4
\$\begingroup\$

I'm an electrical engineering student and I'm currently in In-Plant training in Electricity Company of my country. Recently I visited an electrical grid. I got to know about Earthing transformer. I asked an engineer about how the current flows to the ground in a delta connection as it's not grounded and to flow current there must be a return path to the source. But she didn't give me a satisfying answer, she said there will be current flowing and said that current flows from high potential to low potential (zero potential earth). However I had doubts since that time because unlike in star connection (grounded), there is no path for return current in delta system (with ground I meant).

I know that an earthing transformer creates a virtual ground to current flow in delta system to the virtual ground, so they can measure earth faults for tripping. But it is not clear to me how current flows to the ground if there's no earthing transformer.

I saw in some post saying that this is due to inductive coupling. I'm talking about a 33kV delta system. Since current flows through a closed system, why is this possible?

I'm talking about ungrounded delta system, no neutral, no leg is connected to ground and no earthing transformer. Fully isolated Delta connection. So I want to know how and why current flows to the ground in an earth-fault/1-phase touch earth in a delta-system. In theory it must be zero as there's no return path to the source?

\$\endgroup\$
2
\$\begingroup\$

There is no such thing as an ungrounded system. The delta system is grounded through parasitic capacitance from each phase to ground. This parasitic capacitance appears in the zero sequence network as an impedance (XC) connected to the neutral bus. When you have a phase-earth fault the zero sequence current has to flow through this XC. As such, it is typically very very small.

In the picture i show below i have a one-line drawing showing a source at left (1.0 per unit = nominal voltage) connected to Bus H. The transformer between bus H & L is a wye-delta. So, bus L is an ungrounded delta bus like you are talking about. If we account for the parasitic phase-earth capacitance then we would connect the large XC from the bus to reference bus in each of the three sequence networks (positive, negative, and zero). In my figure i only show it in the zero sequence (bottom) because it is negligible in the positive (top) and negative (middle). Note that ZF = 0 for solid ground fault.

enter image description here

The lady's answer, "current flows from high potential to low potential (zero potential earth)" is just ignorant. Without an earthing source (zig-zag grounding bank etc.) there will be no significant fault current as i describe above. When you get to the point in your studies that you learn/practice symmetrical component analysis you will see this clearly.

I'd recommend Blackburn's book.

russ

\$\endgroup\$
  • 1
    \$\begingroup\$ I see. So in theory there shouldn't be any current but in reality parasitic capacitance creates this. Thanks for the answer, yeah I will get that book. \$\endgroup\$ – Aimkiller Sep 21 '19 at 0:32
  • 1
    \$\begingroup\$ That is correct @Aimkiller. I didn't want to muddy the water, but their is also inter-winding capacitance between the high and low voltage windings of the transformers. This contributes as well, but is negligible in most applications. Here is a paper that describes a case where the inter-winding capacitance (parasitic) is not negligible. russ \$\endgroup\$ – relayman357 Sep 21 '19 at 0:43
2
\$\begingroup\$

In normal configurations a single fault on a delta system will not cause any significant earth fault current. This is useful because the distribution system can tolerate a single fault without interruption to the consumers.

This benefit is of little use if the system does not detect the first fault and fix it. As a result, earth fault detection relays will monitor the phase-earth voltage on each line. (This will create a very weak star/wye point on the system.) When an earth fault is detected on one phase then that phase is deliberately shorted to earth. This can help make fallen wires less likely to electrocute someone. A contact of the relay will signal the fault which can then be investigated, cleared and reset.

If the first fault is not cleared then a second fault will result in high earth current and trip out the system.

\$\endgroup\$
  • \$\begingroup\$ SunnyskyguyEE75 provided me with the right answer for my question. Thank you also for your answer @Transistor \$\endgroup\$ – Aimkiller Sep 20 '19 at 17:11
  • \$\begingroup\$ @Transistor Where did you get your information for your comment: When an earth fault is detected on one phase then that phase is deliberately shorted to earth.? \$\endgroup\$ – relayman357 Sep 20 '19 at 23:54
  • 3
    \$\begingroup\$ @relay, from an ESB Networks (Ireland) MV technician servicing the 10 kV substation at the factory where I work. If I remember correctly they had by regulation to clear the fault within an hour or so or else isolate the line. \$\endgroup\$ – Transistor Sep 21 '19 at 0:10
  • \$\begingroup\$ Thanks @Transistor. In the U.S. distribution systems are operated effectively grounded in vast majority. I know Europeans run resonant grounded systems quite a bit. I had just never heard the application that you described (intentionally shorting the grounded phase). It is interesting - if you run across the regulation that describes it i'd love to hear of it. best regards, russ \$\endgroup\$ – relayman357 Sep 21 '19 at 0:24
1
\$\begingroup\$

Current does not need to flow to ground within a delta transmission system, it is a floating system with a very small capacitive coupling to ground. Please understand that even though a star system usually has a neutral line (3Ø 4-wire), when the load is properly balanced there is no current flowing through this line and it is not needed. United States used to have an ungrounded delta transmission line when there were few customers and shorter lines. This had the advantage of reducing the number of customers lost with a single ground fault and reducing earth-ground current during a fault. In such systems earth ground is reflected through the primary of the source transformer and the secondary of the load transformer.

Distributed power at the customer end in the United States today normally does ground the delta configuration by tapping one phase of each transformer (see fig. 1). This is sometimes used for facilities which need both 3Ø and 1Ø service. Distribution is done by a WYE (Star) configuration.

schematic

simulate this circuit – Schematic created using CircuitLab

This is done primarily to prevent transient voltages, improve the safety of the transmission line, and quickly identify faults.

It is important to know that the system will work fine for delivering power without an earth ground.

\$\endgroup\$
  • 1
    \$\begingroup\$ There are some true statements in your answer, but enough guessing that it is dangerous. Before suggesting the neutral is not needed on 4-wire systems you need to study more. Your comment, For very high voltage applications delta configurations are used is just wrong. You probably assumed this because on high-voltage transmission lines you don't see a neutral. They are not delta configurations. \$\endgroup\$ – relayman357 Sep 21 '19 at 2:16
  • 1
    \$\begingroup\$ The circuit you show (red leg delta) is not used in distribution, it is used in industrial facilities to allow feeding some small amount of lighting loads with grounded neutral for safety. \$\endgroup\$ – relayman357 Sep 21 '19 at 2:17
  • \$\begingroup\$ @relayman357 - Do we know what country he is working in? An ungrounded delta system was originally used here and it behaved exactly as he describes - very low current through a single line ground fault. He's talking about a 33kV system, must be a small country or province. \$\endgroup\$ – Vogon Poet Sep 21 '19 at 2:43
  • \$\begingroup\$ @VogonPoet Thanks for your answer. What my problem was how earth-fault current flows in delta ungrounded system(no leg is grounded I already mentioned it in my OP). As relayman and SunnyskyguyEE75 said, I understood it flows because of parasitic capacitance. So In theory there is zero current, but in reality it's not. \$\endgroup\$ – Aimkiller Sep 21 '19 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.