Balanced Star and Delta Connected Loads

Overview

In this section you will learn how to:

  • analyse phase and line voltage and current relationships for balanced star connected loads,
  • analyse phase and line voltage and current relationships for balanced delta connected loads,
  • calculate phase and line voltages and currents for complex star connected loads,
  • calculate phase and line voltages and currents for complex delta connected loads,
  • calculate active, reactive and apparent power for star connected loads,
  • calculate active, reactive and apparent power for delta connected loads,
  • draw phasor diagrams for balanced three phase loads.
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Turbine

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Balanced Three Phase Loads

Three-phase loads are said to be balanced if the same impedance exists in each phase of the load.  They may be connected either in Delta or Star.  This is shown in Figure 1.

Delta and Star Balanced Load Circuit Diagrams

Figure 1: Delta and Star Balanced Load Circuit Diagrams

© L.Gray UHI

Calculations on Balanced Three Phase Systems

When carrying out calculations on balanced three-phase systems, whether Delta or Star three-wire or four-wire, the procedure is to perform the calculation for one phase only.  The currents and voltages in the other two phases will have the same magnitude but will be shifted in phase by +120˚ or -120˚.

Balanced Delta Load Calculations

We will use phasors for all our calculations. This means using complex numbers in both polar and rectangular form. 

First, the generator phase voltages expressed using phasors.  We saw in Outcome 1 that for a Delta connected generator, line voltage = phase voltage, and each phase voltage is 120° different from the next.  Taking Phase 1 as our reference, for a balanced Delta system with all voltages = V:

VL12=VP1=V⟨0°=V(1+j0) 

VL23=VP2=V⟨120°=V(-12-j32) 

VL31=VP3=V⟨+120°=V(-12+j32) 

Load calculations are best explained with examples.

Example 1 Balanced Delta Load

Three similar coils, each of resistance 7 Ω and inductance 0.03 H are connected in Delta across a UK 415 V, 3-phase supply, as in Figure 2.  Note that the generator configuration is unimportant, as the generator is symmetrical and line voltages are always given (not the phase voltages).

Delta Connected Load diagram

Figure 2: Delta Connected Load

© L.Gray UHI

Determine:

  1. The line voltage
  2. The phase voltage
  3. The phase current
  4. The line current
  5. The power factor
  6. The active, reactive and apparent power taken by each phase of load
  7. The total active, reactive and apparent power taken by the load
  8. Sketch the phasor diagram

View solution

Solution to Example 1 Balanced Delta Load

Consider only Phase 1, as we will have the same answers for Phase 2 and 3, just with phase shifts of -120° and +120° respectively.  Also assume we have taken a “snapshot” of the phasor diagram when Phase 1 voltage is at 0°.

  1. We are told the line voltage in the question: VL=415⟨0°V 

  2. Because the load is Delta connected, Vphase = Vline.  We are given Vline in the question, so VPH=VL=415⟨0°V 

  3. Before we can calculate current, we have to calculate the impedance of the load:

    Remember that pure inductive reactance has a phase shift of +90° and pure capacitive reactance has a phase shift of -90°.

    First the inductive reactance at the UK frequency of 50 Hz:
    XL=2πfL=2π×50×0.03=9.4⟨90°Ω 

    Then the impedance:
    Z=R+jXL=7+j9.4=11.72⟨53.3°Ω 

    Now the phase voltage appears across each phase, so by Ohms Law:
    IPH=VPH/Z=415⟨0°/(11.72⟨53.3°)=35.4⟨-53.3°A 

  4. To find the line current, we need to apply Kirchoff’s Current Law, at junction 1:

    IL1=IP1−IP3 

    Because it is a balanced system IP3 leads IP1 by 120° but is the same magnitude

    IL1 = IP1 − IP3  
    = ( 35.4 ⟨ -53.3° ) − ( 35.4 ⟨ -53.3 + 120° ) = ( 35.4 ⟨ -53.3° ) − ( 35.4 ⟨ 66.7° )
    =(21.156−j28.383)−(14.002+j32.513)=7.154−j60.896=61.315 ⟨ -83.3°A 

    It turns out that this is actually Line Current=3×Phase Current with a -30° phase shift. So this is an easier way to calculate line currents:

    IL1 =3IP1 ⟨ −30° =3×35.4⟨−53.3−30°=61.315⟨−83.3° A 

    This equation works for all three line currents, but don’t forget to apply the -120° to phase current 2, and +120° to phase current 3.

