Circulating fault currents in two-winding and three ...
Transcript of Circulating fault currents in two-winding and three ...
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV 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
Last Updated: Rev 1.1 β 2019-03-11
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Two Winding Transformer Example: DAB - Wye Connection Delta Wye Transformer connection is also explained in detail on our GeneralPAC YouTube channel. Please visit the link below: https://www.youtube.com/watch?v=j5tDh8QKftg&list=PLqJ0Y2s60r-5TG3jkbk9Er5OPI50i37kh
πππ π π π =115ππππ2
600 ππππππ= 22.04Ξ©
First, weβll need to 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.
πππππππ π ππ =πΎπΎπππππππ π πππππππππππππ π ππ
=(115ππππ)2
30 ππππππ= 440.83Ξ©
We need to decide an impedance base for our system. Since the 2-winding transformer is given in a 30MVA base, itβs worthwhile to select 30MVA as our impedance base.
πππ π π π @30ππππππ =πππ π π π πππππππ π ππ
=22.04 Ξ©
440.83 Ξ©= 0.05 ππππ
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 thatβs why weβre interested in the per unit value.
πππ π π π @30ππππππ = 0 + ππ β 0.05 ππππ The resistive component will be ignored. So we will assume the 0.05 pu impedance is fully reactive.
ππππππππβππππππππππππ π π π π π π πππ π π π πππππ π = πππππ π πππππππ π πππ π π π π π π π πππ π π π πππππ π = ππππππ πππππ π πππππππ π πππ π π π π π π π πππ π π π πππππ π
= 0.05 ππππ
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, weβll consider them all equal.
ππππ
(1) = ππππ(2) = 0 + ππ0.10 ππππ
ππππ
(0) = 0 + ππ0.10 ππππ
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 0.10.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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.
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.
TWO WINDING TRANSFORMER
HV: 115 KV X (1) = 10% @30 MVA
LV: 13.8 KV X (1) = 10% @30 MVA
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Building the Short-circuit Model for 2W, DAB-Wye XFMR Now that we have calculated all the impedances to a common 30 MVA base, letβs 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The generator is set at 30 MVA common base. We will assume that all the impedance values are equal to ππ0.05 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. This 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 sequence) and R0 (zero sequence is set to 0. An effective short-circuit 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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 πΌπΌπΏπΏππ with πΌπΌβ ππ both on the HV side, we should expect a 1.73 difference in the magnitude and 30 degrees phase shift in the angle.
Therefore, if we compare the LV line current quantity (πΌπΌπΏπΏππ) with the HV line current quantity (πΌπΌπΏπΏππ), in addition to the magnitude difference which is driven by the transformer ratio, we should expect LV
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπΏπΏππ to lag 30 degrees compared to HV πΌπΌπΏπΏππ. 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 degrees 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 isnβt very clear in Blackburn book but Graingerβs Power Systems Analysis book, makes this very clear in equation 11.20 and 11.88.
LV LINE CURRENT LAGS HV LINE CURRENT BY 30Β°
HV LINE CURRENT LEADS LV LINE CURRENT BY 30Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W Delta-Wye XFMR Zero Sequence Connection According to Blackburn Figure A4.2-1, the zero sequence impedance seen from 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, 3LG Fault at 13.8kV, Analysis @ 13.8kV side Now that we have modeled the system appropriately, letβs 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 βBus Faultβ on secondary 13.8kV 3LG indicates the fault type. Right below the tool bar, there are 3 buttons starting with positive sequence symbol, negative sequence symbol, and zero sequence symbols. There are also buttons for A, B, and C
indicating phase selection ( ). Currently, the positive sequence button is selected so all the values shown are positive sequence (both voltage and current).
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The 3 phase fault on the 13.8kV side produces the following sequence and phase currents.
πΌπΌππ(1) = 8367β 90Β° πΌπΌππ = 8367β 90Β°
πΌπΌππ(2) = 0 πΌπΌπ΅π΅ = 8367β -30Β°
πΌπΌππ(0) = 0 πΌπΌπΆπΆ = 8367β -150Β°
Now that we have reference values, letβs try to hand calculating a 3 phase fault and comparing the hand calculations with the simulation software. We will draw the positive, negative, and zero sequence networks as shown below.
For the positive sequence network, ππ0.05 is clearly the generator impedance while the transformer impedance is clearly ππ0.10. Additionally, we will define our pre-fault voltage as 1pu at 0 degrees. Itβs
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
interesting to note that some books use ππ1.0 ππππ 1β 90Β° for a pre-fault voltage instead of 1β 0Β°. 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 1β 0Β°.
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 ππ0.05 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 β 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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, itβs a one-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 donβt 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.
πΌπΌππ(1) =
1β 0Β°ππ0.15
= -ππ6.667 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255.11ππ
πΌπΌππ
(1) = βππ6.667 β 1255.11 = 8367ππ β -90Β°
Since the impedances were given on a 30MVA base, the result is at a per unit value. Weβll need to convert this to amps by multiplying by the current base. Compare this value with the positive sequence current shown in Aspen OneLiner. The magnitude is the same, however, the angle is 180 degrees out of phase. Thatβs 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 βPhasor #2: LV Secondary 13.8kV β HV Primary 115kVβ
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ
(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
πΌπΌππ
(0) = 0
Now, letβs calculate the phase current values and see if they compare with ASPEN OneLiner. Because three phase faults are balanced, we will use the principles of symmetrical components to calculate the phase B and phase C currents.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) + πΌπΌππ
(0)
Since the sequence components are all referenced to phase A, we can simply add positive, negative, and zero sequence currents to calculate the phase A current (or line A current).
πΌπΌππ = πΌπΌππ(1) + πΌπΌππ
(2) + πΌπΌππ(0)
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0)
π π = 1β 120Β° π π 2 = 1β 240Β°
Phase B current is actually
πΌπΌπ΅π΅ = πΌπΌπ΅π΅(1) + πΌπΌπ΅π΅
(2) + πΌπΌπ΅π΅(0)
Since this is a balanced fault, we can derive πΌπΌπ΅π΅
(1) by simply rotating πΌπΌππ(1) by
240 degrees in the counter-clockwise direction. Similarly, to calculate πΌπΌπ΅π΅
(2) we need to rotate πΌπΌππ
(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 -- πΌπΌπ΅π΅
(0) = πΌπΌππ(0)
πΌπΌπΆπΆ = π π β πΌπΌππ(1) + π π 2 β πΌπΌππ
(2) + πΌπΌππ(0)
π π = 1β 120Β° π π 2 = 1β 240Β°
Phase C current is actually
πΌπΌπΆπΆ = πΌπΌπΆπΆ(1) + πΌπΌπΆπΆ
(2) + πΌπΌπΆπΆ(0)
Since this is a balanced fault, we can derive πΌπΌπΆπΆ
(1) by simply rotating πΌπΌππ(1) by
120 degrees in the counter-clockwise direction. Similarly, to calculate πΌπΌπΆπΆ
(2) we need to rotate πΌπΌππ
(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 -- πΌπΌπΆπΆ
(0) = πΌπΌππ(0)
Now to perform the actual calculations --
πΌπΌππ(1) =
1β 0Β°ππ0.15
= -ππ6.667 ππππ
To summarize, a 3LG fault only has the positive sequence network activated. Therefore, πΌπΌππ
(2) + πΌπΌππ(0) = 0
Phase A current on the 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.08ππππ= 1324.20ππ
πΌπΌππ
(1) = βππ6.667 β 1324.20 = 8367ππ β -90Β°
πΌπΌππ = πΌπΌππ(1) + πΌπΌππ
(2) + πΌπΌππ(0) = πΌπΌππ
(1) = 8367ππ β -90Β°
equals 8367ππ β -90Β° 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 13.8 kV side and toward the source (115 kV side).
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0) πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) = (1β 240Β°) β (8367ππ β -90Β°) = 8367ππ β 150Β°
Since πΌπΌππ
(2)π π ππππ πΌπΌππ(0)are both 0 for a 3
phase fault, the equation simplifies to πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) This matches the ASPEN OneLiner result but with a 180 degrees phase shift.
πΌπΌπΆπΆ = a β πΌπΌππ
(1) + π π 2 β πΌπΌππ(2) + πΌπΌππ
(0) πΌπΌπΆπΆ = a β πΌπΌππ
(1) = (1β 120Β°) β (8367ππ β β 90Β°) = 8367ππ β 30Β°
Since πΌπΌππ
(2)π π ππππ πΌπΌππ(0)are both 0 for a 3
phase fault, the equation simplifies to πΌπΌπΆπΆ = a β πΌπΌππ
(1) This matches the ASPEN OneLiner result but with a 180 degrees phase shift for the reasons explained above.
