Circulating fault currents in two-winding and three ... · 2W Delta-Wye Xfmr Zero Sequence...

Two Winding Transformer Examples Setup

2W, 3LG Fault, Analysis @ 13.8kV side

2W, 3LG Fault, Analysis @ 115kV side 2W, 1LG Fault, Analysis @ 13.8kV side

2W, 1LG Fault, Analysis @ 115kV side

2W, LL Fault, Analysis @ 13.8kV side


Three Winding Transformer Example Setup 3W, 3LG Fault Analysis @ 13.8kV side

3W, 3LG Fault Analysis @ 115kV side 3W, 1LG Fault Analysis @ 13.8kV side

3W, 1LG Fault Analysis @ 115kV side

3W, LL Fault Analysis @ 13.8kV side

3W, LL Fault Analysis @ 115kV side


Symmetrical Fault Currents with Two-Winding and Three-Winding Transformers

In this paper, we will setup a short circuit example for a 2W and 3W Transformer, hand calculate the short circuit phase and sequence qantities, and validate the quantities with a commercial short

circuit software.

By: Abdur Rehman PE Presented by: Abdur Rehman PE & Jason Hall PE

Last Updated: Rev 1.0 – 2018-02-01













Two Winding Transformer Example DAB-Wye Connection The following parameters are given in this example

We need to convert these parameters to something more usable on a common base. Once we have converted the parameters, we need to enter them in a short circuit software for validation purposes.

𝑍𝑍𝑠𝑠𝑠𝑠 =115𝑘𝑘𝑘𝑘2

600 𝑀𝑀𝑘𝑘𝑀𝑀= 22.04Ω

First, we’ll need convert the MVA short-circuit of the generator to an impedance value. Since the generator is connected to the 115kV bus, we can use that as our voltage base to calculate the short circuit impedance.

𝑍𝑍𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒 =𝐾𝐾𝑘𝑘𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒𝑀𝑀𝑘𝑘𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒

=(115𝑘𝑘𝑘𝑘)2

30 𝑀𝑀𝑘𝑘𝑀𝑀= 440.83Ω

We need to decide an impedance base for our system. Since the 2-winding transformer is given in a 30MVA base, it’s worthwhile to select 30MVA as our impedance base

𝑍𝑍𝑠𝑠𝑠𝑠@30𝑀𝑀𝑀𝑀𝑀𝑀=𝑍𝑍𝑠𝑠𝑠𝑠𝑍𝑍𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒

=22.04 Ω

440.83 Ω= 0.05 𝑝𝑝𝑝𝑝

Lastly, we must convert the short-circuit impedance of the generator to a 30 MVA base. Hand calculating short













circuit current is much easier in a per unit system. So that’s why we’re interested in the per unit value.

𝑍𝑍𝑠𝑠𝑠𝑠@30𝑀𝑀𝑀𝑀𝑀𝑀 = 0 + 𝑗𝑗 ∗0.05 𝑝𝑝𝑝𝑝 The resistive component will be ignored. So we will assume the 0.05 pu impedance is fully reactive.

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ℎ𝑟𝑟𝑟𝑟𝑆𝑆𝑟𝑟𝑝𝑝𝑟𝑟 𝑅𝑅𝑅𝑅𝑅𝑅𝑆𝑆𝑅𝑅𝑅𝑅𝑆𝑆𝑆𝑆𝑅𝑅 = 𝑇𝑇𝑟𝑟𝑅𝑅𝑆𝑆𝑟𝑟𝑇𝑇𝑅𝑅𝑆𝑆𝑅𝑅 𝑅𝑅𝑅𝑅𝑅𝑅𝑆𝑆𝑅𝑅𝑅𝑅𝑆𝑆𝑆𝑆𝑅𝑅 = 𝑆𝑆𝑝𝑝𝑆𝑆 𝑇𝑇𝑟𝑟𝑅𝑅𝑆𝑆𝑟𝑟𝑇𝑇𝑅𝑅𝑆𝑆𝑅𝑅 𝑅𝑅𝑅𝑅𝑅𝑅𝑆𝑆𝑅𝑅𝑅𝑅𝑆𝑆𝑆𝑆𝑅𝑅 = 0.05 𝑝𝑝𝑝𝑝

We are also going to assume that the synchronous reactance of the generator is equal to the transient reactive which is equal to the sub transient reactance. Typically, these values are different, however, for simplification purposes, we’ll consider them all equal.

𝑍𝑍𝑇𝑇

(1) =𝑍𝑍𝑇𝑇(2) = 0 + 𝑗𝑗0.10 𝑝𝑝𝑝𝑝

𝑍𝑍𝑇𝑇(0) = 0 + 𝑗𝑗0.10 𝑝𝑝𝑝𝑝

Since the transformer impedance is given as 10% at 30 MVA (positive and zero sequence), and we selected 30 MVA for the generator base, the per unit value is simply .10 We are going to assume that positive sequence impedance and negative sequence impedance are equal. Please also note that 0.10 pu is purely reactive. There is no resistive component for this impedance.

Building the Short-circuit Model for 2W, DAB-Wye XFMR Now that we have calculated all the impedances to a common 30 MVA base, let’s model the system in Aspen Oneliner short-circuit tool so we can validate our hand calculations later.













The generator information was modeled using the information to the left. As shown, the pre-fault per unit voltage is selected as 1 pu and angle is at 0 degrees.













The generator is set at 30 MVA common base. We will assume that all the impedance values are equal to 𝑗𝑗0.05 for simplification purposes.

The transformer information is configured as shown. The primary voltage and secondary voltage as well as transformer rating is appropriately selected. Also note that this is a standard DAB Transformer Connection. Which means that LV line current lags the HV line current by 30 degrees. The 30 MVA base impedances are entered appropriately. Since there is no resistive component given for the impedance, R (positive seq) and R0 (zero sequa) is set to 0. An effective short-ckt impedance is selected for Zg1 (0 + 0j) since the LV Wye winding is grounded. This means that there is no impedance between neutral bushing and ground. If this was an ungrounded transformer, this impedance would be set really high to reflect an effective ungrounded neutral.













2W DAB-Wye Xfmr Current and Voltage Transformation The following section is a brief discussion on current and voltage transformation for a DAB Wye-grounded transformer.

When we compare the HV to LV line current/voltage for the DAB-Wye grounded transformer, we should expect a difference in both magnitude and phase. For example, if we compare 𝐼𝐼𝐿𝐿𝑀𝑀 with 𝐼𝐼∅𝑀𝑀 both on the HV side, we should expect a 1.73 difference in the magnitude and 30 degree phase shift in the angle.

Therefore, if we compare the LV line current quantity (𝐼𝐼𝐿𝐿𝑏𝑏) with the HV line current quantity (𝐼𝐼𝐿𝐿𝑀𝑀), in addition to the magnitude difference which is driven by the transformer ratio, we should expect LV 𝐼𝐼𝐿𝐿𝑏𝑏













to LAG 30 degrees compared to HV 𝐼𝐼𝐿𝐿𝑀𝑀. That is why the DAB-Wye grounded transformer connection is also described as a standard Dy1 connection.

For sequence component, we have to remember that there is a (positive) 30 degree phase shift between positive sequence line current and positive sequence phase current. Similarly, there is an equal amount of phase shift (30 degrees) in the negative direction between the negative sequence line current and negative sequence phase current. This relationship isn’t very clear in Backburn but Grainger’s Power Systems Analysis book makes this very clear in equation 11.20 and 11.88.













2W Delta-Wye Xfmr Zero Sequence Connection According to Blackburn Figure A4.2-1, the zero sequence impedance seen on form the HV, Delta side will effectively be open while the Wye-grounded side is seen as a short with the transformer impedance. This is due to the zero sequence current trap on the Delta side (it does not cause the zero sequence current to flow on the Wye winding). While the grounded-wye causes the short to ground.













