tutorial7_solution.pdf

7/27/2019 tutorial7_solution.pdf

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EEL303: Power Engineering I - Tutorial 7

1. A voltage of 100

2sin(314t + ) is applied to a series RL circuit by closing a switch.R=10ohm and L=0.1H.

(a) What is the maximum possible value of DC component of current that can occur.

(b) If the switch is closed at an instant such that the DC component of current is zero,

what is the instantaneous value of voltage at the time of closing of a switch.(c) If the switch is closed when the instantaneous voltage is zero, find the instantaneous

current 5 seconds later.

[Ans: (a)4.2915A (b)134.75V (c)-4.089A]

Solution:

i(t) =Vmax

|Z

|

sin(t + ) e

RtL sin( )

iss(t) =

Vmax

|Z| sin(t + )

itr(t) =Vmax

|Z| eRtL sin( )

|Z| =

R2 + (L)2 =

(10)2 + (314 0.1)2 = 32.954ohm

= tan1(L

R

)= tan1

(314 0.1

10

)= 72.3350

(a) DC component of current (or) transient term will have its maximum value when( ) =

2and t=0.

Vmax

|Z| =100

2

32.954= 4.2915A

(b) DC component of current will be zero, when ( )At the instant of closing switch (t=0),

v(t) = 100

2sin(72.3350) = 134.75V

(c)v(t) = 100

2sin(314t + ) = 0

= = 314tAt t=5sec, = 314 5 = 1570

i(t) =100

2

32.954

[sin(72.335) e1005sin(1570 72.335)] = 4.089A

Electrical Engineering Dept - IIT Delhi


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I

m = 0.8 j0.6 +j0.25

j0.6(0.55 j6.58) = 1.03 j3.34p.u

4. The one-line diagram of three phase power system is as shown: Each generator is repre-

Figure 1:

sented by an emf behind the sub-transient reactance. All impedances are expressed inp.u on common MVA base. All resistances and shunt capacitances are neglected. Thegenerators are operating on no-load at their rated voltage with their emfs in phase. Athree phase fault occurs at bus 3 through fault impedance Zf=j0.19p.u.

(a) Using Thevenins theorem obtain the impedance to the point of fault and faultcurrent in p.u.

(b) Determine the bus voltages and line currents during fault.

[Ans: (a) Z33=j0.2102; If = -j2.4988 p.u (b) Vf1

= 0.925p.u; Vf2

= 0.925p.u; Vf3

=

0.4748p.u; If12

= 0; If13

= -j1.5007p.u; If33

= -j1.0004p.u]

Solution: z12 = j0.75, z10 = j0.05, z20 = j0.075, z13 = j0.3, z23 = j0.45

y12 = -j1.33, y10 = -j20, y20 = -13.33j, y13 = -3.33j, y23 = -2.22j

YBus =

24.66j 1.33j 3.33j1.33j 16.88j 2.22j3.33j 2.22j 5.55j

YBus =

1 0 0

l21 1 0

l31 l32 1

u11 u12 u13

0 u22 u23

0 0 u33

u11 = -24.66j; u12 = 1.33j; u13 = 3.33j



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l21 u11 = 1.33j = l21 = -0.0539l21 u12 + u22 = -16.88i = u22 = -16.8083jl21 u13 + u23 = 2.22j = u23 = 2.3994jl31 u11 = 3.33j = l31 = -0.1350l31 u12 + l32 u22 = 2.22j = l32=-0.1428l31 u13 + l32 u23 + u33 = -5.55j = u33=-4.7578j

Y = LU

1 0 0

0.0539 1 0

0.1350 0.1428 1

24.66j 1.33j 3.33j

0

16.8083j 2.3994j

0 0 4.7578j

Z13

Z23

Z33

=

0

0

1

1 0 0

0.0539 1 00.1350 0.1428 1

X1

X2

X3

=

0

0

1

X1= 0

0.05X1 + X2 = 0 =

X2=0

X3= 1

24.66j 1.33j 3.33j

0 16.8083j 2.3994j0 0 4.7578j

Z13

Z23

Z33

=

0

0

1

(-4.7578j)(Z33)=1 = Z33= 0.2102j;(

16.8083j)(Z23) + (2.3994j)(Z33) = 0 =

Z23 = 0.03j

(24.66j)(Z13) + (1.33j)(Z23) + (3.33j)(Z33) = 0 = Z13 = 0.03j

If =V03

Z33 + Zf=

1

0.2102j + 0.19j= 2.4988j

Vfi = V

0

i Zi3

Z33 + ZfV03



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Vf1

= V01 Z13

Z33 + ZfV03

= 1 0.03jj0.4002

1 = 0.925p.u

Vf2

= V02

Z23

Z33 + Zf

V03

= 1

0.03j

j0.4002 1 = 0.925p.u

Vf3

= IfZf = (2.4988j)(0.19j) = 0.4748p.u

Ifij =

Vfi Vfjzij

If12

=V

f1 Vf

2

z12=

0.925 0.9250.75j

= 0

If13

=V

f1 Vf

3

z13=

0.925 0.4748j0.3

=

1.5007jp.u

If23

=V

f2 Vf

3

z23=

0.925 0.4748j0.45

= 1.0004jp.u


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tutorial7_solution.pdf