Now to calculate the sequence and phase voltages, we perform the following
Here, we see that positive sequence phase A voltage equal 0 kV which is, what we should expect because the
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(1) = 1β 0Β° β (j0.15) β (βj6.667) = 0pu
ππππ(1) = 0kVβ 0Β°
ππππ = ππππ(1) + ππππ
(2) + ππππ(0) = 0 kVβ 0Β°
short-circuit should drive the voltage to zero. This agrees with the Aspen model.
πππ΅π΅ = π π 2 β ππππ(1) = (1β 240Β°) β (0Vβ 0Β°) = 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 exists 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.
πππΆπΆ = a β ππππ(1) = (1β 120Β°) β (0 kVβ 0Β°) = 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, 3LG Fault at 13.8kV, Analysis @ 115kV side Now, letβs perform the same analysis but on the 115kV bus. ASPEN OneLiner shows the following results. Letβs see if our hand-calculation matches ASPEN OneLiner.
πΌπΌππ(1) =
1β 0Β°ππ0.15
= -ππ6.667 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115 ππππ= 150.613 A
πΌπΌππ
(1) = βππ6.667 β 150.613 = 1004 β -90Β°
πΌπΌππ = πΌπΌππ(1) + πΌπΌππ
(2) + πΌπΌππ(0) = πΌπΌππ
(1) = 1004 β -90Β° Because we know that the HV line current quantities lag 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.
πΌπΌππ(1) = (1004ππ β -90Β°) β (1 β +30Β°)
πΌπΌππ
(1) = (1004ππ β -60Β°)
The current that flows through the 115 kV side is the same as the 13.8 kV side. The only different is that the DAB-Wye-grounded transformer needs to be accounted for β both the magnitude and angle. A 3LG fault only has the positive sequence network activated. Therefore, πΌπΌππ
(2) + πΌπΌππ(0) = 0
If we adjusted only for the magnitude, we should expect the Phase A current to equals 1004ππ β -90Β° on 13.8 kV side. However, this is incorrect because we need to account for the DAB delta transformation. According to ASPEN OneLiner, πΌπΌππ
(1) = 1004ππ β -60Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Because we know that the HV line current quantities LAG the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly by 30 degrees. πΌπΌππ
(1) = (1004ππ β -90Β°) β 1 β +30Β° Which gives us,
πΌπΌππ(1) = (1004ππ β -60Β°)
The hand-calculated value matches ASPEN OneLiner.
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1) + π π β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) = (1β 240Β°) β (1004ππ β -90Β°) β (1 β +30Β°)= (β1004ππ) = (1004ππ β 180Β°)
πΌπΌπ΅π΅ = 1004ππ β 180Β°
Since πΌπΌππ
(2)π π ππππ πΌπΌππ(0)are both 0 for a 3
phase fault, the equation simplifies to πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(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.
πΌπΌπΆπΆ = a β πΌπΌππ
(1) + π π 2 β πΌπΌππ(2) + πΌπΌππ
(0) Since πΌπΌππ
(2)π π ππππ πΌπΌππ(0)are both 0 for a three
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπΆπΆ = a β πΌπΌππ
(1) = (1β 120Β°) β (8367ππ β β 90Β°) β (1 β +30Β°)= 1004ππ β 60Β°
πΌπΌπΆπΆ = 1004ππ β 60Β°
phase fault, the equation simplifies to πΌπΌπΆπΆ = a β πΌπΌππ
(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
πΌπΌππ(1) = βj6.667
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.
πΌπΌππ(1) = βj6.667 β (1β 30Β°) = 6.667β -60Β°
ππππ(1) = 1β 0Β° β (ππ0.05) β (6.667β -60Β°) = 0.73β -13.2Β°
We must be very careful in how we calculate the phase A sequence voltage because of the Delta-AB transformer connection. The πΌπΌππ
(1) was calculated as βj6.667 pu However, this for the LV Side current. Because we must consider the DAB Wye-grounded transformer connection, we need to shift the positive sequence phase current by 30 degrees β as seen on the HV side. This effectively becomes πΌπΌππ
(1) =6.667β -60Β° We need to perform KVL to calculate ππππ
(1) --- which is the source voltage minus the generator impedance (ππ0.05) times the current flowing through the generator impedance which is 6.667β -60Β°.
ππππ(1) = 0.73β -13.2Β°
Since negative and zero sequence networks do not exist, the negative and zero sequence voltages are also zero. Therefore, the phase A voltage
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππ΅π΅πππ π ππ =115kVβ3
ππππ(1) = 0.73β -13.2Β° β
115kVβ3
= 48.51 kVβ -13.2Β°
ππππ
(2) = ππππ(0) = 0
ππππ = ππππ
(1) + ππππ(2) + ππππ
(0) = 48.51 kVβ -13.2Β°
equals ππππ = 48.51 kVβ -13.2Β° The hand-calculated value matches ASPEN OneLiner.
πππ΅π΅ = π π 2 β ππππ
(1) = (1β 240Β°) β 48.51 kVβ -13.2Β°
πππ΅π΅ = 48.51 kVβ -133.2Β°
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 exist 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.
πππΆπΆ = a β ππππ(1) = (1β 120Β°) β 48.51 kVβ -13.2Β°
πππΆπΆ = 48.51 kVβ 106.8Β°
We rotate positive sequence A by 120 degrees in the CCW direction to get phase C voltage. The value calculated matches ASPEN OneLiner.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, 1LG Fault at 13.8kV Bus, Analysis @ 13.8kV side Letβs analyze a 1LG fault (phase A) on the 13.8 k V side of the DAB transformer. According to ASPEN OneLiner, we should get the following results.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The 1LG fault on the 13.8kV side produces the following current quantities are:
πΌπΌππ(1) = 3138β 90Β° πΌπΌππ = 9413β 90Β°
πΌπΌππ(2) = 3138β 90Β° πΌπΌπ΅π΅ = 0
πΌπΌππ(0) = 3138β 90Β° πΌπΌπΆπΆ = 0
The voltage quantities are:
ππππ(1) = 4.980 kV β 0Β° ππππ = 0
ππππ(2) = 2.988 kV β 180Β° πππ΅π΅ = 7.519 kV β -113.4Β°
ππππ(0) = 1.992 kV β 180Β° πππΆπΆ = 7.519 kV β 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 have reference values, letβs 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) =
1β 0Β°ππ0.15 + ππ0.15 + ππ0.10
= βππ2.5ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255ππ
πΌπΌππ
(1) = βππ2.5ππππ β 1255ππ = 3138β -90Β°
The positive sequence current is calculated by dividing the pre-fault voltage (1β 0Β°) by the total impedance of the circuit. The positive sequence impedance is ππ0.05 + ππ0.10 = ππ0.15 The negative sequence impedance is ππ0.05 + ππ0.10 = ππ0.15 The zero sequence impedance is ππ0.10 The positive sequence current matches the result shown in ASPEN OneLiner. Remember, Aspen results are 180 degrees out of phase because itβs viewing the phasors from 13.8kV side looking toward the 115kV side.
πΌπΌππ(2) = βππ2.5 ππππ
πΌπΌππ
(1) = βππ2.5 ππππ β πΌπΌπ΅π΅πππ π ππ = 3138β -90Β°
Since the same positive sequence current flows through both the negative sequence and zero sequence networks for a 1LG fault, πΌπΌππ
(1) = πΌπΌππ(2) = πΌπΌππ
(0)
πΌπΌππ
(0) = βππ2.5 ππππ
πΌπΌππ(0) = βππ2.5 ππππ β πΌπΌπ΅π΅πππ π ππ = 3138β -90Β°
Letβs determine the phase current values for the 1LG fault on the 13.8kV side.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) + πΌπΌππ
(0) = (-ππ2.5ππππ) + (-ππ2.5ππππ) + (-ππ2.5ππππ)= -ππ7.5 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255.109ππ
-ππ7.5 β πΌπΌπ΅π΅πππ π ππ = 9413ππ β -90Β°
Using the principles of symmetrical components, 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Another way of representing phase A current is
3πΌπΌππ(0) = -ππ7.5 ππππ β 9413ππ β -90Β°
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1) + π π β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπ΅π΅ = (1β 240Β°)(-ππ2.5) + (1β 120Β°)(-ππ2.5) + (-ππ2.5) = 0
πΌπΌπ΅π΅ = 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 β this result makes sense because the fault is on phase A, so phase B should have no fault current. This result matches the ASPEN OneLiner simulation.
πΌπΌπΆπΆ = π π β πΌπΌππ(1) + π π 2 β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπΆπΆ = (1β 120Β°)(-ππ2.5) + (1β 240Β°)(-ππ2.5) + (-ππ2.5) = 0
A very similar operation for the phase C current also. We should expect phase C current because this is a 1LG on line A. This result matches the ASPEN OneLiner simulation.
