Short-Transmission Line Solved Problems
Problem-5
A 3-phase line delivers 3600 kW at a p.f. 0·8 lagging to a load. If the sending end voltage is 33 kV, determine
receiving end voltage
line current
transmission efficiency.
The resistance and reactance of each conductor are 5·31 \(\Omega\) and 5.54 \(\Omega\) respectively
Solution-5
Resistance of each conductor, \(R=5 \cdot 31 \Omega\)
Reactance of each conductor, \(X_{L}=5 \cdot 54 \Omega\)
Load power factor, \(\quad \cos \phi_{R}=0 \cdot 8\) (lagging)
Sending end voltage/phase, \(V_{S}=33,000 / \sqrt{3}=19,052 \mathrm{V}\)
- Line current,\[\begin{aligned} I &=\frac{\text { Power delivered / phase }}{\mathrm{V}_{R} \times \cos \phi_{R}}=\frac{1200 \times 10^{3}}{V_{R} \times 0.8} \\ &=\frac{150 \times 10^{5}}{V_{R}} \end{aligned}\]
- , we get, Using approximate expression for\[\begin{aligned} V_{S} & =V_{R}+I R \cos \phi_{R}+I X_{L} \sin \phi_{R} \\ \Rightarrow 19,052 & =V_{R}+\frac{15 \times 10^{5}}{V_{R}} \times 5 \cdot 31 \times 0.8+\\ & \frac{15 \times 10^{5}}{V_{R}} \times 5 \cdot 54 \times 0.6 \\ \Rightarrow V_{R}^{2}-19,052 & V_{R}+1,13,58,000=0 \end{aligned}\]
Solving this equation, we get, \(V_{R}=18,435 \mathrm{V}\)
Line voltage at the receiving end \(=\sqrt{3} \times 18,435=31,930 \mathrm{V}=31 \cdot 93 \mathrm{kV}\)
Line current, \(I=\dfrac{15 \times 10^{5}}{V_{R}}=\frac{15 \times 10^{5}}{18,435}=81 \cdot 36 \mathrm{A}\)
Line losses, \(=3 I^{2} R=3 \times(81 \cdot 36)^{2} \times 5 \cdot 31=1,05,447 \mathrm{W}=105 \cdot 447 \mathrm{kW}\)
Transmission efficiency \(=\dfrac{3600}{3600+105 \cdot 447} \times 100=97 \cdot 15 \%\)
Problem-6
A short 3-\(\phi\) transmission line with an impedance of (6 + j 8) \(\Omega\) per phase has sending and receiving end voltages of 120 kV and 110 kV respectively for some receiving end load at a p.f. of 0·9 lagging. Determine
power output
sending end power factor.
Solution-6
Resistance of each conductor, \(R=6 \Omega\)
Reactance of each conductor, \(X_{L}=8 \Omega\)
Load power factor, \(\cos \phi_{R}=0.9\) lagging
Receiving end voltage/phase, \(\quad V_{R}=110 \times 10^{3} / \sqrt{3}=63508 \mathrm{V}\)
Sending end voltage/phase, \(V_{S}=120 \times 10^{3} / \sqrt{3}=69282 \mathrm{V}\)
Let \(I\) be the load current. Using approximate expression for \(V_{S}\), we get,
\[\begin{aligned} V_{S} &=V_{R}+I R \cos \phi_{R}+I X_{L} \sin \phi_{R} \\ 69282 &=63508+I \times 6 \times 0.9+I \times 8 \times 0.435 \\ 8 \cdot 88 I &=5774 \\ I&=5774 / 8 \cdot 88=650 \cdot 2 \mathrm{A} \end{aligned}\]
- Power output\[=1,11,490 \mathrm{kW}\]
Sending end p.f.
\[\begin{aligned} \cos \phi_{S}&=\dfrac{V_{R} \cos \phi_{R}+I R}{V_{S}}\\ &=\dfrac{63508 \times 0 \cdot 9+650 \cdot 2 \times 6}{69282}=0.88~ \text{lag} \end{aligned}\]
Problem-7
An 11 kV, 3-phase transmission line has a resistance of 1·5 \(\Omega\) and reactance of 4 \(\Omega\) per phase. Calculate the percentage regulation and efficiency of the line when a total load of 5000 kVA at 0.8 lagging power factor is supplied at 11 kV at the distant end.