2W DAB-WYE XFMR, 3LG Fault at 13.8kV, Analysis @ 13.8kV side Now that we have modeled the system appropriately, let’s simulate a 3 phase fault on the 13.8kV bus using Aspen Oneliner. We will then hand calculate the short circuit current to see if it matches the model

At the very bottom center of the window, we can see that “Bus Fault on: … Secondary 13.8kV 3LG” indicating the fault type. Right below the tool bar, there are a 3 button starting with positive sequence symbol, negative sequence symbol, and zero sequence symbol. There are also buttons for A, B, C

indicating phase selection ( ). Currently, the positive sequence button is selected so all the values shown are positive sequence (both voltage and current)













The 3 phase fault on the 13.8kV side produces the following sequence and phase currents.

𝐼𝐼𝑀𝑀(1) = 8367∠90° 𝐼𝐼𝑀𝑀 = 8367∠90°

𝐼𝐼𝑀𝑀(2) = 0 𝐼𝐼𝐵𝐵 = 8367∠-30°

𝐼𝐼𝑀𝑀(0) = 0 𝐼𝐼𝐶𝐶 = 8367∠-150°

Now that we a reference values, let’s try to hand calculate a 3 phase fault and compare the hand calculations with the simulation software. We will draw the positive, negative, and zero sequence networks as show below.

For the positive sequence network, 𝑗𝑗0.05 is clearly the generator impedance while the transformer impedance is clearly 𝑗𝑗0.10. Additionally, we will define our pre-fault voltage as 1pu at 0 degrees. It’s













interesting to note that some books use 𝑗𝑗1.0 𝑟𝑟𝑟𝑟 1∠90° for a pre-fault voltage instead of 1∠0°. This will result in the same fault magnitudes however, there will be a difference in the fault angle. For the purpose of this example, we will use a pre-fault voltage of 1∠0°.

The negative sequence network is like the positive sequence except, it does not have a voltage source. Generators do not generate negative sequence, but negative sequence current flows through their windings – which is why we have the generator impedance 𝑗𝑗0.05 included in the sequence network. It is very common to see similar values for positive and negative sequence networks in the industry as well.

The zero sequence network is very different from the other sequence networks. According to Blackburn, it must satisfy the flow of equal and in phase current in the three phases. Furthermore, transformer and line zero sequence currents are very different than positive and negative sequence currents. Depending on the construction of the transformer, the zero sequence impedance could be 85-90% of the positive sequence impedance – in our model however, we were asked to enter the zero sequence impedance values for the transformer and we entered the same values as the positive sequence impedance.

The effective transmission line impedance is also different. When a 1LG fault occurs, the return path of the fault involves ground wires, the earth, effects of the cable sheath, etc. For positive and negative













sequence networks, it’s a one-way way impedance from one end to another. For this reason, the zero sequence impedance is typically 200-600% of the positive impedance. Please note that we don’t have any transmission lines in our example.

A perfectly balanced 3 phase or 3LG fault only has positive sequence network activated as shown above. The negative and zero sequence values are not activated for a 3LG fault.

𝐼𝐼𝑀𝑀(1) =

1∠0°𝑗𝑗0.15

= -𝑗𝑗6.667 𝑝𝑝𝑝𝑝

𝐼𝐼𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 =30𝑀𝑀𝑘𝑘𝑀𝑀

√3∗ 13.8𝑘𝑘𝑘𝑘= 1255.11𝑀𝑀

𝐼𝐼𝑀𝑀

(1) = −𝑗𝑗6.667∗1255.11 = 8367𝑀𝑀 ∠-90°

Since the impedances were given on a 30MVA base, the result is at a per unit value. We’ll need to convert this to amps by multiplying by the current base. Compare this value with the positive sequence current shown in Aspen Oneline. The magnitude is the same however, the angle is 180 degrees out of phase. That’s because Aspen Oneliner is looking from the 13.8kV side and toward the 115kV side so all the angles are 180 degrees out of phase. This is evident by the window name “Phasor #2: LV Secondary 13.8kV – HV Primary 115kV”













𝐼𝐼𝑀𝑀

(2) = 0

Since the negative and zero sequence networks are not activated for 3 phase faults, the negative and zero sequence currents both equal zero which also matches Aspen Oneliner.

Our hand calculations matches the Aspen Oneliner Model

𝐼𝐼𝑀𝑀

(0) = 0

Now let’s calculate the phase current values and see if they compare with Aspen onliner. Because 3 phase faults are balanced, we will use the principles of symmetrical components to calculate the phase B and phase C currents.

𝐼𝐼𝑀𝑀 = 𝐼𝐼𝑀𝑀

(1) + 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0)

Since the sequence components are all references to phase A, we simply add positive, negative, and zero sequence currents to calculate the phase A current (or line A current).

𝐼𝐼𝑀𝑀 = 𝐼𝐼𝑀𝑀(1) + 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0)













𝐼𝐼𝐵𝐵 = 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀

(1) + 𝑅𝑅 ∙ 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0)

𝑅𝑅 = 1∠120° 𝑅𝑅2 = 1∠240°

Phase B current is actually

𝐼𝐼𝐵𝐵 = 𝐼𝐼𝐵𝐵(1) + 𝐼𝐼𝐵𝐵

(2) + 𝐼𝐼𝐵𝐵(0)

Since this is a balanced fault, we can derive 𝐼𝐼𝐵𝐵

(1) by simply rotating 𝐼𝐼𝑀𝑀(1) by

240 degrees in the counter-clockwise direction.

Similarly, to calculate 𝐼𝐼𝐵𝐵(2) we need to

rotate 𝐼𝐼𝑀𝑀(2) by 120 degrees in the

counter clockwise direction (the direction of system phasor rotation which is ABC in our example). Zero sequence currents have equal magnitudes and equal phase angle therefore -- 𝐼𝐼𝐵𝐵

(0) = 𝐼𝐼𝑀𝑀(0)

𝐼𝐼𝐶𝐶 = 𝑅𝑅 ∙ 𝐼𝐼𝑀𝑀(1) + 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0)

𝑅𝑅 = 1∠120° 𝑅𝑅2 = 1∠240°

Phase C current is actually

𝐼𝐼𝐶𝐶 = 𝐼𝐼𝐶𝐶(1) + 𝐼𝐼𝐶𝐶

(2) + 𝐼𝐼𝐶𝐶(0)

Since this is a balanced fault, we can derive 𝐼𝐼𝐶𝐶

(1) by simply rotating 𝐼𝐼𝑀𝑀(1) by

120 degrees in the counter-clockwise direction. Similarly, to calculate 𝐼𝐼𝐶𝐶

(2) we need to rotate 𝐼𝐼𝑀𝑀

(2) by 240 degrees in the counter clockwise direction. (The direction of system phasor rotation which is ABC in our example). Zero sequence currents have equal magnitudes and equal phase angle therefore -- 𝐼𝐼𝐶𝐶

(0) = 𝐼𝐼𝑀𝑀(0)

Now to perform the actual calculations --


1∠0°𝑗𝑗0.15


To summarize, a 3LG fault only has the positive sequence network activated. Therefore, 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0) = 0

Phase A current on the 13.8kV side equals 8367𝑀𝑀 ∠-90°














√3∗ 13.8𝑘𝑘𝑘𝑘= 1255.11A

𝐼𝐼𝑀𝑀

(1) = −𝑗𝑗6.667∗ 1255.11 = 8367𝑀𝑀 ∠-90°


(2) + 𝐼𝐼𝑀𝑀(0) = 𝐼𝐼𝑀𝑀

(1) = 8367𝑀𝑀 ∠-90°

This matches the Aspen Oneliner result. However, the tool is showing the angle at +90 degrees.

The reason why is because the tool is looking at the phasors from the 13.8kV side and toward the source (115kV side).


(1) + 𝑅𝑅 ∙ 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0) 𝐼𝐼𝐵𝐵 = 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀

(1) = (1∠240°)∗ (8367𝑀𝑀 ∠-90°) = 8367𝑀𝑀 ∠150°

Since 𝐼𝐼𝑀𝑀

(2)𝑅𝑅𝑆𝑆𝑎𝑎 𝐼𝐼𝑀𝑀(0)are both 0 for a 3

phase fault, the equation simplifies to 𝐼𝐼𝐵𝐵 = 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀

(1) This matches the Aspen Oneliner result but with a 180 degree phase shift.

