Three-Phase Circuits

Demonstrative Video

Problem-1: Line and Phase Quantities

Determine the following quantities for the circuit shown

image

Solution-1

The 3 line currents (of both load & source) are: \[\begin{array}{l} \mathbf{I}_{a A}=\frac{\mathbf{V}_{a^{\prime} n}}{Z_{g a}+Z_{1 a}+Z_{A}}=\frac{120 \angle 0^{\circ}}{40+j 30}=\left(2.4 \angle-36.87^{\circ}\right) \mathrm{A} \\ \mathbf{I}_{b B}=\mathbf{I}_{a A}<-120^{\circ}=\left(2.4 \angle-156.87^{\circ}\right) \mathrm{A} \\ \mathbf{I}_{c C}=\mathbf{I}_{a A}<+120^{\circ}=\left(2.4 \angle+83.13^{\circ}\right) \mathrm{A} \end{array}\] The 3 phase voltages of the load are: \[\begin{array}{l} \mathbf{V}_{A N}=\mathbf{I}_{a A} Z_{A}=\left(2.4 \angle-36.87^{\circ}\right)(39+j 28)=\left(115.22 \angle-1.19^{\circ}\right) \mathrm{V}\\ \mathbf{V}_{B N}=\mathbf{V}_{A N} \angle-120^{\circ}=\left(115.22 \angle-121.19^{\circ}\right) \mathrm{V} \\ \mathbf{V}_{C N}=\mathbf{V}_{A N} \angle+120^{\circ}=\left(115.22 \angle+118.81^{\circ}\right) \mathrm{V} \end{array}\]

The 3 line voltages of the load are:

\[\begin{aligned} \mathrm{V}_{A B} &=\left(\sqrt{3} \angle 30^{\circ}\right) \mathbf{{V}_{A N}} \\ &=\left(\sqrt{3} \angle 30^{\circ}\right)\left(115.22 \angle-1.19^{\circ}\right) \\ &=\left(199.58<+28.81^{\circ}\right) \mathrm{V} \end{aligned}\] image

\[\begin{aligned} \mathrm{V}_{B C} &=\mathbf{V}_{A B} \angle-120^{\circ} \\ &=\left(199.58 \angle-91.19^{\circ}\right) \mathrm{V} \\ \mathrm{{V}_{C A}} &=\mathbf{V}_{A B} \angle+120^{\circ} \\ &=\left(199.58 \angle+148.81^{\circ}\right) \mathrm{V} \end{aligned}\]

The 3 phase voltages of the source are: \[\begin{aligned} \mathbf{V}_{a n}=\mathbf{V}_{a^{\prime} \eta}-\mathbf{I}_{a A} Z_{g a} &=120-\left(2.4 \angle-36.87^{\circ}\right)(0.2+j 0.5) \\ =&\left(118.9 \angle-0.32^{\circ}\right) \mathrm{V} \\ \mathbf{V}_{b n}=& \mathbf{V}_{a n} \angle-120^{\circ}=\left(118.9 \angle-120.32^{\circ}\right) \mathrm{V} \\ \mathbf{V}_{c n}=& \mathbf{V}_{a n} \angle+120^{\circ}=\left(118.9 \angle+119.68^{\circ}\right) \mathrm{V} \end{aligned}\] The three line voltages of the source are: \[\begin{aligned} \mathbf{V}_{a b} &=\left(\sqrt{3} \angle 30^{\circ}\right)\mathrm{V_{a n}}=\left(\sqrt{3} \angle 30^{\circ}\right)\left(118.9 \angle-0.32^{\circ}\right) \\ &=\left(205.94 \angle+29.68^{\circ}\right) \mathrm{V} \\ \mathbf{V}_{b c} &=\mathbf{V}_{a b} \angle-120^{\circ}=\left(205.94 \angle-90.32^{\circ}\right) \mathrm{V} \\ \mathbf{V}_{c a} &=\mathbf{V}_{a b} \angle+120^{\circ}=\left(205.94 \angle+149.68^{\circ}\right) \mathrm{V} \end{aligned}\]

