Unit1 And 2 Sample Solutions

Some Solved Questions
Q1. Draw the pu impedance diagram for the power system shown in the figure . Neglect
resistance , and use a base of 100 MV A , 230 kV in 50 Ω line. The ratings of the
generator , motor and transformers are : Generator 40 MV A, 25 kV, X” = 20%
Motor 50 MV A, 11 kV, X” = 30%
Y-Y transformer, 40 MV A, 33 Y-220 Y kV, X = 15%
Y-Δ transformer, 30 MV A, 11 Δ-220 Y kV, X = 15%

Solution: The rectangular blocks are circuit breakers and it is connected at the beginning
and end of each m/c or transformer or transmission line to protect it from high current
due to faults. We don’t have to bother about it. Just neglect it – do not get confused.
Step 1: Identifying the circuits: Let us name it as ckt.1 , ckt(trans line) and ckt 2.
Step 2 : Marking the new bases for each of the 3 ckts.
Ckt(trans line): we should start from this as this is what is given in the question.
And it is 100MVA and 230KV.
Ckt 1: we know the base voltage on the right side of transformer but not left side. The
base on left side (ckt 1) depends on the transformation ratio of Y- Y transformer which is
coming into picture when we are going from transmission line to generator ckt.
Hence VB(ckt 1) / VB(ckt trans line) = 33/220 (since the higher voltage side is always connected
towards the transmission line side)
VB(ckt 1) = 34.5KV
Ckt 2 : Similarly
VB(ckt trans line) / VB(ckt 2) = 220/11
VB(ckt 2) = 11.5KV
Step 3 : Calculating the new reactances
Generator: Xg = 0.2 x (100/40) x (25/34.5)2 = 0.2625pu
Transformer (Y-Y) on the primary side = 0.15 x (100/40) x (33/34.5)2 = 0.343pu
on the secondary side = 0.15 x (100/40) x (220/230)2 = 0.343pu
And we notice that whether we do calculations on primary or secondary side, the pu
value does not change for a transformer.
Transformer (Y- Δ) on the primary side = 0.15 x (100/30) x (220/230)2 = 0.457pu
on the secondary side = 0.15 x (100/30) x (11/11.5)2 = 0.457pu
Motor: 0.3 x (100/50) x (11/11.5)2 = 0.55pu
Transmission line: Actual/ base impedance(since the actual value is given, we can
directly devide by the base value and get the pu value)
Base impedance in the transmission line circuit = 2302/ 100 = 529 ohms
Hence X (line) = 50/529 = 0.0945pu

Q2.The Figure shows the schematic diagram of a radial transmission system. The ratings
and reactances of the various components are shown therein. A load of 60 MW at 0.9
power factor lagging is tapped from the 66 KV substation which is to be maintained at 60
kV . Calculate the terminal voltage of the synchronous machine . represent the
transmission line and the transformers by series reactances only.

60 MW and not 60MV
Solution: Let 100 MVA and 11KV be the base in Generator circuit.
Accordingly the base in transmission line will be 100 MVA, 220 KV.
And in the load circuit, it is 100 MVA and 66KV.
Xt1 = j0.1
Xt2 = j0.08
X line = 150/(2202/100) = j0.31
Iload = 60,000/(sqrt 3 x 60x 0.9) = 641.5A
Base current = 100,000/ (sqrt 3 x 66) = 874.77A
Iload (pu) = 641.5/874.77 = 0.733(-25.84 )A
0

V2(pu) = 60/66 = 0.91(0 )
0

V1 = V2 + Iload(equivalent Reactance)
= 0.91(0 ) + 0.733(-25.84 ) x (j0.1 + j0.31 + j0.08)
0 0

= 1.11(16.85 ) 0

Actual V1 = 1.11 x 11 = 12.2 KV

Q3..A synchronous generator is rated 60 MVA , 11 kV. It has resistance R = 0.1 pu and
a

X = 1.65 pu. It is feeding into an infinite bus bar at 11 kV delivering a current 3.15 kA at
d

0.9 pf lagging.
(a) Determine E and angle δ.
f

(b) Draw a phasor diagram for this operation.
(c) Bus bar voltage falls to 10 kV while the mechanical power input generator and its
excitation remains unchanged. What is the value and pf of the current delivered to the
bus. In this case assume the generator resistance to be negligible.

Solution:
Choose a base of 60MVA, 11kV(as the base MVA to be chosen is not given, we have the
liberty to choose the base same as the rating of the m/c.).
The bus bar voltage is to be maintained at 11KV.
When converted to p.u., it is 11/11(Actual / base) = 1.0(0 )pu
0

Base current = 60/{sqrt(3)x 11} = 3.15 KA
Actual current = 3.15KA
Hence pu Ia = 1(-cos-1 0.9) = 1(-25.84 )pu
0

(a) Ef = Vt + Ia(Ra + jIaXa) = 1.81+j1.438 = 2.312(38.47 ) 0

Hence Actual Ef = 11x2.312 = 25.43 and δ = 38.47 leading
0

(b)

(c) Vt = 10/11 = 0.91pu
Mechanical power input is the input to prime mover(steam) which changes the
active power. Hence P = same as above = Vt x Ia x pf = 1 x 1 x 0.9 = 0.9pu
And P = |Vt| |Ef| sin δ/ Xa (Ef or Excitation is also not changing)
Which gives us δ = 40 0

Hence Ia = (Ef – Vt )/jXa = 1.0436(-30.35)pu

Therefore actual current = 1.0436x3.15 = 3.28 KA and pf = cos30.35 = 0.863 lagging

Q4.Three impedances of 5-j10, 6+j5 and 3+ j15 ohms are connected in star to A, B and C
lines of a 3300 V, 3 phase, 3 wire supply. The phase sequence is ABC. Calculate the line
current Ia.
Solution: this is a case of unbalanced 3 phase star connected load supplied from a
balanced 3 phase supply.
There is 3 wire and hence the phase voltages cannot be taken as balanced(since there is
no neutral connection between supply and load).
Vab = 3300(0 )0

Vbc = 3300(-120 ) 0

Let Va, Vb and Vc be the phase voltages.
Van = (5 – j10)Ia = (5-j10)(Ia1+Ia2+Ia0) = (5-j10)(Ia1+Ia2) (1)
Since Ia + Ib +Ic = In = 3Ia0 = 0 , hence Ia0 = 0(there is no neutral wire and hence zero
sequence currents cannot can flow.)
Vbn = (6+j5)Ib = (6+j5)(Ib1+Ib2) = (6+j5)(a2Ia1 + aIa2) (2)
2
Vcn = (3+j15)Ic = (3+j15)(Ic1+Ic2) == (3+j15)(aIa1 + a Ia2) (3)
(1) – (2) gives us
Van –Vbn = Vab = 3300(0) = (3.67 – j2.3)Ia1 + (12.33 – j12.7)Ia2 (4)
(2) – (3) gives us
Vbn – Vcn = Vbc = 3300(-120 ) = (15.8 – j2.8)Ia1 – (18.84 – j12.8)Ia2
0
(5)
Solving (4) and (5) for Ia1 and Ia2,
Ia1 = 134-j65
Ia2 = 95 +j141
Hence Ia = Ia1+ Ia2 = 241(-18.4 ) A
0

Unit1 And 2 Sample Solutions

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Unit1 And 2 Sample Solutions