Measurement of 3 Phase Reactive Power Using 2 Wattmeters MCQ Quiz - Objective Question with Answer for Measurement of 3 Phase Reactive Power Using 2 Wattmeters - Download Free PDF

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 be employed to measure the power in a 3 phase, three-wire star, or delta connected or balanced or unbalanced load.

In the wattmeter method, the current coils of the wattmeter are connected with any two lines, say R and Y, and the potential coil of each wattmeter is joined on the same line, the third line i.e. B as shown in below figure.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-\phi)\)

\(W_2=V_LI_Lcos(30+\phi)\)

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

Calculation

Given, W₁ = 2400W

W₂ = 1260 W

W = W₁ + W₂

W = 2400 + 1260

W = 3660 W

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

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

Explanation:

In the two-wattmeter method:

• If one wattmeter shows a negative reading, it means the power factor of the load is less than 0.5.
• When the potential terminals of the second wattmeter are interchanged, it shows a positive value, giving the correct magnitude of the power it measures.

Calculation:

The total Power in a three-phase load using two wattmeter method is: P_total = W₁ + W₂
Where: W₁ = 5000 W, W₂ = 1000 W (after correcting the reverse deflection by interchanging potential terminals)
P_total = 5000 + 1000 = 6000 W

Concept:

Power measured by 1- ϕ  wattmeter is given by P = V_pcI_cc cos ϕ 

Where 

V_pc = Voltage of potential coil

I_cc = Current through the current coil

Application:

Here, V_pc = 300 V , I_cc = 5 A , P = 300 W (Full Scale)

300 = V_pc I_cc cos ϕ

300 = 300 x 5 cos ϕ 

cos ϕ  = 0.2

 Important Points

Dynamometer type wattmeter gives more error compared to LPF wattmeter for measuring low power factor. So, the LPF wattmeter is used here for better accuracy.

Concept:

The power triangle is as shown below.

P = Active power (or) Real power in W = Vrms Irms cos ϕ

Q = Reactive power in VAR = Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

S = P + jQ

\(S = \sqrt {{P^2} + {Q^2}}\)

ϕ is the phase difference between the voltage and current

Power factor \(\cos \phi = \frac{P}{S}\)

Calculation:

Given that,

Apparent power (S) = 50 kVA

Reading of wattmeter = Active power (P) = 20 kW

\( \Rightarrow 50 = \sqrt {{{20}^2} + {Q^2}} \)

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)

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

Calculation:

Given that,

\({W_2} = \frac{{{W_1}}}{2}\)

We know that,

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

\( = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - \frac{{{W_1}}}{2}} \right)}}{{\left( {{W_1} + \frac{{{W_1}}}{2}} \right)}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {2{W_1} - {W_1}} \right)}}{{2{W_1} + {W_1}}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 {W_1}}}{{3{W_1}}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ \)

Two Wattmeter Method of Power Measurement

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

In the wattmeter method, the current coils of the wattmeter are connected with any two lines, say R and Y, and the potential coil of each wattmeter is joined on the same line, the third line i.e. B as shown in below figure.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-\phi)\)

\(W_2=V_LI_Lcos(30+\phi)\)

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

Calculation

Given, W₁ = 2400W

W₂ = 1260 W

W = W₁ + W₂

W = 2400 + 1260

W = 3660 W

Given that load is pure inductive – Hence it is zero power factor (lag)

Cos ϕ = 0

⇒ ϕ = 90°

We know that,

\(\begin{array}{l} \phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\\ \Rightarrow 90^\circ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\\ \Rightarrow \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}} = \tan 90^\circ = \infty \end{array}\)

⇒ W₁ + W₂ = 0

Concept:

Power measured by 1- ϕ  wattmeter is given by P = V_pcI_cc cos ϕ 

Where 

V_pc = Voltage of potential coil

I_cc = Current through the current coil

Application:

Here, V_pc = 300 V , I_cc = 5 A , P = 300 W (Full Scale)

300 = V_pc I_cc cos ϕ

300 = 300 x 5 cos ϕ 

cos ϕ  = 0.2

 Important Points

Dynamometer type wattmeter gives more error compared to LPF wattmeter for measuring low power factor. So, the LPF wattmeter is used here for better accuracy.

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.

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)

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

Calculation:

Given that,

\({W_2} = \frac{{{W_1}}}{2}\)

We know that,

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

\( = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - \frac{{{W_1}}}{2}} \right)}}{{\left( {{W_1} + \frac{{{W_1}}}{2}} \right)}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {2{W_1} - {W_1}} \right)}}{{2{W_1} + {W_1}}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 {W_1}}}{{3{W_1}}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ \)

Concept:

The power triangle is as shown below.

P = Active power (or) Real power in W = Vrms Irms cos ϕ

Q = Reactive power in VAR = Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

S = P + jQ

\(S = \sqrt {{P^2} + {Q^2}}\)

ϕ is the phase difference between the voltage and current

Power factor \(\cos \phi = \frac{P}{S}\)

Calculation:

Given that,

Apparent power (S) = 50 kVA

Reading of wattmeter = Active power (P) = 20 kW

\( \Rightarrow 50 = \sqrt {{{20}^2} + {Q^2}} \)

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

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

Calculation:

Given that, line voltage (VL) = 500 V

Cos ϕ = 0.5

Cos ϕ = 0.5 ⇒ ϕ = 60°

W₁ + W₂ = 36 kW

\(\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}\)

\(\Rightarrow \tan 60 = \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}\)

⇒ W₁ – W₂ = 36

Concept:

In a two-wattmeter method,

Total power (P) = P₁ + P₂

The power factor of the system = cos ϕ

\(\tan \phi = \sqrt 3 \frac{{[{P_1} - {P_2}]}}{{\left[ {{P_1} + {P_2}} \right]}}\)

Where P₁ & P₂ are power of wattmeters

Calculation:

P = 30 W, cos ϕ = 0.4

If P₁ and P₂ are two individual wattmeter readings then according to the problem,

P = P₁ + P₂ = 30 kW

Power factor angle ϕ = cos^-1 (0.4) = 66.4°

So, \(\phi = {\tan ^{ - 1}}\left( {\sqrt 3 \frac{{\left( {{P_1} - {P_2}} \right)}}{{\left( {{P_1} + {P_2}} \right)}}} \right)\)

\(66.4^\circ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{P_1} - {P_2}} \right)}}{{30}}} \right)\)

\(\frac{{2.288\; \times \;30}}{{\sqrt 3 }} = {P_1} - {P_2}\)  

39.69 kW = P₁ – P₂

Two Wattmeter Method of Power Measurement

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

In the wattmeter method, the current coils of the wattmeter are connected with any two lines, say R and Y, and the potential coil of each wattmeter is joined on the same line, the third line i.e. B as shown in below figure.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-\phi)\)

\(W_2=V_LI_Lcos(30+\phi)\)

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

Calculation

Given, W₁ = 2400W

W₂ = 1260 W

W = W₁ + W₂

W = 2400 + 1260

W = 3660 W