Power calculations/usage - off grid wedding

[RainorshinePNW]

Hey all- another electrical math question I'm hoping those that are smarter than me in this department can help with.

I'm planning on using my Lightning to supply power for a wedding deep in the national forest. Uses will include mostly catering equipment and lights (LED strings). Total power need is not all that significant, but there are some tricky aspects. Namely the distance from where I need to park the truck to where we'll need power- about 200 ft.

My initial thought was to use the 240v 30amp outlet/circuit in the bed to run 100 ft to a small spiderbox/mobile power distributor (like this), and then from there run 100 ft 120v extension cords. I thought this would deliver more power further than trying to use the (2) 120v 20amp circuits?

Next question is about power usage/consumption. If I add up the watts of everything I need to power (~4000w/4kW), is it simple that multiplied by the number of hours I'll be powering that equipment to get my total kWh used?

I know I'll have about 75% SOC when I reach the destination (at 5000 ft elevation), so trying to understand/make sure I know how much power I'll consume while there, to make sure I have enough left to get back

Thanks!

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[adoublee]

Hey all- another electrical math question I'm hoping those that are smarter than me in this department can help with.

I'm planning on using my Lightning to supply power for a wedding deep in the national forest. Uses will include mostly catering equipment and lights (LED strings). Total power need is not all that significant, but there are some tricky aspects. Namely the distance from where I need to park the truck to where we'll need power- about 200 ft.

My initial thought was to use the 240v 30amp outlet/circuit in the bed to run 100 ft to a small spiderbox/mobile power distributor (like this), and then from there run 100 ft 120v extension cords. I thought this would deliver more power further than trying to use the (2) 120v 20amp circuits?

Next question is about power usage/consumption. If I add up the watts of everything I need to power (~4000w/4kW), is it simple that multiplied by the number of hours I'll be powering that equipment to get my total kWh used?

I know I'll have about 75% SOC when I reach the destination (at 5000 ft elevation), so trying to understand/make sure I know how much power I'll consume while there, to make sure I have enough left to get back

Thanks!

Yes, there is more power inherent to the 30A 240V than two 20A 120V's. Think of the 30A 240V as two 30A 120V's.

Copper is your friend when it comes to distance. The more copper in circuits, the more voltage you will get through to the things you are connecting. If you can learn the wire gauges of potential products you would use, you could use an online voltage drop calculator to see which maintains the highest voltage at the connections. Shoot for not more than 5% drop and you best not be more than about 10% down.

Things I would be thinking about would be 1) Will I be able to balance the load such that there is not more than 3.6kW connected to either one of the two legs of 30A 240V outlet. Also, are there any loads the Lightning will perceive as a ground fault - might try and plug everything in once in advance if possible.

Yes, kW used times hours run is the general formula for kWh you will use. Assume the battery will go down around 10% the kWh rate as the AC power it is telling you it puts out of the outlet. Play with the setting that preserve your driving range in advance, or the truck might shut you down mid ceremony.

Good luck!

 

[Maquis]

Run the 240V circuit in tact as far as you can (full 200’ if possible) and balance the loads on each leg the best you can. That will minimize voltage drop.
It sounds like you don’t have any loads that are voltage-critical, anything with a motor. LED lights should be fine, and worst-case, the heating equipment has to run a little longer to maintain the desired temperature.
Yes, KW x Hrs = kWh.

 

[sotek2345]

Run the 240V circuit in tact as far as you can (full 200’ if possible) and balance the loads on each leg the best you can. That will minimize voltage drop.
It sounds like you don’t have any loads that are voltage-critical, anything with a motor. LED lights should be fine, and worst-case, the heating equipment has to run a little longer to maintain the desired temperature.
Yes, KW x Hrs = kWh.

You are going to have some conversion losses, so to be safe I would do (adds 10%):

KW x Hrs x 1.1 = kWh used

 

[Maquis]

You are going to have some conversion losses, so to be safe I would do (adds 10%):

KW x Hrs x 1.1 = kWh used

That’s a fair point…..I’d hope it’s less than 10%, but better to be on the conservative side.

 

[FlasherZ]

One final note: probably not that great of a deal, but the gauge of your extension cords makes a big difference as your load current goes up. LED lighting performance will not suffer, but with the lower voltage you'll end up with a higher current. Incandescent lamps and motor loads can suffer due to the voltage drops.

4000W should probably be distributed among 4 different cords, minimum 12 AWG (IMO). A lot of 100 ft cords are 14 AWG or even 16 AWG. Your loads will work with thinner cords (14/16), but what will happen is that you will lose power along the cord, P=I^2*R. Resistance of 16 AWG copper is about 0.40 ohms for 100 ft, and since the circuit is 2x that, at 10A you'll end up losing (10*10)*.4 * 2 = 80W along the length of each cord. If you have 4 of those cords, each at 10A, you'll be losing 320W.

That means for a 4800W load equally distributed across 4x16 AWG cords, you'd be losing 1% of your battery capacity every 4 hours just to heating extension cords up.

Compare that to 12 AWG where resistance is 0.159 ohms per 100 ft. You'd only lose 100*100*0.159*2 = ~32W per cord, or 128W across 4 cords, which would give you 10 hours before you'd lose 1%.

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