Calculation of Capacitance MCQ Quiz in తెలుగు - Objective Question with Answer for Calculation of Capacitance - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Inductance per phase per km

\(\begin{array}{l} = 0.2ln\left( {\frac{d}{{0.7788r}}} \right)mH\\ \frac{L}{{ph}} = 0.2ln\left( {\frac{{3.5}}{{0.778 \times 0.75 \times {{10}^{ - 2}}}}} \right) = 1.28mH \end{array}\)

Explanation:

Line Impedance in Transmission Lines

Definition: Line impedance is a term used in electrical engineering to describe the combination of a transmission line's resistance (R), inductance (L), and capacitance (C), which collectively influence the voltage drop along the line and its ability to transfer power efficiently. This concept is crucial in the design and analysis of power transmission systems.

Components of Line Impedance:

• Resistance (R): The opposition to the flow of current in the transmission line, caused by the inherent resistive properties of the conductor material. Resistance leads to power losses in the form of heat.
• Inductance (L): The property of the transmission line that opposes changes in current flow, causing the storage of energy in a magnetic field. Inductance can cause voltage drops and phase shifts between voltage and current.
• Capacitance (C): The ability of the transmission line to store energy in an electric field, due to the potential difference between conductors. Capacitance can influence the charging current and voltage distribution along the line.

Effects of Line Impedance:

• Voltage Drop: As current flows through the transmission line, the impedance causes a voltage drop. This drop is a function of the current and the impedance, leading to a reduction in the voltage delivered at the receiving end.
• Power Losses: The resistive component of impedance results in power losses, primarily in the form of heat. These losses reduce the efficiency of power transmission.
• Power Transfer Capability: The overall impedance affects the line's ability to transfer power efficiently. Higher impedance can limit the amount of power that can be transmitted without excessive voltage drop or power loss.

Importance in Power Transmission:

The concept of line impedance is fundamental in the design and operation of power transmission systems. Understanding and managing impedance is crucial for ensuring efficient and reliable power delivery. Engineers must consider line impedance when determining the appropriate conductor size, material, and configuration to minimize losses and maintain voltage levels within acceptable limits.

Correct Option Analysis:

The correct option is:

Option 4: Line Impedance

This option accurately describes the combination of resistance, inductance, and capacitance in a transmission line, which collectively influence the voltage drop along the line and its ability to transfer power efficiently. Line impedance is a key parameter in the analysis and design of electrical power systems.

Additional Information

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

Option 1: Voltage Level

While the voltage level is an important factor in power transmission, it does not describe the combination of resistance, inductance, and capacitance. Voltage level refers to the potential difference between conductors, which is determined by the power system design and operating requirements.

Option 2: Line Configuration

Line configuration refers to the physical arrangement of conductors and their spacing in a transmission line. While configuration can influence impedance, it does not encompass the combined effects of resistance, inductance, and capacitance.

Option 3: Line Losses

Line losses refer to the power lost during transmission due to resistance and, to a lesser extent, other factors like corona discharge. Although related to impedance, line losses do not fully describe the concept of impedance, which includes inductance and capacitance.

Conclusion:

Understanding line impedance is essential for the efficient design and operation of power transmission systems. The combination of resistance, inductance, and capacitance affects voltage drop, power losses, and the overall ability to transfer power efficiently. By accurately considering these components, engineers can optimize transmission line performance and ensure reliable power delivery.

Concept

Consider two conductors A and B, kept at a distance of 'D'.

In the given figure, C_AN and C_BN represent the line to neutral capacitance.

C_AB represents the line-to-line capacitance.

\(C_{AN}=C_{BN}=2C_{AB}\)

Calculation

Given, \(C_{AB}=1\mu F/km\)

\(C_{AN}=C_{BN}=2(1)\space \mu F/km\)

\(C_{AN}=C_{BN}=2\space \mu F/km\)

Mistake Points Since capacitance is inversely proportional to impedance.

∴ CAN and CBN connected in series are not equal rather the correct expression is:

\(C_{AN}=C_{BN}=2C_{AB}\)

\({D_{ac}} = \sqrt 3 \;D = {D_{a'c'}} = \sqrt 3 \times 5 = 5\sqrt 3 \;m\)

D_aa’ = D_cc’ = D_bb’ = 2D = 10 m

Mutual GMD in hexagonal position.

\({D_m} = {3^{\frac{1}{4}}}\;D\)

\( = 5 \times {3^{\frac{1}{4}}} = 6.58\;m\)

Self GMD in hexagonal position

\({D_s} = {\left( {2\;rD} \right)^{\frac{1}{2}}} = {\left( {0.05} \right)^{\frac{1}{2}}}\)

Phase to neutral capacitance is given by

\({C_n} = ln\frac{{{D_m}}}{{{D_S}}}\)

\({C_n} = \frac{{2\pi\epsilon }}{{ln\left( {\frac{{6.5}}{{{{\left( {0.005} \right)}^{\frac{1}{2}}}}}} \right)}}\)

\(\frac{{0.0242}}{{\log \left( {\frac{4}{r}} \right)}} = 0.01\;\mu F/km\)

\(\Rightarrow \log \left( {\frac{4}{r}} \right) = 2.42\)

⇒ r = 0.015 m

In new configuration, \({D_{eq}} = \sqrt[3]{{4 \times 4 \times 8}} = 5.04\)

