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Power Systems → Transmission Line Performance → Short, Medium & Long Line Models
Last updated 5 September 2026
Question A 3-phase, 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are
0.25 Ω 0.25 \Omega 0.25Ω and
3.925 Ω 3.925 \Omega 3.925Ω respectively. If the voltage at the generator terminal is 17.87 kV, the power factor of the load is
_______ .
Correct answer 0.75 to 0.85
Solution Given:
P 3 ϕ = 3 MW = 3 × 10 6 W P_{3\phi} = 3 \text{ MW} = 3 \times 10^6 \text{ W} P 3 ϕ = 3 MW = 3 × 1 0 6 W V L , load = 17.32 kV ⇒ V p h , load = 17.32 3 = 10 kV = 10000 V V_{L, \text{load}} = 17.32 \text{ kV} \Rightarrow V_{ph, \text{load}} = \frac{17.32}{\sqrt{3}} = 10 \text{ kV} = 10000 \text{ V} V L , load = 17.32 kV ⇒ V p h , load = 3 17.32 = 10 kV = 10000 V V L , gen = 17.87 kV ⇒ V p h , gen = 17.87 3 ≈ 10317.2 V V_{L, \text{gen}} = 17.87 \text{ kV} \Rightarrow V_{ph, \text{gen}} = \frac{17.87}{\sqrt{3}} \approx 10317.2 \text{ V} V L , gen = 17.87 kV ⇒ V p h , gen = 3 17.87 ≈ 10317.2 V R = 0.25 Ω , X = 3.925 Ω R = 0.25 \Omega, X = 3.925 \Omega R = 0.25Ω , X = 3.925Ω Phase current
I = P 3 ϕ 3 V p h , load cos ϕ = 3 × 10 6 3 × 10000 cos ϕ = 100 cos ϕ I = \frac{P_{3\phi}}{3 V_{ph, \text{load}} \cos \phi} = \frac{3 \times 10^6}{3 \times 10000 \cos \phi} = \frac{100}{\cos \phi} I = 3 V p h , load c o s ϕ P 3 ϕ = 3 × 10000 c o s ϕ 3 × 1 0 6 = c o s ϕ 100 Using the phasor relationship for an inductive load:
V p h , gen 2 = ( V p h , load cos ϕ + I R ) 2 + ( V p h , load sin ϕ + I X ) 2 V_{ph, \text{gen}}^2 = (V_{ph, \text{load}} \cos \phi + I R)^2 + (V_{ph, \text{load}} \sin \phi + I X)^2 V p h , gen 2 = ( V p h , load cos ϕ + I R ) 2 + ( V p h , load sin ϕ + I X ) 2 Substituting
I = 100 cos ϕ I = \frac{100}{\cos \phi} I = c o s ϕ 100 :
10317.2 2 = ( 10000 cos ϕ + 100 cos ϕ ⋅ 0.25 ) 2 + ( 10000 sin ϕ + 100 cos ϕ ⋅ 3.925 ) 2 10317.2^2 = (10000 \cos \phi + \frac{100}{\cos \phi} \cdot 0.25)^2 + (10000 \sin \phi + \frac{100}{\cos \phi} \cdot 3.925)^2 10317. 2 2 = ( 10000 cos ϕ + c o s ϕ 100 ⋅ 0.25 ) 2 + ( 10000 sin ϕ + c o s ϕ 100 ⋅ 3.925 ) 2 10317.2 2 = ( 10000 cos ϕ + 25 cos ϕ ) 2 + ( 10000 sin ϕ + 392.5 cos ϕ ) 2 10317.2^2 = (10000 \cos \phi + \frac{25}{\cos \phi})^2 + (10000 \sin \phi + \frac{392.5}{\cos \phi})^2 10317. 2 2 = ( 10000 cos ϕ + c o s ϕ 25 ) 2 + ( 10000 sin ϕ + c o s ϕ 392.5 ) 2 Dividing by
10000 2 10000^2 1000 0 2 :
