UJIAN TENGAH SEMESTER GANJIL UNIVERSITAS BRAWIJAYA MALANG NAMA : SUPARMAN NPM :136060300111003 MK : ANALISA SISTEM TENAGA LANJUT

Question 1: (30 marks)

A. What is the Power mismatch? Discuss and derive the involved equation.

Answer:

It is usual to use the power mismatch as an accurate test. The complex power

mismatch at a node i is the difference between the specified power and the

calculated power where is the voltage evaluated at iteration r

Above equation can be separated into real and imaginary parts and expressed in polar

form or rectangular form as:

It is assumed in the above equations that Vi is the voltage evaluated at iteration r. The most common convergence criterion used in practice is

B. Derive the following Jacobian matric in Newton Raphson method:

Answer:

Equation 1

∑ ( )

Verification:

∑ ( )

( )

( )

Equation 2

( )

Verification:

∑ ( )

( ) ( )

( )

Equation 3

∑ ( )

Verification:

∑ ( )

( ) ( )

( )

Equation 4

( )

Verification:

∑ ( )

( )

( )

Question 2: (20 marks) Discuss briefly the steady state (load flow) modeling of each power system component

involved in the power system.

Answer:

steady state (steady / steady) is the normal condition of post-occurring disorders in an RL circuit with t ≠ 0 and steady state always has the same magnitude but opposite in sign to declare the value zero from existing flows. Picture below is an image of power system modeling and graphic images of steady state conditions:

Picture below is an image of power system modeling

Picture below is graphic images of steady state conditions:

Question 3: (50 marks) Figure 1 shows three phase system that are connected to generators A, B, C and D at 11kV level. The five nodes of the system with the specific ID are given in the figure. The power supplies the load through node 2 at 132kV level. The reactance of the transmission line and component parameters in per unit are shown in the Figure.

a) Change all components in Figure 1 which are given in real parameters to per unit

system. Note: use 100MVA as reference to determine the per unit system. Answer: VBase4 =33 KV VBase5 =33x(11/33)=11 KV VBase3 =11x(11/11)=11 KV VBase2 =33x(11/33)=11 KV VBase1 =11x(11/11)=11 KV ZBase gen 1 = (11 KV) 2/(100 MVA)=1,21 Ω ZBase 2 = (11 KV) 2/(100 MVA)=1,21 Ω ZBase 5 = (11 KV) 2/(100 MVA)=1,21 Ω ZBase 4 = (33 KV) 2/(100 MVA)=10,89 Ω ZBase 1 = (11 KV) 2/(100 MVA)=1,21 Ω

ZGen / Z5 = 1,21 x (11 ) 2/(33) x (200/100)=0,26 pu Z1 = 1,21 x (11 ) 2/(132) x (200/100)=0,16 pu Z2 = 1,21 x (11 ) 2/(11) x (100/100)=1,21 pu Z3 = 1,21 x (11 ) 2/(132) x (150/100)=0,012 pu Z4= 10.89 x (33 ) 2/(33) x (100/100)=10,89 pu Z5 = 0

b) Construct the admittance matrix of the system. Answer:

Y11 Y12 Y13 Y14 Y15

V1

I1

Y21 Y22 Y23 Y24 Y25

V2

I2

Y31 Y32 Y33 Y34 Y35

V3

=

I3

Y41 Y42 Y43 Y44 Y45

V4

I4

Y51 Y52 Y53 Y54 Y55

V5

I5

c) Obtain the Thevenin equivalence circuit at bus 2.

Answer:

V 1

V 2V 3

V 4

V 5I 1

I 2

I 3

I 4I 5

Y 10

Y 40

Y 12

Y 24

Y 23

Y 20

Y 30

Y 45 a Y 45 b Y 45 c

Y 50

d) Assume that the power load required in the system is about 525 MW and all gernerators running on pf 0.9, determine the power flow in the Lines: 1-2, 2-3, 2-4 and calculate the system losses. Answer: