Measurement of Poly Phase Power MCQ Quiz - Objective Question with Answer for Measurement of Poly Phase Power - Download Free PDF

Explanation:

Power Measurement in a 3-Phase Circuit Using Two Wattmeters

Definition: The two-wattmeter method is a commonly used technique for measuring power in a 3-phase circuit. This method requires two wattmeters to be connected to two of the three lines in a 3-phase system. The readings of these wattmeters give the total power in the circuit. The method is especially useful for both balanced and unbalanced loads.

Working Principle: In a 3-phase circuit, the total power can be calculated as the algebraic sum of the readings of the two wattmeters:

Total Power (P) = W₁ + W₂

Where W₁ and W₂ are the readings of the two wattmeters. Depending on the power factor of the circuit, the readings of the wattmeters can be positive, negative, or even zero. The phase angle (φ) between the line voltage and current determines the power factor, and this in turn affects the wattmeter readings.

Given Condition:

In the question, one of the wattmeter readings is positive, and the other is negative. The magnitudes of the readings are different. This condition indicates that the power factor of the circuit is less than 0.5 (lagging).

Analysis:

Let’s analyze the readings of the two wattmeters for various power factors:

• For a power factor of 1 (unity), both wattmeter readings will be positive and equal, as the angle between voltage and current is 0°.
• For a power factor between 0.5 and 1 (lagging), both wattmeter readings will still be positive but unequal. The difference in their magnitudes increases as the power factor decreases.
• For a power factor of 0.5 (lagging), one wattmeter reading becomes zero. This happens because the phase angle φ is 60°, leading to the wattmeter connected to a specific phase showing zero reading.
• For a power factor less than 0.5 (lagging), one wattmeter reading becomes negative. The negative reading occurs because the phase angle φ exceeds 60°, causing the current direction relative to the voltage to result in a negative product in one wattmeter.

In the given scenario, one wattmeter shows a positive reading, and the other shows a negative reading, with different magnitudes. This is a clear indication that the power factor of the circuit is less than 0.5 (lagging).

Conclusion:

The correct answer is:

Option 4: Less than 0.5 (lagging)

This conclusion is based on the characteristic behavior of wattmeters in the two-wattmeter method when the power factor of the circuit is less than 0.5. The negative reading of one wattmeter, combined with the positive reading of the other, confirms this condition.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Unity

If the power factor were unity, both wattmeter readings would be positive and equal. This does not match the given condition where one wattmeter reading is negative, and the magnitudes are different.

Option 2: 2 (lagging)

A power factor of 2 (lagging) is not possible in any electrical circuit, as power factor is defined as the cosine of the phase angle (cos φ), which always lies between -1 and 1. Hence, this option is invalid.

Option 3: 5 (lagging)

Similar to Option 2, a power factor of 5 (lagging) is not physically possible, as power factor values cannot exceed 1. Therefore, this option is also invalid.

Option 4: Less than 0.5 (lagging)

This is the correct answer, as explained above. The negative reading of one wattmeter and the positive reading of the other, with different magnitudes, indicate a power factor less than 0.5 (lagging).

Conclusion:

Understanding the behavior of wattmeters in the two-wattmeter method is essential for determining the power factor of a 3-phase circuit. The key takeaway is that when one wattmeter reading is negative and the other is positive, with different magnitudes, the power factor is less than 0.5 (lagging). This knowledge is crucial for accurately analyzing and diagnosing power systems in electrical engineering applications.

Explanation:

Measurement of Three-Phase Power using the Two Wattmeter Method

Introduction: The two wattmeter method is a widely used technique for measuring the total power in a three-phase system. It is particularly suitable for balanced and unbalanced loads. The method involves the use of two wattmeters connected to the system, and the readings of these wattmeters help determine the total power and the power factor of the system.

Condition: When the two wattmeter readings are equal, the power factor of the circuit is unity (1). This occurs because the phase angle between the line voltage and current in the system is zero, indicating a purely resistive load. Let us delve deeper into the working principle and analysis to understand why this happens.

Working Principle:

In the two wattmeter method:

• Wattmeter 1 measures the power in one phase (P1).
• Wattmeter 2 measures the power in another phase (P2).
• The total power in the system is given by the sum of the readings of the two wattmeters: P = P1 + P2.