Letβs determine the sequence voltages for the 1LG fault on the 13.8kV side.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(1) = 1 β 0Β° β (j0.15)(βj2.5) = 0.625 pu
πππ΅π΅πππ π ππ =13.8kVβ3
ππππ
(1) = (0.625 pu) β πππ΅π΅πππ π ππ = 4.98 kVβ 0Β°
ππππ(1) = 4.98 kVβ 0Β°
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 OneLiner simulation
ππππ(2) = 0 β (j0.15)(βj2.5) = β0.375 pu
ππππ(2) = (β0.375) β οΏ½
13.8ππππβ3
οΏ½ = 2.988 kVβ 180Β°
ππππ
(2) = 2.988 kVβ 180Β°
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 results 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 OneLiner simulation.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(0) = 0 β (j0.10)(βj2.5) = β0.25 pu
ππππ(0) = (β0.25) β οΏ½
13.8ππππβ3
οΏ½ = 1.992 kVβ 180Β°
ππππ
(0) = 1.992 kVβ 180Β°
Similarly, performing KVL around the Zero sequence network results in the following β using the same concept as discussed previously. Note the negative signβ¦ 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 OneLiner simulation.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, 1LG Fault at 13.8kV Bus, Analysis @ 115kV side Now letβs perform the same analysis but on the 115kV bus. ASPEN OneLiner shows the following results.
According to ASPEN OneLiner, the current and voltage values on the 115kV side are expected to be
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(1) = 59.35 kV β -4.0Β° πΌπΌππ
(1) = 377ππ β -60Β° ππππ
(2) = 8.299 kVβ 150Β° πΌπΌππ(2) = 377A β -120Β°
ππππ(0) = 0 πΌπΌππ
(0) = 0 And the phase values on the 115kV side are expected to be
ππππ = 52.02 kV β 0Β° πΌπΌππ = 652 β -90Β° πππ΅π΅ = 66.40 kV β -120Β° πΌπΌπ΅π΅ = 0 πππΆπΆ = 60.50 kV β 108.1Β° πΌπΌπΆπΆ = 652 β 90Β°
Now letβs calculate the sequence current quantities on the 115V side.
From the previous analysis, we know that -ππ2.5 ππππ flows on the LV Side Current of the transformer. However, because we have a DAB connection on the HV side, we need to take that into account.
πΌπΌππ(1) =
1β 0Β°ππ0.40
= -ππ2.5 ππππ (πΏπΏππ πππππππ π πππππππππ π πππ π )
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115 ππππ= 150.613 A
πΌπΌππ
(1) = βππ2.5 β 150.613 = 377 β -90Β° (πΏπΏππ πππππππ π πππππππππ π πππ π ) 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 backward for negative sequence current.
The current that flows through the 115kV side is the same as the 13.8kV side. The only difference is that DAB-Wye-grounded transformer needs to be accounted for β both the magnitude and angle. A 1LG fault has only 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 ππβ -90Β° However, this is incorrect because we need to account for the DAB delta transformation. Because we know that the HV line current quantities LAG the LV line current quantities by 30 degrees, we need to advance the HV quantities accordingly by 30 degrees. πΌπΌππ
(1) = (377 ππ β -90Β°) β 1 β +30Β° Which gives us the πΌπΌππ
(1) =(377 ππ β -60Β°) The hand-calculated value matches ASPEN OneLiner.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) = (377 ππβ -90Β°) β (1 β +30Β°)
πΌπΌππ
(1) = 377 ππβ -60Β° (π»π»ππ πππππππ π πππππππππ π πππ π )
πΌπΌππ(2) = βππ2.5 β 150.613 = 377 β -90Β° (πΏπΏππ πππππππ π πππππππππ π πππ π )
πΌπΌππ
(2) = (377 ππβ -90Β°) β (1 β -30Β°)
πΌπΌππ(2) = 377 ππβ -120Β° (π»π»ππ πππππππ π πππππππππ π πππ π )
The negative sequence current that flows through the 115kV side is the same as the 13.8kV side. The only difference is that 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ
(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 115 kV side are calculated below.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) = 377 ππβ -60Β° + 377 ππβ -120Β°
πΌπΌππ = 652 A β -90Β°
Using the principles of symmetrical components, 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0) πΌπΌπ΅π΅ = (1β 240Β°)(377 ππβ -60Β°) + (1β 120Β°)(377 ππβ -120Β°)
πΌπΌπ΅π΅ = 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.
πΌπΌπΆπΆ = π π β πΌπΌππ
(1) + π π 2 β πΌπΌππ(2)
πΌπΌπΆπΆ = (1β 120Β°)(377 ππβ -60Β°) + (1β 240Β°)(377 ππβ -120Β°)
πΌπΌπΆπΆ = 652 ππβ 90Β°
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 e simulation.
Now to calculate the sequence voltages, we perform the following
The positive sequence current was calculated as
πΌπΌππ(1) = βj2.5
We must be very careful while calculating the phase A sequence voltage because of the Delta AB
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Since we must account for the DAB transformer phase shift on the HV side, we also shift the positive sequence current 30 degrees.
πΌπΌππ(1) = βj2.5 β (1β 30Β°) = 2.5β -60Β°
ππππ(1) = 1β 0Β° β (ππ0.05) β (2.5β -60Β°) = 0.8939β -4.0Β°
πππ΅π΅πππ π ππ =115kVβ3
ππππ(1) = 0.8939β -4.0Β° β
115kVβ3
= 59.35 kVβ -4Β°
transformer connection. The πΌπΌππ
(1) was calculated as βj2.5 pu However, this is for the LV Side Current. 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 β as seen on the HV side. This effectively becomes πΌπΌππ
(1) = 2.5β -60Β° We need to perform KVL to calculate ππππ
(1) --- which is the source voltage minus the generator impedance (ππ0.05) times the current flowing through the generator impedance which is 2.5β -60Β°. We calculate the positive sequence voltage as ππππ
(1) = 0.8939β -4Β° or 59.35 kVβ -4Β° The hand-calculated value matches ASPEN OneLiner.
πΌπΌππ
(1) = βj2.5 Since we must account for the DAB transformer phase shift on the HV side, we also shift the negative sequence current by -30 degrees.
πΌπΌππ(2) = βj2.5 β (1β -30Β°) = 2.5β -120Β°
We must be very careful in how we calculate the phase A sequence negative voltage because of the Delta-AB transformer connection. The πΌπΌππ
(1) was calculated as βj2.5 pu However, this is for the LV Side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(2) = 0 β (ππ0.05) β (2.5β -120Β°) = 0.125β 150Β°
πππ΅π΅πππ π ππ =115kVβ3
ππππ(2) = 0.125β 150Β° β
115kVβ3
= 8.299 kVβ 150Β°
Current. Because we must consider the DAB Wye-grounded transformer connection, we need to shift the negative sequence phase current by -30 degrees β as seen on the HV side. This effectively becomes πΌπΌππ
(1) =2.5β -120Β° We need to perform KVL to calculate ππππ
(1) --- which is the source voltage (we donβt have a source voltage for negative sequence hence 0) minus the generator impedance (ππ0.05) times the current flowing through the generator impedance which is 2.5β -120Β°. We calculate the negative sequence voltage as ππππ
(2) = 0.125β 150Β° or 8.299 kVβ 150Β° The hand-calculated value matches ASPEN OneLiner.
ππππ(0) = 0
Because there is no zero sequence current that flows on the 115kV side, the zero sequence voltage is non-existent. The value calculated matches ASPEN OneLiner.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now to calculate the phase voltage quantities, we perform the following
ππππ = ππππ(1) + ππππ
(2) + ππππ(0)
ππππ = 59.35 kVβ -4Β° + 8.299 kVβ 150Β° = 52.02 kV β 0Β°
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
πππ΅π΅ = π π 2 β ππππ
(1) + π π β ππππ(2) + ππππ
(0) πππ΅π΅ = (1β 240Β°) β 59.35 kVβ -4Β° + (1β 120Β°) β 8.299 kVβ 150Β°
πππ΅π΅ = 66.39 kVβ -120.0Β°
This is the same value as the ASPEN OneLiner simulation.
πππΆπΆ = π π β ππππ(1) + π π 2 β ππππ
(2) + ππππ(0)
πππΆπΆ = (1β 120Β°) β 59.35 kVβ -4Β° + (1β 240Β°) β 8.299 kVβ 150Β°
πππΆπΆ = 60.50 kVβ 108.1Β°
This is the same value as the ASPEN OneLiner simulation.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, LL Fault at 13.8kV Bus, Analysis @ 13.8kV Side Letβs perform a line-to-line (LL) fault on the 13.8 kV bus and analyze the results.
