# Cable Loss (KVAR)

Hi,
I am trying to figure out reactive power loss for a 965 ft 1/0 AWG cable.
I have already calculated the power loss at .3397 kw.
total current&.43 cable resistance=.000168 voltage4.5 kv power factor=.95 The formula for the kw is: ((26.43^2 x .000168) x3) X 965 339.74/1000 = .3397 kw
I have been trying to use the formula Q=(P*sin p.f) 1.73 to calculate the KVAR loss, but I am not coming up with the right answer. The answer should be about 8.19 KVAR.
Any idea?
Thanks,
Troy
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Hi,
I am trying to figure out reactive power loss for a 965 ft 1/0 AWG cable.
I have already calculated the power loss at .3397 kw.
total current&.43 cable resistance=.000168 voltage4.5 kv power factor=.95 The formula for the kw is: ((26.43^2 x .000168) x3) X 965 339.74/1000 = .3397 kw
I have been trying to use the formula Q=(P*sin p.f) 1.73 to calculate the KVAR loss, but I am not coming up with the right answer. The answer should be about 18.19 KVAR.
Any idea?
Thanks,
Troy
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Try the Neher McGrath paper at: http://www.electriciancalculators.com/ampacity/ampacity.htm
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I am still not able to get it. Does anybody know the formula I need to use?
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How many conductors are in the cable? What is the circular mil ohm dc resistance of the conductor at 20 degrees C? What is the ambient temperature? If the cable is run as an individual conductor what is the configuration and distances between it and other cables, and is this a three phase sytem? Is this a shielded conductor? Does it have a concentric neutral? Is this cable in open air or in a duct?
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Who is this, "Jacob Two Two's" cousin ?
Inconsequential...
I trust thou art not working for the mystical land of tomorrows Utility Company.
The Mighty WontVolt
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Troy wrote:

No, the formula for Q at the load is Q=V*A*sqrt(1-pf^2)*1.73. To use 'sin' you need to know the angle between current and voltage (which is arccos(0.95) so another formula would be Q=P*sin(arccos(pf))*1.73). But knowing that sin^2+cos^2=1 makes it possible to find the sin(angle) from the pf directly as I showed.
But this will give you the reactive power of the *load*, not the amount of reactive power (if any) in the *cable*. You can *not* use the real power loss in the cable in the above formula.
To find the reactive power consumed in the cable, you need the cable's reactance. A multi-conductor cable would have some capacitance as well as inductance and resistance. But a cable this short, I doubt there is much reactive losses anyway. Now, if it were a transmission line stretching a few miles, that would be another story.
If you have precision instruments, an experimental method would be to measure the power factor at the load end and then measure it again at the supply end. With that and some math, you can figure out the difference in reactive power through the cable.
The real power losses in such a cable are of interest because they cause heating in the cable. The reactive 'losses' are probably of no consequence.
daestrom
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wrote:

Hi Daestrom,
Thanks for your help! Would it be the shunt capacitance reactance that I need for the calculation? The data sheet I have shows 6691 ohms per kft
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Troy wrote:

Yep. Then you have a 'transmission line' and you can look up formula for a pi configuration for example (1/2 shunt cap, the series inductance, then the other 1/2 shunt cap).
This type of 'four-terminal network' calculation is common for longer transmission lines. In the olden days it was done with some simple graphical methods, but nowadays a computer spreadsheet works just as good.
daestrom
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daestrom wrote:

6691ohms/kft doesn't appear to be a capacitive reactance figure. This would imply that the total line capacitive reactance increases with distance- it doesn't- the susceptance does. What is of more importance at this length is the inductive reactance.
--
Don Kelly
snipped-for-privacy@shawcross.ca
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do people really still use ft as a measurement ? I thought that went out in the 70's ?
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George Vest wrote:

Only in the USA.
--
Don Kelly
snipped-for-privacy@shawcross.ca