power system and planning asst.3 solution

Question1: a)

R=delta f delta P

R1=2000

?

0.1?60

=0.048Hz/MW

R1,pu=R1?500=24Hz/pu

R2=2000

?

0.1?60

=0.192Hz/MW

R1,pu=R1?250=48Hz/pu b)

M=ω?I

0.5?I?ω2=W k0 Thus,

0.5?M?ω=W k0

M=

W k0 0.5?ω

ω=2?π?f

M=

W k0

0.5?2?π?f

=

W k0

π?f

=

21G

π?60

=0.111408G J/rad M pu=

M

2000M

=0.0557 pu

c)

Δf ss=

ΔP D

D+

1

R1+

1

R2

D pu=

D

2000

=

20

2000

=0.01

ΔP D,pu=

ΔP D

2000

=

75

2000

=0.0375

Thus,

Δf ss,pu=

ΔP D,pu

D pu+

1

R1,pu+

1

R2,pu

=

0.0375

0.01+

1

24+

1

48

=0.51724 Hz

Build the block diagram:

And get the results from matlab “scope” for Δω:

From the figure above, the x-axis is the time line from 0-20s and the y-axis the “Δω”results. From this figure, when the curve is stable, the Δω is around -0.52.

d)

The mechanical power inputs to generator1:

From the figure above, the mechanical power inputs to generator 1 is stable around at P mech1=0.0216.

In theoretically:

?P mec?1

pu =?

?f ss

f1pu

=0.02155pu

They are almost the same.

The mechanical power inputs to generator2:

From the figure above, the mechanical power inputs to generator 2 is stable around at P mech2=0.0108.

In theoretically:

?P mec?2

pu =?

?f s s

R f2

pu

=0.01078pu

They are almost the same.

e)

Add the ramping limits by setting a mechanical power rate of change limitation of 10MW/s on generator 1. 10

2000

=0.005.

The block changed to as followed:

And get the results from matlab “scope” for Δω:

The mechanical power inputs to generator1:

The mechanical power inputs to generator2:

Compared the results we gained from part e to the results we gained from part c and d. We can see that the stabilizing values are the same for without adding the ramping limits and adding the ramping limits.

However, for the adding the ramping limits to the generator 1, the system needs more time to reach the stabilizing values.

Question2:

Add the secondary control, ?KI1

S and –KI2

S

which for KI1=0.009 and KI2=0.003to the

Question1 block diagram. And also adding the ΔP load1=0.75*ΔP D,pu, which ΔP D,pu=0.0375 from the question1. Adding the ΔP load2=0.25*ΔP D,pu, which ΔP D,pu=0.0375 from the question1.

The block changed to as followed:

And get the results from matlab “scope” for Δω:

The mechanical power inputs to generator1:

The mechanical power inputs to generator2:

Question3:

The block diagram is the same as the Question2, except to change M’=M*0.8 and D’=D*0.5.

And get the results from matlab “scope” for Δω:

By definition, M=Iω0, with the smaller inertial constant M, it implies the smaller moment of inertia I. The new system should be more vulnerable to disturbance (large M implies large I thus able to endure larger disturbance).

In this case, M′= 80%M,D′= 50%D is suitable. We can see, although with more impact, the value of ?f ss is still in the reasonable range of f0±1Hz and the final steady state of frequency error can be pushed back to 0. Since by definition, D represents the value of how the system responses (changes) when frequency changes, thus smaller D has less impact on the system. For larger D (larger damping coefficient), it implies that the system will returns to steady state without oscillation, yet it requires more time to be stabilized.

We can conclude that M should be larger and D should be smaller in order to stabilize the system.

Question4:

Add the tie line, with reactance 0.1pu, to interconnect the resulting 2 systems from Part2.

The block changed to as followed:

And get the results from matlab “scope” for Δω1:

B1=B2=1

R1

+

1

R2

+D=0.0725pu.

Add B1 and B2 blocks, and then the new block changed to as followed:

And get the results from matlab “scope” for Δω1:

ACE1=?P L1

1

R1+

1

R2+D

2

R1+

2

R2+2D

?

1

R1

+

1

R2

+D

??P L1

2

R1+

2

R2+2D

=?P L1=?P L ACE2= 0,wit??P L2= 0

Appendix: Question1: c)

Question2: e)

Question2:

Question4: Without B:

With B:

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