For a three-phase system with a balanced load, the power factor can be determined using the formula:

Power Factor (cos φ) = (P1 - P2) / (P1 + P2)

Analysis for Unity Power Factor:

When the power factor is unity (1), it implies that the load is purely resistive. In this case, the current and voltage are in phase, and the phase angle φ is zero. Substituting this condition into the formula:

• P1 = P2 (since both wattmeter readings are equal for a purely resistive load).
• cos φ = (P1 - P2) / (P1 + P2).
• cos φ = (P1 - P1) / (P1 + P1) = 0 / 2P1 = 0.

Thus, the power factor of the circuit is unity (cos φ = 1) when the two wattmeter readings are equal. This is a unique condition that arises due to the absence of any phase difference between the voltage and current.

Advantages of Two Wattmeter Method:

• Applicable for both balanced and unbalanced loads.
• Simple and effective for three-phase power measurement.
• Provides a direct way to measure power factor.

Correct Option Analysis:

The correct option is:

Option 2: Unity

This option is correct because the power factor of the circuit is unity when the two wattmeter readings are equal. The condition of equal readings arises when the load is purely resistive, and the current and voltage are in phase.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 0.8 lag

This option is incorrect. When the power factor is 0.8 lag, the phase angle is not zero, and the wattmeter readings will not be equal. Instead, one wattmeter will record a higher value than the other, depending on the phase angle of the load.

Option 3: 0.8 lead

Similar to option 1, this option is also incorrect. A power factor of 0.8 lead indicates a capacitive load, where the current leads the voltage. In this case, the wattmeter readings will again be unequal, and the condition of equal readings will not be satisfied.

Option 4: Zero

This option is incorrect because a power factor of zero implies a completely reactive load (either inductive or capacitive). In such a scenario, the current and voltage are 90° out of phase, and the two wattmeter readings will not be equal. Instead, one wattmeter may show a negative reading depending on the connection.

Conclusion:

The two wattmeter method is a reliable technique for measuring three-phase power and determining the power factor of a circuit. When the two wattmeter readings are equal, it indicates that the power factor of the circuit is unity, corresponding to a purely resistive load. This condition is unique and helps distinguish purely resistive loads from other types of loads in three-phase systems.

Concept

The reactive power in power measurement of a three-phase using 2 wattmeters is given by:

\(Q={\sqrt{3}(W_1-W_2 )}\)

where, W₁ and W₂ are wattmeter readings

Calculation

Given, W₁ = 2000 W

W₂ = 1000 W

\(Q=\sqrt{3}(2000-1000)\)

Q = 1000√3 VAr

Explanation:

In a three-phase system, power measurement using two wattmeters is a common method known as the two-wattmeter method. This method is widely used for measuring power in both balanced and unbalanced three-phase systems. To understand the situation where one of the wattmeters reads zero, we need to delve into the principles of this method and the relationship between the readings and the power factor of the load.

Two-Wattmeter Method:

The two-wattmeter method involves connecting two wattmeters to a three-phase system to measure power. The readings of the two wattmeters (W1 and W2) can be used to determine the total power (P) and the power factor (pf) of the system. The total power is given by the sum of the readings:

P = W1 + W2

The power factor can be determined using the difference between the readings:

pf = cos(Φ) = (W1 - W2) / (W1 + W2)

Where Φ is the phase angle between the line voltage and the line current.

Scenario Analysis:

In the scenario where one of the wattmeters reads zero, let's assume that W1 = 0. This means that the entire power is being measured by the second wattmeter (W2). Therefore, the total power (P) is:

P = W1 + W2 = 0 + W2 = W2

Now, using the formula for power factor:

pf = cos(Φ) = (W1 - W2) / (W1 + W2) = (0 - W2) / (0 + W2) = -W2 / W2 = -1

However, the power factor cannot be -1 because it represents a purely inductive load, which is not possible in this context. Therefore, we need to consider the magnitude of the power factor:

|pf| = |cos(Φ)| = |-1| = 1

This represents a power factor of unity, which means the load is purely resistive. However, this is a theoretical consideration. Practically, the power factor can be determined by the specific configuration of the load and the readings of the wattmeters.