According to ASPEN OneLiner, the current and voltage values on the 115kV side is expected to be
ππππ(1) = 3.98 kV β 0Β° πΌπΌππ
(1) = 4184ππ β 90Β° ππππ
(2) = 3.98 kVβ 0Β° πΌπΌππ(2) = 4184A β -90Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(0) = 0 πΌπΌππ
(0) = 0 And the phase values on the 115 kV side are expected to be
ππππ = 7.98 kV β 0Β° πΌπΌππ = 0 β 0Β° πππ΅π΅ = 3.98 kV β -180Β° πΌπΌπ΅π΅ = 7246 β 0Β° πππΆπΆ = 3.98 kV β -180Β° πΌπΌπΆπΆ = 7246 β 180Β°
Now letβs calculate the sequence current quantities on the 115 V side.
πΌπΌππππ(1) =
1β 0Β°ππ0.05 + j0.10 + j0.10 + j0.05
= -ππ3.333 ππππ (πΏπΏππ πππππππ π πππππππππ π πππ π )
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8 ππππ= 1255.11 A
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 Current. πΌπΌπ π
(1)
= 4184 β -90Β° (πΏπΏππ πππππππ π πππππππππ π πππ π )
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππππ(1) = βππ3.333 β 1255.11 = 4184 β -90Β° (πΏπΏππ πππππππ π πππππππππ π πππ π )
The negative sequence current is simply the positive sequence current but with 180 degrees phase shift.
πΌπΌππππ(2) = 4184 β +90Β° (πΏπΏππ πππππππ π )
The hand-calculated value matches ASPEN OneLiner.
The angles also match β Aspen is showing a transformation from 13.8kV to 115kV sideβ¦ so the angles are 180 degrees out of phase.
To calculate Line A and B current for the sequence currents;
πΌπΌππππ = πΌπΌππππ(1) + πΌπΌππππ
(2) = 0
πΌπΌππππ = π π 2 β πΌπΌππππ(1) + π π β πΌπΌππππ
(2) = (1β 240Β°)(4184 β -90Β°) + (1β 240Β°)(4184 β +90Β°)
= β7246ππ
This matches the ASPEN OneLiner expectations.
πΌπΌπππ π = βπΌπΌππππ = +7246ππ
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 letβs analyze the quantities from a voltage prospective
The positive sequence current was calculated as
πΌπΌππ(1) = βj3.333
To calculate the phase and sequence voltage, we simply take the positive sequence current and multiply by the total positive sequence impedance.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ
(1) = 1β 0Β° β (ππ0.15) β (βj3.333) = 0.50β 0Β° pu
πππ΅π΅πππ π ππ =13.8kVβ3
ππππ(1) = 0.50β 0Β° β
13.8kVβ3
= 3.98 kVβ 0Β°
ππππ
(2) = ππππ(1) = 3.98 kVβ 0Β°
ππππ
(0) = 0
ππππ = ππππ(1) + ππππ
(2) + ππππ(0)
= 3.98 kVβ 0Β° + 3.98 kVβ 0Β° + 0 = 7.96kVβ 0Β°
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 β¦ that gives us positive sequence voltage value. Itβs 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 ππππ 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.
ππππ = π π 2 β ππππ
(1) β π π β ππππ(2) + ππππ
(0) = (1β 240Β°) β 3.98 kV + (1β 120Β°) β 3.98 kV + 0
ππππ = β3.98 kV = 3.98 kVβ 180Β°
πππ π = ππππ = 3.98 kVβ 180Β°
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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
2W DAB-WYE XFMR, 2LG Fault at 13.8kV Bus, Analysis @ 13.8kV Side
Letβs perform a 2LG fault on the 13.8 kV bus and analyze the results.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
According to ASPEN OneLiner, the current and voltage values on the 13.8kV side is expected to be
ππππ(1) = 2.27 kV β 0Β° πΌπΌππ
(1) = 5977ππ β 90Β° ππππ
(2) = 2.27 kV β 0Β° πΌπΌππ(2) = 2391 β -90Β°
ππππ(0) = 2.27 kV β 0Β° πΌπΌππ
(0) = 10758A β -90Β° And the phase values on the 13.8kV side are expected to be
ππππ = 6.83 kV β 0Β° πΌπΌππ = 0 β 0Β° πππ΅π΅ = 0 kVβ 0Β° πΌπΌπ΅π΅ = 9025A β -36.6Β° πππΆπΆ = 0 kVβ 0Β° πΌπΌπΆπΆ = 9025A β -143.4Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now letβs calculate the sequence current quantities on the 13.8 kV side.
πΌπΌππππ(1) =
ππππ
ππ(1) + οΏ½ ππ(2) β ππ(0)
ππ(2) + ππ(0)οΏ½
πΌπΌππππ(1) =
1β 0Β°
ππ0.15 + οΏ½ ππ0.15 β ππ0.10ππ0.15 + ππ0.10οΏ½
= -ππ4.7619 ππππ (πΏπΏππ πππππππ π )
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8 ππππ= 1255.11 A
πΌπΌππππ
(1) = -ππ4.7619 ππππ β 1255.11 = 5977 β -90Β°
For a line to line to ground fault, the positive, negative and zero sequence network diagram are connected in parallel. We shall calculate the positive sequence current for a fault on the 13.8 kV 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.
πΌπΌππππ(1) = -ππ4.7619 ππππ
πΌπΌππππ(2) = βπΌπΌππππ
(1) β οΏ½ππ(0)
ππ(2) + ππ(0)οΏ½
πΌπΌππππ(2) = β(-ππ4.7619) β οΏ½
ππ0.10ππ0.15 + ππ0.10
οΏ½ = ππ1.9047 ππππ
Now, letβs calculate the negative sequence current for a fault on the 13.8 kV 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8 ππππ= 1255.11 A
πΌπΌππππ(2) = ππ1.9047 ππππ β 1255.11 = 2391 β -90Β°
πΌπΌππππ(1) = -ππ4.7619 ππππ
πΌπΌππππ(0) = βπΌπΌππππ
(1) β οΏ½ππ(2)
ππ(2) + ππ(0)οΏ½
πΌπΌππππ(0) = β(-ππ4.7619) β οΏ½
ππ0.15ππ0.15 + ππ0.10
οΏ½ = ππ2.857 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8 ππππ= 1255.11 A
πΌπΌππππ
(0) = ππ2.857 ππππ β 1255.11 = 3586 β 90Β°
Now, letβs 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 β πΌπΌ(0).. so we adjust for the factor of 3, we get
πΌπΌ(0) =10758ππβ 90Β°
3= 3586ππβ 90Β°
This matches our hand calculated values.
Now letβs calculate the phase current quantities on the 13.8kV side.
πΌπΌππππ = πΌπΌππππ
(0) + πΌπΌππππ(1) + πΌπΌππππ
(2) = ππ2.857 ππππ + -ππ4.7619 ππππ + ππ1.9047 ππππ= 0
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 13.8 kV bus. The fault current calculation, according to Blackburn, works like a
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππππ = πΌπΌππππ(0) + π π 2 β πΌπΌππππ
(1) + π π β πΌπΌππππ(2)
= ππ2.857 ππππ + (1β 240Β°) β (-ππ4.7619 ππππ) + (1β 120Β°)β ππ1.9047 ππππ = 7.190β 143Β° ππππ
πΌπΌπππ π = πΌπΌππππ
(0) + π π β πΌπΌππππ(1) + π π 2 β πΌπΌππππ
(2) = ππ2.857 ππππ + (1β 120Β°) β (-ππ4.7619 ππππ) + (1β 240Β°)
β ππ1.9047 ππππ = 7.190β 36.58Β° ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8 ππππ= 1255.11 A
πΌπΌπππ π = 0
πΌπΌππππ = 7.190β 143Β° ππππ β 1255.11A = 9025 β 143Β°ππ
πΌπΌπππ π = 7.190β 36.58Β° ππππ β 1255.11A = 9025 β 36.58Β°ππ
current divider circuit. By following the given equations in Blackburn, the calculated values match the expected values.
Please note that current values in Aspen Oneliner are 180 degrees out of phase.
Now letβs analyze the quantities from a voltage quantities on the 13.8 kV side
ππππ(1) = ππππ
(2) = ππππ(0) = 1β 0Β° β (ππ0.15) β (βj4.7619)
= 0.285β 0Β° pu
πππ΅π΅πππ π ππ =13.8kVβ3
ππππ(1) = 0.285β 0Β° pu β
13.8kVβ3
= 2.27 kVβ 0Β°
ππππ = ππππ(0) + ππππ
(1) + ππππ(2) = 3 β ππππ
(1) = 3 β 2.27 kVβ 0Β°= 6.829kVβ 0Β°
ππππ = ππππ
(0) + π π 2 β ππππ(1) + π π β ππππ
(2) = 0
πππ π = ππππ(0) + π π β ππππ
(1) + π π 2 β ππππ(2) = 0
The hand-calculated value matches ASPEN OneLiner.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Three Winding Transformer Example In this example, the following parameters are given
We need to convert these parameters to something more usable β weβll also need to enter the converted parameters in a short-circuit simulation software.