Problem-2: Complex power and Power Dissipation

The equivalent Y-Y configuration of a circuit is shown. Determine the complex powers provided by the source and dissipated by the line of a a-phase? image

Solution-2

The line current of a-phase can be calculated by the complex power is: \[\begin{aligned} S_{\phi}& =\mathbf{V}_{\phi} \mathbf{I}_{\phi}^{*}\\ \Rightarrow (160+j 120) 10^{3}&=\frac{600}{\sqrt{3}} \mathbf{I}_{a A}^{*} \\ \Rightarrow \mathbf{I}_{a A}& =\left(577.35 \angle-36.87^{\circ}\right) \mathrm{A} \end{aligned}\] The a-phase voltage of the source is: \[\begin{aligned} \mathbf{V}_{a n}& =\mathbf{V}_{A N}+\mathbf{I}_{a A} Z_{1 a} \\ &=600 / \sqrt{3}+\left(577.35 \angle-36.87^{\circ}\right)(0.005+j 0.025)\\ &=\left(357.51 \angle 1.57^{\circ}\right) \mathrm{V} \end{aligned}\]

The complex power provided by the source of aphase is: \[\begin{aligned} S_{a n} &=\mathbf{V}_{a n} \mathbf{I}_{a A}^{*}=\left(357.51 \angle 1.57^{\circ}\right)\left(577.35 \angle 36.87^{\circ}\right) \\ &=\left(206.41 \angle 38.44^{\circ}\right) \mathrm{kV} \mathrm{A} \end{aligned}\] The complex power dissipated by the line of a- phase is: \[\begin{aligned} S_{a A} &=\left|\mathbf{I}_{a A}\right|^{2} Z_{1 a}=(577.35)^{2}(0.005+j 0.025) \\ &=\left(8.50 \angle 78.66^{\circ}\right) \mathrm{kV} \mathrm{A} \end{aligned}\]

Problem-3: Star and Delta Configurations

Each phase of a three-phase alternator, generates a voltage of 3810.5 V and can carry a maximum current of 30 A.

Find the following quantities:

if the alternator is connected in

  1. star configuration

  2. delta configuration

Solution-3

\(\text { Given data: }~~ E_{p h}=3810.5 \mathrm{V} ~~ I_{p h}=30 A\)

Star Delta

$$\begin{aligned} \Rightarrow & \text{Line current}~ \mathrm{I}_{t} =\mathrm{I}_{\mathrm{ph}}=30 \mathrm{A} \\ \Rightarrow & \text{Line voltage} ~\mathrm{E}_{\ell} =\sqrt{3} \mathrm{E}_{\mathrm{ph}}\\ & =6600 \mathrm{V} \\ \Rightarrow & \text{Total}~ \mathrm{kVA} =\sqrt{3} \mathrm{E}_{l} \mathrm{I}_{l} \\ =&\sqrt{3} \times 6600 \times 30 \times 10^{-3} \\ &=342.95 \end{aligned}$$

$$\begin{aligned} \Rightarrow & \mathrm{I}_{\ell}=\sqrt{3} \mathrm{I}_{\mathrm{ph}}=51.96 \mathrm{A} \\ \Rightarrow & \mathrm{E}_{\ell}=3810.5 \mathrm{V} \\ \Rightarrow & \text { Total kVA }=\sqrt{3} \mathrm{E}_{l} \mathrm{I}_{l} \\ =&\sqrt{3} \times 3810.5 \times 51.96 \times 10^{-3} \\ &=342.95 \end{aligned}$$

Problem-4: Balanced 3-phase System

A balanced \(3-\phi\) load connected in star consists of \((6 + j 8)~\Omega\) impedance in each phase. It is connected to a \(3-\phi\) supply of 400 V, 50 Hz.

Find the following quantities:

  1. magnitude of phase current

  2. magnitude of line current

  3. per phase power

  4. total power.

