Problems of power systems. Essay Example

Question 1

= 0.02+j0.04 12Z

= 0.01+j0.03 22Z

= 0.0125+j0.025 23Z = 10-j20 = = 20-j28 = = 15-j15 =

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

= 0.6-j1 = Therefore first iteration =

Per load at bus 3 = -200MW/100 = 2-j0 = 20-j28 =

= 0.8-j1 = Therefore first iteration =

Per load at bus xx = -250MW/100 = 2.5-j0 = 15-j15 =

= 0.833-j1 = Therefore first iteration =

Question 2

= 256.4375kVA Input =

= 0.487 per unit = 0.2 m X

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

= 176.78per unit At base I =

= 0.1506 per unit = ) = th Impedance (Z Current (I) =

=j4.9999979 per unit or 884A Current from generation =

= j0.5471 or 96.71A 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 =1.368sin Power output =

During fault =0.365sin Power output = =1.25sin Power output =Post fault output =

= 1.25, m3 = 0.365, Pm2 = 1.368, Pm1P = 2sin sin = eX= 0.8 p.u; P

= 0.8; sTherefore P = 0.4572rad, = 0.8, thus 2sin

= 0.371rad = 0.8, thus thus 2.44sin 0.371rad= 2.771rad = =  = -0.41059

oCritical clearing angle is 126.94 Using the swing curve above the critical clearing time is 0.41s

Question 5   Let us take lambda to 6.5 and the power will be determined as   ) = 975-(150+83.3+38.89)= 702.81MW1Change in MW (P

Thus the change in lambda

= 2.66 = Change

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

= \$8,228.029 per hour + + Cost =

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

= 0.234 = Change

This means applicable of new lambda for second iteration is 9.16+0.234 = 9.16  ) = 975-(450+325+200) = 0MW2Change in MW (P

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

= \$8,236.25 per hour + + Cost =