πππ π π π =115ππππ2
600 ππππππ= 22.04Ξ©
First, weβll need to convert the MVA short-circuit of the generator to an impedance value. Since the generator is connected to the 115 kV bus, we can simply calculate the impedance value using basic ohms law.
πππππππ π ππ =πΎπΎπππππππ π πππππππππππππ π ππ
=(115ππππ)2
30 ππππππ= 440.83Ξ©
We need to decide an impedance base for our system. Since the 3-winding transformer is given in a 30MVA base, itβs worthwhile to select 30MVA as our impedance base
πππ π π π @30ππππππ =πππ π π π πππππππ π ππ
=22.04 Ξ©
440.83 Ξ©= ππ0.05 ππππ
Lastly, we must convert the short-circuit impedance of the generator to a 30 MVA base.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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 β 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.
πππ»π»ππ = 10% = 0.10ππππ @30ππππππ
πππ»π»ππ = πππππ π = πππππ π @30ππππ = 10% = 0.10ππππ
Given on a 30MVA base. Since we are working on a 30 MVA base, we do not need to convert this.
πππ»π»πΏπΏ = ππππππ = 22.5% = 0.225ππππ @15ππππππ
πππ»π»πΏπΏ = ππππππ = 0.225ππππ β30ππππππ15ππππππ
= 0.45ππππ
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
πππππΏπΏ = 11% = 0.11ππππ @10ππππππ
πππππΏπΏ = πππ π ππ = 0.11ππππ β30ππππππ10ππππππ
= 0.33ππππ
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, letβs model the system in ASPEN OneLiner short-circuit tool.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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, itβs easier to set all the impedances to ππ0.05 as calculated.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The 3 winding transformer is configured as shown. The primary voltage, secondary, and tertiary voltage is selected as the problem statement. The 30 MVA base rating is also selected. The 30 MVA base impedances are entered appropriately (Zps, Zpt, and Zst) for positive sequence short circuit impedance. Since the problem statement did not give the transformer zero sequence impedance values, appropriate values were chosen in its place. An effective infinite impedance is selected for Zg1 since the HV Wye winding is ungrounded connected (99999 + 99999j). An effective grounding impedance (0 + 0j) is selected for Zg2 since the LV Wye winding is grounded.
3W Xfmr Zero Sequence Connection The transformer manufacturer provided impedance values must be converted again to fit the parameters of the equivalent wye-type sequence network which will be used for the sequence impedance network. The following equation is used for this discussion (Equation A4.2-13 to 15 of Blackburn).
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππ»π» =12
(πππ»π»ππ + πππ»π»πΏπΏ β πππππΏπΏ) =12οΏ½πππππ π + ππππππ β πππ π πποΏ½ =
12
(0.10 + 0.45 β 0.33) = 0.11ππππ
ππππ =12
(πππ»π»ππ + πππππΏπΏ β πππ»π»πΏπΏ) =12οΏ½πππππ π + πππ π ππ β πππππποΏ½ =
12
(0.10 + 0.45 β 0.33) = β0.01ππππ
πππΏπΏ =12
(πππ»π»πΏπΏ + πππππΏπΏ β πππ»π»ππ) =12οΏ½ππππππ + πππ π ππ β πππππ π οΏ½ =
12
(0.45 + 0.33 β 0.10) = 0.34ππππ
Itβs important to note that this 3W connection diagram is only valid for the positive and negative sequence impedance networks. The zero sequence connection will look very different due to the HV Wye-ungrounded connection and the LV delta tertiary connection both of which block the zero sequence current flow to ground.
According to Blackburn Figure A4.2-3, the zero sequence connections for the Wye-ungrounded (HV Primary winding) and Delta (LV Tertiary winding) are open. However, the MV secondary winding is closed.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, 3LG Fault at 13.8kV Bus, Analysis @ 13.8kV side Now that we have modeled the system appropriately, letβs simulate a 3 phase fault on the 13.8kV bus using Aspen Oneliner.
At the very bottom center of the window, we can see that βBus Faultβ on secondary 13.8kV 3LG indicates 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, and C
indicating phases. Currently, the positive sequence button is selected ( ) so all the values shown on the window are positive sequence (both voltage and current)
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The 3 phase fault on the 13.8kV side produces the following sequence and phase currents.
πΌπΌππ(1) = 8367β 90Β° πΌπΌππ = 8367β 90Β°
πΌπΌππ(2) = 0 πΌπΌπ΅π΅ = 8367β -30Β°
πΌπΌππ(0) = 0 πΌπΌπΆπΆ = 8367β -150Β°
Now that we have reference values, letβs try to hand calculating a 3 phase fault and comparing the hand calculations with the simulation software. We will draw the positive, negative, and zero sequence networks as shown below.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The three phase fault on the 13.8kV side is fairly simple after drawing the network connection as described in Blackburn chapter 4. Note that a 3 phase fault only has the positive sequence current activated as shown below. There is no negative or zero sequence networks for a 3 phase fault.
πΌπΌππ(1) =
1β 0Β°ππ0.15
= -ππ6.667 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.08ππππ= 1324.20ππ
Since the impedances were given on a 30 MVA base, the result is at a per unit value. Weβll need to convert this to amps by multiplying by the current base.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) = βππ6.667 β 1324.20 = 8367ππ β -90Β°
This matches ASPEN OneLiner with an exception of the angle which is 180 degrees apart.
πΌπΌππ
(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.
πΌπΌππ
(0) = 0
Now, letβs calculate the phase current values and see if they compare with ASPEN OneLiner. Because 3 phase faults are balanced, we will use the principles of symmetrical components to calculate the phase B and phase C currents.
πΌπΌππ
(1) = βππ6.667 β 1324.20 = 8367ππ β -90Β°
This was calculated previously and
shown here for convenience.
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0)
Since πΌπΌππ(2)π π ππππ πΌπΌππ
(0)are both 0 for a 3 phase fault, the equation simplifies to
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1)
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1) = (1β 240Β°) β (8367ππ β β 90Β°)
= 8367ππ β 150Β°
This matches ASPEN OneLiner expectations with an exception of 180 degree out of phase. So if we take -30 degrees (phase B line current as shown below) and add 180 degrees, we get 150 degrees which matches our hand calculations
πΌπΌπΆπΆ = a β πΌπΌππ
(1) + π π 2 β πΌπΌππ(2) + πΌπΌππ
(0)
πΌπΌπΆπΆ = a β πΌπΌππ(1) = (1β 120Β°) β (8367ππ β β 90Β°) = 8367ππ β 30Β°
Since πΌπΌππ(2)π π ππππ πΌπΌππ
(0)are both 0 for a 3 phase fault, the equation simplifies to πΌπΌπΆπΆ = a β πΌπΌππ
(1)
This matches ASPEN OneLiner expectations with an exception of 180 degree out of phase. So if we take -150 degrees (phase C line current as shown below) and add 180 degrees, we get 30 degrees which matches our hand calculations
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, 3LG Fault at 13.8kV Bus, Analysis @ 115kV side Letβs now determine, how voltages and currents should appear on the 115kV side for a 3phase fault on the 13.8kV bus?
According to ASPEN OneLiner, the current and voltages values are expected to be
ππππ(1) = 44.26 ππππ β 0Β° πΌπΌππ
(1) = 1004ππ β -90Β° ππππ
(2) = 0 kV πΌπΌππ(2) = 0 β 0Β°
ππππ(0) = 0 kV πΌπΌππ
(0) = 0 β 0Β°
And the phase values are
ππππ = 44.26ππ β 0Β° πΌπΌππ = 1004ππ β -90Β° πππ΅π΅ = 44.26ππ β β 120Β° πΌπΌπ΅π΅ = 1004ππ β 150Β° πππΆπΆ = 44.26ππ β 120Β° πΌπΌπΆπΆ = 1004ππ β 30Β°
This goes back to our schematic diagram β except, we are calculating the current flow and voltage on the 115kV side.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) =
1β 0Β°ππ0.05
= βj6.667pu
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115ππππ= 150.61ππ
πΌπΌππ
(1) = βj6.667pu β 150.61A = 1004β -90Β°
πΌπΌππ(2) = πΌπΌππ
(0) = 0
πΌπΌππ = πΌπΌππ(1) + πΌπΌππ
(2) + πΌπΌππ(0) = 1004β -90Β°
πΌπΌππ = 1004β -90Β°
The sequence fault current is the same as we calculated previously because the fault location did not change. However, we must calculate the current flow on the 115kV side. For this, we simply multiply with the new current base. Notice how this sequence current value matches the results from ASPEN OneLiner. Also note that negative and zero sequence current is non-existent for a perfect 3 phase fault. Phase A current results in πΌπΌππ =1004β -90Β° which matches ASPEN OneLiner.