𝐼𝐼𝐶𝐶 = a ∙ 𝐼𝐼𝑀𝑀

(1) + 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0) 𝐼𝐼𝐶𝐶 = a ∙ 𝐼𝐼𝑀𝑀

(1) = (1∠120°)∗ (8367𝑀𝑀 ∠−90°) = 8367𝑀𝑀 ∠30°



phase fault, the equation simplifies to 𝐼𝐼𝐶𝐶 = a ∙ 𝐼𝐼𝑀𝑀

(1) This matches the Aspen Oneliner result but with a 180 degree phase shift for the reasons explained above.

Now to calculate the sequence and phase voltages, we perform the following

𝑘𝑘𝑀𝑀(1) = 1∠0°− (j0.15)∗ (−j6.667) = 0pu

Here, we see that positive sequence phase A voltage equal 0 kV which is what we should expect because the short-circuit should drive the voltage to













𝑘𝑘𝑀𝑀

(1) = 0kV∠0°

𝑘𝑘𝑀𝑀 = 𝑘𝑘𝑀𝑀(1) +𝑘𝑘𝑀𝑀

(2) +𝑘𝑘𝑀𝑀(0) = 0 kV∠0°

zero. This agrees with the Aspen model.

𝑘𝑘𝐵𝐵 = 𝑅𝑅2 ∙ 𝑘𝑘𝑀𝑀(1) = (1∠240°)∗ (0V∠0°) = 0kV

As discussed earlier, since this is a balanced three phase fault, all the phase currents will have equal magnitude and displaced by 120 degrees. And since negative and zero sequence does not existing for three phase faults, we rotate the positive sequence voltage by 240 degrees in the CCW direction to get phase B voltage. However, since the positive sequence voltage is zero, the phase B voltage is also zero. The hand-calculated value matches Aspen oneliner.

𝑘𝑘𝐶𝐶 = a ∙ 𝑘𝑘𝑀𝑀(1) = (1∠120°)∗ (0 kV∠0°) = 0kV

We rotate positive sequence A by 120 degrees in the CCW direction to get phase C voltage. But since positive sequence voltage is zero, phase C voltage is also zero. The hand-calculated value matches Aspen oneliner.













2W DAB-WYE XFMR, 3LG Fault at 13.8kV, Analysis @ 115kV side Now let’s perform the same analysis but on the 115kV bus. Aspen Oneliner shows the following results. Let’s see if our hand-calculations matches Aspen Onliner.


1∠0°𝑗𝑗0.15



√3∗ 115 𝑘𝑘𝑘𝑘= 150.613 A

𝐼𝐼𝑀𝑀

(1) = −𝑗𝑗6.667∗ 150.613 = 1004 ∠-90°


(2) + 𝐼𝐼𝑀𝑀(0) = 𝐼𝐼𝑀𝑀

(1) = 1004 ∠-90° Because we know that the HV line current quantities LAGS the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly. For the DAB transformer specifically, we advance 30 degrees forward for positive sequence and 30 degrees backwards for negative sequence.

𝐼𝐼𝑀𝑀(1) = (1004𝑀𝑀 ∠-90°)∗ (1 ∠+30°)

𝐼𝐼𝑀𝑀

(1) = (1004𝑀𝑀 ∠-60°)

The current that flows through the 115kV side is the same as the 13.8kV side. The only different is DAB-Wye-grounded transformer needs to be accounted for – both the magnitude and current. A 3LG fault only has the positive sequence network activated. Therefore, 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0) = 0

If we adjusted only for the magnitude, we should expect the 115kV current to be Phase A current on the 13.8kV side equals 1004𝑀𝑀 ∠-90° However, this is incorrect because we need to account for the DAB delta transformation. According to Aspen Oneliner, 𝐼𝐼𝑀𝑀

(1) = 1004𝑀𝑀 ∠-60°













Because we know that the HV line current quantities LAGS the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly by 30 degrees. 𝐼𝐼𝑀𝑀

(1) = (1004𝑀𝑀 ∠-90°)∗ 1 ∠+30° Which gives us the 𝐼𝐼𝑀𝑀

(1) =(1004𝑀𝑀 ∠-60°) The hand-calculated value matches Aspen oneliner.

𝐼𝐼𝐵𝐵 = 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀(1) + 𝑅𝑅 ∙ 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0)


(1) = (1∠240°)∗ (1004𝑀𝑀 ∠-90°)∗ (1 ∠+30°)= (−1004𝑀𝑀) = (1004𝑀𝑀 ∠180°)

𝐼𝐼𝐵𝐵 = 1004𝑀𝑀 ∠180°



phase fault, the equation simplifies to 𝐼𝐼𝐵𝐵 = 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀

(1) Remember, we have to shift positive sequence forward by 30 degrees because we have a DAB Wye-grounded transformer connection. The hand-calculated value matches Aspen oneliner.


(1) + 𝑅𝑅2 ∙ 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0) Since 𝐼𝐼𝑀𝑀















(1) = (1∠120°)∗ (8367𝑀𝑀 ∠−90°) ∗ (1 ∠+30°)= 1004𝑀𝑀 ∠60°

𝐼𝐼𝐶𝐶 = 1004𝑀𝑀 ∠60°

phase fault, the equation simplifies to 𝐼𝐼𝐶𝐶 = a ∙ 𝐼𝐼𝑀𝑀

(1) Remember, we have to shift positive sequence forward by 30 degrees because we have a DAB Wye-grounded transformer connection. The hand-calculated value matches Aspen oneliner.

Now to calculate the sequence and phase voltages, we perform the following

The positive sequence current was calculated as

𝐼𝐼𝑀𝑀(1) = −j6.667

Since we have to account for the DAB transformer phase shift on the HV side, we have to account for the 30 degrees phase shift.

𝐼𝐼𝑀𝑀(1) = −j6.667∗ (1∠30°) = 6.667∠-60°

𝑘𝑘𝑀𝑀(1) = 1∠0°− (𝑗𝑗0.05)∗ (6.667∠-60°) = 0.73∠-13.2°

We must be very careful in how we calculate the phase A sequence voltage because of the DeltaAB transformer connection. The 𝐼𝐼𝑀𝑀

(1) was calculated as −j6.667 pu However, this for the LV side. Because we must consider the DAB Wye-grounded transformer connection, we need to shift the positive sequence phase current by 30 degrees – as seen on the HV side. This effectively becomes 𝐼𝐼𝑀𝑀

(1) = 6.667∠-60° We need to perform KVL to calculate 𝑘𝑘𝑀𝑀

(1) --- which is the source voltage minus the generator impedance (𝑗𝑗0.05) times the current flowing through the generator impedance which is 6.667∠-60°.

𝑘𝑘𝑀𝑀(1) = 0.73∠-13.2°

Since negative and zero sequence networks do not exist, the negative and zero sequence voltages are also zero. Therefore, the phase A voltage equals 𝑘𝑘𝑀𝑀 = 48.51 kV∠-13.2°













𝑘𝑘𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 =115kV

√3

𝑘𝑘𝑀𝑀(1) = 0.73∠-13.2° ∗

115kV

√3= 48.51 kV∠-13.2°

𝑘𝑘𝑀𝑀

(2) = 𝑘𝑘𝑀𝑀(0) = 0

𝑘𝑘𝑀𝑀 = 𝑘𝑘𝑀𝑀

(1) +𝑘𝑘𝑀𝑀(2) +𝑘𝑘𝑀𝑀

(0) = 48.51 kV∠-13.2°

The hand-calculated value matches Aspen oneliner.

