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Overview

Introduction

Unbalanced loads present a problem in that they can lead to poor performance because of larger current flow in some phases, unexpected neutral current flow, high harmonic content.  Unbalanced motor and transformer loads can fail because of high vibration and overheating.  So steps should always be taken to reduce electrical unbalance, by, for example, ensuring all electrical connections are good and windings are not damaged.   Nonetheless, it is important to be able to calculate the effect of unbalance and this is more difficult than with balanced loads because the neutral voltage and/or current cannot be assumed to be zero.

When performing calculations on balanced three-phase circuits it is usual to work in terms of phase voltages and phase currents and carry out the calculation on one phase as for a single phase circuit.  The solution for the other two phases is then found by taking into account the phase shift of 120° or -120°.

With unbalanced circuits, where, for example, the loads in the three phases are different, the above simplification is impossible, and it is necessary to determine the conditions in each phase separately by applying the laws of networks. 

Unbalanced Delta Connected Load

Assuming the three-phase voltages to be balanced but with the load being different in each phase of the delta, the full line voltage will occur across each phase of the load.  This is shown in Figure 1, with the accompanying phasor diagram in Figure 2.

Figure 1: Delta connected unbalanced load

© L.Gray UHI

Assuming the line voltages to be balanced then, taking the voltage V_L1 as the reference phasor

The phase currents are then given by

N.B. the impedances must be expressed in complex cartesian (rectangular) or polar form.

The line currents are then determined from the phase currents in complex form by the following relationships:

Figure 2: Phasor diagram for an example of a Delta connected unbalanced load

© L.Gray UHI

The Active (P) power can be calculated by adding the power in each phase:

Total Active Power Ptotal = I²_P1 R₁ + I²_P2 R₂ + I²_P3 R₃ (Watts = W)

Example

Three impedances Z_1, Z_2, and Z₃ are delta-connected to a symmetrical three-phase 400 V supply of phase sequence 123

Z₁ = (10 + j0) ohms is connected between lines 1 and 2

Z₂ = (8 + j6) ohms is connected between lines 2 and 3

Z₃ = (5 - j5) ohms is connected between lines 3 and 1

Calculate the:

1. Phase currents
2. Line currents
3. Total active power consumed.
4. Draw the phasor diagram.

View solution

Self Assessment Questions

Now do the questions in the Self Assessment Questions folder for Week 5.

Effect of an Open-Circuited Phase

Suppose that due to a fault, one phase of the delta-connected load becomes open-circuited.  This is shown in Figure 3 where the impedance Z₁ is disconnected.

Figure 3: Delta connected unbalanced load, with phase 1 open circuited

© L.Gray UHI

The current I_P1 now becomes zero

As before, I_P2 = V_L / Z₂

                   I_P3 = V_L/ Z₃

Also I_L1 = I_P1 – I_P3 = - I_P3

        I_L2 = I_P2 – I_P1 = I_P2

        I_L3 = I_P3 – I_P2

Applying this to the previous example we have Figure 4:

Figure 4: Delta connected unbalanced load example, with phase 1 open-circuited

© L.Gray UHI

Using the previous calculations

I_P1 = 0               I_P2 = 40 ⟨ -156.87°                    I_P3 = 56.6 ⟨ 165°

I_L1 = - I_P3 = - 56.6 ⟨ 165° A      

I_L2 = I_P2 = 40 ⟨ -156.87° A                                                                               

I_L3 = I₃ – I₂ = 35.2⟨120.57°       (as before)                                                            

The total Active power now taken by the load is given by:   

40² x 8 + 56.6² x 5 = 12800 +16018 W = 28.8 kW

Self Assessment Questions

Now do the questions in the Self Assessment Questions folder for Week 6.

Unbalanced Star Connected Load Supplied by a Three Phase Four Wire Supply

An unbalanced star-connected load may be connected to a three-phase supply by means of either a three-wire or a four-wire system.  The four-wire system is the easiest to solve and is shown in Figure 5.

Figure 5: Four-wire System Star connected unbalanced load

Assuming that there is no impedance in the neutral (i.e. a short circuit), the voltages at the star point of the supply (o) and the load (o’) must be at the same potential, thus there is no voltage between points o and o’.  The voltages across the loads in each phase are thus equal to the phase voltages of the supply.