1.03172 2 = ( cos ϕ + 0.0025 cos ϕ ) 2 + ( sin ϕ + 0.03925 cos ϕ ) 2 1.03172^2 = (\cos \phi + \frac{0.0025}{\cos \phi})^2 + (\sin \phi + \frac{0.03925}{\cos \phi})^2 1.0317 2 2 = ( cos ϕ + c o s ϕ 0.0025 ) 2 + ( sin ϕ + c o s ϕ 0.03925 ) 2 1.0644 = cos 2 ϕ + 0.005 + 0.00000625 cos 2 ϕ + sin 2 ϕ + 0.0785 tan ϕ + 0.00154 cos 2 ϕ 1.0644 = \cos^2 \phi + 0.005 + \frac{0.00000625}{\cos^2 \phi} + \sin^2 \phi + 0.0785 \tan \phi + \frac{0.00154}{\cos^2 \phi} 1.0644 = cos 2 ϕ + 0.005 + c o s 2 ϕ 0.00000625 + sin 2 ϕ + 0.0785 tan ϕ + c o s 2 ϕ 0.00154 1.0644 = 1 + 0.005 + 0.0785 tan ϕ + 0.001546 cos 2 ϕ 1.0644 = 1 + 0.005 + 0.0785 \tan \phi + \frac{0.001546}{\cos^2 \phi} 1.0644 = 1 + 0.005 + 0.0785 tan ϕ + c o s 2 ϕ 0.001546 0.0594 = 0.0785 tan ϕ + 0.001546 ( 1 + tan 2 ϕ ) 0.0594 = 0.0785 \tan \phi + 0.001546 (1 + \tan^2 \phi) 0.0594 = 0.0785 tan ϕ + 0.001546 ( 1 + tan 2 ϕ ) 0.001546 tan 2 ϕ + 0.0785 tan ϕ − 0.05785 = 0 0.001546 \tan^2 \phi + 0.0785 \tan \phi - 0.05785 = 0 0.001546 tan 2 ϕ + 0.0785 tan ϕ − 0.05785 = 0 Solving for
tan ϕ \tan \phi tan ϕ using the quadratic formula:
tan ϕ ≈ 0.726 \tan \phi \approx 0.726 tan ϕ ≈ 0.726 ϕ = arctan ( 0.726 ) ≈ 35.98 ∘ \phi = \arctan(0.726) \approx 35.98^\circ ϕ = arctan ( 0.726 ) ≈ 35.9 8 ∘ cos ϕ = cos ( 35.98 ∘ ) ≈ 0.809 \cos \phi = \cos(35.98^\circ) \approx 0.809 cos ϕ = cos ( 35.9 8 ∘ ) ≈ 0.809 The power factor is approximately 0.81.
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Correct answer 0.75 to 0.85
Solution Given:
P 3 ϕ = 3 MW = 3 × 10 6 W P_{3\phi} = 3 \text{ MW} = 3 \times 10^6 \text{ W} P 3 ϕ = 3 MW = 3 × 1 0 6 W V L , load = 17.32 kV ⇒ V p h , load = 17.32 3 = 10 kV = 10000 V V_{L, \text{load}} = 17.32 \text{ kV} \Rightarrow V_{ph, \text{load}} = \frac{17.32}{\sqrt{3}} = 10 \text{ kV} = 10000 \text{ V} V L , load = 17.32 kV ⇒ V p h , load = 3 17.32 = 10 kV = 10000 V V L , gen = 17.87 kV ⇒ V p h , gen = 17.87 3 ≈ 10317.2 V V_{L, \text{gen}} = 17.87 \text{ kV} \Rightarrow V_{ph, \text{gen}} = \frac{17.87}{\sqrt{3}} \approx 10317.2 \text{ V} V L , gen = 17.87 kV ⇒ V p h , gen = 3 17.87 ≈ 10317.2 V R = 0.25 Ω , X = 3.925 Ω R = 0.25 \Omega, X = 3.925 \Omega R = 0.25Ω , X = 3.925Ω Phase current
I = P 3 ϕ 3 V p h , load cos ϕ = 3 × 10 6 3 × 10000 cos ϕ = 100 cos ϕ I = \frac{P_{3\phi}}{3 V_{ph, \text{load}} \cos \phi} = \frac{3 \times 10^6}{3 \times 10000 \cos \phi} = \frac{100}{\cos \phi} I = 3 V p h , load c o s ϕ P 3 ϕ = 3 × 10000 c o s ϕ 3 × 1 0 6 = c o s ϕ 100 Using the phasor relationship for an inductive load:
V p h , gen 2 = ( V p h , load cos ϕ + I R ) 2 + ( V p h , load sin ϕ + I X ) 2 V_{ph, \text{gen}}^2 = (V_{ph, \text{load}} \cos \phi + I R)^2 + (V_{ph, \text{load}} \sin \phi + I X)^2 V p h , gen 2 = ( V p h , load cos ϕ + I R ) 2 + ( V p h , load sin ϕ + I X ) 2 Substituting
I = 100 cos ϕ I = \frac{100}{\cos \phi} I = c o s ϕ 100 :
10317.2 2 = ( 10000 cos ϕ + 100 cos ϕ ⋅ 0.25 ) 2 + ( 10000 sin ϕ + 100 cos ϕ ⋅ 3.925 ) 2 10317.2^2 = (10000 \cos \phi + \frac{100}{\cos \phi} \cdot 0.25)^2 + (10000 \sin \phi + \frac{100}{\cos \phi} \cdot 3.925)^2 10317. 2 2 = ( 10000 cos ϕ + c o s ϕ 100 ⋅ 0.25 ) 2 + ( 10000 sin ϕ + c o s ϕ 100 ⋅ 3.925 ) 2 10317.2 2 = ( 10000 cos ϕ + 25 cos ϕ ) 2 + ( 10000 sin ϕ + 392.5 cos ϕ ) 2 10317.2^2 = (10000 \cos \phi + \frac{25}{\cos \phi})^2 + (10000 \sin \phi + \frac{392.5}{\cos \phi})^2 10317. 2 2 = ( 10000 cos ϕ + c o s ϕ 25 ) 2 + ( 10000 sin ϕ + c o s ϕ 392.5 ) 2 Dividing by
10000 2 10000^2 1000 0 2 :
1.03172 2 = ( cos ϕ + 0.0025 cos ϕ ) 2 + ( sin ϕ + 0.03925 cos ϕ ) 2 1.03172^2 = (\cos \phi + \frac{0.0025}{\cos \phi})^2 + (\sin \phi + \frac{0.03925}{\cos \phi})^2 1.0317 2 2 = ( cos ϕ + c o s ϕ 0.0025 ) 2 + ( sin ϕ + c o s ϕ 0.03925 ) 2 1.0644 = cos 2 ϕ + 0.005 + 0.00000625 cos 2 ϕ + sin 2 ϕ + 0.0785 tan ϕ + 0.00154 cos 2 ϕ 1.0644 = \cos^2 \phi + 0.005 + \frac{0.00000625}{\cos^2 \phi} + \sin^2 \phi + 0.0785 \tan \phi + \frac{0.00154}{\cos^2 \phi} 1.0644 = cos 2 ϕ + 0.005 + c o s 2 ϕ 0.00000625 + sin 2 ϕ + 0.0785 tan ϕ + c o s 2 ϕ 0.00154 1.0644 = 1 + 0.005 + 0.0785 tan ϕ + 0.001546 cos 2 ϕ 1.0644 = 1 + 0.005 + 0.0785 \tan \phi + \frac{0.001546}{\cos^2 \phi} 1.0644 = 1 + 0.005 + 0.0785 tan ϕ + c o s 2 ϕ 0.001546 0.0594 = 0.0785 tan ϕ + 0.001546 ( 1 + tan 2 ϕ ) 0.0594 = 0.0785 \tan \phi + 0.001546 (1 + \tan^2 \phi) 0.0594 = 0.0785 tan ϕ + 0.001546 ( 1 + tan 2 ϕ ) 0.001546 tan 2 ϕ + 0.0785 tan ϕ − 0.05785 = 0 0.001546 \tan^2 \phi + 0.0785 \tan \phi - 0.05785 = 0 0.001546 tan 2 ϕ + 0.0785 tan ϕ − 0.05785 = 0 Solving for
tan ϕ \tan \phi tan ϕ using the quadratic formula:
tan ϕ ≈ 0.726 \tan \phi \approx 0.726 tan ϕ ≈ 0.726 ϕ = arctan ( 0.726 ) ≈ 35.98 ∘ \phi = \arctan(0.726) \approx 35.98^\circ ϕ = arctan ( 0.726 ) ≈ 35.9 8 ∘ cos ϕ = cos ( 35.98 ∘ ) ≈ 0.809 \cos \phi = \cos(35.98^\circ) \approx 0.809 cos ϕ = cos ( 35.9 8 ∘ ) ≈ 0.809 The power factor is approximately 0.81.
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