Q1) Differentiate between: Feeder & Distributor ?
Feeder Distributor
The line conduction, which transmits electrical The line driver, which transmits the electric power
power from the substation to the distributor, is from the substation to the service mains conductor,
called a feeder or feeder. is called a distributor.
It is a link (line driver) between the substation and It is a link (line driver) between the supplier and the
the distribution. service mains.
Its design is based on the current-carrying capacity. Its design is based on voltage.
The length of the feeder is usually longer than that The length of the distributor is usually longer than
of the distributor. that of the service main.
The voltage of the feeder is usually higher than that The voltage of the distributor is higher than that of
of the distributor. the service mains.
Tapping is not done by this. This enables fractional removal for the consumers.
Q2) With a neat phasor diagram explain calculation of voltage drop in AC distributors referred to the
receiving end voltage ?
Ans) Calculating voltage drop in AC distributors referred to the
receiving end voltage involves these steps:
1. Gather information: Know the source voltage (Vs), load current
(I), and line impedance (R + jX).
2. Calculate impedance drop: ΔV_impedance = I * Z (where Z = R + jX).
3. Determine power factor (PF).
4. Split impedance into real (R) and reactive (X) components.
5. Calculate real and reactive component losses: ΔV_real = I * R * PF and ΔV_reactive = I * X * √(1 - PF^2).
6. Find total voltage drop: ΔV_total = ΔV_impedance + ΔV_real + ΔV_reactive.
7. Calculate receiving end voltage (Vr): Vr = Vs - ΔV_total.
Ensure Vr remains within acceptable limits for proper load operation.
Q3) Derive equation for volume of conductor required for D.C. 2 wire system with midpoint earthed
with diagram ?
Ans) The equation for the volume of conductor required for a DC 2-
wire system with a midpoint earthed can be derived as follows:
1. Calculate the resistance of the conductor needed to meet the
allowable voltage drop (R_required) based on your system's specifications.
2. Use the formula for resistance: **R = ρ * (L / A)**, where ρ is the
resistivity of the conductor material, L is the length of the conductor,
and A is the cross-sectional area.
3. Rearrange the resistance formula to solve for cross-sectional area
(A): **A = ρ * (L / R_required)**.
4. The volume of conductor required (V_conductor) is the product of the cross-sectional area (A) and the
length (L): **V_conductor = A * L**.>> 5. Substitute the expression for A from step 3 into the equation for
V_conductor: **V_conductor = (ρ * L^2) / R_required**. >> So, the equation for the volume of conductor
required for a DC 2-wire system with a midpoint earthed is: >>**V_conductor = (ρ * L^2) / R_required**
Where:>> - V_conductor is the volume of conductor required (in cubic meters or other appropriate units).
- ρ is the resistivity of the conductor material (in ohm-meter).
- L is the total length of the conductor (in meters).
- R_required is the resistance of the conductor needed to meet the allowable voltage drop (in ohms).
Q4) Explain how Kelvins Law is helpful in deciding the most economical cross section of a conductor
with its limitations ?
Ans) Kelvin's Law, also known as the "Square-Cube Law," is a principle used in electrical engineering to
determine the most economical cross-sectional area of a conductor for a given electrical distribution
system. It helps optimize conductor size while considering both the electrical losses (I^2R losses) and the
cost of the conductor material. 1) Minimizing Conductor Cost: Kelvin's Law states that the electrical
losses in a conductor are proportional to its resistance (R) and inversely proportional to its cross-sectional
area (A). 2) Optimizing Cost vs. Loss Trade-off: By using a larger conductor with lower resistance, you can
reduce the electrical losses in the system. However, this comes at the cost of using more conductor
material, which can be expensive. 3) Meeting Voltage Drop Requirements: Kelvin's Law is particularly
useful in ensuring that the voltage drop in the distribution system remains within acceptable limits. By
choosing the right conductor size, you can control voltage drop and maintain the voltage levels required
for proper equipment operation. Limitations of Kelvin's Law:
1) Linear Assumption: Kelvin's Law assumes a linear relationship between resistance (R) and cross-
sectional area (A). In reality, conductor materials have nonlinear characteristics at high temperatures,
which can affect the accuracy of the calculations. 2) Operating Conditions: The law assumes constant
current and temperature conditions. In practical systems, current may vary, and temperature changes can
occur due to factors like ambient temperature, load fluctuations, and other environmental conditions.