𝑘𝑘𝐵𝐵 = 𝑅𝑅2 ∙ 𝑘𝑘𝑀𝑀

(1) = (1∠240°)∗ 48.51 kV∠-13.2°

𝑘𝑘𝐵𝐵 = 48.51 kV∠-133.2°

As discussed earlier, since this is a balanced three phase fault, all the phase currents will have equal magnitude and displaced by 120 degrees. And since negative and zero sequence does not existing for three phase faults, we rotate the positive sequence voltage by 240 degrees in the CCW direction to get phase B voltage. The value calculated matches Aspen oneliner.

𝑘𝑘𝐶𝐶 = a ∙ 𝑘𝑘𝑀𝑀(1) = (1∠120°)∗ 48.51 kV∠-13.2°

𝑘𝑘𝐶𝐶 = 48.51 kV∠106.8°

We rotate positive sequence A by 120 degrees in the CCW direction to get phase C voltage. The value calculated matches Aspen oneliner.













2W DAB-WYE XFMR, 1LG Fault at 13.8kV Bus, Analysis @ 13.8kV side Let’s analyze a 1LG fault (phase A) on the 13.8k V side of the DAB transformer. According to Aspen Oneliner, we should get the following results.













The 1LG fault on the 13.8kV side produces the following current quantities are:

𝐼𝐼𝑀𝑀(1) = 3138∠90° 𝐼𝐼𝑀𝑀 = 9413∠90°

𝐼𝐼𝑀𝑀(2) = 3138∠90° 𝐼𝐼𝐵𝐵 = 0

𝐼𝐼𝑀𝑀(0) = 3138∠90° 𝐼𝐼𝐶𝐶 = 0

The voltages quantities are:

𝑘𝑘𝑀𝑀(1) = 4.980 kV ∠0° 𝑘𝑘𝑀𝑀 = 0

𝑘𝑘𝑀𝑀(2) = 2.988 kV ∠180° 𝑘𝑘𝐵𝐵 = 7.519 kV ∠-113.4°

𝑘𝑘𝑀𝑀(0) = 1.992 kV ∠180° 𝑘𝑘𝐶𝐶 = 7.519 kV ∠113.4

Remember, the phasor window on the 13.8kV side is looking from the 13.8kV side toward the 115kV side. Aspen Oneliner is unnecessarily adjusting the phase angle by 180 degrees. So the actual values should be 180 degrees out of phase.

Now that we a reference values, let’s try to hand calculate a 1LG fault and compare the hand calculations with the simulation software. We will connect sequence networks as shown below per Blackburn.














1∠0°𝑗𝑗0.15+ 𝑗𝑗0.15 + 𝑗𝑗0.10

= −𝑗𝑗2.5𝑝𝑝𝑝𝑝


√3∗ 13.8𝑘𝑘𝑘𝑘= 1255𝑀𝑀

𝐼𝐼𝑀𝑀

(1) =−𝑗𝑗2.5𝑝𝑝𝑝𝑝∗ 1255𝑀𝑀 = 3138∠-90°

The positive sequence current is calculated by dividing the pre-fault voltage (1∠0°) by the total impedance of the circuit. The positive sequence impedance is 𝑗𝑗0.05 + 𝑗𝑗0.10 = 𝑗𝑗0.15 The negative sequence impedance is 𝑗𝑗0.05 + 𝑗𝑗0.10 = 𝑗𝑗0.15 The zero sequence impedance is 𝑗𝑗0.10 The positive sequence current matches the result shown in Aspen Oneliner. Remember, Aspen results are 180 degrees out of phase because it’s viewing the phasors from 13.8kV side looking toward the 115kV side.

𝐼𝐼𝑀𝑀(2) = −𝑗𝑗2.5 𝑝𝑝𝑝𝑝

𝐼𝐼𝑀𝑀

(1) =−𝑗𝑗2.5 𝑝𝑝𝑝𝑝 ∗𝐼𝐼𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 = 3138∠-90°

Since the same positive sequence current flows through both the negative sequence and zero sequence networks for a 1LG fault, 𝐼𝐼𝑀𝑀

(1) = 𝐼𝐼𝑀𝑀(2) = 𝐼𝐼𝑀𝑀

(0)

𝐼𝐼𝑀𝑀

(0) = −𝑗𝑗2.5 𝑝𝑝𝑝𝑝

𝐼𝐼𝑀𝑀(0) =−𝑗𝑗2.5 𝑝𝑝𝑝𝑝 ∗𝐼𝐼𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 = 3138∠-90°

Let’s determine the phase current values for the 1LG fault on the 13.8kV side.


(1) + 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0) = (-𝑗𝑗2.5𝑝𝑝𝑝𝑝) + (-𝑗𝑗2.5𝑝𝑝𝑝𝑝) + (-𝑗𝑗2.5𝑝𝑝𝑝𝑝)= -𝑗𝑗7.5 𝑝𝑝𝑝𝑝


√3∗ 13.8𝑘𝑘𝑘𝑘= 1255.109𝑀𝑀

-𝑗𝑗7.5∗ 𝐼𝐼𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 = 9413𝑀𝑀 ∠-90°

Using the principles of symmetrical component, we know that phase A current is the sum of positive, negative, and zero sequence currents. This result matches the Aspen Oneliner simulation with an exception of the phase angle which is 180 degrees apart.













Another way of representing phase A current is

3𝐼𝐼𝑀𝑀(0) = -𝑗𝑗7.5 𝑝𝑝𝑝𝑝→ 9413𝑀𝑀 ∠-90°

𝐼𝐼𝐵𝐵 = 𝑅𝑅2∗ 𝐼𝐼𝑀𝑀(1) + 𝑅𝑅 ∗ 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0)

𝐼𝐼𝐵𝐵 = (1∠240°)(-𝑗𝑗2.5) + (1∠120°)(-𝑗𝑗2.5) + (-𝑗𝑗2.5) = 0

𝐼𝐼𝐵𝐵 = 0

To calculate Phase B current, we must shift the positive sequence current by 240 degrees CCW and negative sequence current by 120 degrees CCW per fundamentals of symmetrical components. The total fault current on phase B equals 0 – this result makes sense because the fault is on phase A and B … so phase B should have no fault current. This result matches the Aspen Oneliner simulation.

𝐼𝐼𝐶𝐶 = 𝑅𝑅 ∗ 𝐼𝐼𝑀𝑀(1) + 𝑅𝑅2 ∗ 𝐼𝐼𝑀𝑀

(2) + 𝐼𝐼𝑀𝑀(0)

𝐼𝐼𝐶𝐶 = (1∠120°)(-𝑗𝑗2.5) + (1∠240°)(-𝑗𝑗2.5)+ (-𝑗𝑗2.5) = 0

Very similar operation for the phase C current. We should expect phase C current because this is a 1LG on line A. This result matches the Aspen Oneliner simulation.

Let’s determine the sequence voltages for the 1LG fault on the 13.8kV side.













𝑘𝑘𝑀𝑀(1) = 1 ∠0°− (j0.15)(−j2.5) = 0.625 pu

𝑘𝑘𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 =13.8kV

√3

𝑘𝑘𝑀𝑀

(1) = (0.625 pu)∗𝑘𝑘𝐵𝐵𝑏𝑏𝑠𝑠𝑒𝑒 = 4.98 kV∠0°

𝑘𝑘𝑀𝑀(1) = 4.98 kV∠0°

If we perform KVL around the positive sequence network and calculate the voltage at the point of the fault, we will always find it to equal the pre-fault voltage minus the positive sequence impedance times the positive sequence current. Notice that this is a positive voltage value so the angle is at 0 degrees. This is the same value as the Aspen Onliner simulation

𝑘𝑘𝑀𝑀(2) = 0 − (j0.15)(−j2.5) = −0.375 pu

𝑘𝑘𝑀𝑀(2) = (−0.375)∗ �

13.8𝑘𝑘𝑘𝑘

√3�= 2.988 kV∠180°

𝑘𝑘𝑀𝑀(2) = 2.988 kV∠180°

Similarly, performing KVL around the negative sequence network always gives us zero minus the negative sequence impedance multiplied by the negative sequence current. Note that negative sign which is resulted in this case… this indicates the negative sequence voltage is 180 degrees out of phase compared to the positive sequence voltage. This is the same value as the Aspen Onliner simulation.













