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From Book ContentThe line wavelength may be assumed to be 5000 Km. The practical line loadability may be based on a load angle δ of 35◦ . Assume |VS| = 1.0 pu and |VR| = 0.9 pu. Determine the number of three-phase transmission circuits required for each voltage level. Each transmission tower may have up to two circuits. To limit the corona loss, all 500-kV lines must have at least two conductors per phase, and all 765-kV lines must have at least four conductors per phase. The bundle spacing is 45 cm.

The conductor size should be such that the line would be capable of carrying current corresponding to at least 5000 MVA. Use acsr command in MATLAB to find a suitable conductor size. Following are the minimum recommended spacings between adjacent phase conductors at various voltage levels. (a) Select a suitable voltage level, and conductor size, and tower structure.

Use lineperf program and option 1 to obtain the voltage regulation and transmission efficiency based on a receiving end power of 3000 MVA at 0.8 power factor lagging at the selected rated voltage. Modify your design and select a conductor size for a line efficiency of at least 94 percent for the above specified load. (b) Obtain the line performance including options 4–8 of the lineperf program for your final selection. Summarize the line characteristics and the required line compensation.

For each voltage level SIL is computed from (5.78). From the practical line loadability equation given by (5.97), the real power per circuit is computed, and number of circuits is established and tabulated in the following table. For the 345 kV level we require six three-phase transmission circuit, and for the 500 kV line we need three circuits. Considering the cost of the transmission towers, right of ways and associated equipment, it can be concluded that one circuit at 765 kV is the most economical and practical choice.

6.1. A power system network is shown in Figure 47. The generators at buses 1 and 2 are represented by their equivalent current sources with their reactances in per unit on a 100-MVA base. The lines are represented by π model where series reactances and shunt reactances are also expressed in per unit on a 100 MVA base. The loads at buses 3 and 4 are expressed in MW and Mvar.

(a) Assuming a voltage magnitude of 1.0 per unit at buses 3 and 4, convert the loads to per unit impedances. Convert network impedances to admittances and obtain the bus admittance matrix by inspection. (b) Use the function Y = ybus(zdata) to obtain the bus admittance matrix. The function argument zdata is a matrix containing the line bus numbers, resistance and reactance. (See Example 6.1.) (a) Use Newton-Raphson method and hand calculations to find one of the roots of the polynomial equation.

The conductor size should be such that the line would be capable of carrying current corresponding to at least 5000 MVA. Use acsr command in MATLAB to find a suitable conductor size. Following are the minimum recommended spacings between adjacent phase conductors at various voltage levels. (a) Select a suitable voltage level, and conductor size, and tower structure.

Use lineperf program and option 1 to obtain the voltage regulation and transmission efficiency based on a receiving end power of 3000 MVA at 0.8 power factor lagging at the selected rated voltage. Modify your design and select a conductor size for a line efficiency of at least 94 percent for the above specified load. (b) Obtain the line performance including options 4–8 of the lineperf program for your final selection. Summarize the line characteristics and the required line compensation.

For each voltage level SIL is computed from (5.78). From the practical line loadability equation given by (5.97), the real power per circuit is computed, and number of circuits is established and tabulated in the following table. For the 345 kV level we require six three-phase transmission circuit, and for the 500 kV line we need three circuits. Considering the cost of the transmission towers, right of ways and associated equipment, it can be concluded that one circuit at 765 kV is the most economical and practical choice.

6.1. A power system network is shown in Figure 47. The generators at buses 1 and 2 are represented by their equivalent current sources with their reactances in per unit on a 100-MVA base. The lines are represented by π model where series reactances and shunt reactances are also expressed in per unit on a 100 MVA base. The loads at buses 3 and 4 are expressed in MW and Mvar.

(a) Assuming a voltage magnitude of 1.0 per unit at buses 3 and 4, convert the loads to per unit impedances. Convert network impedances to admittances and obtain the bus admittance matrix by inspection. (b) Use the function Y = ybus(zdata) to obtain the bus admittance matrix. The function argument zdata is a matrix containing the line bus numbers, resistance and reactance. (See Example 6.1.) (a) Use Newton-Raphson method and hand calculations to find one of the roots of the polynomial equation.