DC Distribution MCQ Quiz - Objective Question with Answer for DC Distribution - Download Free PDF

Two-wire DC system with one conductor earthed

Max. voltage between conductors = V_m

Power to be transmitted = P

Load current, \(I_1={P\over V_m}\)

If R₁ is the resistance of each line conductor, then \(R_1={\rho l\over a_1}\)

Line losses, \(W=2I_1^2R_1=2({P\over V_m})^2({\rho l\over a_1})\)

∴ Area of X-section, \(a_1={2P^2\rho l\over WV_m^2}\)

Volume of conductor material required = 2a₁l

\((Volume)_1=2({2P^2\rho l\over WV_m^2})l={4P^2\rho l^2\over WV_m^2}=K\)

Three-wire DC system

Load current, \(I_3={P\over 2V_m}\)

Line losses, \(W=2I_1^2R_1=2({P\over 2V_m})^2({\rho l\over a_3})\)

∴ Area of X-section, \(a_3={P^2\rho l\over 2WV_m^2}\)

Assuming the area of the  X-section of neutral wire to be half that of the outer wire.

Volume of conductor material required = 2.5a3l

\((Volume)_2=2.5({P^2\rho l\over 2WV_m^2})l={2.5P^2\rho l^2\over 2WV_m^2}={5\over 16}(Volume)_1\)

A 3-wire DC distributor, compared to a 2-wire DC distributor with the same maximum voltage to earth, uses only 31.25% of the copper. 

Two-wire DC system with one conductor earthed

Case 1: If the supply voltage of a D.C. 2-wire feeder is V_m

Max. voltage between conductors = Vm

Power to be transmitted = P

Load current, \(I_1={P\over V_m}\)

If R1 is the resistance of each line conductor, then \(R_1={\rho l\over a_1}\)

Line losses, \(W=2I_1^2R_1=2({P\over V_m})^2({\rho l\over a_1})\)

∴ Area of X-section, \(a_1={2P^2\rho l\over WV_m^2}\)

The volume of conductor material required = 2a1l

\((Volume)_1=2({2P^2\rho l^2\over WV_m})l={4P^2\rho l^2\over WV_m^2}=4K\)

Case 2: If the supply voltage of a D.C. 2-wire feeder is increased 100%.

V = 2V_m

Load current, \(I_1={P\over 2V_m}\)

Line losses, \(W=2I_2^2R_2=2({P\over 2V_m})^2({\rho l\over a_2})\)

∴ Area of X-section, \(a_2={P^2\rho l\over 2WV_m^2}\)

The volume of conductor material required = 2a₂l

\((Volume)_2=2({P^2\rho l\over 2WV_m^2})l={P^2\rho l^2\over WV_m^2}=K\)

% change = \({final-initial \over initial}\times 100\)

% change = \({K-4K \over 4K}\times 100\)

The saving in copper is 75%

Explanation:

In a 3-wire DC system, maintaining balanced voltages on either side of the neutral is critical for efficient and stable operation. The device used to achieve this balance is known as a balancer set. Let’s delve deeper into the workings and importance of the balancer set and analyze why it is the correct choice among the options provided.

Balancer Set

Definition: A balancer set in a 3-wire DC system is a device designed to maintain the voltage levels on both sides of the neutral conductor equal. This is essential to ensure that the loads connected to the system receive a stable and balanced voltage, thereby preventing any potential damage or inefficiency in the electrical system.

Working Principle: The balancer set typically consists of two identical machines, usually DC generators or motors, connected in series across the supply voltage. The neutral point is taken from the junction between these two machines. When the system is balanced, the voltages on either side of the neutral are equal. If there is any imbalance due to unequal loading, the balancer set works to redistribute the voltages, ensuring that the voltage drop across each half remains equal.

Advantages:

• Ensures balanced voltages across the system, thereby enhancing the efficiency and lifespan of the electrical components connected to it.
• Helps in maintaining stable operation of the electrical system even under varying load conditions.
• Reduces the risk of electrical faults and potential damage to the system due to voltage imbalances.

Disadvantages:

• Additional complexity and cost due to the inclusion of the balancer set in the system.
• Requires regular maintenance to ensure proper operation and to avoid any potential faults.

