EEL303: Power Engineering I - Tutorial 1
1. (a) If v
i.
ii.
iii.
iv.
= 141.4sin(ωt + 30◦ ) V and i = 11.31cos(ωt − 30◦ ). Find for each
The maximum value
[ Vm = 141.4V , Im = 11.31A]
The RMS value
[Vrms = 100V , Irms = 8A]
The phasor expression in polar and rectangular form [V = 1006 0◦ , I = 86 30◦ ]
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
106 60◦ . 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:
Z = 106 60◦ = 5 + 8.6602i Ω
V = 240
I=
240
= 246 − 60
106 60
φ = tan−1
R = 5Ω
X = 8.6602Ω
S = V I ∗ = 2880 + 4988.30j = P + jQ
4988.3
= 60◦
2880
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 =
2
Vbase
kV A
Xactual =
1.1 × 222
= 1.0648Ω
500
(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:
new
Xpu
100
= 1.1 ×
×
500
22
20
2
= 0.2662p.u.
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 1
4. 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 %
G1
G2
M1
= 0.6862p.u, Xp.u.
= 0.4574p.u, Xp.u.
= 1.0255p.u,
and that for motors is 20 %. [Xp.u.
M2
Xp.u. = 0.6837p.u]
Solution: The p.u. reactances are
2
50
13.2
G1
Xp.u. = 0.15× ×
= 0.6862p.u.
10
13.8
M1
Xp.u.
G2
Xp.u.
2
12.5
50
= 1.0255p.u.
= 0.2× ×
8
13.8
M2
Xp.u.
2
50
13.2
= 0.15× ×
= 0.4574p.u.
15
13.8
2
50
12.5
= 0.2× ×
= 0.6837p.u.
12
13.8
5. 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
G1
corresponding to base values of 200 MVA, 35 kV.
[Xp.u.
= 0.1777p.u,
G2
G3
Xp.u. = 0.08916p.u, Xp.u. = 0.16029p.u]
Solution:
200
×
= 0.1 ×
100
33
35
2
= 0.1777p.u.
G2
Xpu
200
= 0.08 ×
×
150
32
35
2
= 0.08916p.u.
G3
Xpu
200
×
= 0.12 ×
110
30
35
2
= 0.16029p.u.
G1
Xpu
Electrical Engineering Dept - IIT Delhi