[ACADEMIC] Mathcad - Chopping 524: Transients in Power Systems Session 24; Page 2/8 Spring 2018 Im...

ECE 524: Transients in Power Systems

Session 24; /8Spring 2018

Current Chopping

A 480:4160V single phase transformer is rated at 300kVA. The transformer draws a no-loadcurrent of 0.025 p.u. (both magnetization and core losses) at a power factor of 0.12 lagging.The 4160V side is 4 m long and has a shunt parasitic capacitance to ground of 500 pF/m.You can neglect saturation.

(A). When the transformer is de-energized the breaker chops a current of 0.4A (the switchsuccesfully interrupts this current prior to the current 0) on the 4160V side. Assume thetransformer secondary is already open. Find transient voltage resulting from this currentchopping. Neglect core loss resistance.

Define units: kVA kW pu 1 msecs

1000

VHV 4160V 2 VHV 5883.128V

Srated 300kVA

IBHVSrated

VHV IBHV 72.115A Recall this is a single phase transformer so

don't divide by SQRT(3).

Inl 0.025pu pfnl 0.12 lagging

θnl acos pfnl θnl 83.108 deg

Assume that the votlage is at an angle of 0 degrees.

Inl_amps Inl IBHV ej θnl

Inl_amps 0.216 1.79i( ) A

Inl_amps 1.803A 2 Inl_amps 2.55A

Now find the magnetizing inductance and the core loss resistance terms referred to 4160Vside

Ir Re Inl_amps Ir 0.216A

RcVHV

Ir

Rc 19.2284 kΩ



Im Im Inl_amps Im 1.79 A

XmVHV

Im

Xm 2.324 kΩ

LmXm

2 π 60Hz Lm 6.165 H

Total capactiance:

Cg 500pF

m Length 4m

Cg_total Cg Length Cg_total 2000 pF

The resulting circuit is a parallel LC circuit (RLC in part B).

ATPDraw File

4160V RMS

VS V1IV

6.165H2000pF

R=

0 IA

BRK

TimedBreaker

LogicClosed@t0

BRK

6.1

65

2 [H

]

Vcap

0.0

02

[uF

]

PSCAD File

Now find the transient response to the current chopping without the resistance included:

Ichop 0.4A Note that this in an instantaneous current so no SQRT(2)

There will be two parts, first the voltage due to the interrupted inductive current

Z0Lm

Cg_total Z0 55.521 kΩ



Vc0 Ichop Z0 Vc0 22.208 kV

Second there will be an offset due to the initial votlage across the capacitor, but this doesn'tproduce a net dc offset, just an initial point.

θinit acosIchop

2 Inl_amps

θinit 80.974 deg

Vc_init 2 VHV sin θinit Vc_init 5810.278V

However, we really only care about the peak value in this case.

Full response is:

ω01

Lm Cg_total ω0 9005.608

rad

s

f0ω0

2π f0 1433.287

1

s

t 0sec 106sec

1

60Hz

θ acosVc_init

Vc0

θ 74.833 deg

Vcap t( ) Vc0 cos ω0 t θ Vcap 0s( ) 5810.278V

Vcap_pk Vc0 Vcap_pk 22.208 kV

0 5 103 0.01 0.015

2 104

1 104

0

1 104

2 104

Vcap t( )

t



ATP results:

Circuit shown above.ATP Data File

BEGIN NEW DATA CASEC --------------------------------------------------------C Generated by ATPDRAW May, Saturday 17, 2008C A Bonneville Power Administration programC by H. K. Høidalen at SEfAS/NTNU - NORWAY 1994-2006C --------------------------------------------------------C dT >< Tmax >< Xopt >< Copt > 1.E-5 .05 500 1 1 1 1 0 0 1 0C 1 2 3 4 5 6 7 8C 345678901234567890123456789012345678901234567890123456789012345678901234567890/BRANCHC < n1 >< n2 ><ref1><ref2>< R >< L >< C >C < n1 >< n2 ><ref1><ref2>< R >< A >< B ><Leng><><>0 V1 6165.2 0 V1 .002 0/SWITCHC < n 1>< n 2>< Tclose ><Top/Tde >< Ie ><Vf/CLOP >< type > VS V1 -1. .03 .4 1/SOURCEC < n 1><>< Ampl. >< Freq. ><Phase/T0>< A1 >< T1 >< TSTART >< TSTOP >14VS 0 5883.13 60. -1. 1.E3/OUTPUT V1 BLANK BRANCHBLANK SWITCH