Applications: Balancer sets are widely used in industrial and commercial settings where stable and balanced DC supply is critical. They are commonly found in systems with significant and variable loads, such as in manufacturing plants, data centers, and other facilities with sensitive electrical equipment.

Correct Option Analysis:

The correct option is:

Option 4: Balancer set

This option correctly identifies the device used in a 3-wire DC system to maintain balanced voltages on either side of the neutral. The balancer set ensures that the voltage levels remain equal, thereby promoting efficient and stable operation of the system.

Additional Information

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

Option 1: Amplifier

An amplifier is an electronic device used to increase the power, voltage, or current of a signal. While amplifiers are critical in various electrical and electronic applications, they do not serve the purpose of maintaining balanced voltages in a 3-wire DC system. Therefore, this option is incorrect.

Option 2: Booster

A booster, in electrical terms, typically refers to a device used to increase the voltage level in a circuit. Although boosters are important in certain applications to ensure adequate voltage levels, they do not address the issue of balancing voltages on either side of the neutral in a 3-wire DC system.

Option 3: Diverter

A diverter is a device used to redirect the flow of current or voltage in a circuit. While it plays a role in managing electrical pathways, it is not specifically designed to maintain balanced voltages in a 3-wire DC system. Thus, this option is also incorrect.

Conclusion:

Understanding the role of a balancer set in a 3-wire DC system is crucial for maintaining efficient and stable operation. The balancer set ensures that the voltages on either side of the neutral are equal, preventing any imbalance that could lead to inefficiencies or damage. While other devices like amplifiers, boosters, and diverters have their specific applications, they do not fulfill the requirement of balancing voltages in this context.

Concept:

The booster output is given by:

W = V_b × I_fl

where, W = Booster output power

V_b = Booster voltage

I_fl = Full load current

Calculation:

Given, R = 0.5 ohm/km

R_T = L × R

RT = 4 × 0.5 = 2Ω

Vb = I_fl × R_T

Vb = 15 × 2

Vb = 30 V

W = 30 × 15

W = 450 W

Concept:

The booster output is given by:

W = V_b × I_fl

where, W = Booster output power

V_b = Booster voltage

I_fl = Full load current

Calculation:

Given, R = 0.5 ohm/km

R_T = L × R

RT = 4 × 0.5 = 2Ω

Vb = I_fl × R_T

Vb = 15 × 2

Vb = 30 V

W = 30 × 15

W = 450 W

Concept:

On the basis of how DC distributors are fed by the feeders, they are classified as:

→ Distributor fed at one end

→ Distributor fed at both ends

→ Distributor fed at the centre.

→ Ring distributor.

Now the type of distribution given in the question is of type “Distributor fed at one end”.

• In this type of feeding, the distributor is connected to the supply at one end and loads are taken at different points along the length of the distributor.
• In the above figure end P is also called singly fed distributor and loads I₁, I₂ and I₃ tapped off at points Q, R, S respectively.

Points to remember in this type of distribution:

1) The current in the various sections of the distributor away from the feeding point goes on decreasing. Thus the current in the section PQ is more than current in the section QR and the current in the section QR is more than current in section RS.

2) The voltage across the loads away from the feeding point goes on decreasing. Therefore minimum voltage occurs at point S.

3) In case a fault occurs at/on any section of the distributor, the whole distributor will have to be disconnected from the supply mains.

Calculations:

Given- conductor resistance per km = 0.02 Ω

But in 2 wire DC distributor system 2 conductors are present

∴ Resistance per km for 2 wire DC distributor = 0.02 × 2 = 0.04 Ω 

∴ Resistance of section AB = 0.04 × 2 = 0.08 Ω (R_AB)

∴ Resistance of section BC = 0.04 × 2 = 0.08 Ω (R_BC)

Also, I₂ = 200 A, I₁ = 100 A

∴ Current in section AB = I₁ + I₂ = 100 + 200 = 300 A

∴ current in section BC = I₂ = 200 A

i.e. I_AB = 300 A, I_BC = 200 A

Now, Voltage available at load point B

V_B = Voltage at A – Voltage drop in AB

V_B = 500 V – I_AB × R_AB

V_B = 500 V – (300 × 0.08) V

V_B = (500 - 24) V

V_B = 476 V

Now, voltage available at point C

V_C = voltage at B – voltage drop in BC

V_C = 476 V – I_BC × R_BC

V_C = 476 V – (200 × 0.08) V

V_C = 476 V – 16 V

V_C = 460 V

Therefore the voltage available at the farthest point (C) in the system is 460 V.

