A.C. DISTRIBUTOR CALCULATIONS It has already been shown that voltage drop in an inductive circuit is given by I (R cos φ + X sin φ) The total drop will be given by Σ I (R cos φ + X sin φ). Questions on a.c. distributors may be solved in the following three ways : Express voltages, currents and impedances in complex notation and then proceed exactly as in d.c. distributors. Split the various currents into their active and reactive components. Now, the drop in the case of active components will be due to resistance only and in the case of reactive components due to reactance only. Find out these two drops and then add the two to find the total drop.

A.C. DISTRIBUTOR CALCULATIONS In cases where approximate solutions are sufficient, quick results can be obtained by finding the “distribution centre” or centre of gravity of the load.

LOAD DIVISION BETWEEN PARALLEL LINES It is common practice to work two or more cables or overhead lines in parallel when continuity of supply is essential. In the case of a fault developing in one line of cable, the other lines or cables carry the total load till the fault is rectified.” Let us take the case of two lines in parallel and having impedances of and . Their combined impedance is

LOAD DIVISION BETWEEN PARALLEL LINES

SUSPENSION INSULATORS Suspension insulators are used when transmission voltage is high. A number of them are connected in series to form a chain and the line conductor is carried by the bottom most insulator A ‘CAP’ TYPE SUSPENSION INSULATOR

SUSPENSION INSULATORS A ‘cap’ type suspension insulator consists of a single disc-shaped piece of porcelain grooved on the under surface to increase the surface leakage path. A galvanized cast iron cap is cemented at the top of the insulator. In the hollow cavity of the insulator is cemented a galvanized forged steel pin, the lower enlarged end of which fits into the cavity of the steel cap of the lower suspension insulator and forms a ball and socket connection. A string of suspension insulators consists of many units, the number of units depending on the value of the transmission voltage.

SUSPENSION INSULATORS String Efficiency If there are n units in the string, then its efficiency is given by

CALCULATION OF VOLTAGE DISTRIBUTION ALONG DIFFERENT UNITS Fig 41.16

CALCULATION OF VOLTAGE DISTRIBUTION ALONG DIFFERENT UNITS Let, C = capacitance to ground kC = mutual capacitance between units

CALCULATION OF VOLTAGE DISTRIBUTION ALONG DIFFERENT UNITS

CALCULATION OF VOLTAGE DISTRIBUTION ALONG DIFFERENT UNITS

INTERCONNECTORS An interconnector is a tieline which enables two generating stations to operate in parallel. It facilitates the flow of electric power in either direction between the two stations. VOLTAGE DROP OVER THE INTERCONNECTOR Let station 1 supply a current of 1 to station 2 along an interconnector having a resistance of RW and reactance of XTW per phase as shown in Fig.41.17 (a). If the receiving end p.f. is cos φ lagging, then the vector diagram will be as shown in Fig.41.17 (b).

INTERCONNECTORS Fig.41.17

INTERCONNECTORS

SAG AND STRESS ANALYSIS The conductors of a transmission line are attached to suitable insulators carried on supports of wood, iron, steel or reinforced concrete. Obviously, the supports must be strong enough to withstand not only the dead weight of the conductors themselves but also the loads due to ice and sleet that may adhere to them and to wind pressure. Moreover, the minimum factor of safety for the conductors should be 2.0 based on ultimate strength.

SAG AND STRESS ANALYSIS Sag and stresses vary with temperature on account of thermal expansion and contraction of the line conductors. The value of sag as well as tension of a conductor would now be calculated when Supports are at equal levels Supports are at unequal levels

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS Fig.41.18 shows a span of a wire with the two supports at the same elevation and separated by a horizontal distance 2l. It can be proved that the conductor AB forms a catenary with the lowest point O forming the mid-point (where the curve is straight). Fig.41.18

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS Let W be the weight of the wire per unit length and let point O be chosen as the reference point for measuring the co-ordinates of different points on the wire. Consider a point P having co-ordinates of x and y. The tension T at point P can be resolved into two rectangular components and so that . If S is the length of the arc OP, then its weight is WS which acts vertically downward through the centre of gravity of OP. There are four forces acting on OP—two vertical and two horizontal. Since OP is in equilibrium, the net force is zero.

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS . Equating the horizontal and vertical components, we have It may be noted

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT EQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT UNEQUAL LEVELS Fig.41.19 shows a span between two supports A and B whose elevations differ by η, their horizontal spacing being 2l as before. Such spans are generally met with in a hilly country. Let O be the lowest point of the catenary AOB. Obviously, OA is a catenary of half-span and OB of half-span Fig.41.19

SAG AND TENSION WITH SUPPORTS AT UNEQUAL LEVELS From equation

SAG AND TENSION WITH SUPPORTS AT UNEQUAL LEVELS

SAG AND TENSION WITH SUPPORTS AT UNEQUAL LEVELS Having found and , values of and can be easily calculated. It is worth noting that in some cases, may be negative which means that there may be no horizontal point (like point O) in the span. Such a thing is very likely to happen when the line runs up a steep mountain side

EFFECT OF WIND AND ICE It is found that under favourable atmospheric conditions, quite an appreciable thickness of ice is formed on transmission lines. The weight of ice acts vertically downwards i.e. in the same direction as the weight of the conductor itself as shown in Fig 41.20 Fig 41.20

EFFECT OF WIND AND ICE If is the weight of ice per unit length of the conductor and the force per unit length exerted by the wind, then total weight of the conductor per unit length is

EFFECT OF WIND AND ICE Note. If P is the wind pressure per unit projected area of the conductor, then wind load or force per unit length of the ice-covered conductor is Ice is in the form of a hollow cylinder of inner diameter = D and outer diameter = (D + 2R). Hence, volume per unit length of such a cylinder is

EFFECT OF WIND AND ICE If ρ is the ice density, then weight of ice per unit length of the conductor is When ice and wind loads are taken into account, then the approximate formulae Here ‘d’ represents the slant sag in a direction making an angle of θ with the vertical. The vertical sag would be = d cos θ.

EFFECT OF WIND AND ICE Similarly, formulae become