    Summarising our Answers so far:

    VL1 = VP1 = 415 ⟨ 0° V

    IP1 = 35.4 ⟨ - 53.3° A

    IL1 = 61.315 ⟨ -83.3° A

    VL2 = VP2 = 415 ⟨ -120° V

    IP2 = 35.4 ⟨ -173° A

    IL2 = 61.315 ⟨ -203.3° A

    VL3 = VP3 = 415 ⟨ +120° V

    IP3 = 35.4 ⟨ 66.7° A

    IL3 = 61.315 ⟨ 36.7° A

    These will be used to draw our phasor diagram.

  5. Remembering our Single Phase AC, we know that Power Factor (PF) = the cosine of the angle between the voltage and current. Using our Phase 1 values, we have  PF = cos(53.3°) = 0.598 lagging (because of – sign).

  6. Again, from Single Phase AC theory:

    Active Power, P = VI x Power Factor.  Calculating for Phase 1:

    P = 415 x 35.4 x 0.598 = 8785 W (Watts)

    Reactive Power Q = VI x sin(angle between voltage and current)

    Q = 415 x 35.4 x sin(53.3°) = 11779 VAr (Volts-Amps Reactive)

    Apparent Power = VI

    S = 415 x 35.4 = 14691 VA (Volt-Amps)

  7. Since there are 3 phases, the total power taken by the load is 3 x the power taken by each phase:

    Total Active Power = 3 x 8785 W = 26.355 kW

    Total Reactive Power = 3 x 11779 VAr = 35.337 kVAr

    Total Apparent Power = 3 x 14691 VA = 44.073 kVA

     

    We can check these last answers using, e.g.:

    Active Ptotal=3VLILcosθ=3×415×61.315×0.598=26.355 kW 

    Apparent Stotal=Ptotal2+Qtotal2=263552+353372=44.083 kVA (allowing for rounding errors) 

  8. Here is the phasor diagram (drawn using Matlab© R2020b). Note that the voltage scale is smaller than the current scale.

    phasor diagram

    © L.Gray UHI

 

Self Assessment Question 1 Balanced Delta Load

Three similar coils, each of resistance 10 Ω and inductance 0.07 H are connected in Delta across a 400 V, 3-phase supply, at 60 Hz. 

Delta Connected Load diagram

Figure 2: Delta Balanced Load

© L.Gray UHI

Determine:

  1. The line voltage
  2. The phase voltage
  3. The phase current
  4. The line current
  5. The power factor
  6. The active, reactive and apparent power taken by each phase of load
  7. The total active, reactive and apparent power taken by the load
  8. Sketch the phasor diagram

View solution

Solution to Self Assessment Question 1

  1. We are told the line voltage in the question: VL=400⟨0°V 

  2. Because the load is Delta connected, Vphase = Vline.  We are given Vline in the question, so VPH=VL=400⟨0°V 

  3. Calculate the impedance of the load:

    First the inductive reactance at 60 Hz:

    XL=2πfL=2π×60×0.07=26.4⟨90°Ω 

    Then the impedance

    Z=R+jXL=10+j26.4=28.230⟨69.254°Ω 

    Now the phase voltage appears across each phase, so by Ohms Law:

    IPH = VPH / Z = 400 ⟨ 0° / ( 28.230 ⟨ 69.254° ) = 14.169 ⟨ − 69.254° A  

  4. To find the line current, we need to apply Kirchoff’s Current Law, at junction 1:

    IL1 = IP1 − IP3  

    Because it is a balanced system IP3 leads IP1 by 120° but is the same magnitude

    IL1 = IP1 − IP3  
    = ( 14.169 ⟨ -69.254° ) − ( 14.169 ⟨ -69.254 + 120° ) = ( 14.69 ⟨ -69.254° ) − ( 14.169 ⟨ 50.746° )
    =(5.019−j13.250)−(8.966+j10.971)=-3.947−j24.221=24.540 ⟨ −99.255°A

