Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this document? Why not share!

2,360 views

Published on

DC distributed load(Uniformly loaded), fed at one end or both ends

No Downloads

Total views

2,360

On SlideShare

0

From Embeds

0

Number of Embeds

2

Shares

0

Downloads

62

Comments

1

Likes

1

No embeds

No notes for slide

- 1. Attention Students: We missed this in a hurry: Please go through this DC DISTRIBUTED LOAD 1. Fed at one end (Uniformly loaded) : Let ‘L’m be the length of distributor and r ohm be the resistance/m Run. Consider a point C on the distributor at a distance x metres from the feeding pointA. The current at point C is = iL – ix = i(L-x) Amp. This is true and it can be understood if we consider the follwing example: Current flowing in DE will 10x5 – 10x2 = 30A
- 2. Now consider a small length dx near point C. Its resistance is rdx and the voltage drop over length dx is dv = i(L-x)rdx = ir(L – x ) Total voltage drop in the feeder uptopoint C v= = ir(Lx – x2/2) The voltage drop upto point B can be obtained by putting x = l VAB = ir(L2 – L2/2) = irL2/2 =( iL)(rL)/2 = IR/2 Loss = = = i2rL3/3 Watts 2. DC distributed load fed at both the ends(VA = VB)(Uniformly loaded) Current supplied from each feeding point = iL/2 Consider a point C at a distance x meters from the feding point A. Then current at point C is = iL/2 – ix = i(L/2 – x )
- 3. Consider a small length dx near point C. Its resistance is rdx and the voltage drop over length dx is dv = i(L/2 - x)rdx Voltage drop upto point C = (ir/2)(Lx –x2) Obviously point of minimum potential will be the mid point. Maximum voltage drop (when x =L/2) = ir/2(L2/2 – L2/4) = irL2/8 = iLrL/8 = I R/8 Minimum potential (at mid- point) = V – IR/8 Power loss in loss in the distributor fed from both the ends be calculated firstly for half of the line and then can be doubled. Loss =2 ∫ ( L / 2 − x)( L / 2 − x )iirdx = i2rL3/12 Watts Problems : 1. A 2 wire dc distributor 200m long is uniformly loaded with 2A/m. Resistance of single wire is 0.3 ohm/km. If the distributor is fed at one end, calculate: (i) The voltage drop upto a distance of 150m from the feeding point. (ii) The maximum voltage drop Solution: I = 2A/m, r = 2x0.3/1000 = 0.0006 ohm/m (i) Vx = ir (Lx – L2/2) = 2x0.0006(200x150 – 1502/2) = 22.5V (ii) I = iL = 400A; R = rL = 0.12 Total voltage drop = IR/2 = 24V 2. A 2 wire dc distributor cable 1000m long is loaded with 0.5A/m. Resistance of each conductor is 0.05 ohm/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220V. what is the minimum voltage and when it occurs.
- 4. Solution: Current loading: i = 0.5A/m r = 2x0.05x1000 = 0.1x10-3 ohm L = 1000m I = iL = 500A R = rL = 0.1 ohm] Maximum voltage drop = IR/8 = 500x0.1/8 = 6.25V Minimum voltage occurs at the mid point and its value is = V –IR/8 = 220 – 6.25 = 213.75V

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

2. A 2 wire dc distributor cable 1000m long is loaded with 0.5A/m. Resistance of each conductor is 0.05 ohm/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220V. what is the minimum voltage and when it occurs.