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=26.43
cable resistance=.000168
voltage=34.5 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
Reply to
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=26.43 cable resistance=.000168 voltage=34.5 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
Reply to
Troy
Try the Neher McGrath paper at:
formatting link
Reply to
Gerald Newton
I am still not able to get it. Does anybody know the formula I need to use?
Reply to
Troy
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
Reply to
daestrom
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
Reply to
Troy
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?
Reply to
Gerald Newton
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
Reply to
The Mighty WontVolt
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
Reply to
daestrom
do people really still use ft as a measurement ? I thought that went out in the 70's ?
Reply to
George Vest
Only in the USA.
Reply to
Don Kelly
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.
check your data.
Reply to
Don Kelly

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