    It turns out that this is actually Line Current=3×Phase Current with a -30° phase shift. So this is an easier way to calculate line currents:

    IL1 = 3 IP1 ⟨ − 30°   =3×14.169⟨−69.254−30°=24.541⟨−99.254° A 

    This equation works for all three line currents, but don’t forget to apply the -120° to phase current 2, and +120° to phase current 3.

    Summarising our Answers so far:

    VL1 = VP1 = 400 ⟨ 0° V

    IP1 = 14.169 ⟨ - 69.254° A

    IL1 = 24.54 ⟨ -99.254° A

    VL2 = VP2 = 400 ⟨ -120° V

    IP2 = 14.169 ⟨ -189.254° A

    IL2 = 24.54 ⟨ -219.254° A

    VL3 = VP3 = 400 ⟨ +120° V

    IP3 = 14.169 ⟨ 50.746° A

    IL3 = 24.54 ⟨ 20.746° A

    These will be used to draw our phasor diagram.

  5. Remembering our Single Phase AC, we know that Power Factor (PF) = the cosine of the angle between the voltage and current. Using our Phase 1 values, we have  PF = cos(69.254°) = 0.354 lagging (because of – sign).

  6. From Single Phase AC theory, calculating for Phase 1:

    P = VI x PF = 400 x 14.169 x 0.354 = 2006 W

    Q = VIsinφ = 400 x 14.169 x sin(69.254°) = 5300 VAr

    S = VI = 5668 VA

  7. Since there are 3 phases, the total power taken by the load is 3 x the power taken by each phase:

    Total Active Power = 3 x 2006 W = 6018 W

    Total Reactive Power = 3 x 5300 VAr = 15.9 kVAr

    Total Apparent Power = 3 x 5668 VA = 17.0 kVA

     

    We can check these last answers using, e.g.:

    Active Ptotal=3VLILcosθ=3×400×24.54×0.354=6018 W 

    Apparent Stotal=Ptotal2+Qtotal2=60182+159002=17 kVA 

    Here is a circuit diagram proving these to calculations to be correct (allowing for rounding errors):

    circuit diagram
    © L.Gray UHI

  8. Here is the phasor diagram:

    Phasor diagram

    © L.Gray UHI

    Now do the questions in the Self-Assessment Questions folder, for Week 3.

 

Example 2 Balanced Star Load

Three similar coils, each of resistance 7 Ω and inductance 0.03 H are connected in Star across a UK 415 V, 3-phase supply, as shown in Figure 3.  Note that the generator configuration is unimportant, as the generator is symmetrical and line voltages are always given (not the phase voltage

Star Load diagram

© L.Gray UHI

Determine:

  1. The line voltage
  2. The phase voltage
  3. The phase current
  4. The line current
  5. The power factor
  6. The active, reactive and apparent power taken by each phase of load
  7. The total active, reactive and apparent power taken by the load
  8. Sketch the phasor diagram

View solution

Solution to Example 2, Balanced Star Load

Consider only Phase 1, as we will have the same answers for Phase 2 and 3, just with phase shifts of -120° and +120° respectively.  Also assume we have taken a “snapshot” of the phasor diagram when line voltage VL12 is at 0°.

  1. We are told the line voltage in the question: VL12 = 415⟨0° V

  2. For a Star connnected load, VL12 = VP1 – VP2. In outcome 1, we saw a phasor diagram for the Star connected generator which showed the results of this calculation (Figure 4)

    phasor diagram

    Figure 4 Phasor Diagram for Balanced Star Connected Load or Generator
    © L.Gray UHI

    This shows the relationship between phase and line voltages in a balanced Star connected system, and we can see that VL12=3×VP1⟨30° 

    So our Phase 1 Voltage, VP1=VL123⟨−30° 

    so VP1=4153⟨−30°=239.6⟨−30°V

    Note that the choice of reference phasor is completely arbitrary. We could have taken our phasor “snapshot” when VP1 was along the positive x axis, in which case VL12 would have had a phase angle of +30° (at that point in time).