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) = (1β 240Β°) β (1004 β 150Β°) = 1004A β 150Β° phase B current also matches ASPEN
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = 1004A β 150Β°
OneLiner at πΌπΌπ΅π΅ = 1004A β 150Β°
πΌπΌπΆπΆ = a β πΌπΌππ(1) = (1β 120Β°) β (1004 β 150Β°) = 1004A β 30Β°
Phase C current also matches ASPEN OneLiner at πΌπΌπΆπΆ = 1004A β 30Β°
Now to calculate the sequence and phase voltages:
ππππ
(1) = 1β 0Β° β (j0.05) β (βj6.667) = 0.667pu
πππ΅π΅πππ π ππ =115kVβ3
ππππ(1) = 0.667pu β
115kVβ3
= 44.28kVβ 0Β°
ππππ
(2) = ππππ(0) = 0
ππππ = ππππ
(1) + ππππ(2) + ππππ
(0) = 44.28kVβ 0Β°
Since negative and zero sequence networks do not exist, the negative and zero sequence voltages are also zero. Therefore, the phase A voltage is simply ππππ = 44.28kVβ 0Β°
πππ΅π΅ = π π 2 β ππππ
(1) = (1β 240Β°) β (44.28kVβ 0Β°)= 44.28kVβ 240Β°
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
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππ΅π΅ = 44.28kVβ 240Β°
since negative and zero sequence do not exist 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.
πππΆπΆ = a β ππππ(1) = (1β 120Β°) β (44.28kVβ 0Β°)
= 44.28kVβ 120Β°
πππΆπΆ = 44.28kVβ 120Β°
We rotate positive sequence A by 120 degrees in the CCW direction to get phase C voltage. The value calculated matches ASPEN OneLiner.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, 1LG Fault at 13.8kV Bus Analysis @ 13.8kV side We will build the concepts of analyzing 3Phase faults here.
First, letβs evaluate our expectations based on the ASPEN OneLiner simulation. If we place a 1LG fault on the 13.8 kV bus, this is what we should expect.
Notice that the bottom center indicates βBus Fault on: β¦ 13.8 kV 1LG Type Aβ β which is a Phase A to Ground fault.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
According to ASPEN OneLiner, the current and voltage values on the 13.8 kV side is expected to be
ππππ(1) = 6.070kV β 0Β° πΌπΌππ
(1) = 1993ππ β 90Β° ππππ
(2) = 1.897kVβ -180Β° πΌπΌππ(2) = 1993ππ β 90Β°
ππππ(0) = 4.173kV β 180Β° πΌπΌππ
(0) = 1993ππ β 90Β°
And the phase and line current quantities are
ππππ = 0V β 0Β° πΌπΌππ = 5978 β 90Β° πππ΅π΅ = 9.316kV β -132.2Β° πΌπΌπ΅π΅ = 0 β 0Β° πππΆπΆ = 9.317kVππ β 132.2Β° πΌπΌπΆπΆ = 0 β 0Β°
Letβs now hand calculate these values to see if we match β but we first start by drawing the sequence network diagram for a ground fault. According to Blackburn, the positive, negative, and zero sequence networks are connected in series for a 1LG fault as shown below.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) =
1β 0Β°ππ0.15 + ππ0.15 + ππ0.33
= βππ1.587ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255ππ
πΌπΌππ
(1) = βππ1.587ππππ β 1255ππ = 1991ππ β -90Β°
The positive sequence current is calculated by dividing the pre-fault voltage (1β 0Β°) by the total impedance of the circuit. The positive sequence impedance is ππ0.05 + ππ0.11 + (βππ0.10) = ππ0.15 The negative sequence impedance is ππ0.05 + ππ0.11 + (βππ0.10) = ππ0.15 The zero sequence impedance is ππ0.34 + (βππ0.10) = ππ0.33
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
The positive sequence current matches the result shown in ASPEN OneLiner with an exception of the 180 degrees difference in the angle.
πΌπΌππ(2) = βππ1.587ππππ
πΌπΌππ
(1) = βππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 1991ππ β -90Β°
Since the same positive sequence current flows through both the negative sequence and zero sequence networks for a 1LG fault, πΌπΌππ
(1) = πΌπΌππ(2) = πΌπΌππ
(0) This also agrees with Aspen Oneliner, with an exception of the angle.
πΌπΌππ
(0) = βππ1.587ππππ
πΌπΌππ(0) = βππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 1991ππ β -90Β°
Letβs determine the phase current values for the 1LG fault on the 13.8 kV side.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) + πΌπΌππ
(0)
= (-ππ1.587ππππ) + (-ππ1.587ππππ)+ (-ππ1.587ππππ)
πΌπΌππ = -ππ4.762 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255.109ππ
-ππ4.762 β πΌπΌπ΅π΅πππ π ππ = 5978ππ β -90Β°
Another way of representing phase A current is
3πΌπΌππ(0) = -ππ4.762 ππππ β 5978ππ β -90Β°
Using the principles of symmetrical components, 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.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1) + π π β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπ΅π΅ = (1β 240Β°)(βj1.587) + (1β 120Β°)(βj1.587) + (βj1.587)
πΌπΌπ΅π΅ = 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 β 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.
πΌπΌπΆπΆ = π π β πΌπΌππ(1) + π π 2 β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπΆπΆ = (1β 120Β°)(βj1.587) + (1β 240Β°)(βj1.587) + (βj1.587)
= 0
πΌπΌπΆπΆ = 0
Letβs determine the sequence voltages for the 1LG fault on the 13.8kV side.
ππππ(1) = 1 β 0Β° β (j0.15)(βj1.587) = +0.76195 pu
πππ΅π΅πππ π ππ =13.8kVβ3
(0.76195 ) β πππ΅π΅πππ π ππ = 6.07kVβ 0Β°
If we perform KVL around the positive sequence network and calculate the voltage at the point of the fault, we will always find it equal to 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 Oneliner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(2) = 0 β (j0.15)(βj1.587) = β0.23805
ππππ(2) = (β0.23805) β οΏ½
13.8ππππβ3
οΏ½ = 1.897kVβ 180Β°
Similarly, performing KVL around the negative sequence network always gives us zero minus the negative sequence impedance multiplied by the negative sequence current. Note the negative sign, the negative sign 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 OneLiner simulation
ππππ(0) = 0 β (j0.33)(βj1.587) = β0.5237
ππππ(0) = (β0.5237) β οΏ½
13.8ππππβ3
οΏ½ = 4.172kVβ 180Β°
Similarly, performing KVL around the Zero sequence network results in the following β using the same concept as discussed previously. Note that negative signβ¦ 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 Oneliner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now letβs calculate the phase voltage on the 13.8 kV side using the sequence components
ππππ = ππππ(1) + ππππ
(2) + ππππ(0)
ππππ = 6.07kVβ 0Β° + 1.897kVβ 180Β° + 4.127kVβ 180 = 0
ππππ = 0
When we combine the positive, negative, and zero sequence voltages to calculate the phase A voltage at the point of the fault, we effectively calculate 0V. This makes sense because at the point of the fault, we should expect the voltage to drive to 0. This is the same value as the ASPEN OneLiner simulation
πππ΅π΅ = π π 2 β ππππ
(1) + π π β ππππ(2) + ππππ
(0)
πππ΅π΅ = (1β 240) β 6.07kVβ 0Β° + (1β 120)1.897kVβ 180Β°+ 4.127kVβ 180
πππ΅π΅ = 9.32kVβ -132.2Β°
Calculating the phase B voltage, we effectively get a slight increase in voltage compared to nominal values. This is the same value as the ASPEN OneLiner simulation
πππΆπΆ = π π β ππππ
(1) + π π 2 β ππππ(2) + ππππ
(0)
Calculating the phase C voltage, we effectively get a slight increase in voltage
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππΆπΆ = (1β 120) β 6.07kVβ 0Β° + (1β 240) β 1.897kVβ 180Β°+ 4.127kVβ 180
πππΆπΆ = 9.32kVβ 132.2Β°
compared to nominal values. Notice the 180 degrees difference in the phase angle compared to the phase B voltage. This is the same value as the ASPEN OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, 1LG Fault at 13.8kV Bus Analysis @ 115kV side Finally, letβs calculate the sequence and phase voltages on the 115 kV side for the same fault.
