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Synchronous Machines Study 3

The document describes a short circuit test on a synchronous machine model in PSCAD. It applies a short circuit, analyzes the simulation results, and validates that the theoretical time constants match those demonstrated in the simulation graphs.

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0% found this document useful (0 votes)
90 views6 pages

Synchronous Machines Study 3

The document describes a short circuit test on a synchronous machine model in PSCAD. It applies a short circuit, analyzes the simulation results, and validates that the theoretical time constants match those demonstrated in the simulation graphs.

Uploaded by

Panu Mark II
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Synchronous Machines Study 3

A Short Circuit Test on the Machine Model


Motivation:

This example demonstrates the classical short circuit test of a synchronous machine. The associated
discussion of the simulation results serves as a validation of the PSCAD model.

System Overview:

The circuit diagram of this example is shown in Figure 1.

Figure 1: Short Circuit Test Set Up

To conduct the SC test, the machine has to be running in steady state in open circuit conditions. This is
achieved by adjusting the phase angle and magnitude of the machine voltage with respect to the source
voltage so that the current in the machine is zero (negligible) in steady state.

Voltage magnitude and phase of the infinite source is 230.0 kV and 0.0 degree, respectively. Same
quantities for the machine are 13.8 kV and –31.08 degrees (includes phase shift by transformer and
interface, ∆t=50µs). Field voltage necessary to produce 1.0 PU terminal voltage on the open circuited
machine is 1.0 PU. These initial conditions give open circuit condition for the machine.

The machine is run at constant speed by locking the rotor (Enab = 0) at synchronous speed. Thus, there
are no prime mover dynamics involved. The exciter dynamics are also eliminated by feeding a constant
voltage (Ef=1.0 PU) to the exciter. Machine saturation is disabled. The ideal transformer is simply a ratio
changer with negligible leakage (0.005 PU) reactance and no saturation. These simplifications allow us to
focus on the machine dynamics better.

Rev.1 Page 1
Synchronous Machines Study 3

The relevant section of the machine parameters is shown in Figure 2.

Figure 2: The 13.8 kV. 120 MVA Generator Parameters

A short circuit is applied at 0.5056 seconds (time 0.5056 seconds is chosen just for convenience so that
the Phase A current does not have a DC component during the SC test).

Analysis and Simulation Results:

Let us validate the model by comparing the theoretical time constants for the given machine parameters
with the time constants demonstrated by the simulation graphs. For details refer to [1].

Sub Transient time constant

The sub transient component of short circuit current should decay with the sub transient (or damper)
time constant (Td”) given by the following equation,

 Xd __   0.280 
Td "     Tdo __     0.039  34.7 ms (1)
 Xd _   0.314 

Thus, the sub transient effects will be seen for only about two cycles.

Transient time constant

The transient component should decay with the transient time constant (Td’)

 Xd _   0.314 
Td '     Tdo _     6.55  2.03 s (2)
 Xd   1.014 

The subtransient and transient time constants can be seen from the expanded view of the Phase A fault
current in Figure 4 below.

Rev.1 Page 2
Synchronous Machines Study 3

Figure 3: Fault Current (Isc)

Figure 4: Decay of sub-transient and transient components of the Phase A fault current

Field current decay time constant

The time constant of the field current to decay to its pre–fault value is also Td” (if a constant field
voltage, as we have applied in this case). This can be verified from the field current plot in Figure 5.

After the sub–transient effects have disappeared, but the transient component is still present, the
magnitude of the field current is given by,

 Xd   1.014 
I 'fo     I fo     1  3.23 PU (3)
 Xd _   0.314 

Rev.1 Page 3
Synchronous Machines Study 3

The initial DC component of field current is approximately the midpoint of the first cycle which is about
3.2 PU as shown in Figure 5 and agrees with the above calculation. With a fixed field voltage, the field
current will return to its pre–fault value in steady state. The decay of the field current during the
transient period is given by,

t t
 
I 'f  I fo  I 'fo  I fo  e Td '
= 1  3.23  1  e
Td '
(4)

Thus, after a time period equal to Td’, the field current will decay to around 37% of its initial value.

I 'f Td '  1  3.23  1  0.37  1.825 PU

Figure 5: Field Current Response

From Figure 5, it can be seen that If reached approximately 1.825 PU after about 2.0 seconds from the
fault inception. This agrees with the theoretical calculation of Td’. An exponential curve (ExpCurve) with
a time constant of 2.03 seconds is superimposed on If to show that If indeed decays with this time
constant. If SC is not ideal, but has a resistance (the fault resistance in our case can be considered
negligible), this time constant could further be reduced.

Moreover, the transient and sub–transient components of current are only different by about 12% as
shown below,

I a"  Xd _   0.314 
    1.12
I a'  Xd __   0.280 

With the fast decay rate of Ia”, this difference is difficult to observe. Hence, from Figure 4 it can be
noticed that the sub–transient and transient currents are almost of the same magnitude (Ia" = 25.35 kA
and Ia'= 22.6 kA).

Rev.1 Page 4
Synchronous Machines Study 3

Another calculation that can be verified is the ratio of the sub–transient component Ia” to the steady
state fault current Ia. Note that with constant field excitation, we have,

I a"  Xd   1.014 
    3.6
I a  Xd __   0.280 

From the top plot of Figure 3 (Phase A fault current), we obtain a ratio value of 3.57 (= 25.0/7.0), which
is close to the value calculated using the above equation.

The steady state Ia of 7.0 kA also exactly matches the calculated value as seen in Figure 6.

Figure 6: Steady State Fault Current

The armature time constant, Ta (0.278 s), is the decay time constant of the fundamental frequency
component of If (on the stator side this is the time constant at which the DC component and the second
harmonic component of stator current decay). This time constant is estimated in Figure 7. The initial
peak–to–peak magnitude of the 60 Hz component after sub–transient influence disappeared is about
3.0 PU. The 60 Hz component reached 37% of this value (1.1 PU) in about 0.270 seconds. This value
closely agrees with the given value for parameter Ta.

Figure 7: Decay of the fundamental frequency component in the field current

Rev.1 Page 5
Synchronous Machines Study 3

Discussion:

- The theoretical results and the simulation results are very close. Therefore, the model of the
synchronous machine is accurately represented in PSCAD.

PSCAD:

Refer to PSACD case: SM_study_03.pscx

References:

[1] C.V. Jones, The Unified Theory of Electrical Machines, Plenum Press, N.Y., 1967, Chapter 20

Based on the original document for PSCAD v.2. by: Dr Om Nayak and Dr. Ani Gole.
Prepared by: Dharshana Muthumuni

Copyright © 2018 Manitoba Hydro International Ltd. All Rights Reserved.

Rev.1 Page 6

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