How do you calculate 11kv line losses?

How do you calculate 11kv line losses?

Example:

  1. 11 KV Distribution Line have following parameter.
  2. Main length of 11 KV Line is 6.18 Kms.
  3. Total nos.
  4. 25KVA Transformer Iron Losses = 100 W, Copper Losses= 720 W, Average LT Line Loss= 63W.
  5. 63KVA Transformer Iron Losses = 200 W, Copper Losses= 1300 W, Average LT Line Loss= 260W.

What is the equation for power Ohm’s law?

Power and Joule’s Law Power is directly linked to Ohm’s by Joule’s law, which says that the heat produced in resistance is proportional to the square of the current flowing through it over a given time. We can express this as P=V*I and because V=I*R, we get P = I*I*R or P=I2R.

Which is the correct equation for OHM’s law?

However, these power equations are so commonly associated with the Ohm’s Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm. REVIEW: Power measured in watts, symbolized by the letter “W”. Joule’s Law: P = I2R ; P = IE ; P = E2/R

How to calculate power loss in transmission cables?

Current in the transmission cables = P/V. *Power loss is not equal to V^2/R because this V does not represent potential drop across the cables. If the cables are connected to your home then Power = V^2 / (Resistance of cables + Resistance of your house) Hence Power loss at the cables = I^2 R or (P/V)^2 R. Aug 15, 2011.

How is power loss related to power supply efficiency?

If a resistor dissipates .25 W of power then a ½ W rating must be used. Power supply efficiency is the ratio of output power (POUT) to the input power (PIN). Since input power is always more than the output power because the internal circuit to operate the power supply, there is a power loss.

How to calculate the power of a lamp?

Now that we know the current, we can take that value and multiply it by the voltage to determine power: This tells us that the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat. Let’s try taking that same circuit and increasing the battery’s voltage to see what happens.