Problems of power systems. Essay Example

Question 1

= 0.02+j0.04 12Z

= 0.01+j0.03 22Z

= 0.0125+j0.025 23Z

problems of power systems. = 10-j20problems of power systems. 1
=

problems of power systems. 2 = 20-j28problems of power systems. 3
=

problems of power systems. 4 = 15-j15problems of power systems. 5
=

Per load at bus 2 = -400MW/100 = 4-j0

= 0.6-j1problems of power systems. 7
= problems of power systems. 6
Therefore first iteration =

Per load at bus 3 = -200MW/100 = 2-j0

problems of power systems. 8 = 20-j28problems of power systems. 9
=

= 0.8-j1problems of power systems. 11
= problems of power systems. 10
Therefore first iteration =

Per load at bus xx = -250MW/100 = 2.5-j0

problems of power systems. 12 = 15-j15problems of power systems. 13
=

= 0.833-j1problems of power systems. 15
= problems of power systems. 14
Therefore first iteration =

Question 2

= 256.4375kVAproblems of power systems. 16
Input =

= 0.487 per unitproblems of power systems. 17
= 0.2 m X

= 2.5 x0.487 = 1.2186 per unitm X0.25

= 176.78per unitproblems of power systems. 18
At base I =

= 0.1506 per unitproblems of power systems. 20
= problems of power systems. 19
) = th Impedance (Z

problems of power systems. 21
Current (I) =

=j4.9999979 per unit or 884Aproblems of power systems. 22Current from generation =

= j0.5471 or 96.71Aproblems of power systems. 23
For each motor =

When there is a fault at d

Through generation I= 884A

Through motors I = -j5.0+2(-j0.5471)=-j6.0942 per unit or 1077.33A

Fault at p

Through generation I= 884A

Through motor I = 96.71A

Fault at R

Through generation I= 3(96.71A) = 290.13A

Through motor I = 96.71A

Question 3

=613.1/100=6.13 times the pickup.p,actual/If,maxThis would have a ratio of I

=0.135+0.3=0.435 seconds.CT+TP=TB=0.135 seconds. So we need R1 to operate in TPFrom the time-overcurrent curves, we see that R2 (set at TDS of ½) will operate in about T

=0.435, then move right until we get to the vertical line corresponding to 6.13 times pickup. The intersection corresponds to desired TDS curve.B=0.135 point on the ordinate until we get to TPBecause CT ratio and Tap are the same for R1, we can use the 6.13 times pickup for it as well. Move up from the T

Question 4

Pre-fault condition

problems of power systems. 25 =1.368sinproblems of power systems. 24Power output =

During fault

problems of power systems. 27 =0.365sinproblems of power systems. 26Power output =

problems of power systems. 29 =1.25sinproblems of power systems. 28Power output =Post fault output =

= 1.25, m3 = 0.365, Pm2 = 1.368, Pm1P

problems of power systems. 32 = 2sinproblems of power systems. 31sinproblems of power systems. 30 = eX= 0.8 p.u; P

= 0.8; sTherefore P

problems of power systems. 33

= 0.4572rad, problems of power systems. 35
= 0.8, thus problems of power systems. 342sin

= 0.371radproblems of power systems. 37
= 0.8, thus thus problems of power systems. 362.44sin

problems of power systems. 380.371rad= 2.771radproblems of power systems. 40
= problems of power systems. 39
=

problems of power systems. 41

problems of power systems. 42= -0.41059

oCritical clearing angle is 126.94

problems of power systems. 43

Using the swing curve above the critical clearing time is 0.41s

Question 5

problems of power systems. 44

problems of power systems. 45

problems of power systems. 46

Let us take lambda to 6.5 and the power will be determined as

problems of power systems. 47

problems of power systems. 48

problems of power systems. 49

) = 975-(150+83.3+38.89)= 702.81MW1Change in MW (P

Thus the change in lambda

= 2.66problems of power systems. 51
= problems of power systems. 50
Change

This means applicable of new lambda for second iteration is 6.5+2.66 = 9.16

problems of power systems. 52

problems of power systems. 53

problems of power systems. 54

) = 975-(482.5+305.8+186.7)= 0MW1Change in MW (P

= 186.7MW3 = 305.8MW and P2Since the units of production are now P1 = 482.5MW, P

problems of power systems. 55

problems of power systems. 56

problems of power systems. 57

problems of power systems. 60
+ problems of power systems. 59+problems of power systems. 58
Cost =

problems of power systems. 63
+ problems of power systems. 62
+ problems of power systems. 61
Cost =

= $8,228.029 per hourproblems of power systems. 66
+ problems of power systems. 65
+ problems of power systems. 64
Cost =

as 3 and P2, P1b) If we look at the constrains imposed on P

problems of power systems. 67

problems of power systems. 68

problems of power systems. 69

Since one generator exceeds the limit we reduce it to the maximum value which is 450MW and continue with iteration as

) = 975-(450+305.8+186.7) = 32.5MW2Change in MW (P

Thus the other generators whose production is known will iterate using change of lambda as

= 0.234problems of power systems. 71
= problems of power systems. 70
Change

This means applicable of new lambda for second iteration is 9.16+0.234 = 9.16

problems of power systems. 72

problems of power systems. 73

) = 975-(450+325+200) = 0MW2Change in MW (P

= 200MW3 = 325MW and P2Since the units of production are now P1 = 450MW, P

problems of power systems. 74

problems of power systems. 75

problems of power systems. 76

problems of power systems. 79
+ problems of power systems. 78+problems of power systems. 77
Cost =

problems of power systems. 82
+ problems of power systems. 81
+ problems of power systems. 80
Cost =

= $8,236.25 per hourproblems of power systems. 85
+ problems of power systems. 84
+ problems of power systems. 83
Cost =

  1. Hadi Saadat, “Power System Analysis”, McGraw-Hill, 2004

  2. John J. Grainger and William D. Stevenson, Jr. “Power System Analysis”, McGraw-Hill, 1994