𝑘𝑘𝑀𝑀(0) = 0 − (j0.10)(−j2.5) =−0.25 pu

𝑘𝑘𝑀𝑀(0) = (−0.25)∗ �

13.8𝑘𝑘𝑘𝑘

√3�= 1.992 kV∠180°

𝑘𝑘𝑀𝑀(0) = 1.992 kV∠180°


(1) + 𝑘𝑘𝑀𝑀(2) = 0

𝑘𝑘𝐵𝐵 = 𝑘𝑘𝑀𝑀(0) + 𝑅𝑅2 ∙ 𝑘𝑘𝑀𝑀

(1) + 𝑅𝑅 ∙ 𝑘𝑘𝑀𝑀(2) = 7.5𝑘𝑘𝑘𝑘∠-113 .4°

𝑘𝑘𝐶𝐶 = 𝑘𝑘𝑀𝑀(0) + 𝑅𝑅 ∙ 𝑘𝑘𝑀𝑀

(1) + 𝑅𝑅2 ∙ 𝑘𝑘𝑀𝑀(2) = 7.5𝑘𝑘𝑘𝑘∠113 .4°

Similarly, performing KVL around the Zero sequence network results in the following – using the same concept as discussed previously. Note that negative sign… this indicates the Zero sequence voltage is 180 degrees out of phase compared to the positive sequence voltage and in phase compared to the negative sequence voltage. This is the same value as the Aspen Onliner simulation.













2W DAB-WYE XFMR, 1LG Fault at 13.8kV Bus, Analysis @ 115kV side Now let’s perform the same analysis but on the 115kV bus. Aspen Oneliner shows the following results.

According to Aspen oneliner, the current and voltages values on the 115kV side is expected to be

𝑘𝑘𝑀𝑀(1) = 59.35 kV ∠-4.0° 𝐼𝐼𝑀𝑀

(1) = 377𝑀𝑀 ∠-60°













𝑘𝑘𝑀𝑀(2) = 8.299 kV∠150° 𝐼𝐼𝑀𝑀

(2) = 377A ∠-120° 𝑘𝑘𝑀𝑀

(0) = 0 𝐼𝐼𝑀𝑀(0) = 0

And the phase values on the 115kV side are expected to be

𝑘𝑘𝑀𝑀 = 52.02 kV ∠0° 𝐼𝐼𝑀𝑀 = 652 ∠-90° 𝑘𝑘𝐵𝐵 = 66.40 kV ∠-120° 𝐼𝐼𝐵𝐵 = 0 𝑘𝑘𝐶𝐶 = 60.50 kV ∠108.1° 𝐼𝐼𝐶𝐶 = 652 ∠90°

Now let’s calculate the sequence current quantities on the 115V side.

From the previous analysis, we know that -𝑗𝑗2.5 𝑝𝑝𝑝𝑝 flows on the LV side of the transformer. However, because we have a DAB connection on the HV side, we need to take that into account.


1∠0°𝑗𝑗0.40

= -𝑗𝑗2.5 𝑝𝑝𝑝𝑝 (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)


√3∗ 115 𝑘𝑘𝑘𝑘= 150.613 A

𝐼𝐼𝑀𝑀

(1) = −𝑗𝑗2.5∗ 150.613 = 377 ∠-90° (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅) Because we know that the HV line current quantities LAGS the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly. For the DAB transformer specifically, we advance 30 degrees forward for positive sequence current and 30 degrees backwards for negative sequence current.

The current that flows through the 115kV side is the same as the 13.8kV side. The only different is DAB-Wye-grounded transformer needs to be accounted for – both the magnitude and current. A 1LG fault only has positive sequence, negative sequence, and zero sequence networks activated. If we adjusted only for the magnitude, we should expect the 115kV current to be Phase A current on the 13.8kV side equals 377 𝑀𝑀∠-90° However, this is incorrect because we need to account for the DAB delta transformation. Because we know that the HV line current quantities LAGS the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly by 30 degrees. 𝐼𝐼𝑀𝑀

(1) = (377 𝑀𝑀 ∠-90°)∗ 1 ∠+30° Which gives us the 𝐼𝐼𝑀𝑀

(1) =(377 𝑀𝑀 ∠-60°) The hand-calculated value matches Aspen oneliner.













𝐼𝐼𝑀𝑀(1) = (377 𝑀𝑀∠-90°) ∗ (1 ∠+30°)

𝐼𝐼𝑀𝑀

(1) = 377 𝑀𝑀∠-60° (𝐻𝐻𝑘𝑘 𝑟𝑟𝑇𝑇𝑎𝑎𝑅𝑅)

𝐼𝐼𝑀𝑀(2) = −𝑗𝑗2.5∗ 150.613 = 377 ∠-90° (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)

𝐼𝐼𝑀𝑀

(2) = (377 𝑀𝑀∠-90°)∗ (1 ∠-30°)

𝐼𝐼𝑀𝑀(2) = 377 𝑀𝑀∠-120° (𝐻𝐻𝑘𝑘 𝑟𝑟𝑇𝑇𝑎𝑎𝑅𝑅)

The negative sequence that flows through the 115kV side is the same as the 13.8kV side. The only different is DAB-Wye-grounded transformer needs to be accounted for. Remember, we must shift negative sequence backwards by 30 degrees because we have a DAB Wye-grounded transformer connection. The hand-calculated value matches Aspen oneliner.













𝐼𝐼𝑀𝑀

(0) = 0

Because we have a Delta transformer on the HV side, we know that zero sequence current is trapped within the delta and does not pass on the HV side. The hand-calculated value matches Aspen oneliner.

The line current on the 115kV side are calculated below.


(1) + 𝐼𝐼𝑀𝑀(2) = 377 𝑀𝑀∠-60° + 377 𝑀𝑀∠-120°

𝐼𝐼𝑀𝑀 = 652 A ∠-90°

Using the principles of symmetrical component, we know that phase A current is the sum of positive, negative, and zero sequence currents. However, since the zero sequence current is blocked, we omit the zero sequence current from our calculations. This matches the Aspen Oneliner result.













𝐼𝐼𝐵𝐵 = 𝑅𝑅2∗ 𝐼𝐼𝑀𝑀

(1) + 𝑅𝑅 ∗ 𝐼𝐼𝑀𝑀(2) + 𝐼𝐼𝑀𝑀

(0) 𝐼𝐼𝐵𝐵 = (1∠240°)(377 𝑀𝑀∠-60°) + (1∠120°)(377 𝑀𝑀∠-120°)

𝐼𝐼𝐵𝐵 = 0

To calculate Phase B current, we must shift the positive sequence current by 240 degrees CCW and negative sequence current by 120 degrees CCW per fundamentals of symmetrical components. Again, we omit the zero-sequence current because it does not flow through the 115kV side. This result matches the Aspen Oneliner simulation.

𝐼𝐼𝐶𝐶 = 𝑅𝑅 ∗ 𝐼𝐼𝑀𝑀

(1) + 𝑅𝑅2∗ 𝐼𝐼𝑀𝑀(2)

𝐼𝐼𝐶𝐶 = (1∠120°)(377 𝑀𝑀∠-60°) + (1∠240°)(377 𝑀𝑀∠-120°)

𝐼𝐼𝐶𝐶 = 652 𝑀𝑀∠90°

To calculate Phase C current, we must shift the positive sequence current by 120 degrees CCW and negative sequence current by 240 degrees CCW per fundamentals of symmetrical components. Again, we omit the zero-sequence current because it does not flow on the 115kV side. This result matches the Aspen Oneliner simulation.

Now to calculate the sequence voltages, we perform the following


𝐼𝐼𝑀𝑀(1) = −j2.5

We must be very careful in how we calculate the phase A sequence voltage because of the DeltaAB transformer













Since we must account for the DAB transformer phase shift on the HV side, we must account for the 30 degrees phase shift for the positive sequence current.