According to ASPEN OneLiner, the current and voltage values on the 115kV side expected to be
ππππ(1) = 61.13 kV β 0Β° πΌπΌππ
(1) = 239ππ β -90Β° ππππ
(2) = 5.270 kVβ -180Β° πΌπΌππ(2) = 239ππ β -90Β°
ππππ(0) = 0 kV πΌπΌππ
(0) = 0 β 90Β°
And the phase values on the 115kV side are expected to be
ππππ = 55.85 kV β 0Β° πΌπΌππ = 478 β -90Β° πππ΅π΅ = 63.92 kV β -115.9Β° πΌπΌπ΅π΅ = 239 β 90Β° πππΆπΆ = 63.92 kV β 115.9Β° πΌπΌπΆπΆ = 239 β 90Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now letβs hand-calculate the phase currents on the 115 kV side
πΌπΌππ
(1) = βππ1.587ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115ππππ= 150.613ππ
πΌπΌππ
(1) = βππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 239A β -90Β°
The positive sequence current that flows through the 13.8kV side is the same as the 115kV side. The only difference is the base. This matches the ASPEN OneLiner result.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ
(2) = βππ1.587ππππ
πΌπΌππ(2) = βππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 239A β -90Β°
The negative sequence current has the same result. This matches the ASPEN OneLiner result.
πΌπΌππ(0) = 0
The zero sequence current will certainly be different as a result of the HV Wye-ungrounded connection. On the zero sequence network diagrams for 1LG fault, notice there is no zero sequence current flow through the 115 kV side. Thatβs precisely because of the Wye-ungrounded connection which blocks zero sequence current. This matches the ASPEN OneLiner result.
The line current on the 115kV side are calculated below.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) = (-ππ1.587ππππ) + (-ππ1.587ππππ) = -ππ3.174 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115ππππ= 150.613ππ
Using the principles of symmetrical components, we know that phase A current is the sum of positive, negative, and zero sequence currents. However, since the zero sequence current is
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ = -ππ3.174 β πΌπΌπ΅π΅πππ π ππ = 478 β -90Β°
blocked, we omit the zero sequence current from our calculations. This matches the ASPEN OneLiner result.
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0)
πΌπΌπ΅π΅ = (1β 240Β°)(βj1.587) + (1β 120Β°)(βj1.587)= +ππ1.587ππππ
πΌπΌπ΅π΅ = +ππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 239.02ππ β 90Β°
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.
πΌπΌπΆπΆ = π π β πΌπΌππ(1) + π π 2 β πΌπΌππ
(2)
πΌπΌπΆπΆ = (1β 120Β°)(βj1.587) + (1β 240Β°)(βj1.587)= +ππ1.587ππππ
πΌπΌπΆπΆ = +ππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 239.02ππ β 90Β°
Now letβs calculate the sequence voltage on the 115kV side.
ππππ(1) = 1 β 0Β° β (ππ0.05)(βj1.587) = 0.92065 pu
πππ΅π΅πππ π ππ =115kVβ3
ππππ
(1) = (0.92065 ) β πππ΅π΅πππ π ππ = 61.13kVβ 0Β°
If we perform KVL around the positive sequence network and calculate the voltage at the point of the fault, we will always find it equal to the pre-fault voltage minus the positive sequence impedance times the positive sequence current. Since, we want to know the positive sequence current only on the 115kV side, we need to select the
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
appropriate positive sequence impedance, which is ππ0.05 This is the same value as the ASPEN OneLiner simulation
ππππ(2) = 0 β (j0.05)(βj1.587) = β0.07935
ππππ
(2) = (β0.07935) β πππππππ π ππ = 5.27 kVβ 180Β°
Similarly, performing KVL around the negative sequence network always gives us zero minus the negative sequence impedance multiplied by the negative sequence current. We just need to select the appropriate impedance this time since we want the negative sequence voltage. Note that the negative sign 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 OneLiner simulation
ππππ(0) = 0
The zero sequence voltage will be zero since the Wye-ungrounded connection blocks the zero sequence current on the HV side. This is the same value as the ASPEN OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now letβs calculate the phase voltages on the 115kV side.
ππππ = ππππ(1) + ππππ
(2) + +ππππ(0)
ππππ = 61.13kVβ 0Β° + 5.27 kVβ -180Β° = 55.85 kV β 0Β°
ππππ = 55.85 kV β 0Β°
We must combine the positive, negative, and zero sequence voltages to calculate the phase A voltage seen on the 115 kV. 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
πππ΅π΅ = π π 2 β ππππ
(1) + π π β ππππ(2)
πππ΅π΅ = (1β 240Β°) β 61.13kVβ 0Β° + (1β 120Β°) β 5.27 kVβ -180Β°
πππ΅π΅ = 63.92 kVβ -115.9Β°
This is the same value as the ASPEN OneLiner simulation
πππΆπΆ = π π β ππππ
(1) + π π 2 β ππππ(2) + ππππ
(0) πππΆπΆ = (1β 120Β°) β 61.13kVβ 0Β° + (1β 240Β°) β 5.27 kVβ -180Β°
Calculating the phase C voltage, we effectively get a slight increase in voltage compared to nominal values. Notice the 180 degrees difference in the phase angle compared to the phase B
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππΆπΆ = 63.92 kVβ 115.9Β°
voltage. This is the same value as the ASPEN OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, LL Fault at 13.8kV Bus Analysis @ 13.8 kV side So far, we have performed 3LG and 1LG fault analysis for 3W transformers. Letβs simulate a LL fault and determine
According to ASPEN OneLiner, the current and voltages values on the 13.8 kV side is expected to be
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(1) = 3.984 kV β 0Β° πΌπΌππ
(1) = 4184 ππ β 90Β° ππππ
(2) = 3.984 kVβ 0Β° πΌπΌππ(2) = 4184 β β 90Β°
ππππ(0) = 0kV πΌπΌππ
(0) = 0 β 90Β°
And the phase values on the 13.8kV side are expected to be
ππππ = 7.967 kV β 0Β° πΌπΌππ = 0 β 0Β° πππ΅π΅ = 3.984 kV β 180Β° πΌπΌπ΅π΅ = 7246 β 0Β° πππΆπΆ = 3.984 kV β 180Β° πΌπΌπΆπΆ = 7246 β 180Β°
Now letβs hand-calculate the phase currents on the 13.8 kV side
πΌπΌππ(1) =
1β 0Β°ππ0.15 + ππ0.15
= βππ3.333 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 13.8ππππ= 1255.11 ππ
πΌπΌππ
(1) = βππ3.333 ππππ β πΌπΌπ΅π΅πππ π ππ = 4184ππ β -90Β°
The positive sequence current is calculated by dividing the pre-fault voltage (1β 0Β°) by the total impedance of the circuit. Itβs important to remember that positive and negative sequence currents are in parallel for LL fault and zero sequence is not activated. The positive sequence impedance is ππ0.05 + ππ0.11 + (βππ0.10) = ππ0.15 The negative sequence impedance is ππ0.05 + ππ0.11 + (βππ0.10) = ππ0.15 The positive sequence current matches the result shown in ASPEN OneLiner with an exception of the angle which is 180 degrees out of phase.
πΌπΌππ(2) = βπΌπΌππ
(1) = + ππ3.333 pu
πΌπΌππ(0) = ππ3.333 ππππ β πΌπΌπ΅π΅πππ π ππ = 4184ππ β 90Β°
Since the same positive sequence current and negative sequence current flow to the faulted area (in parallel) and since positive and negative sequence impedance are equal, we should expect
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌππ(1) = βπΌπΌππ
(2) with equal magnitudes but 180 degrees out of phase. The negative sequence current matches the result shown in ASPEN OneLiner with an exception of the angle which is 180 degrees out of phase.
πΌπΌππ
(0) = 0
Letβs determine the phase current values for the LL fault on the 13.8 kV side.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) + πΌπΌππ
(0) = (4184ππ β -90Β°) + (4184ππ β 90Β°) + 0= 0
πΌπΌππ = 0
Using the principles of symmetrical component, we know that phase A current is the sum of positive, negative, and zero sequence current. The result is zero and it makes sense because the fault is between line B and line C. So line A does not have any fault current. This result matches the ASPEN OneLiner simulation with an exception of the phase angle which is 180 degrees apart.
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ(1) + a β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπ΅π΅ = (1β 240Β°)(4184ππ β -90Β°) + (1β 120Β°)(4184ππ β +90Β°)
+ 0 = 7246β 0Β°
πΌπΌπ΅π΅ = 7246β -180Β°
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. This result matches the ASPEN OneLiner simulation with an exception to the angle which is 180 degrees out of phase.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = a β πΌπΌππ(1) + π π 2 β πΌπΌππ
(2) + πΌπΌππ(0)
πΌπΌπΆπΆ = (1β 120Β°)(4184ππ β -90Β°) + (1β 240Β°)(4184ππ β +90Β°)
+ 0 = 7246β 180Β°
πΌπΌπ΅π΅ = 7246β 0Β°
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. The result also makes sense because
πΌπΌπΆπΆ = βπΌπΌπ΅π΅ This result matches the ASPEN OneLiner simulation with an exception to the angle which is 180 degrees out of phase.
Letβs determine the sequence voltages for the LL fault on the 13.8 kV side.
