Test: DC Distribution System - Electrical Engineering (EE) MCQ

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.

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

Maximum voltage drop

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

Types Of DC Distributors:

• DC distributors are usually classified on the basis of the way they are fed by the feeders. Following are the four types of DC distributors.
  • Distributor fed at one end
  • Distributor fed at both ends
  • Distributor fed at the center
  • Ring distributor
     

Distributor Fed At One End:

• In this type, the distributor is connected to the supply at one end and loads are tapped at different points along its length.
• The following figure shows the single-line diagram of a distributor fed at one end. It is worth noting that -

• The current in various sections of the distributor away from the feeding point goes on decreasing.
• From the above figure, the current in section DE is less than the current in section CD and likewise.
• The voltage also goes on to decrease away from the feeding point. In the above figure, the voltage at point E will be minimum.
• In case of a fault in any section of the distributor, the whole distributor will have to be disconnected from the supply. Thus, the continuity of supply is interrupted.

Distributor Fed At Both Ends:

• In this type, the distributor is connected to supply at both ends, and voltages at feeding points may or may not be equal.
• The minimum voltage occurs at some load point which is shifted with the variation of load on different sections of the distributor.

• If a fault occurs at any feeding point, continuity of the supply is ensured from the other feeding point.
• If a fault occurs on any section of the distributor, continuity of the supply is ensured on both sides of the fault with respective feeding points.
• The conductor cross-section area required for a doubly fed distributor is much less than that required for a distributor fed at one end.

Distributor Fed At The Center:

• As the name implies, the distributor is supplied at the center point.
• Voltage drop at the farthest ends is not as large as that would be in a distributor fed at one end.

Ring Main DC Distributor:

• In this type, the distributor is in the form of a closed ring and fed at one point.
• This is equivalent to a straight distributor fed at both ends with equal voltages.

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

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:

• 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.

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

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

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:

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

Concept:

• For the same conductor length, the same amount of power, same losses and same maximum voltage to earth, 3 wire DC system requires a minimum conductor area.
• For transmitting the same amount of power at the same voltage, a three-phase transmission line requires less conductor material than a single-phase line; The three-phase transmission system is so cheaper
• For a given amount of power transmitted through a system, the three-phase system requires conductors with a smaller cross-sectional area; This means a saving of copper and thus the original installation costs are less.
   

Important Points:

•  Below is given the table which shows the ratio of conductor-material in any system compared with that in the corresponding 2-wire DC system.
• Cos φ is the power factor in an AC system.

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:

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 = 200/400 = 0.5I₁

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 =