Note:

There are a few advantages of other types of the distribution system.

• In this type of distribution, if a fault occurs on any feeding point of the distributor or on any section of the distributor, the continuity of the supply is maintained from the other operating feeding point.
• Also the area of cross-section required for doubly-fed distributors is much less than that of a singly fed distributor.

Fig. Distributor fed at the center

• Distributor fed at the center is equivalent to two singly fed distributors, each distributor having a common feeding point and length equal to half of the total length.
• In-ring main distribution, the distributor is in the form of a closed ring. It is equivalent to a straight distributor fed at both ends with equal voltages, where the two ends being brought together to form a closed ring.
• The distributor ring may be fed at one or more than one point.

We Have,

Resistance of cable is 0.5 ohm/km

Hence, total Resistance of 4 km long cable (R) = 0.5 × 4 = 2 Ω

Load Current (I) = 15 A

So total Voltage drop in the cable = IR = 15 × 2 = 30 volts

Total Power Loss (P) = I²R

⇒ P = I2R = 15² × 2 = 225 × 2 = 450 W

Concept: 

Total current supplied by the distributor, I = 800 × 1 = 800A

Total resistance of the distributor , R = 2 × 0.05 × 0.8 = 0.08 ohm

For a uniformly loaded DC-distributed wire fed from both sides with equal voltages, the V_min occurs at mid-point( x = l/2).

So the maximum voltage drop is at the mid-point:

Calculation:

\(V_{drop}=\frac{i.r×(l-x)^2}{2}\)  = \(\frac{i.r×(l-l/2 )^2}{2}\) = \(\frac{i.l×r.l}{2}\)  

Maximum voltage drop = \(\frac{IR}{8}\) = \(\frac{800×0.08}{8}\) = 8 volts

Concept:

The booster output is given by:

W = V_b × I_fl

where, W = Booster output power

V_b = Booster voltage

I_fl = Full load current

Calculation:

Given, R = 0.5 ohm/km

R_T = L × R

RT = 4 × 0.5 = 2Ω

Vb = I_fl × R_T

Vb = 15 × 2

Vb = 30 V

W = 30 × 15

W = 450 W

Consider 200 V (Fig i) and 400 V (Fig ii) system as shown below"

We have,

P₁ = V₁I₁ = 200I₁  and P₂ = V₂I₂ = 400I₂ 

As the same power is delivered in both cases,

Therefore,

P₁ = P₂

200I₁ = 400I₂ 

I₂ = \(\frac{200}{400} = 0.5\)I₁

Now, power loss in 200V system (W₁) = 2I₁²R₁

And, power loss in 400V system ( W₂) = 2I₂²R₂  =  2(0.5I₁)²R₂ = 0.5I12R2

As power loss in the two cases is the same, W₁ = W₂

2I₁²R₁ = 0.5I₁²R₂;

⇒ R₂/R₁ = 4

We know resistance is inversely proportional to area.

Therefore:  A₁/A₂ = 4,  Also, ratio of volume ; v₁/v₂ = 4

⇒ v₁ = 4v₂ 

(Note: 'v' abbreviated for volume)

Hence,

Percentage Saving in copper = \(\frac{V_{1}-V_{2}}{V_{1}}\) × 100 = \(\frac{4v_2-v_2}{4v_2}\times 100 =75\%\)

Explanation:

On the basis of how DC distributors are fed by the feeders, they are classified as:

→ Distributor fed at one end

→ Distributor fed at both ends

→ Distributor fed at the centre.

→ Ring distributor.

Now the type of distribution given in the question is of type “Distributor fed at one end”.