Solution-4

image

\[\begin{aligned} E_{ph}& =400 / \sqrt{3}=230.94 \mathrm{V}\\ Z&=(6+j 8) \Omega=10 \angle 53.13^{\circ} \Omega \\ \mathrm{I}_{\mathrm{ph}}&=230.94 / 10=23.094 \mathrm{A} \\ \end{aligned}\]

\[\begin{aligned} \mathrm{I}_{\ell}&=\mathrm{I}_{\mathrm{ph}}=23.094 \mathrm{A}\\ P_{1-\phi}&=E_{p h} I_{p h} \cos \theta\\ =&230.94 \times 23.094 \times \cos 53.13^{\circ}\\ & =3200 \mathrm{W} \end{aligned}\]

\[\begin{aligned} \mathrm{P}_{\mathrm{T}}&=\sqrt{3} \mathrm{E}_{l} \mathrm{I}_{l} \cos \theta \\ =&\sqrt{3} \times 400 \times 23.094 \times \cos 53.13^{\circ}\\ &=9600 \mathrm{W} \\ \text{or}~\mathrm{P}_{\mathrm{T}}&=3 \mathrm{P}=9600 \mathrm{W} \end{aligned}\]

image

Problem-5: Phase Sequence

Determine the phase sequence of the set of voltages given by \[\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =400 \cos \left(\omega \mathrm{t}-10^{\circ}\right) \mathrm{V} \\ \mathrm{v}_{\mathrm{BN}} & =400 \cos \left(\omega \mathrm{t}-250^{\circ}\right) \mathrm{V} \\ \mathrm{v}_{\mathrm{CN}} & =400 \cos \left(\omega \mathrm{t}-130^{\circ}\right) \mathrm{V} \end{aligned}\]

Solution-5

\[\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =400 \cos \left(\omega \mathrm{t}-10^{\circ}\right) \mathrm{V} \\ \mathrm{v}_{\mathrm{BN}} & =400 \cos \left(\omega \mathrm{t}-250^{\circ}\right) \mathrm{V} \\ \mathrm{v}_{\mathrm{CN}} & =400 \cos \left(\omega \mathrm{t}-130^{\circ}\right) \mathrm{V} \end{aligned}\] Given voltage phasors are:

\[\begin{aligned} \mathrm{V}_{\mathrm{AN}}=282.8 \angle-10^{\circ} \\ \mathrm{V}_{\mathrm{BN}}=282.8 \angle-250^{\circ}\\ \mathrm{V}_{\mathrm{CN}}=282.8 \angle-130^{\circ} \end{aligned}\]

image Note: \(\mathrm{V}_{\mathrm{C}}\) is lagging \(\mathrm{V}_{\mathrm{A}}\). Hence phase sequence is \(\mathrm{ACB}\)

Problem-6: Balanced System & Phase Sequence

In a three-phase balanced supply, voltage \(\mathrm{V}_{\mathrm{C}}=110 \angle 65^{\circ} \mathrm{V}\).

Taking the phase sequence as \(\mathrm{ABC}\), find

Solution-6

\[\begin{aligned} \mathrm{V_A}& =\mathrm{Vc} \times 1 \angle-120^{\circ}=110 \angle-55^{\circ} \mathrm{V} \\ \mathrm{V_B}& =\mathrm{Vc} \times 1 \angle 120^{\circ}=110 \angle 185^{\circ} \mathrm{V} \\ \mathrm{V_C}& =110 \angle 65^{\circ} \mathrm{V} \end{aligned}\]

image

\[\begin{aligned} \mathrm{V_{AB}}& =\mathrm{V_A}-\mathrm{V_B}=190.5 \angle-25^{\circ} \mathrm{V} \\ \mathrm{V_{BC}}& =\mathrm{V_B}-\mathrm{V_C}=190.5 \angle-145^{\circ} \mathrm{V} \\ \mathrm{V_{CA}}&=\mathrm{V_C}-\mathrm{VA}=190.5 \angle 95^{\circ} \mathrm{V} \end{aligned}\]

Problem-7: Unbalanced Delta System

A \(3-\phi\) 3-wire \(240~\mathrm{V}\) ABC system has a delta-connected load with impedances \[\mathrm{Z}_{\mathrm{AB}}=10 \angle 0^{\circ} \Omega ;~ \mathrm{Z}_{\mathrm{BC}}=10 \angle 30^{\circ} \Omega;~ \mathrm{Z}_{\mathrm{CA}}=15 \angle-30^{\circ} \Omega\] Taking \(\mathrm{V_{BC}}\) as reference, determine


image
\(\checkmark\) draw the phasor diagram also.

