3-Phase AC Circuits 2

LessonAbout 1.5 h2 of 2 in 3P AC Circuits

3P AC Circuits 2

3-phase power measurement is the process of determining total active power in three-phase systems using one or more electrodynamometer wattmeters.

1. Power Measurement Using Wattmeter

Wattmeter is an electrodynamic instrument designed to measure active electric power in watts.

  • Current coil has low resistance and connects in series with line
  • Potential coil has high resistance and connects across load voltage
  • Deflection angle is proportional to average active power

Wattmeter Power Reading

\[ W = V I \cos\theta \]

Where: \(W\) = wattmeter reading (W), \(V\) = potential coil voltage (V), \(I\) = current coil current (A), \(\theta\) = phase angle between \(V\) and \(I\) (\(^\circ\))

EE Board Exam October 1980

Example. Three equal impedances of \(8 + j7\ \Omega\) are connected in delta across a \(240\ \text{V}\), 3-phase line. Find the reading of a single-phase wattmeter with its current coil in line a and potential coil across lines b and c.

Solution.

\[ Z = 8 + j7 = 10.63\angle 41.186^\circ\ \Omega \qquad\qquad I_a = \sqrt{3}\left(\dfrac{240}{10.63}\right) = 39.104\ \text{A} \]
\[ W = V_{bc} I_a \cos(120^\circ - 71.186^\circ) = 240(39.104)\cos 48.814^\circ = 6180\ \text{W} \]

Answer: \(6180\ \text{W}\)

2. 3-Phase Power Measurement with Balanced Load Using Single Wattmeter

Single-wattmeter method measures total 3-phase active power in balanced loads using one wattmeter and special potential connections.

2.1 Neutral Point Available

Neutral-point method connects the wattmeter potential coil directly between one line conductor and the system neutral.

  • Measures single-phase active power
  • Total 3-phase power equals three times reading
  • Requires accessible 4-wire neutral conductor

Single Wattmeter with Neutral Available

\[ P_{\text{total}} = 3 W \]

Where: \(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)

2.2 Potential-Lead Shift Method

Potential-lead shift method connects current coil in line 1 and shifts potential lead sequentially to lines 2 and 3.

  • First position gives reading \(W'\)
  • Second position gives reading \(W''\)
  • Total active power is sum of two readings

Potential-Lead Shift Method

\[ P_{\text{total}} = W' + W'' \]

Where: \(P_{\text{total}}\) = total active power (W), \(W'\) = first position reading (W), \(W''\) = second position reading (W)

Example. A wattmeter connected in line 1 of a \(230\ \text{V}\) 3-phase system reads \(5760\ \text{W}\) when connected to line 2 and \(3380\ \text{W}\) when shifted to line 3. Calculate the line current.

Solution.

\[ \theta = \tan^{-1}\left[\sqrt{3}\left(\dfrac{5760 - 3380}{5760 + 3380}\right)\right] = 24.276^\circ \]
\[ I_L = \dfrac{5760 + 3380}{\sqrt{3}(230)\cos 24.276^\circ} = 25.168\ \text{A} \]

Answer: \(25.168\ \text{A}\)

2.3 Artificial Neutral Method

Artificial neutral method creates a star point \(N\) using three equal non-inductive resistors \(R\) connected across lines.

  • Star point potential equals true load neutral potential
  • Potential coil connects between line 1 and artificial neutral \(N\)
  • Total power equals three times wattmeter reading

Artificial Neutral Method

\[ P_{\text{total}} = 3 W \]

Where: \(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)

Example. An artificial neutral is formed using \(88.8\ \Omega\) resistors across a \(400\ \text{V}\) 3-phase line. If a wattmeter reads \(900\ \text{W}\), determine line current.

Solution.

\[ I_1 = \dfrac{W}{V_{13}\cos 30^\circ} = \dfrac{900}{400\cos 30^\circ} = 2.6\ \text{A} \]

Answer: \(2.6\ \text{A}\)

2.4 T-Method

T-method connects potential coil between line 1 and junction of two equal resistors \(R\) placed across lines 2 and 3.

  • Creates midpoint potential reference between lines 2 and 3
  • Wattmeter registers exactly half of total 3-phase power
  • Total active power equals twice wattmeter reading

T-Method Power Formula

\[ P_{\text{total}} = 2 W \]

Where: \(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)

Example. In the T-method of measuring 3-phase power, a \(7.5\ \text{hp}\) motor operating at \(88\%\) efficiency is supplied by a 3-phase line. Calculate the wattmeter reading.

