QUESTION : In Example 11.2, find the voltage between the two ends of either wire

Question

QUESTION: In Example 11.2, find the voltage between the two ends of either wire - that is , the voltage between power plant and city across either wire, not the voltage between the two wires at the generator output. Answer for both choices of transmission voltage.

---Example 11.2 (Background to help answer question above): Suppose the power plant of example 11.1 sends its power out over a two-wire transmission line like that in figure 11.3, and each wire has a resistance of 1.52 ?. What percentage of the plant's 1,244 MW output is lost in transmission if the voltage between the ttwo wires of the transmission line is (a) 115 kV, the lowest voltage used for long distance transmission in North America, or (b) 345 kB, typical of longer transmission lines?

Power plant in example 11.1 produces 1,244 MW of electric power from its single generator who's output voltage is 22 kV

11.1 ELECTRICITY/ 297 wer to r lost to Power plant V Transmission line City trans- n the where R is the wire' the same as the generator power plant, transmission line ower dissipated in the wire: and end user-in this case a the same resistance and car ice this value. But here's the more complex, but DC trans- FIGURE 11.3 s resist-Simplified diagram showing a city. Most AC systems carry "three-phase power" and are loss scales as the square of mission uses simple two-wire systems as shown here. the transmission line drops power output P IV-set, wer-with a low voltage V low current I, or somewhere ission is IR, we're clearly igher voltage V. That's the night wonder why we don't I. We could-but then we losses rather than the qua- current. More important, ductor-both of which are re determined by a mix of on handling high yaltage

Explanation / Answer

given output voltage of the generator Vo = 22,000 V

Power Po = 1244 MW

a. transmission line voltage V = 115 kV

resistance of each wire , R = 1.52 ohm

power lost = i^2*R

i = Po/V

hence

dP = Po^2*R/V^2

hence

dP/Po = Po*R/V^2

hence

dP/Po = 1244*10^6*1.52/115,000^2 = 14.2977693 %

b. V = 345 kV

dP/Po = 14.297769*(115/345)^2 = 1.588641041 %

c. voltage difference across the tow ends of a single wire = dP*R/V

for V = 115 kV

dV = 2350.90139252 V

for V = 345 kV

dV = 87.0704223 V

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.