Solution-7

$$\begin{aligned} \mathrm{V}_{\mathrm{BC}}&=240 \angle 0^{\circ} \mathrm{V}\\ \mathrm{V} \mathrm{CA} & =240 \angle-120^{\circ} \mathrm{V}\\ \mathrm{V}_{\mathrm{AB}}&=240 \angle-240^{\circ} \mathrm{V} \end{aligned}$$

$$\begin{aligned} \mathrm{Z}_{\mathrm{AB}}&=10 \angle 0^{\circ} \Omega \\ \mathrm{Z}_{\mathrm{BC}}&=10 \angle 30^{\circ} \Omega\\ \mathrm{Z}_{\mathrm{CA}}&=15 \angle-30^{\circ} \Omega \end{aligned}$$

Phase currents are: \[\begin{aligned} \mathrm{I_{BC}}&=\frac{\mathrm{V}_{\mathrm{BC}}}{\mathrm{Z}_{\mathrm{BC}}}=\frac{240 \angle 0^{\circ}}{10 \angle 30^{\circ}}=24 \angle-30^{\circ} \mathrm{A}=(20.7846-\mathrm{j} 12) \mathrm{A}\\ \mathrm{I_{CA}}& = \frac{\mathrm{V}_{\mathrm{CA}}}{\mathrm{Z}_{\mathrm{CA}}}=\frac{240 \angle-120^{\circ}}{15 \angle-30^{\circ}}=16 \angle-90^{\circ} \mathrm{A}=(0-\mathrm{j} 16) \mathrm{A}\\ \mathrm{I_{AB}}&==\frac{\mathrm{V}_{\mathrm{AB}}}{\mathrm{Z}_{\mathrm{AB}}}=\frac{240 \angle-240^{\circ}}{10 \angle 0^{\circ}}=24 \angle-240^{\circ} \mathrm{A}=(-12+\mathrm{j} 20.7846) \mathrm{A} \end{aligned}\]

Line currents are obtained by applying \(\mathrm{KCL}\) at junction \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). Thus \[\begin{aligned} \mathrm{I_A}& =\mathrm{I}_{\mathrm{AB}}-\mathrm{I}_{\mathrm{CA}}=(-12+\mathrm{j} 20.7846)-(-\mathrm{j} 16)\\ & =(-12+\mathrm{j} 36.7846)=38.6925 \angle 108.07^{\circ} \mathrm{A}\\ \mathrm{I_B}&=\mathrm{I}_{\mathrm{BC}}-\mathrm{I}_{\mathrm{AB}}=(32.7846-\mathrm{j} 32.7846)=46.3644 \angle-45^{\circ} \mathrm{A}\\ \mathrm{I_C}&=\mathrm{I_{CA}}-\mathrm{I_{BC}}=(-20.7846-\mathrm{j} 4)=21.166 \angle-169.11^{\circ} \mathrm{A} \end{aligned}\]

\[\begin{aligned} \mathrm{R}_{\mathrm{AB}}&=10 ~\Omega \\ \mathrm{R}_{\mathrm{BC}}&=10 \cos 30^{\circ}=8.6603 ~\Omega \\ \mathrm{R}_{\mathrm{CA}}&=15 \cos 30^{\circ}=12.9904 ~\Omega \end{aligned}\]

Real power consumed by the load : \[\begin{aligned} &= \left(24^{2} \times 10\right)+\left(24^{2} \times 8.6603\right)+\left(16^{2} \times 12.9904\right)\\ &=14073.8 \mathrm{W}=14.0738 \mathrm{kW} \end{aligned}\]

image

Problem-8: Unbalanced Star System

A \(3-\phi\), 4 -wire, \(208~\mathrm{V}\) system has a star connected load with impedances

\[\mathrm{Z}_{\mathrm{A}}=6 \angle 0^{\circ} \Omega ;~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ} \Omega;~ \mathrm{Z}_{\mathrm{C}}=5 \angle 45^{\circ} \Omega\]

