If 80 kw is to be transmitted over two 0.055 ohms lines, estimate how much power is saved if the voltage is stepped up from 120v to 1200v and then down again, rather than simply transmitting at 120v. Assume the transformers are each 99% efficient
To solve this problem, we need to compare the power loss in both scenarios: transmitting at 120V and transmitting at 1200V.
1. Transmitting at 120V:
The power transmitted is 80 kW (given) = 80,000 W.
The resistance of each line is 0.055 ohms (given), so the total resistance is 2 * 0.055 = 0.11 ohms.
Using the formula P = I^2 * R, we can find the current, I.
80,000 = I^2 * 0.11
I^2 = 80,000 / 0.11
I^2 = 727,272.73
I = √(727,272.73) ≈ 853.56 A
The power loss is given by P_loss = I^2 * R = (853.56)^2 * 0.11 ≈ 80,000 W (approximately).
2. Transmitting at 1200V:
Now, since the voltage is stepped up to 1200V, the current will decrease as per the formula P = IV. We'll assume the power is still 80 kW.
80,000 = 1200 * I
I = 80,000 / 1200
I ≈ 66.67 A (approximately)
The power loss in this scenario is P_loss = I^2 * R = (66.67)^2 * 0.11 ≈ 500 W (approximately).
Therefore, the power saved by stepping up the voltage and then stepping it down again is approximately 80,000 - 500 = 79,500 W.