Unbalanced Three-Phase Systems Full Analysis

A special technique for handling unbalanced three-phase systems is the method of symmetrical components, which is beyond the scope of this text. We will learn this in the near future.

Make sure to read “what a three-phase circuit” first.

After we learn about three phase circuit, we will learn:

  1. Balanced three phase voltage
  2. Balanced three phase power
  3. Unbalanced three phase power
  4. Three phase power measurement

Unbalanced Three-Phase Systems

Learning and understanding three-phase systems would be incomplete without learning and analyzing unbalanced three-phase systems.

An unbalanced three-phase system is not a rare thing in power transmission and distribution.

When we are dealing with either:

  1. Balanced three-phase system
  2. Unbalanced three-phase system

We need to know what caused them to go “balanced” or “unbalanced”.

There are two causes of this unbalanced system:

  1. The voltage sources are not equal in magnitude and/or have differences in phase angle from each other phase.
  2. The load impedances are unequal from each other.

In balanced system, we always have:

  1. Equal voltage sources in magnitude and phase angle. For example, a three-phase system with a voltage source at 120V and 50Hz frequency for each phase.
  2. Equal load impedance. For example, a three-phase system with only resistance loads or inductive loads or capacitive loads with the same value for all lines.

Thus,

An unbalanced system is due to unbalanced voltage sources or an unbalanced load.

There is also one thing to remember that an unbalanced three-phase voltage source is a very rare phenomenon.

Analyzing unbalanced three-phase systems will take a lot of time. Hence, in this post we will assume that every circuit we use has balanced voltage sources and unbalanced load impedances.

Read also : maximum average power transfer formula

Unbalanced Load in Three-Phase System

A three-phase system is balanced if all the line loads are equal to each other. If one of the loads is increased, then it will be an unbalanced system.

Why?

Because that line will draw more current than the other two.

Effects of Unbalanced Three-Phase System

  • Increased heat by three-phase motors.
  • Reduced lifetime of machine by increased heat.
  • Power losses I2R increased.
  • Motor drives become unreliable.

Properties of Unbalanced Three-Phase System

  • The three-phase waveform is disturbed.
  • The line currents are not equal to each other.
  • Neutral wire is needed.
  • Higher power losses.

How to Solve Unbalanced Three-Phase System

Even if we are dealing with three-phase systems, we can still use mesh and nodal analysis. Of course KCL and KVL are also useful.

Observe the circuit in Figure.(1) below. Here we have an star-connected four-wire unbalanced three-phase system consists of:

  • Balanced three-phase voltage source (not drawn in the circuit)
  • Unbalanced star-connected load impedances (Z1, Z2, and Z3).
unbalanced three-phase system
Figure 1. Unbalanced three-phase Y-connected load

Since we already set the load impedances are unbalanced, all the ZA, ZB, and ZC are unequal.

Using Ohm’s law, we get the line currents as:

(1)   \begin{align*}I_a=\frac{V_{AN}}{Z_A} \\I_b=\frac{V_{BN}}{Z_B} \\I_c=\frac{V_{CN}}{Z_C}\end{align*}

In a balanced three-phase system, current produced in the neutral line should be zero. But here, the current in the neutral line is not zero.

Applying KCL at node N gives the neutral line current as

(2)   \begin{align*}I_n=-(I_a+I_b+I_c)\end{align*}

If we are dealing with a three-phase system which doesn’t have a neutral line, we can still use mesh analysis to find line currents Ia, Ib, and Ic. This includes the delta-star, star-delta, and delta-delta three-phase (and three-wire) system.

If we are talking about long distance power distribution and transmission, we are dealing with multiple three-wire systems with earth as the neutral conductor.

To calculate power in an unbalanced three-phase system requires that we find the power in each phase.

The total power is not simply three times the power in one phase but the sum of the powers in the three phases.

Unbalanced Three-Phase Systems Problem Examples

1. The unbalanced star-load of Figure.(1) has balanced voltage sources 100 V and the a c b sequence. Calculate the line currents and the neutral current, if known

ZA = 15 Ω
ZB = 10 + j5 Ω
ZC = 6 − j8 Ω.