ATP Simulation Results

(f ile Prob1A.pl4; x-v ar t) v :V1 30.5 31.6 32.7 33.8 34.9 36.0*10-3

-30

-20

-10

0

10

20

30

*103The overvoltage in the simulationresults is very dependent on wherethe current is actually chopped.Unless 0.4A falls on a timestep thecurrent chopped will be somewhatlarger and the resulting voltage willalso be larger, with some caseshaving Vpk over 23kV.PSCAD results are similar and arenot shown.

Vt0 5811.9V Vpk 22.211kV



(B). Repeat part (B) with core loss resistance included.

Now include the impact of the TRV reduction resistor:

α1

2 Rc Cg_total α 13001.572

1

s

Recall that:

ω0 9005.608rad

s

ω0 α so the system is overdamped

ωd ω02

α2 ωd 9377.627i

rad

s

fdωd

2 π fd 1.492i kHz

General homogeneous solution for an underdamped system:

vc t( ) A1 es1 t

A2 es2 t

=

s1 α α2

ω02 s1 3623.944

1

s

s2 α α2

ω02 s2 22379.199

1

s

tvc t( )

d

ds1 A1 e

s1 t s2 A2 e

s2 t=

We need to apply the following intial conditions to find A1 and A2

vc 0sec( ) Vc_init= where:

θinit acosIchop

2 Inl_amps

θinit 80.974 deg

Vc_init 2 VHV sin θinit 90deg θnl Vc_init 5657.537 V

i 0sec( ) Ichop= where: Ichop 0.4A



Therefore

vc 0sec( ) A1 A2=

Now take the derivative of the voltage expression :

tvc 0 sec( )

d

d

ic 0sec( )

Cg_total= s1 A1 s2 A2=

where:ic 0sec( )

Cg_total

i 0sec( ) iR 0sec( )

Cg_total

=

i 0sec( ) Ichop=and:

iR 0sec( )Vc_init

Rc=

In this case:

Ichop

Cg_total2 10

8V

s

Use a solve block to find A1 and A2

Initial Guesses

A1 0V A2 0V

Given

A1 A2 Vc_init=

s1 A1 s2 A2Ichop

Cg_total

=

As Find A1 A2 As

17414.381

11756.844

V

A1 As0 A1 17414.381 V

A2 As1 A2 11756.844V



Then we get

vcap t( ) A1 es1 t

A2 es2 t

Now to find the peak value, we want to take the derivative and set it to zero, with another solveblock.

dvcap t( ) s1 A1 es1 t

s2 A2 es2 t

tt 0.1msec

Given

dvcap tt( ) 0V

s=

tpeak Find tt( ) tpeak 0.076 msec

vcap tpeak 11.076 kV vcap 0sec( ) 5657.537 V

0 1 103 2 10

31.5 10

4

1 104

5 103

0

vcap t( )

t

(f ile Prob1B.pl4; x-v ar t) v :V1 23.5 24.0 24.5 25.0 25.5 26.0*10-3

-12

-10

-8

-6

-4

-2

0

*103

Vt0 5656.6 V

Vpk 11.008 kV

Good match with analyticalresults.



(C). Repeat (A) and (B) if the breaker cleared at the natural current zero.

(f ile Prob1A.pl4; x-v ar t) v :V1 0 10 20 30 40 50*10-3

-6000

-4000

-2000

0

2000

4000

6000

Note that thereis now noovervoltage.

(f ile Prob1B.pl4; x-v ar t) v :V1 0 10 20 30 40 50*10-3

-6000

-4000

-2000

0

2000

4000

6000

[ACADEMIC] Mathcad - Chopping 524: Transients in Power Systems Session 24; Page 2/8 Spring 2018 Im...

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Transcript of [ACADEMIC] Mathcad - Chopping 524: Transients in Power Systems Session 24; Page 2/8 Spring 2018 Im...