8
\]
\[
I_{\text{line}} = \frac{4000 \times 10^3}{\sqrt{3} \times 11 \times 10^3 \times 0.8}
\]
\[
I_{\text{line}} = \frac{4000}{\sqrt{3} \times 11 \times 0.8}
\]
\[
I_{\text{line}} \approx \frac{4000}{15.184} \approx 263.3 \, \text{A}
\]

### Step 2: Calculate the Total Impedance of the Line
The total impedance of the line \( Z_{\text{line}} \) is the series combination of the resistance and reactance for the total length of the line:
\[
Z_{\text{line}} = (R \times L) + j(X \times L)
\]
\[
Z_{\text{line}} = (0.2 \times 4) + j(0.5 \times 4)
\]
\[
Z_{\text{line}} = 0.8 + j2.0 \, \Omega
\]

### Step 3: Calculate the Voltage Drop in the Line
The voltage drop across the line is calculated by:
\[
V_{\text{drop}} = I_{\text{line}} \times Z_{\text{line}}
\]
\[
V_{\text{drop}} = 263.3 \times (0.8 + j2.0)
\]
First, calculate the magnitude of the impedance:
\[
|Z_{\text{line}}| = \sqrt{(0.8)^2 + (2.0)^2} = \sqrt{0.64 + 4} = \sqrt{4.64} \approx 2.15 \, \Omega
\]

The voltage drop is then:
\[
V_{\text{drop}} = 263.3 \times 2.15 \approx 565.1 \, \text{V}
\]

### Step 4: Calculate the Received End Voltage
The voltage at the receiving end \( V_r \) is the supply voltage minus the voltage drop:
\[
V_r = V_s - V_{\text{drop}}
\]
\[
V_r = 11,000 \, \text{V} - 565.1 \, \text{V} = 10,434.9 \, \text{V}
\]

Thus, the voltage at the receiving end is approximately \( 10.43 \, \text{kV} \).

### Step 5: Calculate the Efficiency of the Line
The efficiency \( \eta \) of the line is given by:
\[
\eta = \frac{\text{Power received at the load}}{\text{Power supplied}} \times 100
\]

The power supplied is \( 4000 \, \text{kW} \), and the power received at the load is the real power minus the losses in the line. The power loss in the line is given by:
\[
\text{Power loss} = I_{\text{line}}^2 \times R_{\text{total}}
\]
Where:
- \( R_{\text{total}} = R \times L = 0.2 \times 4 = 0.8 \, \Omega \)

Thus:
\[
\text{Power loss} = (263.3)^2 \times 0.8 \approx 55,533.6 \, \text{W} = 55.53 \, \text{kW}
\]

Now, calculate the efficiency:
\[
\eta = \frac{4000 - 55.53}{4000} \times 100 = \frac{3944.47}{4000} \times 100 \approx 98.61\%
\]

### Final Answer:
- **Received end voltage**: \( 10.43 \, \text{kV} \)
- **Efficiency of the line**: \( 98.61\% \)
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