Taking the voltage \(\mathrm{V_{AN}}\) as reference, determine line and neutral currents. Draw the phasor diagram. Also calculate the real power consumed by the load.

image

Solution-8

\(\mathrm{Z}_{\mathrm{A}}=6 \angle \mathrm{0}^{\circ} \Omega ;~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ} \Omega\);   \(\mathrm{Z}_{\mathrm{C}}=5 \angle 45^{\circ} \Omega\)

Taking the phase sequence as ABC,

$$\begin{aligned} \mathrm{V}_{\mathrm{AN}}&=\frac{208}{\sqrt{3}} \angle 0^{\circ} V\\ &=120.08 \angle 0^{\circ} \mathrm{V}\\ \mathrm{V}_{\mathrm{BN}}&=120.08 \angle-120^{\circ} \mathrm{V}\\ \mathrm{V}_{\mathrm{CN}}&=120.08 \angle-240^{\circ} \mathrm{V} \end{aligned}$$

$$\begin{aligned} \mathrm{I}_{\mathrm{A}}&=\frac{\mathrm{V}_{\mathrm{AN}}}{\mathrm{Z}_{\mathrm{A}}}=\frac{120.08}{6 \angle 0^{\circ}}\\ &=20.01 \angle 0^{\circ} \mathrm{A}\\ \mathrm{I}_{\mathrm{B}}&=\frac{\mathrm{V}_{\mathrm{BN}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{120.08 \angle-120^{\circ}}{6 \angle 30^{\circ}}\\ &=20.01 \angle-150^{\circ} \mathrm{A}\\ \mathrm{I}_{\mathrm{C}}&=\frac{V_{C N}}{Z_{C}}=\frac{120.08 \angle-240^{\circ}}{5 \angle 45^{\circ}}\\ &=24.01 \angle 75^{\circ} \mathrm{A} \end{aligned}$$

$$\begin{aligned} I_{N}&=-\left(I_{A}+I_{B}+I_{C}\right)\\ &=(-8.89-j 13.19) A\\ &=15.91 \angle-124^{\circ} A \end{aligned}$$

image

Resistances in different phases are:

\(\mathrm{R}_{\mathrm{A}}=6 \Omega ~~ \mathrm{R}_{\mathrm{B}}=6 \cos 30^{\circ}=5.1962 \Omega ~~ \mathrm{R}_{\mathrm{C}}=3.53 \Omega\)

Real power consumed by the load \[\begin{aligned} =& \left(20.01^{2} \times 6\right)+\left(20.01^{2} \times 5.19\right)+\left(24.01^{2} \times 3.53\right) \\ =&6524.58~ \mathrm{W}= 6.5246 ~\mathrm{kW} \end{aligned}\]

Problem-9: Unbalanced Systems

A \(3-\phi\), 3-wire, 208 V, ACB system has star connected load with impedances \(\mathrm{Z}_{\mathrm{A}}=6 \angle 0^{\circ} \Omega ; ~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ} \Omega\) and \(\mathrm{Zc}=5 \angle 45^{\circ} \Omega .\) Taking \(\mathrm{V}_{\mathrm{BC}}\) as reference, Determine

  1. line currents

  2. the voltage across load impedances

image

Solution-9

$$\begin{aligned} \mathrm{V}_{\mathrm{BC}}&=208 \angle \mathrm{0}^{\circ} \mathrm{V} \\ \quad \mathrm{V}_{\mathrm{CA}}&=208 \angle 120^{\circ} \mathrm{V} \\ \quad \mathrm{V}_{\mathrm{AB}}&=208 \angle-120^{\circ} \mathrm{V} \end{aligned}$$

$$\begin{aligned} Z_{A}&=6 \Omega \\ Z_{B}&=(5.1962+j 3) \Omega \\ Z_{C}&=(3.5355+j 3.5355) \Omega \end{aligned}$$

