Solution Manual - Distribution System Modeling and Analysis 3rd Kersting

Problem
2.1




Problem
2.2
Tap




#1 #2 #3 #4 #5 #6 #7 #8

,Problem
2.3
N1 5000' N2 2500' N3



T1 T2



1 2 3 4 5 6 7 8 9 10 11 12


Problem 2.3.1-2.3.9 Answers for 2.3.5, 2.3.6, and 2.3.9 are kW demands




Problem 2.3.j

j ⋅ 0deg
VN1 := 2500⋅ e pf := .95 kVAT1 := 25 kVAT2 := 37.5 kVhi := 2.4 Vlow := 240

j ⋅ 40deg j ⋅ 50deg
zline := 0.306 + 0.6272j zpuT1 := 0.018 ⋅ e zpuT2 := 0.02⋅ e


DN1N2 := 5000 DN2N3 := 2500

2
kVhi ⋅ 1000
ZbaseT1 := ZbaseT1 = 230.4000
kVAT1

ZT1 := zpuT1⋅ ZbaseT1 ZT1 = 3.1769 + 2.6658j

2
kVhi ⋅ 1000
ZbaseT2 := ZbaseT2 = 153.6000
kVAT2

ZT2 := zpuT2⋅ ZbaseT2 ZT2 = 1.9746 + 2.3533j

DN1N2
zN1N2 := zline⋅ zN1N2 = 0.2898 + 0.5939j
5280
DN2N3
zN2N3 := zline⋅ zN2N3 = 0.1449 + 0.2970j
5280

Note: The voltage drops will be computed for a "worst case" situation. For each segment or transformer
the maximum kVA demand on that segment or transformer will be used to compute the voltage drop to
the remote end. This remote end voltage will then be assumed to be the voltage at that node when the
maximum diversified demand downstream occurs.

, 57.89 j ⋅ acos( pf ) arg ( kVADemandN1)
kVADemandN1 := ⋅e kVADemandN1 = 60.9368 = 18.1949
pf deg
⎯
kVADemandN1 arg ( IN1N2)
IN1N2 := IN1N2 = 24.3747 = −18.1949
V N1 deg
1000
arg ( VN2)
VN2 := VN1 − zN1N2⋅ IN1N2 VN2 = 2488.7963 = −0.2658
deg



22.71 j ⋅ acos( pf ) arg ( kVADemandT1)
kVADemandT1 := ⋅e kVADemandT1 = 23.9053 = 18.1949
pf deg


⎯
kVADemandT1 arg ( IT1)
IT1 := IT1 = 9.6052 = −17.9290
VN2 deg
1000

arg ( VT1)
VT1 := VN2 − IT1⋅ ZT1 VT1 = 2451.9975 = −0.6196
deg


⎛ Vlow ⎞
VlowT1 := VT1⋅ ⎜ ⎟ VlowT1 = 245.1998
⎝ kVhi⋅ 1000 ⎠

41.56 j ⋅ acos( pf ) arg ( kVADemandN2)
kVADemandN2 := ⋅e kVADemandN2 = 43.7474 = 18.1949
pf deg

⎯
kVADemandN2 arg ( IN2N3)
IN2N3 := IN2N3 = 17.4989 = −18.1949
V N1 deg
1000
arg ( VN3)
VN3 := VN2 − zN2N3⋅ IN2N3 VN3 = 2484.7879 = −0.3619
deg

arg ( VT2)
VT2 := VN3 − IN2N3⋅ ZT2 VT2 = 2439.4506 = −1.0341
deg

⎛ Vlow ⎞
VlowT2 := VT2⋅ ⎜ ⎟ VlowT2 = 243.9451
⎝ kVhi⋅ 1000 ⎠
Problem
2.4
j ⋅ 0deg
VN1 := 2500⋅ e pf := .95 kVAT1 := 25 kVAT2 := 37.5 kVhi := 2.4 Vlow := 240

j ⋅ 40deg j ⋅ 50deg
zline := 0.306 + 0.6272j zpuT1 := 0.018 ⋅ e zpuT2 := 0.02⋅ e

, DN1N2 := 5000 DN2N3 := 2500

2
kVhi ⋅ 1000
ZbaseT1 := ZbaseT1 = 230.4000
kVAT1

ZT1 := zpuT1⋅ ZbaseT1 ZT1 = 3.1769 + 2.6658j

2
kVhi ⋅ 1000
ZbaseT2 := ZbaseT2 = 153.6000
kVAT2

ZT2 := zpuT2⋅ ZbaseT2 ZT2 = 1.9746 + 2.3533j

DN1N2
zN1N2 := zline⋅ zN1N2 = 0.2898 + 0.5939j
5280
DN2N3
zN2N3 := zline⋅ zN2N3 = 0.1449 + 0.2970j
5280


kWDemandN1 := 72.43 pf := 0.95


kVAT1 := 25 kVAT2 := 37.5

kWDemandN1 j ⋅ acos( pf ) arg ( kVADemandN1)
kVADemandN1 := ⋅e kVADemandN1 = 76.2421 = 18.1949
pf deg

kVAtotal := kVAT1 + kVAT2

kWDemandN1
AF :=
kVAtotal

AF = 1.1589

kWDemandT1 := AF⋅ kVAT1 kWDemandT1 = 28.9720


kWDemandT2 := AF⋅ kVAT2 kWDemandT2 = 43.4580

kWDemandT1 j ⋅ ( acos( pf ) ) arg ( kVADemandT1)
kVADemandT1 := ⋅e kVADemandT1 = 30.4968 = 18.1949
pf deg

kWDemandT2 j ⋅ ( acos( pf ) ) arg ( kVADemandT2)
kVADemandT2 := ⋅e kVADemandT2 = 45.7453 = 18.1949
pf deg

Note: For all segment and transformer currents, for the constant current model the allocated kVA will be used
along with the Node N1 voltage to compute the currents.


⎯
kVADemandN1 arg ( IN1N2)
IN1N2 := IN1N2 = 30.4968 = −18.1949
V N1 deg
1000