\(\Rightarrow C = \frac{{0.0242}}{{\log \left( {\frac{{5.04}}{{0.015}}} \right)}} = 0.0096\;\mu F/km\)

Equivalent equilateral spacing is given by

\(d = \sqrt {{d_{12}}{d_{23}}{d_{31}}} = \sqrt[3]{{2 \times 2 \times 4}} = 2.52m\) 

\(r = \frac{{2.52}}{2} = 6.25 \times {10^{ - 3}}m\)

Line to neutral capacitance is given by,

\({C_{an}} = \frac{{2\pi {\epsilon_0}}}{{\ln \left( {\frac{d}{r}} \right)}} = \frac{{2\pi \times 8.854 \times {{10}^{ - 12}}}}{{\ln \left[ {\frac{{2.52}}{{6.25 \times {{10}^{ - 3}}}}} \right]}}\)  = 9.2728 × 10^-12 F/m

For the line length of 100 km

C_an = 9.27 × 10^-12 × 100 × 10³ = 9.2728 × 10^-7 F = 0.92728 × 10^-6 F

Charging current \({I_C} = \frac{{{V_{Ph}}}}{{{X_C}}} = \frac{{6600}}{{\sqrt 3 }} \times 2\pi fc\)

\(= \frac{{66000}}{{\sqrt 3 }} \times 2\pi \times 50 \times .92728 \times {10^{ - 6}}\) 

Reactive power demand by capacitance

Q₁ = P Tan ϕ₁

cos ϕ₁ = 0.85, ϕ₁ = 31.788

Tan ϕ₁ = 0.6197

∴ Q₁ = 150 × 0.6197 = 92.96 × 10³ VAR

For raising the P.F from 0.85 lag to unity, the capacitor bank will supply the required VAR to load

\(\therefore C' = \frac{{{Q_1}}}{{\omega {V^2}}} = \frac{{{Q_1}}}{{2\pi f \times {V^2}}} = \frac{{92.96 \times {{10}^3}}}{{2\pi \times 50 \times {{\left( {15 \times {{10}^3}} \right)}^2}}}\)

\(= \frac{{92.96 \times {{10}^3}}}{{2\pi \times 50 \times 15 \times 15 \times {{10}^6}}}\)

= 1.316 μF

Now the capacitance per phase required will be (in μF),

C = C' / 3

C = 0.438 μF

\({\rm{Mutual\;GMD}},{{\rm{D}}_{\rm{m}}} = \sqrt[3]{{8 \times 8 \times 16}} = 10.079{\rm{\;m}}\)

\({\rm{Self\;GMD}},{\rm{\;}}{{\rm{D}}_{\rm{s}}} = \sqrt { 60 \times {{10}^{ - 2}} \times \frac{5}{2} \times {{10}^{ - 2}}} = 0.1225{\rm{\;m}}\)

\({\rm{Capacitance\;per\;km}} = \frac{{{{10}^{ - 6}}}}{{18\ln \frac{{{{\rm{D}}_{\rm{m}}}}}{{{{\rm{D}}_{\rm{s}}}}}}}\)

\(= \frac{{{{10}^{ - 6}}}}{{18\ln \left( {\frac{{10.079}}{{0.1225}}} \right)}} = 0.0125 \times {10^{ - 6}}{\rm{F}}\)

The charging current per km \(= \frac{{240 \times {{10}^3}}}{{\sqrt 3 }} \times 2{\rm{\pi }} \times 50 \times 0.0125 \times {10^{ - 6}}\)

= 0.548 A

The mutual GMD of the conductor

\( = \sqrt[3]{{2 \times 2 \times 4}} = 2.519\;m\)

Radius of conductor = 0.3 cm

Capacitance per phase per meter \( = \frac{{2\pi {\epsilon_0}}}{{ln\frac{{2.519}}{{0.3 \times {{10}^{ - 2}}}}}} = 8.25pF/m\)

= 8.25 × 10-9 F/km

The charging current \( = \frac{{132 \times {{10}^3}}}{{\sqrt 3 }} \times 8.25 \times {10^{ - 9}} \times 2\pi \times 50 = 0.197\;A\)

The capacitance per phase is given by

\(\begin{array}{l} C = \frac{{2\pi\epsilon {_o}}}{{ln\frac{d}{r}}}\;F/metre\\ = \frac{{2\pi \times 8.85 \times {{10}^{ - 12}}}}{{ln\frac{d}{r}}}\\ = \frac{{{2\pi\times8.85\times{10}^{ - 12}}}}{{\;ln\;\frac{{\sqrt[3]{{6 \times 6 \times 12}}}}{{20 \times {{10}^{ - 3}}}}}}\\ = 27.48 \times {10^{ - 12}}\;F/m\;\; \end{array}\)

for 200 km, \(C = 27.48 \times {10^{ - 12}} \times 200 \times 10^3 = 5.49\;\mu F\) 

\(\omega L = \frac{1}{{3\omega C}}\)

\(\Rightarrow L = \frac{1}{{3{\omega ^2}C}} = \frac{1}{{3 \times {{\left( {2\pi \times 50} \right)}^2} \times 5.49 \times {{10}^{ - 6}}}}\)

\(\Rightarrow \;L = 0.614\;H\)

MVA rating of the suppressor coil is –

\(= \frac{{{V^2}}}{{3\omega L}} = \frac{{{{66}^2}\times10^6}}{{3 \times 2\pi \times 50 \times 0.614}} = 7.523\;MVA\)