<content> <p>Complex & Polar</p> </content>

e1bc36df-ab2b-4d43-9dc8-3ce215c62f51 40 Norm Schutzkus 4 5 false false 0 0 decimal

&[DATE] &[TIME] - &[FILENAME]

z Conjugate z Re i z Im * - 11 line :

mag ang ∠ mag ang deg * cos i ang deg * sin * + (* :

z1 2.236 63.435 ∠ :

x ∠ x abs x arg 21 mat :

j 1 - sqrt :

z1 1 1.9999 i * +

z1 Re 1

z1 abs 2.236

z1 arg deg 63.435

z1 Im 1.9999

z1 arg rad 1.1071

z1 ∠ 1 el 2.24

z1 ∠ 2 el ° 63.43

Transient Overvoltages
Transient overvoltages are due to network ringing at the beginning of the transient. Temporary overvoltages are caused at steady state due to reactive power unbalance in the system. Consider the single phase representation of a power system, shown in Figure 3.1.2. For the purpose of studying the two different types of overvoltages, the system in front of the load bus can be represented by a Thevenin source. The load at the bus is represented by a series RL equivalent. The bus capacitor accounts for the line charging variables and for additional compensation to compensate for the load reactance.
Although in a practical problem a more complex system representation is required, the single phase example below demonstrates the conditions under which transient and temporary overvoltages manifest and their main characteristics.

ϕ.S 0 deg * :

Consider the following system parameters,

f 60 Hz * :

ω 2 π * f * :

L 0.050 H * :

R 5.0 Ω * :

C 0.000013 F * :

ϕ 30 deg * :

P 200 W * :

V.m 200 V * :

t V.S 2 sqrt V.m * ω t * ϕ.S + cos * :

The load equivalent circuit can be found as

Q P ϕ tan * : 115.47 W *

load reactive power

Z.L V.m 2^P 2^Q 2^+ sqrt / : 173.205 Ω *

R.L Z.L ϕ cos * : 150 Ω *

L.L Z.L ϕ sin * ω / : 0.23 H *

The system operates at steady state, prior to opening the breaker at t = 0. The system steady state prior to the load rejection is calculated as

Z.L R.L j ω * L.L * + :

Z1 1 j ω * C * 1 Z.L / + / :

Zt R j ω * L * + Z1 + :

V.ps V.m j ϕ.S * exp * :

Therefore, the line current is

I.line V.ps Zt / : 1.0446 0.2977 i * + (A *

I.line ∠ 1 el 1.086 A *

I.line ∠ 2 el ° 15.9

and the voltage at the load bus is

V.load V.ps Z1 Zt / * :

V.load ∠ 1 el 201.505 V *

V.load ∠ 2 el ° 6 -

After the opening of the breaker, the equivalent circuit is reduced to that of Figure 3.1.1. The system response can be obtained following similar steps as in the case of a second order circuit. This section demonstrates how a numerical integration method can be used to obtain the system response. With reference to Section 3.1.2, we define the following functions providing the derivatives of the inductor current and capacitor voltage respectively.

I.L V.C t DI.L R I.L * - V.C - t V.S + L / :

I.L V.C t DV.C I.L C / :

Define the integration step.

dt 0.1 ms * :

h1 1.5 dt * :

h2 0.5 dt * :

T 0.09 s * :

The initial conditions can be found from the pre-disturbance steady state.

I.L 1 el 2 sqrt I.line ∠ 1 el * I.line ∠ 2 el cos * :

V.C 1 el 2 sqrt V.load ∠ 1 el * V.load ∠ 2 el cos * :

The system transient overvoltage depends on the instant of switching. This time can be adjusted by adjusting the source phase angle. Assume

ϕ.S 0 deg * :

The system numerical solution is obtained as

I.L 1 el 0 A * :

V.C 1 el 0 V * :

N T dt / Floor : 900

I.L 2 el 0 A * :

V.C 2 el 0 V * :

t 1 el 0 s * :

t 2 el 0 s * :

so, you don't have one starting point, you have two: one and the next in the discretization scheme. This is because the differential system is with second derivatives, coupled one.