    Remember also that although we have calculated line and phase voltage for one phase, the other two phases are just displaced by -120 and +120 degrees, respectively, if the system is balanced.

  3. Calculate the impedance of the load:

    XL=2πfL=2π×50×0.03=9.425⟨90°Ω 

    Then the impedance:

    Z=R+jXL=7+j9.425=11.740⟨53.398°Ω 

    Now the phase voltage appears across each phase, so by Ohms Law:

    IPH= VPH / Z = 239.6 ⟨ −30°/ ( 11.74 ⟨ 53.398° ) = 20.409⟨−83.398°A  

  4. For a Star connected load the phase current = the line current.

    So IL=20.409⟨-83.398°A

    Summarising our Answers so far:

    VL12 = 415 ⟨ 0° V

    VP1 = 239.6 ⟨ - 30° V

    IP1 = IL1 = 20.409 ⟨ -83.398° A

    VL23 = 415 ⟨ -120° V

    VP1 = 239.6 ⟨ -150° V

    IP2 = IL2 = 20.409 ⟨ -203.398° A

    VL31 = 415 ⟨ +120° V

    VP1 = 239.6 ⟨ 90° V

    IP3 = IL3 = 20.409 ⟨ 36.602° A

    These will be used to draw our phasor diagram.

  5. e) Remembering our Single Phase AC, we know that Power Factor (PF) = the cosine of the angle between the voltage and current. Using our Phase 1 values, we have

    -83.398 – (-30) = -53.398

    PF = cos(53.398°) = 0.596 lagging (because of – sign).

  6. Again, from Single Phase AC theory: Calculating for Phase 1:

    P = VI x PF = 239.6 x 20.409 x 0.596 = 2914 W

    Q = VIsinΦ = 239.6 x 20.409 x sin(53.398°) = 3926 VAr

    S = VI = 239.6 x 20.409 = 4890 VA

  7. Since there are 3 phases, the total power taken by the load is 3 x the power taken by each phase:

    Total Active Power = 3 x 2914 W = 8.742 kW

    Total Reactive Power = 3 x 3926 VAr = 11.778 kVAr

    Total Apparent Power = 3 x 4890 VA = 14.670 kVA

     

    We can check these last answers using:

    Ptotal=3VLILcosθ=3×415×20.409×0.596=8743 W (the same, allowing for rounding errors). Notice that this equation works for both Star and Delta connected loads.  

    Stotal=Ptotal2+Qtotal2=87422+117782=14.670 kVA 

    Here is a circuit diagram proving these to calculations to be correct (allowing for rounding errors):

    Star Connection diagram

    © L.Gray UHI

  8. Here is the phasor diagram (drawn using Matlab© R2020b). Note that the voltage scale is smaller than the current scale.

    phasor diagram

    © L.Gray UHI

 

Self Assessment Question 2 Balanced Star Load

Three similar coils, each of resistance 10 Ω and inductance 0.07 H are connected in Delta across a 400 V, 3-phase supply, at 60 Hz. 

example diagram

Determine:

  1. The line voltage
  2. The phase voltage
  3. The phase current
  4. The line current
  5. The power factor
  6. The active, reactive and apparent power taken by each phase of load
  7. The total active, reactive and apparent power taken by the load
  8. Sketch the phasor diagram

View solution

Solution to Self Assessment Question 2 Balanced Star Load

  1. We are told the line voltage in the question: VL12=400⟨0°V 

  2. For a Star connected load:

    Phase 1 Voltage, VP1=VL123⟨-30°=4003⟨-30°=230.94⟨-30° 

    The other two phases are just displaced by -120 and +120 degrees, respectively, if the system is balanced.

