POWER SYSTEM PLANNING AND DESIGN (2180903)
ASSIGNMENT-1
2) Explain the factor consist for choice and spacing between conductor for transmission line design.
3) Explain corona effect and losses with standard formula.
4) Derive sag tension relation for span, width tower at each end located at equal level.
5) Explain the different types of extra high voltage tower.
6) What is the limitation of HVAC transmission line?
(7) Explain the use bundled conductors.
(8) Explain the insulation co-ordination.
ASSIGNMENT-2
(1) Explain the following distribution system with fig.
(1)Radial system (2) Parallel pr loop system (3) Network or grid system
(2) What is lamp flicker? What are its causes? What type of loads are
responsible for it ? how can it be reduced?
(3) Discuss Kelvin’s law the find the most economical conductor size.
what are the limitations of this law?
(4) State the different types of arrangements for distribution system and
also the advantages and disadvantages of radial secondary distribution system.
(5) Explain the lamp flickers, its causes and remedies.
(6) Discuss in details the steps in planning and designing electrical distribution schemes.
(7) List and explain the types of primary distribution system.
(8) Discuss briefly the design consideration in distribution system. Define and explain terms feeder, Discuss and service mains.
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(9) Compare radial, rind and grid distribution system. State their
application.
10) Discuss classification of distribution system with neat diagrams. What are the advantages and dis-advantages of each?
Ex-1 A Single phase a.c, distributor 500m long has a total loop impedance of (0.02+j0.04) and is fed from one end at 250V.If is loaded as under (1) 50A at unity power factor ,200m from the feeding point.(2)100A at 0.8 power factor lag.300m from the feeding point.(3)50 A at 0.6 power factor lag, at the far end.
We will split the currents into their active and reactive components as under :
50 × 1 = 50 A ; 100 × 0.8 = 80 A; 50 × 0.6 = 30 A
These are shown in(a). The reactive or wattless components are
50 × 0 = 0 ; 100 × 0.6 = 60 A ; 50 × 0.8 = 40 A
These are shown in Fig.(b). The resistances and reactances are shown in their respective figures.
First Method
Current in section AD (Fig.)
for sum of the three load currents.
∴ current in AD
= 50 + 100 (0.8 − j 0.6) + 50 (0.6 − j 0.8)
= 160 − j100
Impedance of section AD
= (200/500) (0.02 + j 0.04)
= (0.008 + j 0.016) W
Second Method
We will split the currents into their active and reactive components as under :
50 × 1 = 50 A ; 100 × 0.8 = 80 A; 50 × 0.6 = 30 A
These are shown in(a). The reactive or wattless components are
50 × 0 = 0 ; 100 × 0.6 = 60 A ; 50 × 0.8 = 40 A
These are shown in Fig.(b). The resistances and reactances are shown in their respective figures.
Drops due to active components of currents are given by taking moments
= 50 × 0.008 + 80 × 0.012 + 30 × 0.02 = 1.96 V
Drops due to reactive components
= 60 × 0.024 + 40 × 0.04 = 3.04 Total drop = 1.96 + 3.04 = 5 V
This is approximately the same as before.
Third Method
The centre of gravity (C.G.) of the load is at the following distance from the feeding end
Ex-2 A 3 core distribution cable is 300m long and supplies a load of 100 KW at 440 volts at 0.9 p.f. lag for 3000hours in a year. T unit and he cable cost including installation is rs(13a +32)per meter where “a” is the cross section area of each conductor in sq. cm. cost of energy wasted is 12paisa per unit and the rate of interest and depreciation is 15%. The resistance per km of the conductor of 1cm2 cross-section is 0.213ohm. Find the most economical cross-section of the distributor cable.
Ex-3 A 2 wire dc distributor AB is 300m long and is fed at both the ends A and B .The distributor supplies a uniformly distributed load of 0.25A/m and concentrated loads of 40A at C and 60 A at D. The distance AC and BD are 120m each. The loop resistance of the distributor is 0.1 ohm/100m. Both A and B are maintained at 300V. Find the currents fed at A and B and the potential at point C and D.
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