For an existing ac transmission line, the string efficiency is 80%, if dc voltage is supplied for the same set up, the string efficiency will be
Since there are no currents flowing in any part of the insulator circuit(assuming infinite leakage resistance),the voltage across each of the equal capacitance units is equal, i.e. uniform voltage distribution exists,the string efficiency becomes 100%
strain insulator is used for low voltage line < 11KV
What is the value of safety factor of pin type insulator?
Safety factor of insulator=Puncture strain⁄ Flashover voltage.For pin type insulator the safety factor is 10.
A 4 insulator string has string efficiency of 80% . 45 kV is applied to the string. What is voltage across lowest unit?
Let voltage across lowest unit be V then,45⁄(4*V)=80⁄100 or,V=(45*100)⁄(4*80)=14.06 V.
The electric breakdown strength of a material depends on its
The electric breakdown strength of a material depends on its composition thickness and moisture electric current .
For most of the solid substance, the value of dielectric constant is
DIELECTRIC CONSTANT of solid materials lie from 0 to 10. examples- Bakelite 4.4 to 5.4, cellulose ascetate 3.3 to 3.9, glass 7.6 to 8 etc.
At which point of a conductor embedded in a slot does the maximum temperature occur?
Maximum temperature occur at the centre of a conductor which is embedded in a slot.
Sag depends upon
Sag of transmission line depends upon weight of conductor, length of span, tensile strength of conductor, temperature etc.
An overhead line crossing a road supported by two towers of 50m& 70m from road level with span 250 m. The weight of conductor is 1.5 Kg/m & working tension is 1500 Kg. Determine clearance from road level.
C=t⁄W=1500⁄1.5=1000 m,h=(70-50)=20 m,a=l⁄2-(C*h)⁄250=45 m,sag=a2=1.01 m,clearance=(50-1.01)=48.9 m
What type of ladder is used for overhead lines work?
dry wooden ladder is used for better insulation purpose
The parameters of transposed overhead transmission line are given as self inductor Xs = 0.4 Ω ⁄ Km and mutual reactance Xm = 0.1 Ω ⁄ Km. The positive sequence reactance X1 and zero sequence reactance X0, respectively in Ω ⁄ Km are
For transmission line Positive sequence impedance(X1) = (Xs - Xm) = (0.4 − 0.1) = 0.3 Ω ⁄ Km. Zero sequence impedance (X0) = Xs + 2Xm = 0.4 + 2*0.1 = 0.6 Ω ⁄ Km.
What is the sag for a span 400 m? If the ultimate tensile of consideration strength of conductor is 6000 kgf. The weight of conductor is 550 kgf/m & safety factor is 2.
Maximum working tension(H) = ultimate tensile strength ⁄ safety factor = 6000 ⁄ 2 = 3000 Kgf. l(Span) = 400 m, weight of conductor (wc) = 550 kgf/km, Sag=Wc*l2 ⁄ 8 H = 0.55*(400)2 ⁄ (8*3000) = 3.6 m..
string efficiency is defined as
String efficiency= voltage across the string/(n x voltage across the disc nearer to conductor).
As because of shunt capacitance effect between insulators and tower (as in case of disc type insulators) the voltage across each insulator is non-uniformly distributed. If shunt capacitance is reduced then voltage across insulators are uniformly distributed. Then string efficiency increases.
When transformers or switchgears are to be installed in a line, the poles used are
Usually these H-type substations are at road side. In general substations consists of transformers, switch gear equipment etc. In these substations the transformers ratings will be 200KVA but not beyond. These H type substations are also called as pole mounted substations.
The electrical power transmitted by a line is at the speed of light, so for line frequency 50 Hz, length of the line should be
λ = v/f = 3X108/50=6 X 106 m= 6000 km
String efficiency can be improved by which method?
Using guard ring and grading insulator helps in equalizing the potential across various units of string.
Silicon content in electric sheet steel is limited to 5% as it
Silicon in electric steel sheet makes steel brittle and also punching becomes difficult. So it is limited to 5%.
A generating station has a maximum demand 2 MW and generates 8.76 million unit annually the load factor of the station is ____.
Load factor = Avg. Load / Max. Load X 100
Avg. Load = 3.76 X 106 / 8760 = 1000 units
Load factor = 100 / 2000 X 1000 = 50% = 0.5
The ratio of average load to maximum load is called
Load factor = average load / maximum demand. The ratio of average load to the maximum demand during a given period known as load factor.i.e. Load factor = Average load ⁄ Maximum demand. If the plant is in operation for T hours, Load factor = (Average load x T)(Maximum demand x T) = Units generated in T hours; Maximum demand x T hours. The load factor may be daily load factor, monthly load factor or annual load factor if the time period considered is a day or month or year. Load factor is always less than 1 because average load is smaller than the maximum load.