  3. Calculate the impedance of the load:

    XL=2πfL=2π×60×0.07=26.389⟨90°Ω 

    Then the impedance:

    Z=R+jXL=10+j26.389=28.22⟨69.246°Ω 

    Now the phase voltage appears across each phase, so by Ohms Law:

    IPH= VPH / Z = 230.94 ⟨ − 30° / ( 28.22 ⟨ 69.246° ) = 8.184⟨−99.246°A  

  4. For a Star connected load the phase current = the line current.

    So IL=8.184⟨-99.246°A

    Summarising our Answers so far:

    VL12 = 400 ⟨ 0° V

    VP1 = 230.94 ⟨ - 30° V

    IP1 = IL1 = 8.184 ⟨ -99.246° A

    VL23 = 400 ⟨ -120° V

    VP1 = 230.94 ⟨ -150° V

    IP2 = IL2 = 8.184 ⟨ -219.246° A

    VL31 = 400 ⟨ +120° V

    VP1 = 230.94 ⟨ 90° V

    IP3 = IL3 = 8.814 ⟨ 20.754° A

    These will be used to draw our phasor diagram.

  5. Remembering our Single Phase AC, we know that Power Factor (PF) = the cosine of the angle between the voltage and current. Using our Phase 1 values, we have

    -99.246 – (-30) = -69.246

    PF = cos(69.246°) = 0.354 lagging (because of – sign).

  6. Again, from Single Phase AC theory, calculating for Phase 1:

    P = VI x PF = 230.94 x 8.184 x 0.354 = 669 W

    Q = VIsinΦ = 230.94 x 8.184 x sin(69.246°) = 1767 VAr

    S = VI = 230.94 x 8.184 = 1890 VA

  7. Since there are 3 phases, the total power taken by the load is 3 x the power taken by each phase:

    Total Active Power = 3 x 669 W = 2007 W

    Total Reactive Power = 3 x 1767 VAr = 5301 VAr

    Total Apparent Power = 3 x 1890 VA = 5670 VA

     

    We can check these last answers using:

    Ptotal=3VLILcosθ=3×400×8.184×0.354=2007 W

    Stotal=Ptotal2+Qtotal2=20072+53012=56.68 kVA (allowing for rounding errors) 

  8. Here is the phasor diagram (drawn using Matlab© R2020b). Note that the voltage scale is smaller than the current scale

    phasor diagram

    © L.Gray UHI

    Now do the questions in the Self-Assessment Questions folder for Week 4.

 

Star-Delta Starting

You may have heard of Star-Delta starting of three phase induction motors.  The analysis of Star and Delta connected loads gives us some idea of how this works:

With Star-Delta starting, the motor’s stator is initially connected in Star configuration when the motor is started.  Remember this is when each stator winding has one end connected to a common point.  In this configuration the line voltage is much larger than the phase voltage, but the line and phase currents are the same.

Once the motor reaches a certain speed, usually about 80% of its final operating speed, a three pole – double throw (3PDT) centrifugal switch will automatically change the stator connection from Star to Delta.   Because of the motor’s rotating inertia, the change is quick enough to cause no observable speed difference (even for the short period of time that the stator current is totally disconnected).  In Delta configuration the line current is 3 times the phase current.

Comparing the current of a Delta connected load to the current of a Star connected load, we get:

IL(delta) IL(star)= 3 V L Z V L 3Z so IL(star)=13IL(delta) 

So starting in Star configuration has two differences compared to starting in Delta (or Direct-On-Line = DOL) configuration:

  1. The starting current is 3 times lower. Motors draw their highest current when they are not rotating, so this is a big advantage as reduces inrush current and improves efficiency.
  2. Because motor torque is proportional to current, the starting torque is 3 times lower. This can be a disadvantage if the motor is driving a high load from start up.

Conclusion

In this section you have learnt how to analyse star and delta connected three-phase balanced loads.  You have learnt how to calculate line and phase voltages and line and phase currents for both resistive and complex (including capacitance and/or inductance) loads.   You have also learnt how to draw phasor diagrams and how to calculate active, reactive and apparent power for these loads. 

Finally, you have seen how star-delta starting can be used for three-phase motors.  In the next section, Outcome 3, you will learn how to analyse three-phase systems with unbalanced loads.

© University of the Highlands and Islands