MVA MW := Tpu 1 := - University of Idaho · Title: Mathcad - HW08 Solution V14.xmcd Author: joel...

ECE 404Introduction to Power Systems

HW #8 Chapter 3: 62Chapter 4: 18, 19, 20, & 32

j 1−:= MVA MW:= pu 1:= Tpu 1:=

3.62 A three-phase transformer is rated 1 GW, 13.8kV Δ/345 kV Y with Zeq = j 0.10 per unit. Thetransformer high-voltage winding has plus or minus 10% taps. The system base quantitiesare:

Sbase_sys 500 MVA⋅:=

Vbase_XLL 13.8 kV⋅:=

Vbase_HLL 345 kV⋅:=

Determine the per-unit equivalent circuit for the following tap settings:

a. Rated tap andb. + 10% tap (providing 10% voltage increase for the high-voltage winding.)

Zbase_sysHVbase_HLL

2

Sbase_sys:= Zbase_sysH 238.05 Ω=

Zeq j 0.10⋅ Tpu⋅:=

Sbase_T 1000 MVA⋅:=

Zbase_TVbase_HLL

2

Sbase_T:= Zbase_T 119.025 Ω=

a.

at13.8 kV⋅

345 kV⋅:= at 0.040=

bVbase_XLLVbase_HLL

:= b 0.040=

catb

:= c 1=

C:\JoeLaw\Classes\ECE404\Homework\HW08\HW08 Solution V14.xmcd

of 6 October 3, 2008



Zpu_sysZeq Zbase_T⋅

Zbase_sysH:= Zpu_sys 0.05j pu⋅=

+VH-

Zeq

+VX-

IHIH

b.

atp1013.8 kV⋅

1.1 345⋅ kV⋅:= atp10 0.036=

catp10

b:= c 0.909=

Yeq1

Zpu_sys:=

1Yeq

0.05j pu⋅=Yeq 20j− pu⋅=

c Yeq⋅ 18.182j− pu⋅=1

c Yeq⋅0.055j pu⋅= inductor

1 c−( )Yeq 1.818j− pu⋅=1

1 c−( ) Yeq⋅0.55j pu⋅= inductor

c2 c−( )Yeq 1.653j pu⋅=1

c2 c−( ) Yeq⋅0.605j− pu⋅= capacitor

+VH-

+VX-

IHIXcYeq

(1-c)Yeq (c2-c)Yeq

+VH-

+VX-

IHIX





For fun determine the admittance looking in the low side with the high side open circuited.

Yseriesc Yeq⋅ c2 c−( )⋅ Yeq⋅

c Yeq⋅ c2 c−( ) Yeq⋅+:= Yseries 1.818j pu⋅=

1 c−( ) Yeq⋅ 1.818j− pu⋅=

Yparallel Yseries 1 c−( ) Yeq⋅+:= Yparallel 0 pu⋅=

4.18 A 230-kV, 60 Hz, three-phase completely transposed overhead line has one ACSR954-kcmil conductor per phase and flat horizontal phase spacing, with 8 m between adjacentconductors.

Determine the inductance in H/m and the inductive reactance in Ω/km.

Looking at Table A.4: Cardinal has 954-kcmil.

r' 0.0403 ft⋅:= r' 1.228 cm⋅=

D12 8 m⋅:= D13 16 m⋅:= D23 8 m⋅:=

Ds3 D12 D13⋅ D23⋅:= Ds 10.079 m⋅=

La 2 10 7−⋅

Hm⋅ ln

Dsr'

⎛⎜⎝

⎞⎟⎠

⋅:= La 1.342 10 6−×

Hm⋅=

Xa 2 π⋅ 60⋅ Hz⋅ La⋅:= Xa 0.506Ω

km⋅=





4.19 Rework Problem 4.18 if the phase spacing between adjacent conductors is:

a. increased by 10% to 8.8 m,b. decreased by 10% to 7.2 m.

Compare the results with those of Problem 4.18

a.

D12i 8.8 m⋅:= D13i 17.6 m⋅:= D23i 8.8 m⋅:=

Dsi3 D12i D13i⋅ D23i⋅:= Dsi 11.087 m⋅=

Lai 2 10 7−⋅

Hm⋅ ln

Dsir'

⎛⎜⎝

⎞⎟⎠

⋅:= Lai 1.361 10 6−×

Hm⋅=

Lai La−

La1.42 %⋅=

Xai 2 π⋅ 60⋅ Hz⋅ Lai⋅:= Xai 0.513Ω

km⋅=

Xai Xa−

Xa1.42 %⋅=

b.

D12d 7.2 m⋅:= D13d 14.4 m⋅:= D23d 7.2 m⋅:=

Dsd3 D12d D13d⋅ D23d⋅:= Dsd 9.071 m⋅=

Lad 2 10 7−⋅

Hm⋅ ln

Dsdr'

⎛⎜⎝

⎞⎟⎠

⋅:= Lad 1.321 10 6−×

Hm⋅=

Lad La−

La1.57− %⋅=

Xad 2 π⋅ 60⋅ Hz⋅ Lad⋅:= Xad 0.498Ω

km⋅=

Xad Xa−

Xa1.57− %⋅=





4.20 Calculate the inductive reactance in Ω/km of a bundled, 60 Hz, three-phase completelytransposed overhead line having three ACSR 1113-kcmil conductors per bundle, with 0.5mbetween conductors in the bundle. The horizontal spacing between bundle centers are 10,10, and 20 m

d

D6 D6

dd

d 0.5 m⋅:= D6 10 m⋅:=

Looking at Table A.4: Finch has 1113-kcmil.

Calculate the GMR

r'6 0.0435 ft⋅:= r'6 1.326 cm⋅=

DSL3

r'6 d2⋅:= DSL 0.149 m=

Calculate the GMD

D126 D6:= D136 2 D6⋅:= D236 D6:=

Ds63 D126 D236⋅ D136⋅:= Ds6 12.599 m=

Calculate the inductive reactance

XL6 2 π⋅ 60⋅ Hz⋅ 2⋅ 10 7−⋅

Hm⋅ ln

Ds6DSL

⎛⎜⎝

⎞⎟⎠

⋅:= XL6 0.335Ω

km⋅=





4.32 Calculate the capacitance-to-neutral in F/m and admittance-to-neutral in S/km for thesingle-phase line in Problem 4.8. Neglect the effect of the earth plane.

(4.8 A 60 Hz single-phase, two-wire overhead line has solid cylindricalcopper conductors with 1.5 cm diameter. The conductors arearranged in a horizontal configuration with 0.5 m spacing.)

D7

Can Ca’n

a a’

D7 0.5 m⋅:=

r71.52

cm⋅:=

r7 0.75 cm⋅=

ε0 8.854 10 12−×

Fm⋅=

Can2 π⋅ ε0⋅

lnD7r7

⎛⎜⎝

⎞⎟⎠

:= Can 13.247 10 12−×

Fm⋅=

Yan j 2⋅ π⋅ 60⋅ Hz⋅ Can⋅:= Yan 4.994j 10 6−×

Skm⋅=

OR

Zan1

j 2⋅ π⋅ 60⋅ Hz⋅ Can⋅:= Zan 2.002j− 108

× Ω m⋅⋅=

Yan1

Zan:= Yan 4.994j 10 6−

×S

km⋅=



MVA MW := Tpu 1 := - University of Idaho · Title: Mathcad - HW08 Solution V14.xmcd Author: joel...

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