ππππ(1) = 1 β 0Β° β (j0.15)(βj3.333) = 0.50 pu
πππ΅π΅πππ π ππ =13.8kVβ3
(0.50) β πππ΅π΅πππ π ππ = 3.984 kVβ 0Β°
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 OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(2) = 0 β (j0.15)(βj3.333) = β0.50 pu
We must be very careful hereβ¦ since the positive and negative sequence are in parallel, the sequence voltages should be equal to each other. But performing our normal calculations gives us a voltage thatβs 180 degrees out of phase.
ππππ(2) = ππππ
(1) = 0.50
ππππ(2) = (0.50) β οΏ½
13.8ππππβ3
οΏ½ = 3.984 kVβ 0Β°
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 OneLiner simulation
ππππ(0) = 0
Zero sequence voltage is not activated for phase to phase faults so this value is zero. This is the same value as the ASPEN OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
Now letβs calculate the phase voltage on the 13.8kV side using the sequence components
ππππ = ππππ(1) + ππππ
(2) + ππππ(0)
ππππ = 3.984 kVβ 0Β° + 3.984 kVβ 0Β° = 7.968 kVβ 0Β°
(Nominal voltage)
When we combine the positive, negative, and zero sequence voltages to calculate the phase A voltage at the point of the fault, we effectively calculate 1 pu. This makes sense for phase A voltage because this is the un-faulted phase. This is the same value as the ASPEN OneLiner simulation
πππ΅π΅ = π π 2 β ππππ
(1) + π π β ππππ(2) + ππππ
(0) πππ΅π΅ = (1β 240) β (3.984 kVβ 0Β°) + (1β 120) β (3.984 kVβ 0Β°)
= 3.984 kVβ 180Β°
πππ΅π΅ = 3.984 kVβ 180
Calculating the phase B voltage, we effectively get half the nominal voltage which makes because this is a phase to phase fault on phase B, C. This is the same value as the ASPEN OneLiner simulation
πππΆπΆ = π π β ππππ
(1) + π π 2 β ππππ(2) + ππππ
(0) πππΆπΆ = (1β 120) β (3.984 kVβ 0Β°) + (1β 240) β (3.984 kVβ 0Β°)
= 3.984 kVβ 180Β°
πππΆπΆ = 3.984 kVβ 180Β°
Calculating the phase B voltage, we effectively get half the nominal voltage which makes because this is a phase to phase fault on phase B, C. This is the same value as the ASPEN OneLiner simulation
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, LL Fault at 13.8kV Bus Analysis @ 115kV side
According to ASPEN OneLiner, the current and voltages values on the 115 kV side is expected to be
ππππ(1) = 55.33 kV β 0Β° πΌπΌππ
(1) = 502 A β -90Β° ππππ
(2) = 11.07 kVβ 0Β° πΌπΌππ(2) = 502 A β 90Β°
ππππ(0) = 0 kV πΌπΌππ
(0) = 0 β 90Β°
And the phase and line quantities on the 115kV side are expected to be
ππππ = 66.40 kV β 0Β° πΌπΌππ = 0 β 0Β° πππ΅π΅ = 50.71 kV β -130.9Β° πΌπΌπ΅π΅ = 870 β -180Β° πππΆπΆ = 50.71 kV β 130.9Β° πΌπΌπΆπΆ = 870 β 0Β°
Now letβs hand-calculate the phase currents on the 115kV side
πΌπΌππ
(1) = βππ3.333 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115ππππ= 150.613ππ
πΌπΌππ
(1) = βππ1.587ππππ β πΌπΌπ΅π΅πππ π ππ = 502A β -90Β°
The positive sequence current that flows through the 13.8 kV side is the same as the 115kV side. The only difference is the base. This matches the ASPEN OneLiner result
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
.
πΌπΌππ
(2) = +ππ3.333ππππ
πΌπΌππ(2) = +ππ3.333 ππππ β πΌπΌπ΅π΅πππ π ππ = 502A β 90Β°
The negative sequence current has the same result except the angle is 180 degrees out of phase. This matches the ASPEN OneLiner result.
πΌπΌππ(0) = 0
We should not expect any zero sequence current since the sequence network is not activated. This matches the ASPEN OneLiner result.
The line current on the 115 kV side are calculated below.
πΌπΌππ = πΌπΌππ
(1) + πΌπΌππ(2) = (-ππ3.333 ππππ) + (ππ3.333 ππππ) = 0 pu
πΌπΌππ = 0
Using the principles of symmetrical components, we know that phase A current is the sum of positive, negative, and zero sequence currents. However, since the zero sequence is not activated, the result simply equal 0. This makes sense because this a phase B to C fault we should expect no fault current to flow on line a. This matches the ASPEN OneLiner result.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πΌπΌπ΅π΅ = π π 2 β πΌπΌππ
(1) + π π β πΌπΌππ(2) + πΌπΌππ
(0)
πΌπΌπ΅π΅ = (1β 240Β°)(-ππ3.333 ππππ) + (1β 120Β°)(+ππ3.333 ππππ)= β5.772 ππππ
πΌπΌπ΅π΅πππ π ππ =30ππππππ
β3 β 115ππππ= 150.613ππ
πΌπΌπ΅π΅ = βππ5.7622 ππππ β πΌπΌπ΅π΅πππ π ππ = 870ππ β -180Β°
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 exist This result matches the ASPEN OneLiner simulation.
πΌπΌπΆπΆ = π π β πΌπΌππ
(1) + π π 2 β πΌπΌππ(2) + πΌπΌππ
(0)
πΌπΌπ΅π΅ = (1β 240Β°)(-ππ3.333 ππππ) + (1β 120Β°)(+ππ3.333 ππππ)= 5.772 ππππ
πΌπΌπ΅π΅ = βππ5.7622 ππππ β πΌπΌπ΅π΅πππ π ππ = 870ππ β 180Β°
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 exist This result matches the ASPEN OneLiner simulation.
Now letβs calculate the sequence voltage on the 115kV side.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(1) = 1 β 0Β° β (ππ0.05)(βj3.333) = 0.83335 pu
πππ΅π΅πππ π ππ =115kVβ3
ππππ
(1) = (0.83335 ) β πππ΅π΅πππ π ππ = 55.33 kVβ 0Β°
If we perform KVL around the positive sequence network and calculate the voltage at the point of the fault, we will always find it equal to the pre-fault voltage minus the positive sequence impedance times the positive sequence current. Since we want to know the positive sequence current only on the 115 kV side, we need to select the appropriate positive sequence impedance which is j0.05 This is the same value as the ASPEN OneLiner simulation
ππππ(2) = 0 β (j0.05)(βj3.333) = β0.16665
ππππ
(2) = (β0.16665) β πππππππ π ππ = 11.07 kVβ 180Β°
Similarly, performing KVL around the negative sequence network always gives us zero minus the negative sequence impedance multiplied by the negative sequence current. We just need to select the appropriate impedance this time since we want the negative sequence voltage. 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 OneLiner simulation except the angle which is 180 degrees out of phase.
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
ππππ(0) = 0
The zero sequence voltage will be zero since the Wye-ungrounded connection blocks the zero sequence current on the HV side. This is the same value as the ASPEN OneLiner simulation
Now letβs calculate the phase voltages on the 115 kV side.
ππππ = ππππ(1) + ππππ
(2) + ππππ(0)
ππππ = 55.33 kVβ 0Β° + 11.07 kVβ 180Β° = 44.26 kV β 0Β°
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 ππππ is a very different value compared to ASPEN OneLiner. Please note that this is the un-faulted phase.
πππ΅π΅ = π π 2 β ππππ
(1) + π π β ππππ(2)
πππ΅π΅ = (1β 240Β°) β 55.33 kVβ 0Β° + (1β 120Β°) β 11.07 kVβ 180Β°
πππ΅π΅ = 61.61 kVβ -111.04Β°
Both the πππ΅π΅, and πππΆπΆ hand calculated quantities are very different than ASPEN OneLiner. The reason for this difference is because we calculated the negative sequence voltage with an angle of 180 degrees instead of 0 degrees (ππππ
(2) = 11.07 kVβ 180Β°). This makes a big difference with the unbalanced phasor additions.
πππΆπΆ = π π β ππππ
(1) + π π 2 β ππππ(2) + ππππ
(0) πππΆπΆ = (1β 120Β°) β 55.33 kVβ 0Β° + (1β 240Β°) β 11.07 kVβ 180Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
πππΆπΆ = 61.61 kVβ +111.04Β°
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
3W XFMR, 2LG Fault at 13.8kV Bus Analysis @ 13.8kV side
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
2W, 2LG 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
2W, 2LG Fault, Analysis @ 13.8kV side
References Grainger, John J., et al. Power System Analysis. McGraw-Hill Education, 2016.
Blackburn, Lewis J. βProtective Relaying: Principles and Applications, Fourth Edition.β CRC Press
Jason Hall PE, Steve Cresse PE, Phil Tibbits PE, Casey Harmin PE β co-workers