How Do You Calculate Line Voltage and Current in Delta Connections?

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Discussion Overview

The discussion revolves around calculating line voltage, line current, kVA rating, and kW for a three-phase delta-connected generator based on given parameters such as full load current, voltage, and power factor. The scope includes technical calculations and clarifications related to electrical engineering principles.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Post 1 presents calculations for line voltage, line current, kVA, and kW but expresses uncertainty about the correct substitutions in the formulas.
  • Post 2 states that line voltages should equal phase voltages in a delta connection.
  • Post 3 questions the appropriateness of using line current and line voltage in the kVA and power calculations.
  • Posts 4 and 5 provide alternative calculations, asserting that line voltage and line current should be 240V and 100A, respectively, and suggest that the question may be poorly worded.
  • Post 6 raises a question about the origin of the 100A current measurement, asking whether it refers to the generator windings or the line.
  • Post 8 discusses the implications of assuming identical impedances across phases and the need to consider power phasors for total kVA and kW calculations.
  • Post 9 references textbook definitions to assert that line voltage equals phase voltage in delta connections and provides multiple methods to calculate kVA and kW, seeking validation of these approaches.
  • Post 11 expresses agreement with the calculations presented by others.

Areas of Agreement / Disagreement

Participants express differing views on the correct values for line voltage and current, with some asserting that line voltage should be 240V while others calculate it as 138.5V. There is also uncertainty regarding the interpretation of the current measurement and its source. Overall, multiple competing views remain without a consensus.

Contextual Notes

Some participants note that the question may be poorly worded, which could contribute to the confusion regarding the definitions and calculations involved. There is also a lack of clarity on whether the current measurement refers to the generator windings or the line.

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Homework Statement



each phase of a three phase delta connected generator supplies a full load current of 100A at a voltage of 240V and at a power factor of 0.6 lagging.

Find:
a) line voltage
b) line current
c) the kVA rating
d) the KW

2. The attempt at a solution

a) Line voltage = Phase Voltage (in delta)

VP= VL/1.732
VP= 240/1.732
VP= 138.5V

Therefore the line voltage is 138.5 volts.

b) Phase current at full load= 100A

Line current= Phase current x 1.732

IL= (100)(1.732)
IL= 173.2A

Therefore the line current is 173.2 amps.

c) kVA= (1.732)(VL)(IL)/1000

=(1.732)(240)(173.2)/1000
= 700VA

d) P= (1.732)(VL)(IL)(0.6)

= (1.732)(240)(173.2)(0.6)
= 43kW

How do these answers look? I am a little unsure of which voltages/currents to substitue into the kVA and P formulas.
Thanks,
 
Last edited:
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I may be a bit rusty at power circuits, but aren't the line voltages equal to the phase voltages for a delta-connected generator?
 
^ Yes. What I am unsure of is question c and d. I am not exactly sure which values to substitute into kVA= (1.732)(VL)(IL)/1000 and P= (1.732)(VL)(IL)(0.6). I used the line current and line voltage for both. Is this right? (assuming that the line current I calculated is correct to begin with.)
 
Hi JGrecs, my answers.

a)240 volts
b)100 amps
c)240*100*1.732=41.6KVA
d)240*100*1.732*0.6 =25KW

The thing here is that the question IMO isn't worded very well so that might be half the problem.

Remember when connected in delta the line and phase voltages are the same, while phase and line currents will be different.

When connected in star line and phase voltage will be different, while line and phase currents will be the same.

Is there an actual answer included with the question?
 
Last edited:
Aussielec said:
Hi gneill, my answers.

a)240 volts
b)100 amps
c)240*100*1.732=41.6KVA
d)240*100*1.732*0.6 =25KW

The thing here is that the question IMO isn't worded very well so that might be half the problem.

Remember when connected in delta the line and phase voltages are the same, while phase and line currents will be different.

When connected in star line and phase voltage will be different, while line and phase currents will be the same.

Is there an actual answer included with the question?

No answers provided. You say for b), that the line current is 100A. Does this mean that the Load current=Line current for delta connections? Also for question a), If I am not mistaken, the phase voltage is calculated Vload/1.732 which comes out to 138.5V. So shouldn't VLine be 138.5V?
 
Last edited:
When connected in delta line and phase current will be different. But this what I can't understand- Where is this 100 amps measured from? The generator windings or the line?

The question reads. "supplies a full load current of 100A"? But from where?
 
Aussielec said:
When connected in delta line and phase current will be different. But this what I can't understand- Where is this 100 amps measured from? The generator windings or the line?

The question reads. "supplies a full load current of 100A"? But from where?

each phase of a three phase delta connected generator supplies a full load current of 100A

The Windings?

100A per phase/winding
 
Last edited:
As I said, I'm a bit rusty on multiphase power system measurement/specification conventions. But if we were to assume that the load consists of identical impedances across the phases supplies, and each is drawing 100A with a power factor of 0.6, then we could determine the appropriate impedance (pf corresponds to 53.13° lag angle for the current) and create a Spice model:

attachment.php?attachmentid=56738&stc=1&d=1363319778.gif

[note that the ground is for the convenience of the simulation software]

Each load will draw its 100A, and each phase voltage source will produce 100A, all shifted according to the phase angles of the supplies.

In order to answer the given problem, I would first have to recall or find out: does a single impedance and the power it dissipates characterize the system, or should the power phasors of each load be summed to find total kVA and kW numbers?
 

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Ok,

A) According to textbook: VLoad=VPhase=VLine (Delta)

Therefore VLine in this case = 240V

B) According to textbook: ILine=IPhase X 1.732 (Delta)

IPhase=100A (Would you agree)?

Therefore ILine= 173.2A

c)

kVA= kW/PF

kVA= 43/0.6
kVA= 72kVA

or

kVA= 240 X 100 X 1.732
kVA= 72kVA

or

kVA= 240 X 1.732 X 3
kVA= 72kVAD) Textbook: P= 1.732 X VLine X ILine X PF or P= 3 X VPhase X IPhase X PF

P= 1.732 X 240V X 173.2A X 0.6
P= 43kW

or

P= 3 X 240V X 100A x 0.6
P= 43kW These should all be correct. Any objections brehs?
 
Last edited:
  • #10
Thanks for the help, Gneill & Aussielec.
 
  • #11
Your results look okay to me!
 

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