ι 2 N range t ι 1 + el t 1 el N ι - 1 + (dt * + : I.L ι 1 + el I.L ι el h1 I.L ι el V.C ι el t ι el DI.L * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DI.L * - eval : V.C ι 1 + el V.C ι el h1 I.L ι el V.C ι el t ι el DV.C * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DV.C * - eval : 31 line for

Note: 3 issues for using range interval instead for loop

First, you can't assing a matrix with indexes, like in mathcad

Second, equations are coupled, so, even maybe you can do that for a small number of intervals in SMath semi symbolic engine, it's impractical. Notice that IT needs values for VC, for the iteration.

Third ... I can't remember, but guess that there are another somewhere there.

t ms / V.C V / augment

Power electronic equipment:
This type of equipment inherently generates current harmonics due to the switching action of the power converters. The order and amplitude of the current harmonics depend on the type of converter and they are called characteristic harmonics. For the normal operation of this equipment, harmonic filters are installed on the system bus to absorb the device characteristic harmonics. However, the converter may produce harmonics of uncharacteristic order. Since filtering is not provided for these orders, interactions may occur between the converter current injections and the system harmonic impedance. Another condition that can produce harmonic interactions is temporary system frequency
change, which results in the detuning of the converter harmonic filters.

A harmonic current source is injected into the load bus. This source represents the harmonics of the transformer magnetizing current immediately following the load rejection.

t I.T 0.8 A * 2 ω * t * cos * 0.4 A * 3 ω * t * cos * + :

the capacitor voltage is given by

I.L V.C t DV.C I.L t I.T - C / :

dt 0.1 ms * :

h1 1.5 dt * :

h2 0.5 dt * :

T 0.09 s * :

N T dt / Floor : 900

ι 2 N range :

I.L 2 el 0 A * :

V.C 2 el 0 V * :

t 1 el 0 s * :

t 2 el 0 s * :

ι 2 N range t ι 1 + el t 1 el N ι - 1 + (dt * + : I.L ι 1 + el I.L ι el h1 I.L ι el V.C ι el t ι el DI.L * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DI.L * - eval : V.C ι 1 + el V.C ι el h1 I.L ι el V.C ι el t ι el DV.C * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DV.C * - eval : 31 line for

t ms / V.C V / augment

Breaker Recovery Voltage
Part of the transient phenomena studies includes the study of the recovery voltage appearing across a circuit breaker during opening operation. Excessive recovery voltage may impair the breaker's ability to interrupt the current. Proper circuit representation of the system can provide an estimate of the recovery voltage.

With reference to Figure 3.1.3, the fault occurs at the load side of the breaker. The fault current is essentially limited by the system inductance. As soon as the contacts of the circuit breaker open, an arc will develop which will sustain the fault current. Upon extinction of the arc, the fault current will be interrupted near its natural zero crossing. As soon as the fault current has been interrupted, the capacitance on the other side of the breaker will charge to the source voltage through the system inductance. Due to the resonance in the system, the capacitor voltage and, therefore, the breaker recovery voltage will overshoot reaching its first maximum value. The rate of change of the recovery voltage at the initial system ringing and its maximum value will stress the breaker insulation. The arc may reignite and the fault current may restrike if the breaker has not built sufficient insulation withstand capability at the moment of the recovery.

The analysis of the recovery voltage can commence at the moment of fault current interruption. With respect to the source voltage, this happens at a phase angle equal to the system impedance angle. Consider the following parameters.

ϕ ω L * R / atan :

f 60 Hz * :

ω 2 π * f * :

L 0.015 H * :

R 1.0 Ω * :

C 1 μF * :

V.m 200 V * :

t V.S 2 sqrt V.m * ω t * ϕ - cos * :

dt 10 μs * :

T 5 ms * :

I.L V.C t DI.L R I.L * - V.C - t V.S + L / :

I.L V.C t DV.C I.L C / :

h1 1.5 dt * :

h2 0.5 dt * :

N T dt / Floor : 500

ι 2 N range :

I.L 2 el 0 A * :

V.C 2 el 0 V * :

t 1 el 0 s * :

t 2 el 0 s * :

ι 2 N range t ι 1 + el t 1 el N ι - 1 + (dt * + : I.L ι 1 + el I.L ι el h1 I.L ι el V.C ι el t ι el DI.L * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DI.L * - eval : V.C ι 1 + el V.C ι el h1 I.L ι el V.C ι el t ι el DV.C * + h2 I.L ι 1 - el V.C ι 1 - el t ι 1 - el DV.C * - eval : 31 line for

t ms / V.C V / augment