Three Phase Wye Delta Transformer Feeding Single Phase Load

In summary, the "Three Phase Wye Delta Transformer Feeding Single Phase Load" discusses the configuration and operation of a three-phase transformer setup where a wye (Y) connection on the primary side and a delta (Δ) connection on the secondary side are utilized to supply single-phase loads. This configuration allows for efficient voltage transformation and distribution, enabling the use of three-phase systems to power single-phase devices. The article highlights key aspects such as load balancing, phase shifting, and the benefits of using transformers to ensure stable voltage supply and minimize losses in electrical systems.
  • #1
BuzzBar
3
2
Three wye-delta (120:480) is feeding single phase load as per diagram. How will current be distributed in the delta winding when single phase is connected between line A and B ?

When I solve by hand I get different results when simulating in Multisim.
Hand calculation - Winding A will be in parallel with winding B&C (which are in series and B+C=-A). This will cause the current to be split evenly between windings A (50A) and B&C (50A).

However when I use Multisim currents are winding A (66A) and B&C(33A).
Which calculation is correct? I don't understand how Multsim is solving winding A to carry double the current in winding B&C,

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  • #2
Welcome to PF.

Is this for schoolwork? If so, I can move it to the schoolwork forums for you. Also, can you clarify on your diagram which are "line" A/B/C and "winding" A/B/C? Thanks.
 
  • #3
I haven't actually tried to solve this yet, IDK, maybe later. But I suspect that you might be adding up the currents algebraically, without regard to their phase difference. In fact, they will add like vectors with 120o angles. The problem is well defined with that "phasor" provision. Frankly, it's just a book keeping exercise of KVL/KCL with out of phase sinusoids.

Please let us know what confuses you about the analysis by hand. Write out the equations of the voltages and currents in the secondary circuit (I know, it is a PITA, that's why I haven't done it for you, yet, LOL). Sorry if I'm not giving you credit for knowing about vectors/phasors in circuits. Often there is a bit of guessing required about the OP's background knowledge.
 
  • #4
berkeman said:
Welcome to PF.

Is this for schoolwork? If so, I can move it to the schoolwork forums for you. Also, can you clarify on your diagram which are "line" A/B/C and "winding" A/B/C? Thanks.
Hi, yes its is for school. I added line and winding currents designation to the diagram.

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  • #5
DaveE said:
I haven't actually tried to solve this yet, IDK, maybe later. But I suspect that you might be adding up the currents algebraically, without regard to their phase difference. In fact, they will add like vectors with 120o angles. The problem is well defined with that "phasor" provision. Frankly, it's just a book keeping exercise of KVL/KCL with out of phase sinusoids.

Please let us know what confuses you about the analysis by hand. Write out the equations of the voltages and currents in the secondary circuit (I know, it is a PITA, that's why I haven't done it for you, yet, LOL). Sorry if I'm not giving you credit for knowing about vectors/phasors in circuits. Often there is a bit of guessing required about the OP's background knowledge.
Hi, I rearranged the delta circuit and treat each winding as as voltage source. I use VAB for reference and consider it to be at angle 0. When I add VBC and VCA as vector sum I get them to be equal to -VAB. So from this I have two equal voltage sources, so current should be dived equally between two voltage source branches,
 

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  • #6
The circuit is overdefined by having three voltage sources, with a closed delta that must sum to zero. Something must give.

The voltage of the delta secondary, relative to ground is floating, unspecified.
 
  • #7
You have two voltage sources across a single load. One has double the source impedance of the other unless otherwise defined. It should be clear after stating that assumption.
 
  • #8
It is obvious that Multisim has used the equation ## I_{AC}+I_{CB}+ I_{BA} = 0A ## instead of ## I_{BA} = \frac{I_{A}}{2} ## has been used by @BuzzBar.
The system is unbalanced and in my opinion the Multisim approach is wrong.
 

FAQ: Three Phase Wye Delta Transformer Feeding Single Phase Load

What is a Three Phase Wye Delta Transformer?

A Three Phase Wye Delta Transformer is a type of transformer configuration that connects three-phase electrical systems. In this setup, one side (the primary side) is connected in a Wye (Y) configuration, while the other side (the secondary side) is connected in a Delta (Δ) configuration. This arrangement allows for efficient voltage transformation and can adapt three-phase power to single-phase loads.

How does a Wye Delta Transformer work when feeding a single-phase load?

When feeding a single-phase load, the Wye Delta Transformer takes three-phase input and transforms it to a lower voltage suitable for the single-phase load. The Wye configuration on the primary side provides a neutral point, which can be used for grounding and stabilizing the voltage. The Delta configuration on the secondary side allows for the distribution of the three-phase power across the single-phase load, ensuring that the load receives the appropriate voltage and current.

What are the advantages of using a Wye Delta Transformer for single-phase loads?

Using a Wye Delta Transformer for single-phase loads offers several advantages, including improved voltage regulation, reduced harmonic distortion, and enhanced system stability. Additionally, this configuration allows for the connection of multiple single-phase loads while maintaining balanced three-phase power, which can improve overall system efficiency and reduce the risk of overloads.

What are the potential issues when using a Wye Delta Transformer with single-phase loads?

Potential issues include unbalanced loads, which can lead to overheating and reduced efficiency. If the single-phase load is significantly imbalanced, it may cause voltage fluctuations and affect the performance of the transformer. Additionally, if the transformer is not properly sized for the load, it can lead to increased losses and reduced lifespan of the transformer.

How do you calculate the capacity of a Wye Delta Transformer for a single-phase load?

To calculate the capacity of a Wye Delta Transformer for a single-phase load, you need to determine the total power requirement of the single-phase load in kilowatts (kW). You can then convert this to kilovolt-amperes (kVA) by dividing the kW by the power factor (PF) of the load: kVA = kW / PF. Ensure that the transformer capacity exceeds this value to accommodate inrush currents and any potential load increases.

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