𝐼𝐼𝑀𝑀(1) =−j2.5∗ (1∠30°) = 2.5∠-60°

𝑘𝑘𝑀𝑀(1) = 1∠0°− (𝑗𝑗0.05)∗ (2.5∠-60°) = 0.8939∠-4.0°


√3

𝑘𝑘𝑀𝑀(1) = 0.8939∠-4.0°∗

115kV

√3= 59.35 kV∠-4°

connection. The 𝐼𝐼𝑀𝑀

(1) was calculated as −j2.5 pu However, this for the LV side. Because we must consider the DAB Wye-grounded transformer connection for HV calculations, we need to shift the positive sequence phase current by 30 degrees – as seen on the HV side. This effectively becomes 𝐼𝐼𝑀𝑀


(1) --- which is the source voltage minus the generator impedance (𝑗𝑗0.05) times the current flowing through the generator impedance which is 2.5∠-60°. We calculate the positive sequence voltage as 𝑘𝑘𝑀𝑀

(1) = 0.8939∠-4° or 59.35 kV∠-4° The hand-calculated value matches Aspen oneliner.

𝐼𝐼𝑀𝑀

(1) = −j2.5 Since we must account for the DAB transformer phase shift on the HV side, we must account for the (negative) 30 degrees phase shift for the negative sequence current.

𝐼𝐼𝑀𝑀(2) = −j2.5∗ (1∠-30°) = 2.5∠-120°

We must be very careful in how we calculate the phase A sequence negative voltage because of the DeltaAB transformer connection. The 𝐼𝐼𝑀𝑀

(1) was calculated as −j2.5 pu However, this for the LV side. Because













𝑘𝑘𝑀𝑀(2) = 0 − (𝑗𝑗0.05)∗ (2.5∠-120°) = 0.125∠150°


√3

𝑘𝑘𝑀𝑀(2) = 0.125∠150°∗

115kV

√3= 8.299 kV∠150°

we must consider the DAB Wye-grounded transformer connection, we need to shift the negative sequence phase current by -30 degrees – as seen on the HV side. This effectively becomes 𝐼𝐼𝑀𝑀


(1) --- which is the source voltage (we don’t have a source voltage for negative sequence hence 0) minus the generator impedance (𝑗𝑗0.05) times the current flowing through the generator impedance which is 2.5∠-120°. We calculate the negative sequence voltage as 𝑘𝑘𝑀𝑀

(2) = 0.125∠150° or 8.299 kV∠150° The hand-calculated value matches Aspen oneliner.

𝑘𝑘𝑀𝑀(0) = 0

Because there is no zero sequence current that flow on the 115kV side, the zero sequence voltage is non-existent. The value calculated matches Aspen oneliner.













Now to calculate the phase voltages quantities, we perform the following


(2) +𝑘𝑘𝑀𝑀(0)

𝑘𝑘𝑀𝑀 = 59.35 kV∠-4° + 8.299 kV∠150° = 52.02 kV ∠0°

We must combine the positive, negative, and zero sequence voltages to calculate the phase A voltage seen on the 115kV. Note that zero sequence voltage is essentially 0 so we can omit this from our calculations This is the same value as the Aspen Oneliner simulation

𝑘𝑘𝐵𝐵 = 𝑅𝑅2 ∗ 𝑘𝑘𝑀𝑀

(1) + 𝑅𝑅 ∗ 𝑘𝑘𝑀𝑀(2) +𝑘𝑘𝑀𝑀

(0) 𝑘𝑘𝐵𝐵 = (1∠240°)∗ 59.35 kV∠-4°+ (1∠120°)∗ 8.299 kV∠150°

𝑘𝑘𝐵𝐵 = 66.39 kV∠-120.0°

This is the same value as the Aspen Oneliner simulation

𝑘𝑘𝐶𝐶 = 𝑅𝑅 ∗ 𝑘𝑘𝑀𝑀(1) + 𝑅𝑅2 ∗ 𝑘𝑘𝑀𝑀

(2) +𝑘𝑘𝑀𝑀(0)

𝑘𝑘𝐶𝐶 = (1∠120°)∗ 59.35 kV∠-4°+ (1∠240°)∗ 8.299 kV∠150°

𝑘𝑘𝐶𝐶 = 60.50 kV∠108.1°

This is the same value as the Aspen Onliner simulation













2W DAB-WYE XFMR, LL Fault at 13.8kV Bus, Analysis @ 13.8kV Side Let’s perform a LL fault on the 13.8kV bus and analyze the results.

According to Aspen oneliner, the current and voltages values on the 115kV side is expected to be

𝑘𝑘𝑀𝑀(1) = 3.98 kV ∠0° 𝐼𝐼𝑀𝑀

(1) = 4184𝑀𝑀 ∠90° 𝑘𝑘𝑀𝑀

(2) = 3.98 kV∠0° 𝐼𝐼𝑀𝑀(2) = 4184A ∠-90°













𝑘𝑘𝑀𝑀(0) = 0 𝐼𝐼𝑀𝑀

(0) = 0 And the phase values on the 115kV side are expected to be

𝑘𝑘𝑀𝑀 = 7.98 kV ∠0° 𝐼𝐼𝑀𝑀 = 0 ∠0° 𝑘𝑘𝐵𝐵 = 3.98 kV ∠-180° 𝐼𝐼𝐵𝐵 = 7246 ∠0° 𝑘𝑘𝐶𝐶 = 3.98 kV ∠-180° 𝐼𝐼𝐶𝐶 = 7246 ∠180°

Now let’s calculate the sequence current quantities on the 115V side.

𝐼𝐼𝑓𝑓𝑏𝑏(1) =

1∠0°𝑗𝑗0.05+ j0.10 + j0.10 + j0.10

= -𝑗𝑗3.333 𝑝𝑝𝑝𝑝 (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)


√3∗ 13.8 𝑘𝑘𝑘𝑘= 1255.11 A

𝐼𝐼𝑓𝑓𝑏𝑏

(1) = −𝑗𝑗3.333∗1255.11 = 4184 ∠-90° (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)

For a line to line fault, the positive and negative sequence network diagram is connected in parallel. The fault current is essentially the pre fault voltage divided by the total positive and negative sequence impedance. We adjust by the base current and get the following current for positive sequence phase a current on the LV side.

𝐼𝐼𝑅𝑅(1) = 4184 ∠-90° (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)

The hand-calculated value matches













The negative sequence current is simply the positive sequence current but with 180 degree phase shift.

𝐼𝐼𝑓𝑓𝑏𝑏(2) = 4184 ∠+90° (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)

Aspen oneliner.

The angles also match – Aspen is showing a transformation from 13.8kV to 115kV side… so the angles are 180 degrees out of phase.

To calculate Line A and B current for the sequence currents;

𝐼𝐼𝑓𝑓𝑏𝑏 = 𝐼𝐼𝑓𝑓𝑏𝑏(1) + 𝐼𝐼𝑓𝑓𝑏𝑏

(2) = 0

𝐼𝐼𝑓𝑓𝑏𝑏 = 𝑅𝑅2 ∗ 𝐼𝐼𝑓𝑓𝑏𝑏(1) + 𝑅𝑅 ∗𝐼𝐼𝑓𝑓𝑏𝑏

(2) = (1∠240°)(4184 ∠-90°) + (1∠240°)(4184 ∠+90°)

=−7246𝑀𝑀

This matches the Aspen Oneliner expectations.

𝐼𝐼𝑓𝑓𝑠𝑠 =−𝐼𝐼𝑓𝑓𝑏𝑏 = +7246𝑀𝑀

Line C current is essentially Line B current but in the other direction. This makes sense because Line C and Line B current should be equal but in opposite direction.

This matches the Aspen Oneliner expectations.

Now let’s analyze the quantities from a voltage prospective


𝐼𝐼𝑏𝑏(1) = −j3.333

To calculate the phase and sequence voltage, we simply take the positive sequence current and multiply by the total positive sequence impedance.













