In a typical power system, a low voltage line is considered as resistive network. Please help me to understand with appropriate equations.
Thanks!!
The spacing between the wires is less, typically around 30 cms in low voltage lines which results in a low reactance value (around 0.3 ohms/km) but the resistance is quite high (a typical low voltage ACSR conductor "rabbit" has around 0.5 ohms/ km).
L= do/di (variation of flux with respect to variation of current) and C= dq/dv (variation of charge with respect to variation of voltage). As the supply voltage is low, capecitive reactance(Xc) offered could be neglected, and air the insulation (bare conductor). Reluctance offered to flux development, is high(low permitivity), and low voltage lines are mainly low power domestic lines with low current carrying capacity hence inductive reactance(Xl) offered is also neglected, or line could be considered as resistive line only and (Z=R)...
Higher the dia of conductor, higher is the cross section, lower is resistance . In High voltage lines or EHV lines dia is high, current capacity is high and resistance per km is low ( as low as .03 ohms per km of 400kV line)). Due to economic reasons one can't use it when power carrying capacity is moderate. We use thinner dia conductor whose resistance is higher.
Inductance is a function of dia of conductor and phase spacing or rather the ratio of it. Inductance is much lower in Distribution lines than the EHV lines.
X/R for 400 kV lines could be 10 , but for distibution lines it could be less than one.
There is no advantage in spending money on thicker conductor to reduce resistance and no use in increasing spacing at extra cost of tower and insulators to increase reactance.
It is therefore an issue of economics.
The resistance can be calculated as
r0 [ohm/km]=(rho [ohm x mm2 / m] x length [10^3 m / km]) / s [mm2]
rho_Al = 1/34 ohm x mm2 / m
rho_Cu = 1/57 ohm x mm2 / m
So, the smaller the cross sectional area the higher the resistance, while the reactance does not change too much.
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