andrew.d.zachar(at)gmail.
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 Posted: Mon Jan 10, 2011 2:48 pm Post subject: Very Simple Question about landing lights, wire size, and ci |
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Hey, everyone.
I have a pretty simple (ha!) question regarding the set up of a landing light arrangement for my in-work RV-7.
I've just ordered a set of 2 55W HID lights (http://www.ddmtuning.com/Products/Apexcone-Raptor-HID-Kit) and I'm starting to think about wiring for one light in each wing (or wingtip). (Note: The lights haven't arrived, and the instructions there might give some guidance, but I thought I'd ask anyway...)
Using page 1-3 of the 'connection as a guide, I'm going to assume these lights are really designed for 13.0 volts. Nominally, My 55W lights will pull A=W/V or 55/13.0 or 4.23A. I see from the wiring table later that 22 gauge will hold 5 A and has a resistance of 16.1 Ohms per 1000 feet. At 24 feet, I get 0.3864 Ohms.
To get the voltage drop in the wire... Volts = 4.23 Amps x .3864 Ohms = 1.63 V, Assuming the alternator is regulated to 13.8 V (still following page 1-3), the light is getting 13.8V - 1.6V = 12.2V, I'm losing 4.23A x 1.6V = 6.8 Watts lost through the wire. 12% loss.
I have a feeling the startup loads are a little higher than 4.23A, so let's bump up to 20 gauge wire.
(10.2 Ohms per 1000 feet. At 24 feet, I get 0.2448 Ohms. Voltage drop = 4.23 Amps x .2448 Ohms = 1.04 V, Assuming 13.8 V (still following page 1-3), the light is getting 13.8V - 1.04V = 12.76V. I'm losing 4.23A x 1.04V = 4.4 Watts lost through the wire. 8% loss)
Finally with 18 gauge wire, 6.39 Ohms per 1000 feet. At 24 feet, I get 0.15336 Ohms. Voltage drop = 4.23 Amps x .15336 Ohms = .65 V, Assuming 13.8 V (still following page 1-3), the light is getting 13.8V - 0.65V = 13.15V. I'm losing 4.23A x .65V = 2.8 Watts lost through the wire for 5% loss.)
That seems like a nice small loss. So, I'll go with 18 gauge wire (not that much heavier than 22ga, which could technically hold the 4.23A, but maybe not with startup loads.
So now I have two runs of 18 ga wire each carrying about 5 amps. I think the default answer of the list will be to separately protect the circuits with their own fuses in a fuse-block, but, for the sake of discussion, let's pretend I really want to run them through a singe breaker on the panel (or a breaker switch)...again, just for discussion.
At 4.23 Amps per side, I could use a fuse or circuit breaker (per the table) for 18 ga, which is 10A. But I may trip the breaker if warm-up current is higher than 4.23A+4.23A. If I bump up the breaker to 12.5A or 15A, I'm now not adequately (more rise in wire temp before tripping) the 18 ga wire. Right?
So, if I really want to run them both through the same breaker or switch-breaker, I need to bump up the wire size so the warm-up loads (is assuming 1.5x the steady state load a good estimate for startup loads?) of ~7A x 2 lights or 14A can be handled. Does that push me to a 15A breaker and therefore 14 ga wire?
Am I on the right track here? I seem to be leading myself to 20 ga (7A) or 18 ga (10A) wire with independent fuse protection, but a DPST switch controlling both, but I'm curious about the logic behind combining them in a breaker anyhow.
Thoughts? Is this the correct process for wire and circuit protection sizing? (Add up all the grouped loads for a given protection device (if you must combine), then size the wire appropriately for the size of protection device being used?)
As always, thanks in advance.
--
Andrew Zachar
andrew.d.zachar(at)gmail.com (andrew.d.zachar(at)gmail.com)
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