Solution-7
Resistance of each conductor, \(R=1 \cdot 5 \Omega\)
Reactance of each conductor, \(X_{L}=4 \Omega\)
- Receiving end voltage/phase,\[V_{R}=\dfrac{11 \times 10^{3}}{\sqrt{3}}=6351 \mathrm{V}\]
- Load power factor,\[\cos \phi_{R}=0.8 \text { lagging }\]
- Load current,\[\begin{aligned} I &=\frac{\text { Power delivered in } \mathrm{kVA} \times 1000}{3 \times V_{R}} \\ &=\frac{5000 \times 1000}{3 \times 6351}=262.43 \mathrm{A} \end{aligned}\]
- Using the approximate expression for\[\begin{aligned} V_{S} &=V_{R}+I R \cos \phi_{R}+I X_{L} \sin \phi_{R} \\ &=6351+262.43 \times 1 \cdot 5 \times 0.8+262 \cdot 43 \times 4 \times 0.6\\ &=7295 \cdot 8 \mathrm{V} \end{aligned}\]
- % regulation\[=\frac{V_{S}-V_{R}}{V_{R}} \times 100=\frac{7295 \cdot 8-6351}{6351} \times 100=14 \cdot 88 \%\]
- Line losses\[=3 I^{2} R=3 \times(262.43)^{2} \times 1 \cdot 5=310 \times 10^{3} \mathrm{W}=310 \mathrm{kW}\]
Output power \(=5000 \times 0.8=4000 \mathrm{kW}\)
Input power \(=\text { Ouput power }+\text { line losses }=4000+310=4310 \mathrm{kW}\)
- Transmission efficiency\[=\dfrac{\text { Output power }}{\text { Input power }} \times 100=\frac{4000}{4310} \times 100=92 \cdot 8 \%\]
Problem-8
A 3-phase, 50 Hz, 16 km long overhead line supplies 1000 kW at 11 kV, 0·8 p.f. lagging. The line resistance is 0·03 \(\Omega\) per phase per km and line inductance is 0·7 mH per phase per km. Calculate the sending end voltage, voltage regulation and efficiency of transmission.
Solution-8
Resistance of each conductor, \(R=0.03 \times 16=0.48 \Omega\)
Reactance of each conductor, \(X_{L}=2 \pi f L \times 16=2 \pi \times 50 \times 0.7 \times 10^{-3} \times 16=3 \cdot 52 \Omega\)
Receiving end voltage/phase, \(V_{R}=\dfrac{11 \times 10^{3}}{\sqrt{3}}=6351 \mathrm{V}\)
Load power factor, \(\cos \phi_{R}=0.8\) lagging
- Line current\[I=\frac{1000 \times 10^{3}}{3 \times V_{R} \times \cos \phi}=\frac{1000 \times 10^{3}}{3 \times 6351 \times 0.8}=65.6~ \mathrm{A}\]
Sending end voltage/phase,
\[\begin{aligned} V_{S}&=V_{R}+I R \cos \phi_{R}+I X_{L} \sin \phi_{R}\\ & =6351+65 \cdot 6 \times 0 \cdot 48 \times 0 \cdot 8+65 \cdot 6 \times 3 \cdot 52 \times 0 \cdot 6 \\ &=6515~ \mathrm{V} \end{aligned}\]- % Voltage regulation\[\begin{aligned} &=\frac{V_{S}-V_{R}}{V_{R}} \times 100\\ &=\frac{6515-6351}{6351} \times 100=2 \cdot 58 \% \end{aligned}\]
- Line losses\[=3 I^{2} R=3 \times(65 \cdot 6)^{2} \times 0.48=6 \cdot 2 \times 10^{3} \mathrm{W}=6 \cdot 2 ~\mathrm{kW}\]
- Input power\[=\text { Output power }+\text { Line losses }=1000+6 \cdot 2=1006 \cdot 2~ \mathrm{kW}\]
- Transmission efficiency\[=\frac{\text { Output power }}{\text { Input power }} \times 100=\frac{1000}{1006 \cdot 2} \times 100=99 \cdot 38 \%\]