Solution.

\[ P_{\text{in}} = \dfrac{7.5(0.746)}{0.88} = 6.358\ \text{kW} \]
\[ W = \dfrac{P_{\text{in}}}{2} = \dfrac{6.358}{2} = 3.179\ \text{kW} \]

Answer: \(3.179\ \text{kW}\)

2.5 CT Method

CT method utilizes two current transformers with secondaries paralleled in difference configuration through one wattmeter current coil.

  • Current coil carries geometric difference of CT currents
  • Single wattmeter reads total 3-phase power directly
  • Suitable for high-voltage and high-current power measurement

CT Method Power Formula

\[ P_{\text{total}} = W \]

Where: \(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)

3. n-Phase Power Measurement Using n Number of Wattmeters

n-wattmeter method measures active power in an \(n\)-phase \(n\)-wire system by placing one wattmeter current coil in each phase line.

  • All potential coil leads join at common arbitrary junction \(O\)
  • Valid for both balanced and unbalanced polyphase loads
  • Total power equals algebraic sum of all \(n\) readings

n-Phase Power with n Wattmeters

\[ P_{\text{total}} = W_1 + W_2 + \dots + W_n \]

Where: \(P_{\text{total}}\) = total active power (W), \(W_1, W_2, W_n\) = individual wattmeter readings (W)

EE Board Exam April 1993

Example. An unbalanced 3-phase load has Phase I at \(120\ \text{V}, 100\ \text{A}, 1.0\ pf\), Phase II at \(100\ \text{V}, 230\ \text{A}, 0.8\ pf\), and Phase III at \(110\ \text{V}, 85\ \text{A}, 0.77\ pf\). Calculate total power.

Solution.

\[ P_{\text{total}} = 120(100)(1) + 100(230)(0.8) + 110(85)(0.77) = 37599.5\ \text{W} = 37.6\ \text{kW} \]

Answer: \(37.6\ \text{kW}\)

4. n-Phase Power Measurement Using n-1 Number of Wattmeters

Blondel’s Theorem states that total power in any \(n\)-wire system is measured using \(n-1\) wattmeters by placing common reference point \(O\) on one conductor.

  • Named after French engineer André Blondel
  • Potential coil voltage on reference line becomes zero
  • Eliminates one wattmeter without affecting total power accuracy

Blondel’s Theorem (n-1 Wattmeters)

\[ P_{\text{total}} = W_1 + W_2 + \dots + W_{n-1} \]

Where: \(P_{\text{total}}\) = total active power (W), \(W_1, W_2\) = readings of remaining \((n-1)\) wattmeters (W)

Example. Two wattmeters connected using line c as common potential point measure an unbalanced load. If \(W_1 = 4140(60)\cos 30^\circ\ \text{W}\) and \(W_2 = 4140(78.4)\cos 26^\circ\ \text{W}\), find total power.

Solution.

\[ W_1 = 4140(60)\cos 30^\circ = 215.12\ \text{kW} \qquad\qquad W_2 = 4140(78.4)\cos 26^\circ = 291.73\ \text{kW} \]
\[ P_{\text{total}} = W_1 + W_2 = 215.12 + 291.73 = 506.85\ \text{kW} \]

Answer: \(506.85\ \text{kW}\)

5. Two-Wattmeter Reading When Load Is Balanced

Two-wattmeter method measures active power in 3-wire 3-phase circuits using two wattmeters connected across lines.

  • \(W_1\) reads \(V_L I_L \cos(30^\circ - \theta)\)
  • \(W_2\) reads \(V_L I_L \cos(30^\circ + \theta)\)
  • Total active power \(P = W_1 + W_2\)
  • Total reactive power \(Q = \sqrt{3}(W_1 - W_2)\)

Two-Wattmeter Readings (Balanced Load)

\[ W_1 = V_L I_L \cos(30^\circ - \theta) \]
\[ W_2 = V_L I_L \cos(30^\circ + \theta) \]
\[ P_{\text{total}} = W_1 + W_2 \qquad\qquad Q_{\text{total}} = \sqrt{3}(W_1 - W_2) \]

Where: \(W_1, W_2\) = wattmeter readings (W), \(V_L\) = line voltage (V), \(I_L\) = line current (A), \(\theta\) = load power factor angle (\(^\circ\)), \(P_{\text{total}}\) = total active power (W), \(Q_{\text{total}}\) = total reactive power (VAR)

EE Board Exam April 1984

Example. A balanced 3-phase load draws \(75\ \text{A}\) line current at \(230\ \text{V}\) line-to-line at \(0.848\) lagging power factor. Find two-wattmeter readings.