𝑘𝑘𝑏𝑏(1) = 1∠0°− (𝑗𝑗0.15)∗ (−j3.333) = 0.50∠0° pu


√3

𝑘𝑘𝑏𝑏(1) = 0.50∠0°∗

13.8kV

√3= 3.98 kV∠0°

𝑘𝑘𝑏𝑏

(2) = 𝑘𝑘𝑏𝑏(1) = 3.98 kV∠0°

𝑘𝑘𝑏𝑏(0) = 0

𝑘𝑘𝑏𝑏 = 𝑘𝑘𝑏𝑏

(1) +𝑘𝑘𝑏𝑏(2) +𝑘𝑘𝑏𝑏

(0) = 3.98 kV∠0°+ 3.98 kV∠0° + 0 = 7.96kV∠0°

This gives us the voltage drop across the generator and transformer impedance. Then we take the pre fault voltage and subtract by the voltage drop … that gives us positive sequence voltage value. It’s interesting to note that for a line to line fault, the positive sequence voltage is 0.50 per unit. The same is true for the negative sequence voltage. This allows us to mentally predict the voltage levels for any line to line fault. The phase 𝑘𝑘𝑏𝑏 voltage is simply the combination of all three sequence component. We get full rated voltage. This makes sense because the fault was on line B and line C. Line A was unaffected. The hand-calculated value matches Aspen oneliner.

𝑘𝑘𝑏𝑏 = 𝑘𝑘𝑏𝑏

(0) + 𝑅𝑅2 ∙ 𝑘𝑘𝑏𝑏(1) ∗ 𝑅𝑅 ∙ 𝑘𝑘𝑏𝑏

(2) = (1∠240°) ∙ 3.98 kV + (1∠120°) ∙ 3.98 kV+ 0

𝑘𝑘𝑏𝑏 = −3.98 kV = 3.98 kV∠180°

𝑘𝑘𝑠𝑠 = 𝑘𝑘𝑏𝑏 = 3.98 kV∠180°

For a phase to phase fault, we see that the faulted phase has a voltage depression of 50% Due to the characteristics of a phase to phase fault, both of the faulted phases will have the same phase voltage. The value calculated matches Aspen oneliner.













2W DAB-WYE XFMR, 2LG Fault at 13.8kV Bus, Analysis @ 13.8kV Side

Let’s perform a 2LG fault on the 13.8kV bus and analyze the results.













According to Aspen oneliner, the current and voltages values on the 13.8kV side is expected to be

𝑘𝑘𝑀𝑀(1) = 2.27 kV ∠0° 𝐼𝐼𝑀𝑀

(1) = 5977𝑀𝑀 ∠90° 𝑘𝑘𝑀𝑀

(2) = 2.27 kV ∠0° 𝐼𝐼𝑀𝑀(2) = 2391 ∠-90°

𝑘𝑘𝑀𝑀(0) = 2.27 kV ∠0° 𝐼𝐼𝑀𝑀

(0) = 10758A ∠-90° And the phase values on the 13.8kV side are expected to be

𝑘𝑘𝑀𝑀 = 6.83 kV ∠0° 𝐼𝐼𝑀𝑀 = 0 ∠0° 𝑘𝑘𝐵𝐵 = 0 kV∠0° 𝐼𝐼𝐵𝐵 = 9025A ∠-36.6° 𝑘𝑘𝐶𝐶 = 0 kV∠0° 𝐼𝐼𝐶𝐶 = 9025A ∠-143.4°













Now let’s calculate the sequence current quantities on the 13.8kV side.


𝑘𝑘𝑓𝑓

1 + � 𝑍𝑍(2) ∙ 𝑍𝑍 (0)

𝑍𝑍 (2) + 𝑍𝑍 (0)�


1∠0°

1 + � 𝑗𝑗0.15 ∙ 𝑗𝑗0.10𝑗𝑗0.15 + 𝑗𝑗0.10�= -𝑗𝑗4.7619 𝑝𝑝𝑝𝑝 (𝐿𝐿𝑘𝑘 𝑆𝑆𝑇𝑇𝑎𝑎𝑅𝑅)


√3∗ 13.8 𝑘𝑘𝑘𝑘= 1255.11 A


(1) = -𝑗𝑗4.7619 𝑝𝑝𝑝𝑝∙ 1255.11 = 5977 ∠-90°

For a line to line to ground fault, the positive and negative and zero sequence network diagram are connected in parallel. We shall calculate the positive sequence current for a fault on the 13.8kV bus. The fault current calculation, according to Blackburn, works like a current divider circuit. By following the given equations in Blackburn, the calculated values match the expected values.

𝐼𝐼𝑓𝑓𝑏𝑏(1) = -𝑗𝑗4.7619 𝑝𝑝𝑝𝑝

𝐼𝐼𝑓𝑓𝑏𝑏(2) = −𝐼𝐼𝑓𝑓𝑏𝑏

(1) ∙ �𝑍𝑍 (0)

𝑍𝑍 (2) + 𝑍𝑍 (0)�

𝐼𝐼𝑓𝑓𝑏𝑏(2) = −(-𝑗𝑗4.7619) ∙ �

𝑗𝑗0.10𝑗𝑗0.15 + 𝑗𝑗0.10

� = 𝑗𝑗1.9047 𝑝𝑝𝑝𝑝


√3∗ 13.8 𝑘𝑘𝑘𝑘= 1255.11 A

Now, let’s calculate the negative sequence current for a fault on the 13.8kV bus. The fault current calculation, according to Blackburn, works like a current divider circuit. By following the given equations in Blackburn, the calculated values match the expected values.













𝐼𝐼𝑓𝑓𝑏𝑏(2) = 𝑗𝑗1.9047 𝑝𝑝𝑝𝑝∙ 1255.11 = 2391 ∠-90°

𝐼𝐼𝑓𝑓𝑏𝑏(1) = -𝑗𝑗4.7619 𝑝𝑝𝑝𝑝

𝐼𝐼𝑓𝑓𝑏𝑏(0) = −𝐼𝐼𝑓𝑓𝑏𝑏

(1) ∙ �𝑍𝑍 (2)

𝑍𝑍 (2) + 𝑍𝑍 (0)�

𝐼𝐼𝑓𝑓𝑏𝑏(0) = −(-𝑗𝑗4.7619) ∙ �

𝑗𝑗0.15𝑗𝑗0.15 + 𝑗𝑗0.10

� = 𝑗𝑗2.857 𝑝𝑝𝑝𝑝


√3∗ 13.8 𝑘𝑘𝑘𝑘= 1255.11 A


(0) = 𝑗𝑗2.857 𝑝𝑝𝑝𝑝 ∙ 1255.11 = 3586 ∠90°

Now, let’s calculate the zero sequence current for a fault on the 13.8kV bus. The fault current calculation, according to Blackburn, works like a current divider circuit. By following the given equations in Blackburn, the calculated values match the expected values. Aspen calculates the zero sequence current as 3 ∙ 𝐼𝐼(0).. so we adjust for the factor of 3, we get

𝐼𝐼(0 ) =10758𝑀𝑀∠90°

3= 3586𝑀𝑀∠90°

This matches our hand calculated values.

Now let’s calculate the phase current quantities on the 13.8kV side.