\(\left[\begin{array}{cc}Z_{A}+Z_{B} & -Z_{B} \\ -Z_{B} & Z_{B}+Z_{c}\end{array}\right]\left[\begin{array}{l}I_{1} \\ I_{2}\end{array}\right]=\left[\begin{array}{c}V_{A B} \\ V_{B C}\end{array}\right]\) \(\left[\begin{array}{cc}11.1962+j 3 & -5.1962-j 3 \\ -5.1962-j 3 & 8.7317+j 6.5355\end{array}\right]\left[\begin{array}{l}I_{1} \\ I_{2}\end{array}\right]=\left[\begin{array}{c}208 \angle-120^{\circ} \\ 208 \angle 0^{\circ}\end{array}\right]\)

Determinant of the coefficient matrix \(= 90.9323 \angle 48.580\)

\(\left[\begin{array}{l}\mathrm{I}_{1} \\ \mathrm{I}_{2}\end{array}\right]=\frac{1}{90.9323 \angle 48.58^{\circ}}\left[\begin{array}{cc}8.7317+\mathrm{j} 6.5355 & 5.1962+\mathrm{j} 3 \\ 5.1962+\mathrm{j} 3 & 11.1962+\mathrm{j} 3\end{array}\right]\left[\begin{array}{c}208 \angle-120^{\circ} \\ 208 \angle 0^{\circ}\end{array}\right]\) \(=\left[\begin{array}{c}-3.6097-\mathrm{j} 22.9799 \\ 11.7974-\mathrm{j} 23.7446\end{array}\right]=\left[\begin{array}{l}23.2617 \angle-98.93^{\circ} \\ 26.5139 \angle-63.58^{\circ}\end{array}\right]\)

\(\left[\begin{array}{l}\mathrm{I}_{1} \\ \mathrm{I}_{2}\end{array}\right]=\left[\begin{array}{l}-3.6097-\mathrm{j} 22.9799 \\ 11.7974-\mathrm{j} 23.7446\end{array}\right]\)

From the mesh currents, line currents can be calculated:

\(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{1}=(-3.6109-\mathrm{j} 22.9797) \mathrm{A}=23.2617 \angle-98.93^{\circ} \mathrm{A}\) \(\mathrm{I}_{B}=\mathrm{I}_{2}-\mathrm{I}_{1}=(15.4071-\mathrm{j} 0.7647) \mathrm{A}=15.426 \angle-2.84^{\circ} \mathrm{A}\) \(\mathrm{I_C}=-\mathrm{I}_{2}=(-11.7974+\mathrm{j} 23.7446) \mathrm{A}=26.5139 \angle 116.42^{\circ} \mathrm{A}\)

Voltages across the loads can be calculated as:

\(\mathrm{V}_{\mathrm{AO}}=\mathrm{Z}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=(-21.6654-\mathrm{j} 137.8782) \mathrm{A}=139.5702 \angle-98.93^{\circ} \mathrm{V}\) \(\mathrm{V}_{\mathrm{BO}}=\mathrm{Z}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}=(82.3503+\mathrm{j} 42.2497) \mathrm{A}=92.556 \angle 27.16^{\circ} \mathrm{V}\) \(\mathrm{V}_{\mathrm{CO}}=\mathrm{Z}_{\mathrm{C}} \mathrm{I_C}=(-125.6599+\mathrm{j} 42.2404) \mathrm{A}=132.5695 \angle 161.42^{\circ} \mathrm{V}\)

Problem-10: Power measurement by two-watt meter

The power input to a \(2000 ~V, 50 \mathrm{Hz},\) 3-phase motor is measured by two wattmeters which indicate \(100 \mathrm{kW}\) and \(300 \mathrm{kW}\) respectively. Calculate

  1. the input power

  2. the power factor

  3. the line current.