Hence we have:

$I_{L1}=I_{P1}=\frac{V_{P1}}{Z_1}$          $I_{L2}=I_{P2}=\frac{V_{P2}}{Z_2}$           $I_{L3}=I_{P3}=\frac{V_{P3}}{Z_3}$ 

The neutral current is given by I_N = I_L1 + I_L2 + I_L3

If one phase becomes open-circuited with a four-wire system the current in that phase will become zero and the currents in the other two phases will remain unchanged.

Example of an Unbalanced 4-Wire Star Connected Load

Three impedances Z_R = (4 + j3) ohms.  Z_Y = (8 + j6) ohms and Z_B = (6 + j4.5) ohms are star-connected across a four-wire 400 V supply                      

Determine:

1. The current in each line
2. The neutral current
3. The Active power taken by the load
4. Draw the phasor diagram

View solution

Self Assessment Questions

Now do the questions in the Self Assessment Questions folder for Weeks 7 and 8.

Unbalanced Star connected load supplied by a three phase three wire supply

The circuit for a three-wire system is shown in Figure 6.  Here the star point of the load is not connected to the star point of the supply and there is thus no neutral connection.  This means therefore that the two star points (O and O’) will not necessarily be at the same potential.

Figure 6: 3-wire System, Unbalanced Star connected load

© L.Gray UHI

In order to determine the current in each phase of the load it is first necessary to find the voltages across each of the loads.  Equating the voltages around the loop ORO’O we have.

V_1O = V_1O’ + V_O’O

V_1O’ = V_1O – V_O’O

Now

I_P1 = V_1O’ /Z₁ = V_{1O ‘}Y₁ = (V_1O – V_O’O) Y₁    Remembering Y (admittance) = 1/Z

Similarly         

I_P2 = (V_2O – V_O’O) Y₂      

and       

I_P3 = (V_3O – V_O’O) Y₃

Also the sum of the currents at o’ is zero therefore

I_P1 + I_P2 + I_P3 = 0

And so

(V_1O – V_O’O)Y₁ + (V_2O – V_O’O)Y₂ + (V_3O – V_O’O) Y₃= 0

V_1O Y₁ + V_2OY₂ + V_3OY₃ – V_O’O (Y₁ + Y₂ + Y₃) = 0

$V_{\text{o'o}}=\frac{V_{10}{Y}_1+V_{20}{Y}_2+V_{30}{Y}_3}{Y_1+Y_2+Y_3}$ 

Then

V_1O’ = V_1O – V_O’O

V_2O’ = V_2O – V_O’O

V_3O’ = V_3O - V_O’O

And

I_P1 = I_L1 =V_1O’ / Z₁

I_P2 = I_L2 = V_2O’ / Z₂

I_P3 = I_L3 = V_3O’ / Z₃

If one phase becomes open-circuited then the current in that phase will fall to zero.  The currents in the other two phases will be equal in magnitude but opposite in sign and given by the line voltage divided by the sum of the impedances in those two phases. For example if Z₁ becomes open-circuited then:

I_P1 = I_L1 = 0

I_P2 = - I_P3 = V₂₃ / (Z₂ + Z₃)   and of course, I_L2 = I_P2 and I_L3 = I_P3

Example

Loads of 2 ohms, (2 + j2) ohms and –j5 ohms are connected in star to the L1, L2 and L3 outputs of a three-phase 433-V, 50-Hz three-wire symmetrical system.

Determine

1. The potential of the load star point with respect to the supply neutral
2. The load phase voltages
3. The line currents
4. The Active power drawn by the load.   

View solution

Self Assessment Questions

Now do the questions in the Self Assessment Questions folder for Weeks 9 and 10.

Conclusion

In this section you have learnt how to analyse systems with unbalanced loads.  You have done this with Delta, 3-wire Star, and 4 wire Star connected unbalanced loads.  You have also learnt how to calculate the Active power consumed by unbalanced loads.  In the next section, Outcome 4, you will learn how this power can be measured using Wattmeters.