In a balanced three-phase system, if one of the wattmeters reads zero, it indicates that the load has a power factor of 0.5. This situation arises because the angle between the voltage and the current is 60 degrees (cos(60) = 0.5). Therefore, the correct power factor in this scenario is:

Correct Option: 0.5

The correct answer is option 3, which indicates that the power factor of the three-phase load is 0.5.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Zero

If the power factor were zero, it would imply that the load is purely reactive (either inductive or capacitive), and no real power is being consumed. This would mean that both wattmeters should read zero, which is not the case here.

Option 2: Unity

A unity power factor (pf = 1) indicates a purely resistive load where the voltage and current are in phase. In such a case, both wattmeters would have equal readings, and neither would read zero. Therefore, this option is incorrect.

Option 4: 0.707

This value corresponds to a power factor angle of 45 degrees (cos(45) = 0.707). For this power factor, the readings of the two wattmeters would be different but not zero. Therefore, this option is incorrect.

Conclusion:

Understanding the two-wattmeter method and the relationship between the wattmeter readings and the power factor is crucial for accurately determining the power factor of a three-phase load. In this case, when one of the wattmeters reads zero, the power factor of the load is 0.5, which corresponds to an angle of 60 degrees between the voltage and the current. This analysis helps in correctly identifying the operating conditions of the three-phase system.

Two Wattmeter Method of Power Measurement

The Wattmeter Method can measure the power in a 3-phase, three-wire star, delta-connected, balanced, or unbalanced load.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

where, \(ϕ=tan^{-1}({\sqrt{3}(W_1-W_2)\over W_1+W_2})\)

Calculation

Given, cos ϕ = 0.5

ϕ = 60° 

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_1=V_LI_Lcos(30+60)\)

\(W_1=V_LI_Lcos(90)=0\)

The value of other wattmeter is:

\(W_2=V_LI_Lcos(30-60)\)

\(W_1=V_LI_Lcos(-30)={\sqrt{3}\over 2}W\)

Observation Table

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, W1 = 2 W2

\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)

⇒ ϕ = 30°

Power factor = cos 30° = 0.866

Important Point:

According to Blondell’s theorem, the no. of watt meters to be required to measure the total power in n-phase system is either N (or) N–1.

When a separate neutral wire is available in the system the no. of watt meters to be required is N.

When the neutral wire is not available in the system, then the no. of watt meters to be required is (N–1).

Power is measured according to Blondel's theorem

One line is acting as a common line for the return path. Hence the minimum number of wattmeters required is 2.

Additional Information

Measurement of polyphase power: 3 - ϕ

1) For 3ϕ – 4 wire → neutral is present

2) As per Blondel’s theorem, if neutral is not present for the ‘n’ wire system, then the ‘n-1’ number of wattmeter’s are required for the ‘n’ wire system.

Concept:

Wattmeter reads the power and it is given by

P = VPC ICC cos ϕ

VPC is the voltage across pressure coil

ICC is current flows through the current coil

ϕ is the phase angle between VPC and ICC

Application:

The circuit representation of the given question is as shown below.

• The potential coil is connected across Z2.
• It reads the voltage across Z2 only.
• So, Wattmeter reads only power consumed by Z2.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q=√{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{√{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, W1 = 157.74 W

W2 = 100 W

W₁ - W₂ = 57.74 W

Using above formula,

Reactive Power (Q) = √3 × 57.74 = 100 VAr

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, 

Reversing the connections

W1 = 4 kW, W2 = - 2 kW

Total reactive power (Q) = √3[4 - (- 2)] = 6 √3 kVAR

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{4} - {(-2)}} \right)}}{{{4} - {2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{6\sqrt 3}}{{{2}}}} \right)\)

ϕ = 79.10^o

cosϕ = 0.2 (approx)

Important Points 

Power measurement using two wattmeter method

According to Blondel's Theorem, for 'n' phase power measurement, the no. of wattmeters required is 'n-1'.

Therefore, for the 3-phase circuit, 3-1 = 2 wattmeters are required.

The total 3ϕ power is the sum of the power measured by individual wattmeters.