• In this type of feeding, the distributor is connected to the supply at one end and loads are taken at different points along the length of the distributor.
• In the above figure end P is also called singly fed distributor and loads I1, I2 and I3 tapped off at points Q, R, S respectively.

Points to remember in this type of distribution:

1) The current in the various sections of the distributor away from the feeding point goes on decreasing. Thus the current in the section PQ is more than current in the section QR and the current in the section QR is more than current in section RS.

2) The voltage across the loads away from the feeding point goes on decreasing. Therefore minimum voltage occurs at point S.

3) In case a fault occurs at/on any section of the distributor, the whole distributor will have to be disconnected from the supply mains.

Concept:

In a uniformly loaded distributor fed at one end, the maximum total voltage drop = IR/2

In a uniformly loaded distributor fed at both ends, the maximum total voltage drop = IR/8

The maximum voltage drop in the case of uniformly loaded distributor fed at both ends is one-fourth of the maximum voltage drop in the case of uniformly loaded distributor fed at one end.

Calculation:

Length of distributor = 200 m = 0.2 km

Current supplied by distributor = 2 amperes/meter

Total current supplied by distributor (I) = 200 × 2 = 400 A

The resistance of single wire = 0.3 Ω/km

Total resistance = 0.3 × 0.2 = 0.06 Ω

For Two wire, R = 0.06 × 2 Ω

Maximum voltage drop \( = \frac{{IR}}{2} = \frac{1}{2} × 400 × 0.06\times 2 = 24\;V\)

The DC transmission systems are classified as:

DC two-wire

DC two-wire with mid-point earthed

DC three-wire

(i) DC two-wire:

• In the 2-wire DC the system, one is the outgoing wire called positive wire and the other is the return wire called negative wire as shown in Figure below. 
• The return wire is also connected to earthing.

(ii) DC two-wire with mid-point earthed:

• The two-wire DC system with mid-point earthed is shown in the figure below.
•  The maximum voltage between any conductor and earth is Vm and the maximum voltage between conductors is 2V_m.
• It increases the voltage level in comparison to two-wire systems.

(iii) DC three-wire:

In a 3-wire DC system, there are two outers and a middle wire which is connected to earth.

When the load is balanced, the current in the neutral wire is zero. 

DC 3- wire system is shown in figure below.

Conclusion:

From the above concept, it is clear that the DC four wires with line earthed are not considered for DC transmission.

The DC transmission systems are classified as:

DC two-wire

DC two-wire with mid-point earthed

DC three-wire

(i) DC two-wire:

• In the 2-wire DC the system, one is the outgoing wire called positive wire and the other is the return wire called negative wire as shown in Figure below. 
• The return wire is also connected to earthing.

(ii) DC two-wire with mid-point earthed:

• The two-wire DC system with mid-point earthed is shown in the figure below.
•  The maximum voltage between any conductor and earth is Vm and the maximum voltage between conductors is 2Vm.
• It increases the voltage level in comparison to two-wire systems.

(iii) DC three-wire:

In a 3-wire DC system, there are two outers and a middle wire which is connected to earth.

When the load is balanced, the current in the neutral wire is zero. 

DC 3- wire system is shown in figure below.

Conclusion:

From the above concept, it is clear that the DC four wires with line earthed are not considered for DC transmission.

The correct answer is option 3):(50%)

Concept:

 Let L = Length of each conductor in metre

σ = Current density in A/m²

P = Power supplied in watts

A = Area of Conductor

The volume of Cu required for both conductors is V = 2 × A × L

Calculation:

Given

V = 220 V

Supply Current per feeder conductor

I = P / 220

The area of conductor required

A = \( I\over σ\)  =  \(P\over 220σ\)

The volume of Cu required for both conductors is

 = 2 × A × L

Vol₁= \({2 × P\times L \over 200\sigma }\)

if voltage doubled 

Volume of Cu required for both conductors is

 Vol2= 2 × A× L

= \({2 × P\times L \over 440\sigma }\)

\({Vol_1 - Vol_2 \over Vol_1} \times 100 = { {2 × P\times L \over 220\sigma } - { 2P\times L \over 440\sigma } \over {2 × P\times L \over 220\sigma }} \times 100\)

= 50%