Solution.

\[ \theta = \cos^{-1}(0.848) = 32^\circ \]
\[ W_1 = (230)(75)\cos(30^\circ - 32^\circ) = 17.24\ \text{kW} \qquad\qquad W_2 = (230)(75)\cos(30^\circ + 32^\circ) = 8.10\ \text{kW} \]

Answer: \(17.24\ \text{kW}, 8.10\ \text{kW}\)

6. Power Factor Computation Using Two-Wattmeter

Power factor angle \(\theta\) is determined directly from the ratio or difference of two wattmeter readings \(W_1\) and \(W_2\).

  • At \(pf = 1.0\), \(W_1 = W_2\)
  • At \(pf = 0.5\), \(W_2 = 0\)
  • At \(pf < 0.5\), \(W_2\) is negative (current coil reversed)

Power Factor Angle Formula

\[ \tan\theta = \sqrt{3}\left(\dfrac{W_1 - W_2}{W_1 + W_2}\right) \]
\[ \theta = \tan^{-1}\left[\sqrt{3}\left(\dfrac{W_1 - W_2}{W_1 + W_2}\right)\right] \]
\[ pf = \cos\theta \]

Where: \(\theta\) = power factor angle (\(^\circ\)), \(W_1, W_2\) = wattmeter readings (W), \(pf\) = load power factor

Example. Two wattmeters read \(4800\ \text{W}\) and \(400\ \text{W}\) after reversing its current coil on a 3-phase load. Solve for load power factor.

Solution.

\[ \theta = \tan^{-1}\left[\sqrt{3}\left(\dfrac{4800 - (-400)}{4800 + (-400)}\right)\right] = 63.963^\circ \]
\[ pf = \cos(63.963^\circ) = 0.439 \]

Answer: \(0.439\)

7. Formula Summary

Formula NameFormulaVariables
Wattmeter Power Reading\(W = V I \cos\theta\)\(W\) = power (W), \(V\) = voltage (V), \(I\) = current (A), \(\theta\) = phase angle (\(^\circ\))
Single Wattmeter with Neutral Available\(P_{\text{total}} = 3 W\)\(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)
Potential-Lead Shift Method\(P_{\text{total}} = W' + W''\)\(P_{\text{total}}\) = total active power (W), \(W', W''\) = readings (W)
Artificial Neutral Method\(P_{\text{total}} = 3 W\)\(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)
T-Method Power Formula\(P_{\text{total}} = 2 W\)\(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)
CT Method Power Formula\(P_{\text{total}} = W\)\(P_{\text{total}}\) = total active power (W), \(W\) = wattmeter reading (W)
n-Phase Power with n Wattmeters\(P_{\text{total}} = W_1 + W_2 + \dots + W_n\)\(P_{\text{total}}\) = total active power (W), \(W_1, W_2, W_n\) = readings (W)
Blondel’s Theorem (n-1 Wattmeters)\(P_{\text{total}} = W_1 + W_2 + \dots + W_{n-1}\)\(P_{\text{total}}\) = total active power (W), \(W_1, W_2\) = readings (W)
Two-Wattmeter Readings (Balanced Load)\(W_1 = V_L I_L \cos(30^\circ - \theta)\)
\(W_2 = V_L I_L \cos(30^\circ + \theta)\)
\(P_{\text{total}} = W_1 + W_2\)
\(Q_{\text{total}} = \sqrt{3}(W_1 - W_2)\)		\(W_1, W_2\) = wattmeter readings (W), \(V_L\) = line voltage (V), \(I_L\) = line current (A), \(\theta\) = power factor angle (\(^\circ\)), \(P_{\text{total}}\) = active power (W), \(Q_{\text{total}}\) = reactive power (VAR)
Power Factor Angle Formula\(\tan\theta = \sqrt{3}\left(\dfrac{W_1 - W_2}{W_1 + W_2}\right)\)
\(\theta = \tan^{-1}\left[\sqrt{3}\left(\dfrac{W_1 - W_2}{W_1 + W_2}\right)\right]\)
\(pf = \cos\theta\)		\(\theta\) = power factor angle (\(^\circ\)), \(W_1, W_2\) = wattmeter readings (W), \(pf\) = power factor