\(W=W_1+W_2\)

The reading of the wattmeters is given by:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

The total power is given by:

\(W_1+W_2=\sqrt{3}V_LI_Lcosϕ\)

Calculation

The value of phase angle (ϕ) is given by:

\(ϕ=tan^{-1}[{\sqrt{3}(W_1-W_2)\over W_1+W_2}]\)

If W₁ = W₂

\(ϕ=tan^{-1}(0)\)

ϕ = 0°

Important Points

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Explanation:

W₁ = -ve, W₂ = +ve

|W₁| ≠ |W₂|, then the powere factor range is given as

The range of ϕ is, 60 < ϕ < 90

Now the range of power factor is, 0.5 > cos ϕ > 0.

The correct answer is option 4):

Concept:

Two wattmeter method:

• The two-wattmeter method uses two voltage measurements referenced to the same phase (line) and the two currents flowing into that phase. The assumption is that the three-phase system is balanced, i.e., the summation of all voltages = 0 V and the summation of all currents = 0 A.
• This is true if there is no leakage current from neutral to ground.

Reading of both wattmeters are

W₁ = √3 V_pI_pcos (30 - ϕ )

W₂ = √3 V_pI_pcos (30 + ϕ)

where V_p → Voltage across pressure coil I

p → Current through coil ϕ → Phase angle

\(\mathrm{Power\, Factor} = \cos \phi = \cos\left [ \tan^{-1}\sqrt{3}\left ( \frac{w_{1}-w_{2}}{w_{1}+w_{2}} \right ) \right ]\)

• The result can be summarised as At zero power factor lagging, W1 reads the positive value and W2 reads the negative value.
• For power factors lying between 0 to 0.5 lagging, W1 reads the positive value and W2 reads the negative value.
• At 0.5 power factor lagging, W1 reads the positive value and W2 reads the zero value.Hence option 4 is correct
• For power factors lying between 0.5 to 1 lagging, W1 reads the positive value and W2 reads the positive value.
• At unity power factor lagging, W1 reads the positive value and W2 reads the positive value. 

Electrodynamic type wattmeter

• A wattmeter consists of a current and pressure coil.
• The current coil is connected in series with the load while the pressure coil is connected in parallel with the load.
• Ideally, the current in the pressure coil is in phase but practically the current lags the applied voltage.
• This is due to the presence of inductance in the pressure coil.
• If there was no inductance, the current in the pressure coil will be in phase with the applied voltage, the wattmeter will be free from the effect of power factor and frequency variations.

Concept:

Power in a three-phase circuit is,

\(P = √ 3 {V_L}{I_L}\cos \phi\)

Two wattmeter method:

The connection diagram using wattmeters as shown below.

\({W_1} = {I_R}{V_{RB}}\cos \left( {{I_R} \ ^\wedge{V_{RB}}} \right)\)

\({W_2} = {I_Y}{V_{YB}}\cos \left( {{I_Y}^\wedge{V_{YB}}} \right)\)

From the phasor diagram,

\({I_R}^\wedge{V_{RB}} = 30 - \phi \)

\({I_Y}^\wedge{V_{YB}} = 30 + \phi \)

\({W_1} = {I_R}{V_{RB}}\cos \left( {30 - \phi } \right)\)

\(\Rightarrow {W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)\)

\({W_2} = {I_Y}{V_{YB}}\cos \left( {30 + \phi } \right)\)

\(\Rightarrow {W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

\({W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - \phi } \right) + \cos \left( {30 + \phi } \right)} \right]\)

\(= √ 3 {V_L}{I_L}\cos \phi\)

⇒ Total three-phase power \( = {\rm{\;}}{{\rm{W}}_1} + {{\rm{W}}_2} = √ 3 {V_L}{I_L}\cos \phi \)

Total three-phase power is the sum of two wattmeters.

\({W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)\)

\({W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

\({W_1} - {W_2} = √ 3 {V_{ph}}{I_{ph}}\sin \phi\)

\(√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\sin \phi \)

Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)

Reactive power is equal to √3 times the difference between the readings of the two wattmeters.

\({W_1} + {W_2} = 3{V_{ph}}{I_{ph}}\cos \phi\)

\(√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\cos \phi \)

\(\Rightarrow \phi = {\tan ^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Power factor \(= cos\;\phi\)

\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)

Calculation:

Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)

= √3 (20 - 10) = 17.32 kVar