1

UNIFORMLY LOADED DISTRIBUTOR FED AT ONE END

Figure (a) represents a distributor of length l meters which is uniformly loaded with i amp per unit length. In order to determine the drop in such a feeder, consider a length dx of the feeder at P which is at a distance x from end A. Let r be the resistance (in ) of both go and return of the distributor per unit length.

First Method

The resistance of the length, = . (1)

This portion of dx of the distributor will carry a current which is to be tapped in the length PB of

the distributor.

So the current in portion = ( ) (2)

Voltage drop in length = ( ) (3)

Therefore, the total drop up to the point P of the distributor,

0= ( )

0 (4)

Or = ( 2

2) (5)

A B

Fig. (b)

dx

P

x

lil i(l-x)

2

If we consider x=l

Then = (2 2

2) =

2

2 (6)

Or =1

2()() =

1

2 (7)

Where I is the total current and R is the total resistance.

UNIFORMLY LOADED DISTRIBUTOR FED AT BOTH ENDS

Let us consider a distributor (Fig. c) of length l which is fed at both ends.

Consider a small length dx of the distributor. Let P be the point of minimum potential. The load

current due to length = . , where i is the current in unit length.

The voltage drop in length AP = (). () = .

0=

2

2

0 (1)

Similarly voltage drop in PB = .

0=

()2

2 (2)

Since VA voltage drop in AP = VB voltage drop in PB

2

2=

()2

2 (3)

Elemental current

Total resistance up to x

3

Example: A distributor 500 m long has a total resistance of 0.08 and is loaded as shown in figure

d. A voltage of 200 V is maintained at feeding points A and B. In addition to the concentrated

loads it has been uniformly loaded at the rate of 1 A/m. Calculate:

i. Minimum potential point

ii. The value of minimum potential

iii. The current supplied by each end

Solution: From observation it seems that E be the point of minimum voltage.

So, let E be the minimum potential point.

Resistance per meter, =0.08

500= 0.00016 /

Resistance of section

= 0.00016 100 = 0.016

= 0.00016 100 = 0.016

= 0.00016 50 = 0.008

= 0.00016 150 = 0.024

= 0.00016 100 = 0.016

In this case, drop in AE = drop in BE

Or [drop in AE (due to concentrated load+ due to uniformly distributed load)]

= [drop in BE (due to concentrated load+ due to uniformly distributed load)]

[( + 150) 0.016 + ( + 70) 0.016 + 0.008 +

10.00016(250)2

2]

= [(50 ) 0.024 + (150 ) 0.016 +1 0.00016 (250)2

2]

A B

C

200 V

D E F100 m 100 50 150 100

x+150 50-x

80 A

200 V

70 A 50 A 100 A

x+70 x 150-x

Fig. (d)

4

Or, (0.016 + 0.016 + 0.008) + 2.4 + 1.12 = 3.6 0.04

Or, = 1

Hence E is the actual minimum potential point.

Current supplied by end A = 250 +(150+1) = 401 Amp

Current supplied by end B = 250 +(150-1) = 399 Amp

Voltage drop in section BE = (149 0.016 + 49 0.024) +10.00016(250)2

2=

3.56 + 5 = 8.56

Voltage at E= 200-8.56= 191.49 V

RING DISTRIBUTOR

A ring distributor is a distributor which is arranged to form a closed circuit and it can have one or several feeding points. The chief interest in the ring distributor lies in the economy in wire which can be effected by its use and by the proper choice in the number of feeding points.

In order to know the current distribution and the voltage at different tapping points, the distributor can be assumed to be cut into two sections A and spread out. Then it can be treated as a feeder fed at both ends.

5

RING DISTRIBUTOR WITH INTERCONNECTION

Sometimes to minimize the voltage drop between two points an interconnection is provided between two points as shown. The current in different section of the ring main is obtained by the use of Kirchhoffs current law.

3 = ( ) + ( 5 4) + ( 1 2)

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Example: A 2 wire distributor ABCDEFA is in the form of a ring main and is fed at point A and is loaded as follows: 20 A at B, 50 A at C, 25 A at D, 40 A at E and 30 A at F. The resistance of the sections (go & return) are AB=0.08 , BC=0.1 , CD=0.12 , DE=0.14 , EF=0.09 , and FA=0.16 . Find the load point at where the potential difference is minimum. If an interconnection having a resistance of 0.1 be connected between the feeding A and D i.e. at the minimum potential point, find the minimum voltage drop in the system at the point at which it now occurs.

Let D be the point of minimum potential point.

The resultant voltage drop in the closed loop ABCDEF is zero.

Therefore we can write,

0.16 + 0.09( 30) + 0.14( 70) 0.12(95 ) 0.1(145 )

0.08(165 ) = 0

7

0.69 = 51.6

= 74.8 Amp

Hence, it can be seen that the point D receives the current from both directions. So, D is the point of minimum potential. Now let the points A and D are interconnected, then the new current distribution is shown in Figure h.

Considering the loop AFEDA,

0.16 + 0.09( 30) + 0.14( 70) 0.1 = 0

Or 0.39 0.1 12.5 = 0

Or = 3.9 125 (1)

Take the loop ADCBA,

0.1 0.12(95 ) 0.1(145 ) 0.08(165 ) = 0

8

0.4 + 0.3 = 39.1

Or = 97.75 0.75 (2)

From equation (1) and (2), 3.9 125 = 97.75 0.75

Or = 48

From equation (2), = 97.75 0.75 = 97.75 0.75 48

= 61.75

It is seen that the new minimum potential point is E and current in section FE= 48-30= 18 A

Therefore, maximum voltage drop = 0.16 48 + 0.09 18 = 9.3

Example: A two wire dc network is represented in the following figure. The feeding point is P and is at 250 V. Calculate the voltage at point Q, R, and S. Resistances represented are for go and return.

9

Total load current supplying from P = 100+100 + 70 = 270 Amps

Current distribution is shown in by arrow in the figure.

Taking the loop SPQ,

0.1 (270 ) 0.075() 0.25( ) + 0.25(70 + ) = 0

Or,

= 6.75 + 5 + 445 (1)

Taking the loop PQR,

0.075 + 0.2 0.2(100 ) 0.08 = 0

Or,

= 0.9375 + 5 250 (2)

From (1) and (2)

6.75 + 5 + 445 = 0.9375 + 5 250

= 90.6

Taking the loop PRS,

0.08 + 0.4(100 ) 0.5(200 ) 0.1(270 ) = 0

Or,

1.08 + 0.9 + 0.1 = 167

Or,

= 154.5 0.835 0.925 (3)

From (1) and (3),

6.75 + 5 + 445 = 154.5 0.835 0.925

Or, 6.6575 + 5.835 = 290.5

Or, 6.6575 90.6 + 5.835 = 290.5

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Or, = 48.7

Putting the value of x and z in (1),

= 6.75 90.6 + 5 48.7 + 445

= 76.5

= 250 90.6 0.075 = 243.8

= 250 76.5 0.08 = 243.88

= 250 [270 (90.6 + 76.5)] 0.1 = 250 10.3 = 239.7