3) Material Costs: While Kelvin's Law helps optimize conductor size for electrical losses, it doesn't account
for fluctuations in material costs. The cost of copper or aluminum, which are commonly used as
conductors, can vary over time, impacting the economic considerations.
Q5) Explain interconnected supply system with neat diagram ?
Ans) An interconnected supply system, or grid, is a large network of
power generation sources, transmission lines, and distribution
systems that work together to deliver electricity over a wide area.
It ensures reliability by sharing resources, enhances efficiency, and
can span multiple regions or countries. Challenges include
complexity and security concerns, but it forms the backbone of
modern electrical power distribution.
1) Multiple Generation Sources: An interconnected supply system
typically incorporates various types of power generation sources,
including power plants such as fossil fuel-based, nuclear, hydro
electric, wind, and solar farms. 2) Transmission Network: The
system includes an extensive network of high-voltage transmission lines that transport electricity from
power generation facilities to various load centers. 3) Load Centers: Within the interconnected supply
system, there are areas known as load centers, where electricity demand is concentrated.
4) Distribution Network: At the local level, the transmission lines are connected to distribution networks
that serve homes, businesses, and industries.
Q6) State & explain factors affecting soil resistivity ?
Ans) Soil resistivity is a critical parameter in electrical engineering, especially in grounding and earthing
systems. It measures the resistance of the soil to the flow of electrical current. Several factors affect soil
resistivity: 1. Soil Type and Composition: - Different types of soil have varying resistivity values. For
example, clay-rich soils tend to have lower resistivity, while sandy soils have higher resistivity.
2. Moisture Content: - Soil resistivity is highly dependent on moisture content. Wet or saturated soils
have lower resistivity, while dry soils have higher resistivity. - Seasonal changes in moisture levels can
lead to variations in soil resistivity, which must be considered in grounding system design.
3. Temperature: - Soil temperature affects resistivity. As temperature increases, resistivity generally
decreases, and vice versa.- Seasonal temperature variations can impact soil resistivity, so measurements
should be taken under representative conditions. 4. Salinity and Chemicals: - The presence of salts and
certain chemicals in the soil can significantly influence resistivity. These substances increase the soil's
conductivity, reducing resistivity. - Soil pollution or contamination can alter its resistivity as well.
5. Pressure and Compaction: - The pressure on soil and its compaction level can affect resistivity.
Compacted soil typically has higher resistivity compared to loose, less compacted soil.
6. Depth: - Soil resistivity tends to decrease with depth because deeper layers often have higher
moisture content, which lowers resistivity. - Grounding systems that extend deeper into the ground can
achieve lower resistance values.
Q7) State the types of bus bar systems & explain duplicate bus bar system with diagram.
Ans) There are two main types of bus bar systems:
1. Isolated Bus Bar System: In this system, each bus bar is electrically isolated from the others, allowing
for individual control and protection. It is commonly used in critical applications where fault isolation is
crucial. 2. Non-Isolated Bus Bar System: In this system, multiple bus bars are connected electrically, and
they share the same voltage potential. It is used in less critical applications and is more cost-effective but
lacks fault isolation capabilities.
Duplicate bus bar system - A duplicate bus bar system is a redundancy arrangement in electrical
substations or distribution systems. It involves the duplication of critical components, such as bus bars, to
ensure continued operation in case of a fault or maintenance. If one set of bus bars or components fails,
the duplicate set takes over, minimizing downtime and ensuring uninterrupted power supply. This
redundancy enhances system reliability but increases construction and maintenance costs.
Q8) As per IS 3043 draw and explain plate electrode system of earthing ?
Ans) A plate electrode system of earthing involves
burying a metal plate (usually copper or aluminum) in
the ground. This buried plate serves as an electrode to
provide a low-resistance path for electrical current to
dissipate safely into the earth.