Solution:

Using Equation.(1), the line currents are

    \begin{align*}&I_a=\frac{100\angle{0^o}}{15}=6.67\angle{0^o} A\\&I_b=\frac{100\angle{120^o}}{10+j5}=\frac{100\angle{120^o}}{11.18\angle{26.56^o}}=8.94\angle{93.44^o} A\\&I_c=\frac{100\angle{-120^o}}{6-j8}=\frac{100\angle{-120^o}}{10\angle{-53.13^o}}=10\angle{-66.87^o} A\\\end{align*}

Using Equation.(2), the current in the neutral line is

    \begin{align*}I_n&=-(I_a+I_b+I_c)\\&=-(6.67-0.54+j8.92+3.93-j9.2)\\&=-10.06+j0.28=10.06\angle{178.4^o}A\end{align*}

2. Observe the circuit in Figure.(2) below. Here we have an unbalanced circuit with star-star connection three-wire. Find:
(a) the line currents
(b) the total complex power absorbed by the load
(c) the total complex power supplied by the source.

unbalanced three-phase system

Solution:

(a) We use mesh analysis to find the required currents. For mesh 1,

    \begin{align*}120\angle{-120^o}-120\angle{0^o}+(10+j5)I_1-10I_2=0\end{align*}

Or

(2.1)   \begin{align*}(10+j5)I_1-10I2=120\sqrt{3}\angle{30^o}\end{align*}

For mesh 2,

    \begin{align*}120\angle{120^o}-120\angle{-120^o}+(10-j10)I_2-10I_1=0\end{align*}

Or

(2.2)   \begin{align*}-10I_1+(10-j10)I_2=120\sqrt{3}\angle{-90^o}\end{align*}

Equations.(2.1) and (2.2) form a matrix equation:

    \begin{align*}\begin{bmatrix}10+j5 & -10 \\-10 & 10-j10\end{bmatrix}\begin{bmatrix}I_1 \\ I_2\end{bmatrix}=\begin{bmatrix}120 \sqrt{3}\angle{30^o} \\ 120 \sqrt{3}\angle{-90^o}\end{bmatrix}\end{align*}

The determinants are

    \begin{align*}\Delta&=\begin{vmatrix}10+j5 & -10 \\-10 & 10-j10\end{vmatrix}=50-j50=70.71\angle-45^o\\\Delta_1&=\begin{vmatrix}120\sqrt{3}\angle30^o & -10 \\120\sqrt{3}\angle-90^o & 10-j10\end{vmatrix}=207.85(13.66-j13.66)\\&=4015\angle-45^o\\\Delta_2&=\begin{vmatrix}10+j5 & 120\sqrt{3}\angle30^o \\-10 & 120\sqrt{3}\angle-90^o\end{vmatrix}=207.85(13.66-j5)\\&=3023.4\angle-20.1^o\\\end{align*}

The mesh currents are

    \begin{align*}I_1&=\frac{\Delta_1}{\Delta}=\frac{4015.23\angle-45^o}{70.71\angle-45^o}=56.78A\\I_2&=\frac{\Delta_2}{\Delta}=\frac{3023.4\angle-20.1^o}{70.71\angle-45^o}=42.75\angle24.9^oA\end{align*}

The line currents are

    \begin{align*}&I_a=I_1=56.78A\\&I_b=I_2-I_1=38.78+j18-56.78=25.46\angle135^oA\\&I_c=-I_2=42.57\angle-155.1^oA\\\end{align*}

(b) We can now calculate the complex power absorbed by the load.
For phase A,

    \begin{align*}S_A=|I_a|^2Z_A=(56.78)^2(j5)=j16,120VA\end{align*}

For phase B,

    \begin{align*}S_B=|I_b|^2Z_B=(25.46)^2(10)=6480VA\end{align*}

For phase C,

    \begin{align*}S_C=|I_c|^2Z_C=(42.75)^2(-j10)=-j18,276VA\end{align*}

The total complex power absorbed by the load is

    \begin{align*}S_L=S_A+S_B+S_C=6480-j2156VA\end{align*}

(c) We check the result above by finding the power supplied by the source. For the voltage source in phase a,

    \begin{align*}S_a=-V_{an}I^*_a=-(120\angle0^o)(56.78)=-6813.6VA\end{align*}

For the source in phase b,

    \begin{align*}S_b&=-V_{bn}I^*_b=-(120\angle-120^o)(25.46\angle-135^o)\\&=-3055.2\angle105^o=790-j2951.1VA\end{align*}

For the source in phase c,

    \begin{align*}S_c&=-V_{cn}I^*_c=-(120\angle120^o)(42.75\angle155.1^o)\\&=-5130\angle275.1^o=-456.03+j5109.7VA\end{align*}

The total complex power supplied by the three-phase source is

    \begin{align*}S_s=S_a+S_b+S_c=-6480+j2156VA\end{align*}

showing that Ss + SL = 0 and confirming the conservation principle of ac power.

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