Solution-10

  • Given : \(\quad \mathrm{W}_{1}=100 ~\mathrm{kW}\) and \(\mathrm{W}_{2}=300~ \mathrm{kW}\)

  • Input power: \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=400~ \mathrm{kW}\)

  • Power factor angle

    \[\begin{aligned} \quad \tan \theta& =\sqrt{3} \frac{\mathrm{W}_{2}-\mathrm{W}_{1}}{\mathrm{W}_{1}+\mathrm{W}_{2}}\\ &=\sqrt{3} \times \frac{200}{400}=0.8660\\ \theta& =40.89^{\circ} \\ \cos \theta&=0.7559 \end{aligned}\]

  • Power factor \(=0.7559\) lagging

  • Three-phase power, \(P=\sqrt{3} V_{l} I_{l} \cos \theta\)

  • Line current, \(\mathrm{I}_{l}=\frac{400 \times 10^{3}}{\sqrt{3} \times 2000 \times 0.7559}=152.76 \mathrm{A}\)

Problem-11: Power measurement by two-watt meter

Two watt meters are connected to measure the power in a 3-phase 3-wire balanced load. Determine the total power and power factor, if the two watt-meters read

  1. \(1000 \mathrm{W}\) each, both positive

  2. \(1000 \mathrm{W}\) each of opposite sign.

Solution-11

  • Case-1: Given data: \(\quad \mathrm{W}_{1}=1000~ \mathrm{W} ; \quad \mathrm{W}_{2}=1000 \mathrm{W}\)

    • Total power \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=2000 \mathrm{W}\)

    • \(\tan \theta=\sqrt{3} \frac{\mathrm{W}_{2}-\mathrm{W}_{1}}{\mathrm{W}_{1}+\mathrm{W}_{2}}=0\)

    • Power factor angle \(\theta=0\)

    • Power factor \(=\cos 0^{\circ}=1.0\)

  • Case-2: Given data: \(\quad \mathrm{W}_{1}=-1000 \mathrm{W} ; \quad \mathrm{W}_{2}=1000 \mathrm{W} ; \quad\)

    • Total power \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=0\)

    • \(\tan \theta=\sqrt{3} \frac{W_{2}-W_{1}}{W_{1}+W_{2}}=\infty ;\)

    • Power factor angle \(\theta=90^{\circ}\)

    • Power factor = 0 lagging.

Problem-12: Power measurement by two-watt meter

Across \(400 \mathrm{V}, 3\) -phase supply mains, a star-connected balanced load of \((16+\mathrm{j} 12) \Omega\) impedance is connected.

  1. Taking \(V_a\) as reference, determine the line currents and the power absorbed by the load

  2. If two wattmeters are used to measure the power, what will be the readings of the wattmeters?

image

Solution-12

  • \(\quad V_{A}=\frac{400}{\sqrt{3}} \angle 0^{\circ}=230.94 \angle 0^{\circ}~V ~~ Z=(16+j 12)=20 \angle 36.87^{\circ} \Omega\)

  • Line currents are: \[\begin{aligned} \mathrm{I}_{\mathrm{A}}&=\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{Z}}=11.54 \angle-36.87^{\circ} \mathrm{A} \\ \mathrm{I}_{\mathrm{B}}&=11.54 \angle-156.87^{\circ} \mathrm{A}\\ \mathrm{I}_{\mathrm{C}}&=11.54 \angle 83.13^{\circ} \mathrm{A} \end{aligned}\]

  • Total power \[P=\sqrt{3} V_{l} I_{l} \cos \theta=\sqrt{3} \times 400 \times 11.54 \cos 36.87^{\circ}=6400 \mathrm{W}\]

  • Watt meter readings: \[\begin{aligned} \quad \mathrm{W}_{1}&=\sqrt{3} \mathrm{V}_{\text {ph }} \mathrm{I}_{\text {ph }} \cos \left(\theta+30^{\circ}\right)\\ &=\sqrt{3} \times 230.94 \times 11.54 \cos 66.87^{\circ}=1814.35 \mathrm{W}\\ \mathrm{W}_{2}&=\sqrt{3} \mathrm{V}_{\mathrm{ph}} \mathrm{I}_{\mathrm{ph}} \cos \left(\theta-30^{\circ}\right)\\ & =\sqrt{3} \times 230.94 \times 11.54 \cos 6.87^{\circ}=4585.64 \mathrm{W} \end{aligned}\]