Power Systems Test 3

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Power Systems Test 3

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Question 1
1 / -0

A 400 KV transmission line having line having line capacitance of 12.68 nF/km and line inductance 1.2 mH/km ignoring the length of the line, its ideal power transfer capability in MW is

A
62.01 MW

B
528.6 MW

C
520.10 MW

D
584.9 MW

Solution

\({\rm{Z}} = \sqrt {\frac{{\rm{L}}}{{\rm{C}}}} = \sqrt {\frac{{1.2 \times {{10}^{ - 3}}}}{{12.68 \times {{10}^{ - 9}}}}} = 307.63{\rm{\;\Omega }}\)
Ideal power transfer capability
\({\rm{P}} = \frac{{{{\rm{V}}^2}}}{{\rm{Z}}} = \frac{{{{\left( {400} \right)}^2}}}{{307.63}} = 520.10{\rm{\;MW}}\)
Question 2
1 / -0

The per unit impedance of a short transmission line is j 0.04. The line has a load current of (1 + j0.8) pu with a receiving end voltage of 1∠0°pu. The average reactive power flow over the line is _

A
-0.76 pu

B
-1.52 pu

C
1.52 pu

D
-3.48pu

Solution

The sending end voltage is, V_S = V_R + IZ = 1∠0°+ (1 + j0.8) (j0.04) = 0.9688 ∠2.36° pu
Reactive Power flowing is given as,
\({Q_{avr}}=\frac{Q_S+Q_R}2 = \frac{1}{{2X}}\left( {V_S^2 - V_R^2} \right) = \frac{1}{{2\left( {0.04} \right)}}\left( {{{0.9688}^2} - {1^2}} \right) = - 0.76\;pu\)
Question 3
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A load of 15 MW at a power factor of 0.8 lagging can be delivered by a three-phase transmission line having conductors each of resistance 1 Ω per kilometer. The voltage at the receiving end is to be 132 kV and the loss in the transmission is to be 5%. The length of the transmission line is _____ (in km)

A
36-38

B
364-381

C
367-386

D
369-382

Solution

\(P = \sqrt 3 \;{V_L}{I_L}\cos \phi \)
\(\Rightarrow 15 \times {10^6} = \sqrt 3 \times 132 \times {10^3} \times {I_L} \times 0.8\)
⇒ I_L = 82 A
Line losses = 5% of power delivered
= 0.05 × 15 × 10⁶ = 750 kW
Let R is the resistance of one conductor
Line losses = 3I²R
⇒ 750 × 10³ = 3 × (82)² × R
\(\Rightarrow R = \frac{{750\; \times \;{{10}^3}}}{{3\; \times \;{{\left( {82} \right)}^2}}} = 37.18\;{\rm{\Omega }}\)
Resistance of each conductor per km is 1 Ω
Length of the line = 37.18 km
Question 4
1 / -0

A three phase, 50 Hz, 16 km long overhead line supplies 1000 kW at 11 kV, 0.8 pf lagging. The line resistance is 0.03 Ω per phase per km and line inductance is 0.7 mH per phase per km. The percentage voltage regulation is _______

A
2.5-2.7

B
2.54-2.71

C
2.57-2.76

D
2.59-2.72

Solution

Resistance of each inductor (R) = 0.03 × 16 = 0.48 Ω
Reactance of each conductor (X_L) = 2π × 50 × 0.7 × 10^-3 × 16
= 3.52 Ω
Receiving end voltage/phase, \({V_R} = \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} = 6350.8\;V\)
Load power factor, cos ϕ_R = 0.8 lagging
P = 3 V I cos ϕ
\(\Rightarrow 1000 \times {10^3} = 3 \times \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} \times I \times 0.8\)
⇒ I = 65.6 A
Sending end voltage,
V_s = V_R + I R cos ϕ_R + I × sin ϕ_R
\(= \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} + \left( {65.6 \times 0.48 \times 0.8} \right) + \left( {65.6 \times 3.52 \times 0.6} \right)\)
= 6514.6 V
% voltage regulation \(\frac{{{V_s} - {V_R}}}{{{V_R}}} \times 100\)
\(= \frac{{6514.6 - 6350.8}}{{6350.8}} \times 100\)
= 2.58%
Question 5
1 / -0

An 11 kV ( Receiving end) , 3 - phase transmission line has resistance of 4 Ω per phase. The efficiency of the line when supplying the load of 4MW at 0.8 lagging power factor, is nearly

A
99%

B
95%

C
82%

D
90%

Solution

Voltage (V) = 11 kV
Load (P₀) = 4 MW
Power factor = cos ϕ = 0.8 lag
Receiving end current,
\({I_r} = \frac{{4 \times {{10}^6}}}{{\sqrt 3 \times 11 \times {{10}^3} \times 0.8}} = 262.43\;A\)
Total line losses which are occurring in resistance only.
P_loss = 3 I²_r R
= 3 (262.43)² (4)
= 0.826 MW
Input power (P_in) = P₀ + P_loss
⇒ P_in = 4 MW + 0.826 MW = 4.826 MW
\(Efficiency\left( \eta \right) = \frac{{{P_{out}}}}{{{P_{in}}}} = \frac{4}{{4.826}} \times 100 = 82.8\%\)