𝐼𝐼𝑓𝑓𝑏𝑏 = 𝐼𝐼𝑓𝑓𝑏𝑏(0) + 𝐼𝐼𝑓𝑓𝑏𝑏

(1) + 𝐼𝐼𝑓𝑓𝑏𝑏(2)

= 𝑗𝑗2.857 𝑝𝑝𝑝𝑝 + -𝑗𝑗4.7619 𝑝𝑝𝑝𝑝 + 𝑗𝑗1.9047 𝑝𝑝𝑝𝑝= 0

𝐼𝐼𝑓𝑓𝑏𝑏 = 𝐼𝐼𝑓𝑓𝑏𝑏(0) + 𝑅𝑅2 ∙ 𝐼𝐼𝑓𝑓𝑏𝑏

(1) + 𝑅𝑅 ∙ 𝐼𝐼𝑓𝑓𝑏𝑏(2)

= 𝑗𝑗2.857 𝑝𝑝𝑝𝑝 + (1∠240°) ∙ (-𝑗𝑗4.7619 𝑝𝑝𝑝𝑝) + (1∠120°)∙ 𝑗𝑗1.9047 𝑝𝑝𝑝𝑝 = 7.190∠143° 𝑝𝑝𝑝𝑝

For a line to line to ground fault, the positive and negative and zero sequence network diagram are connected in parallel. We shall calculate the positive sequence current for a fault on the 13.8kV bus. The fault current calculation, according to Blackburn, works like a current divider circuit. By following the given equations in Blackburn, the calculated values













𝐼𝐼𝑓𝑓𝑠𝑠 = 𝐼𝐼𝑓𝑓𝑏𝑏(0) + 𝑅𝑅 ∙ 𝐼𝐼𝑓𝑓𝑏𝑏

(1) + 𝑅𝑅2 ∙ 𝐼𝐼𝑓𝑓𝑏𝑏(2)

= 𝑗𝑗2.857 𝑝𝑝𝑝𝑝 + (1∠120°) ∙ (-𝑗𝑗4.7619 𝑝𝑝𝑝𝑝) + (1∠240°)∙ 𝑗𝑗1.9047 𝑝𝑝𝑝𝑝 = 7.190∠36 .58° 𝑝𝑝𝑝𝑝


√3∗ 13.8 𝑘𝑘𝑘𝑘= 1255.11 A

𝐼𝐼𝑓𝑓𝑅𝑅 = 0

𝐼𝐼𝑓𝑓𝑏𝑏 = 7.190∠143° 𝑝𝑝𝑝𝑝 ∙ 1255.11A = 9025 ∠143°𝑀𝑀

𝐼𝐼𝑓𝑓𝑠𝑠 = 7.190∠36.58° 𝑝𝑝𝑝𝑝 ∙ 1255.11A = 9025 ∠36.58°𝑀𝑀

match the expected values.

Please note that current values in Aspen Oneliner are 180 degrees out of phase.

Now let’s analyze the quantities from a voltage quantities on the 13.8kV side

𝑘𝑘𝑏𝑏(1) =𝑘𝑘𝑏𝑏

(2) = 𝑘𝑘𝑏𝑏(0) = 1∠0°− (𝑗𝑗0.15)∗ (−j4.7619)

= 0.285∠0° pu


√3

𝑘𝑘𝑏𝑏(1) = 0.285∠0° pu∗

13.8kV

√3= 2.27 kV∠0°

𝑘𝑘𝑏𝑏 = 𝑘𝑘𝑏𝑏(0) +𝑘𝑘𝑏𝑏

(1) +𝑘𝑘𝑏𝑏(2) = 3 ∙ 𝑘𝑘𝑏𝑏

(1) = 3 ∙ 2.27 kV∠0°= 6.829kV∠0°

𝑘𝑘𝑏𝑏 = 𝑘𝑘𝑏𝑏(0) + 𝑅𝑅2 ∙ 𝑘𝑘𝑏𝑏

(1) + 𝑅𝑅 ∙ 𝑘𝑘𝑏𝑏(2) = 0

𝑘𝑘𝑠𝑠 = 𝑘𝑘𝑏𝑏

(0) + 𝑅𝑅 ∙ 𝑘𝑘𝑏𝑏(1) + 𝑅𝑅2 ∙ 𝑘𝑘𝑏𝑏

(2) = 0

The hand-calculated value matches Aspen oneliner.













Three Winding Transformer Example In this example, the following parameters are given

We need to convert these parameters to something more usable – we’ll also need to enter the converted parameters in a short-circuit simulation software.

𝑍𝑍𝑠𝑠𝑠𝑠 =115𝑘𝑘𝑘𝑘2

600 𝑀𝑀𝑘𝑘𝑀𝑀= 22.04Ω

First, we’ll need convert the MVA short-circuit of the generator to an impedance value. Since the generator is connected to the 115kV bus, we can simply calculate the impedance value using basic ohms law.

𝑍𝑍𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒 =𝐾𝐾𝑘𝑘𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒𝑀𝑀𝑘𝑘𝑀𝑀𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒

=(115𝑘𝑘𝑘𝑘)2

30 𝑀𝑀𝑘𝑘𝑀𝑀= 440.83Ω

We need to decide an impedance base for our system. Since the 3-winding transformer is given in a 30MVA base, it’s worthwhile to select 30MVA as our impedance base

𝑍𝑍𝑠𝑠𝑠𝑠@30𝑀𝑀𝑀𝑀𝑀𝑀 =𝑍𝑍𝑠𝑠𝑠𝑠𝑍𝑍𝑏𝑏𝑏𝑏𝑠𝑠𝑒𝑒

=22.04 Ω

440.83 Ω= 𝑗𝑗0.05 𝑝𝑝𝑝𝑝

Lastly, we must convert the short-circuit impedance of the generator to a 30 MVA base.













Typical three-winding transformer impedances are given in H, M, and L or high, medium, and low windings. The leakage impedance is given in ZHM, ZHL, and ZML – at different MVA and voltage ratings. For convenience, P, S, and T values are also used to signify primary, secondary, and tertiary. In this example, we will setup the transformer impedance like below.

𝑍𝑍𝐻𝐻𝑀𝑀 = 10% = 0.10𝑝𝑝𝑝𝑝 @30𝑀𝑀𝑘𝑘𝑀𝑀

𝑍𝑍𝐻𝐻𝑀𝑀 = 𝑍𝑍𝑝𝑝𝑠𝑠 =𝑍𝑍𝑝𝑝𝑠𝑠@30𝑀𝑀𝑀𝑀 = 10% = 0.10𝑝𝑝𝑝𝑝

Given on a 30MVA base. Since we are working on a 30 MVA base, we do not need to convert this.

𝑍𝑍𝐻𝐻𝐿𝐿 = 𝑍𝑍𝑝𝑝𝑝𝑝 = 22.5% = 0.225𝑝𝑝𝑝𝑝 @15𝑀𝑀𝑘𝑘𝑀𝑀

𝑍𝑍𝐻𝐻𝐿𝐿 = 𝑍𝑍𝑝𝑝𝑝𝑝 = 0.225𝑝𝑝𝑝𝑝∗30𝑀𝑀𝑘𝑘𝑀𝑀15𝑀𝑀𝑘𝑘𝑀𝑀

= 0.45𝑝𝑝𝑝𝑝

Given on a 15MVA base. Since we are working on a 30 MVA base, we need to convert this value to a common MVA base as shown

𝑍𝑍𝑀𝑀𝐿𝐿 = 11% = 0.11𝑝𝑝𝑝𝑝 @10𝑀𝑀𝑘𝑘𝑀𝑀

𝑍𝑍𝑀𝑀𝐿𝐿 = 𝑍𝑍𝑠𝑠𝑝𝑝 = 0.11𝑝𝑝𝑝𝑝∗30𝑀𝑀𝑘𝑘𝑀𝑀10𝑀𝑀𝑘𝑘𝑀𝑀

= 0.33𝑝𝑝𝑝𝑝

Given on a 10MVA base. Since we are working on a 30 MVA base, we need to convert this value to a common MVA base as shown.

Building the 3W XFMR Short-circuit Model Now that we have calculated all the impedances to a common 30 MVA base, let’s model the system in Aspen Oneliner short-circuit tool.













The generator information was modeled using the information to the left. As shown, the pre-fault per unit voltage is 1 and angle is at 0 degrees. The generator is set at 30 MVA common base. Although the tool is set to use the sub transient impedance for fault calculations, it’s easier to set all the impedances to 𝑗𝑗0.05 as calculated.











3W, LL Fault Analysis @ 115kV

Circulating fault currents in two-winding and three ... · 2W Delta-Wye Xfmr Zero Sequence...

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