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tutorial on Power Engineering, Papers of Electric Machines

Indian Institute of TechnologyElectric Machines

Questions and answer on Power engineering

Typology: Papers

2020/2021

Uploaded on 03/01/2021

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EEL303: Power Engineering I - Tutorial 1
1. (a) If v= 141.4sin(ωt + 30◦) V and i= 11.31cos(ω t −30◦). Find for each
i. The maximum value [ Vm= 141.4V,Im= 11.31A]
ii. The RMS value [Vrms = 100V,Irms = 8A]
iii. The phasor expression in polar and rectangular form [V= 10060◦,I= 8630◦]
iv. Is the circuit inductive or capacitive
Solution:
v= 141.4Cos(ωt −60◦)i= 11.31Cos(ωt −30◦)
The current leads voltage by 30◦. Hence, the circuit is capacitive.
2. A single phase AC voltage of 240 V is applied to a series circuit whose impedance is
10660◦. Find R, X, P, Q and the power factor of the circuit. [R=5 Ω, X=8.6602 Ω,
I=246−60◦, P= 2.880 kW, Q= 4.9883 kVAR, p.f=0.5]
Solution:
V= 240 Z= 10660◦= 5 + 8.6602iΩR= 5Ω X= 8.6602Ω
I=240
10660 = 246−60 S=V I∗= 2880 + 4988.30j=P+j Q
φ=tan−14988.3
2880 = 60◦Cosφ =Cos60 = 0.5
3. (a) A generator is rated 500 MVA, 22 kV. Its Y connected windings have a reactance
of 1.1 p.u. Find the ohmic value of the reactance. [Xactual=1.0648 Ω ]
Solution:
Xp.u. =Xactual
Xbase
Xbase =V2
base
kV A Xactual =1.1×222
500 = 1.0648Ω
(b) The above generator is in a circuit for which bases are specified as 100 MVA, 20
kV. Find the new p.u. value of the reactance of generator. [Xpu=0.2662 p.u ]
Solution:
Xnew
pu = 1.1×100
500 ×22
202
= 0.2662p.u.
Electrical Engineering Dept - IIT Delhi
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EEL303: Power Engineering I - Tutorial 1

  1. (a) If v = 141. 4 sin(ωt + 30◦) V and i = 11. 31 cos(ωt − 30 ◦). Find for each

i. The maximum value [ Vm = 141. 4 V , Im = 11. 31 A] ii. The RMS value [Vrms = 100V , Irms = 8A] iii. The phasor expression in polar and rectangular form [V = 100^6 0 ◦, I = 8^6 30 ◦] iv. Is the circuit inductive or capacitive

Solution: v = 141. 4 Cos(ωt − 60 ◦) i = 11. 31 Cos(ωt − 30 ◦)

The current leads voltage by 30◦. Hence, the circuit is capacitive.

  1. A single phase AC voltage of 240 V is applied to a series circuit whose impedance is 10 6 60 ◦. Find R, X, P, Q and the power factor of the circuit. [R=5 Ω, X=8.6602 Ω, I=24^6 − 60 ◦, P= 2.880 kW, Q= 4.9883 kVAR, p.f=0.5]

Solution:

V = 240 Z = 10^6 60 ◦^ = 5 + 8. 6602 i Ω R = 5Ω X = 8.6602Ω

I =

= 24^6 − 60 S = V I∗^ = 2880 + 4988. 30 j = P + jQ

φ = tan−^1

= 60◦^ Cosφ = Cos60 = 0. 5

  1. (a) A generator is rated 500 MVA, 22 kV. Its Y connected windings have a reactance of 1.1 p.u. Find the ohmic value of the reactance. [Xactual=1.0648 Ω ]

Solution:

Xp.u. =

Xactual Xbase

Xbase =

V (^) base^2 kV A

Xactual =

1. 1 × 222

(b) The above generator is in a circuit for which bases are specified as 100 MVA, 20 kV. Find the new p.u. value of the reactance of generator. [Xpu=0.2662 p.u ]

Solution: Xpunew = 1. 1 ×

×

= 0. 2662 p.u.

Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 1

  1. Two generators at 10 MVA, 13.2 kV and 15 MVA, 13.2 kV are connected in parallel to a bus bar. They feed supply to two motors of inputs 8 MVA and 12 MVA respectively. The operating voltage of motors is 12.5 kV. Assuming base quantities as 50 MVA and 13.8 kV, draw the reactance diagram. The percentage reactance for generators is 15 % and that for motors is 20 %. [X p.u.G^1 = 0. 6862 p.u, X p.u.G^2 = 0. 4574 p.u, X p.u.M^1 = 1. 0255 p.u, X p.u.M^2 = 0. 6837 p.u]

Solution: The p.u. reactances are

X p.u.G^1 = 0. 15 ×

×

= 0. 6862 p.u. X p.u.G^2 = 0. 15 ×

×

= 0. 4574 p.u.

X p.u.M^1 = 0. 2 ×

×

= 1. 0255 p.u. X p.u.M^2 = 0. 2 ×

×

= 0. 6837 p.u.

  1. Three generators are rated as follows: G1 - 100 MVA, 33 kV, 10 % X; G2 - 150 MVA, 32 kV, 8 % X; G3 - 110 MVA, 30 kV, 12 % X. Determine reactance of generators corresponding to base values of 200 MVA, 35 kV. [X p.u.G^1 = 0. 1777 p.u, X p.u.G^2 = 0. 08916 p.u, X p.u.G^3 = 0. 16029 p.u]

Solution: X puG^1 = 0. 1 ×

×

= 0. 1777 p.u.

X puG^2 = 0. 08 ×

×

= 0. 08916 p.u.

X puG^3 = 0. 12 ×

×

= 0. 16029 p.u.

Electrical Engineering Dept - IIT Delhi