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Problems of power systems. Essay Example
 Category:Engineering and Construction
 Document type:Math Problem
 Level:Undergraduate
 Page:1
 Words:567
Question 1
= 0.02+j0.04 _{12}Z
= 0.01+j0.03 _{22}Z
= 0.0125+j0.025 _{23}Z
= 10j20
=
= 20j28
=
= 15j15
=
Per load at bus 2 = 400MW/100 = 4j0
= 0.6j1
=
Therefore first iteration =
Per load at bus 3 = 200MW/100 = 2j0
= 20j28
=
= 0.8j1
=
Therefore first iteration =
Per load at bus xx = 250MW/100 = 2.5j0
= 15j15
=
= 0.833j1
=
Therefore first iteration =
Question 2
= 256.4375kVA
Input =
= 0.487 per unit
= 0.2 _{m }X
= 2.5 x0.487 = 1.2186 per unit_{m } X0.25
= 176.78per unit
At base I =
= 0.1506 per unit
=
) = _{th }Impedance (Z
Current (I) =
=j4.9999979 per unit or 884ACurrent 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}/I_{f,max}This would have a ratio of I
=0.135+0.3=0.435 seconds._{CT}+T_{P}=T_{B}=0.135 seconds. So we need R1 to operate in T_{P}From the timeovercurrent 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 T_{P}Because 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
Prefault condition
=1.368sinPower output =
During fault
=0.365sinPower output =
=1.25sinPower output =Post fault output =
= 1.25, _{m3} = 0.365, P_{m2} = 1.368, P_{m1}P
= 2sinsin = _{e}X= 0.8 p.u; P
= 0.8; _{s}Therefore P
= 0.4572rad,
= 0.8, thus 2sin
= 0.371rad
= 0.8, thus thus 2.44sin
0.371rad= 2.771rad
=
=
= 0.41059
^{o}Critical 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)= 702.81MW^{1}Change 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)= 0MW^{1}Change in MW (P
= 186.7MW_{3 }= 305.8MW and P_{2}Since the units of production are now P1 = 482.5MW, P
+ +
Cost =
+
+
Cost =
= $8,228.029 per hour
+
+
Cost =
as _{3 } and P_{2}, P_{1}b) 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) = 32.5MW^{2}Change 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) = 0MW^{2}Change in MW (P
= 200MW_{3 }= 325MW and P_{2}Since the units of production are now P1 = 450MW, P
+ +
Cost =
+
+
Cost =
= $8,236.25 per hour
+
+
Cost =

Hadi Saadat, “Power System Analysis”, McGrawHill, 2004

John J. Grainger and William D. Stevenson, Jr. “Power System Analysis”, McGrawHill, 1994