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I have no background of electrical engineering so if my question sounds stupid then accept my apology in advance. I want to calculate the power loss during the transmission. I am using the simple formula which is

$$P_{loss} = Resistance \cdot I^2$$

Where P is power loss
Resistance is in Ohms
and I is the current

Replacing the I by P / V Where V is the voltage now the formula looks like

$$P_{loss} = Resistance \cdot \Big( \frac{P}{V} \Big)^2$$

If the value of resistance is 5.24 Ohms and voltage is 50 kV and the value of P is 940 kW then I get the value of P(loss) = 1852 W which does not seem intuitive.

Looking for your guidance (if i am calculating wrong). Accept my apology in advance if my question bothers you.

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  • \$\begingroup\$ 1852 watts per kilometre seems like a valid result. Please note this is true only for unity power factor. Otherwise the losses will be higher. Formulae for 3-phase line loss can be found here \$\endgroup\$ – venny Aug 19 '14 at 9:10
  • \$\begingroup\$ @venny i consider the value of R as 0.2 ohms per km so the distance between two points are 26.2 km so in order to get the total resistance i multiplied 26.2 by 0.2 which is equal to 5.24 ohms. Is it the right way? or shall i calculate the loss by putting the value of R by 0.2 and then multiply the result with the distance to get the total loss? \$\endgroup\$ – user2293224 Aug 19 '14 at 10:24
  • \$\begingroup\$ you can do both, the result will be same. The loss is actually 1852 watts, not 1852 watts per km as stated in my previous comment. \$\endgroup\$ – venny Aug 19 '14 at 10:37
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=> Short answer: Sounds reasonable.

Common approach would be: Try to approach this way: What causes the power loss if you transmit power along a line?

For simple, resistive loads (based on Ohms law: \$U = R \cdot I\$ and \$P = U \cdot I\$)

  • The current flowing through the line (which is the current that you may use as well to power a device connected at the end of the line):

$$P_{loss} = R_{line} \cdot I_{line}^2$$

or

  • The voltage drop along the line (from where you feed in the power [start] to where you extract it [usually end of the line] - NOT the voltage between line and ground, which I assume you 50kV are)

$$P_{loss} = \frac{V_{along\;line}^2} {R_{line}}$$

However, since you seem only to know the load and line-to-ground voltage, using

$$P = I_{line} \cdot V_{line,gnd}$$

to substitute the current is correct. The resistance appears reasonable for a longer medium voltage line and < 1 % ohmic losses is not that bad.

If you are interested, take a look at this page:

AC Transmission Line Losses, by Curt Harting

For AC: Use RMS values. If you have a multiple phase system, make sure you dont confuse line-to-line and line-to-ground voltages... Also the skin effect has to be taken into account.

General advice: If you are referring to values per length, use a dash: [P'] = W/m and also use them if you write something like "P = ...".

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  • \$\begingroup\$ It seems that substituting a permuted form of the same equation into itself would not yield any new information. I have always used P = IV to calulate energy/time of a circuit component using the voltage across and current through it. So naturally, I would think of the line as a large (long) resistor of vale 5.25 Ohms. Since we do not know the voltage across it, nor the current through it, how can we confidently calculate the power consumed by it? \$\endgroup\$ – sherrellbc Aug 19 '14 at 12:23
  • \$\begingroup\$ .. The subtitution of current using I = P/V assumes the line resistance is disspiating 940kW with 50kV across it with 18.8A through it; that is not true. Unless there is some trickery here I am not seeing this does not make much sense. \$\endgroup\$ – sherrellbc Aug 19 '14 at 12:27
  • \$\begingroup\$ @sherrellbc Do you have any suggestions?? \$\endgroup\$ – user2293224 Aug 19 '14 at 23:13
  • \$\begingroup\$ First step would be to clearly name all known values: What do you know about the line (resistance, line-ground voltage) and its load (voltage = line-ground voltage of line, power consumption 940 kW?). The most important thing is to get a good overview of what is given. With a given load and voltage, calculating the current makes sense and so does calculating power losses if resistance of the line is given and current equals load current. Hope this helped... \$\endgroup\$ – molotovsoda Aug 20 '14 at 8:19

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