1) Plate Electrode: A metal plate, typically made of
Materials like copper or aluminum, is buried vertically
in the ground. This plate acts as the grounding electrode,
providing a pathway for electrical current to safely
dissipate into the earth. 2) Installation: The plate is
buried in a trench, with a portion exposed above the
ground. The exposed portion is usually connected to
the elect rical system that requires grounding.
3) Soil Contact: The effectiveness of the plate
electrode system depends on the quality of contact
between the metal plate and the surrounding soil.
Good contact ensures a low- resistance pathway for
electrical current to flow into the ground.
4) Size Matters: The size of the plate electrode is
crucial. Larger plates provide more surface area
for contact with the soil, resulting in lower resistance
and better grounding performance.
5) Low Resistance Path: The primary purpose of the
plate electrode system is to establish a low-resistance
path for fault currents, such as those from electrical faults or lightning strikes.
Q9) Explain steps in designing of an earthing grid of substation with reference to IEEE standard ?
Ans) Designing an earthing grid for a substation is a critical aspect of ensuring electrical safety and system
reliability. The design process often follows IEEE Standard 80, which provides guidelines for grounding in
substations and other facilities. Here are the key steps in designing an earthing grid for a substation with
reference to IEEE Standard 80.
1) Gather Site Data: Collect information on substation layout, soil resistivity, fault currents, and safety
requirements. 2) Define Objectives: Clearly state grounding goals, such as personnel safety and
equipment protection. 3) Calculate Fault Currents: Determine maximum fault currents to size
conductors. 4) Measure Soil Resistivity: Measure soil resistivity at multiple points on the site.
5) Determine Design Parameters: Define permissible voltages, fault clearing time, and ground-fault
current path impedance. 6) Select Grounding Electrodes: Choose appropriate electrodes based on soil
resistivity and performance criteria. 7) Calculate Grid Dimensions: Determine grid layout to meet
grounding resistance and minimize voltages.
Q10) Explain the terms touch potential and step potential with relevant equivalent circuits ?
Ans) Touch potential refers to the voltage difference or electrical
potential between an energized object or component and a
grounded surface or object that a person may simultaneously
touch. In electrical systems, when a fault or malfunction occurs,
it can create hazardous conditions where the touch potential
can expose individuals to electric shock. Touch potential is a
critical consideration in electrical safety and grounding design
to minimize the risk of electrical accidents.
Step potential is a term used in electrical safety to describe a
potential difference that can occur between a person's feet
when they are in contact with the ground in the vicinity of an
electrical fault. It is an important consideration in situations
where electrical faults can lead to electrical shock hazards, particularly in high-voltage environments. Step
potential is essential to understand because it helps assess the risk to individuals who may need to move
away from a faulted area safely.
Q11) State the symbols used & function of the following equipments:
i) Lightening arrester - A lightning arrester, also known as a surge arrester
or lightning rod, is a protective device designed to divert and safely
dissipate the high voltage and current associated with lightning strikes,
thereby protecting electrical systems and buildings from damage due to lightning-induced surges.
ii) Isolator with earth blade - An isolator with an earth blade, commonly known as a disconnect switch
/with an earth or grounding blade, is an electrical switch used to physically disconnect electrical
equipment or circuits from power sources.
iii) Air blast CB with current tripping - An air blast circuit breaker (CB) with current tripping is an electrical
switching device that uses a high-pressure air blast to extinguish electrical arcs during circuit interruption.
Current tripping means that it responds to excessive current levels to open the circuit.
iv) Three phase transformer with no load tap changer- A three-phase transformer with no-load tap
changer is an electrical device that transfers electrical energy between three-phase circuits without a tap
changer feature for adjusting the voltage ratio under no-load conditions.
v) Current Transformer -: A current transformer is an instrument transformer used to step down high
currents in power systems to a level suitable for measurement and protection purposes. It provides a
proportional secondary current (typically 5A) when a primary current flows through it.
vi) Potential Transformer- A potential transformer is an instrument transformer used to step down high
voltages in power systems to a level suitable for measurement and protection. It provides a proportional
secondary voltage (typically 120V or 240V) when a primary voltage is applied across its terminals,
