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Old 24-04-2020, 06:48   #1
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Dump load theory question

I've proceeded to attempt to build my own dump load resistor from an oven element. Is there anyone who can check and confirm my math? ��

Oven element
230v 3500w 15 ohms. 50" total length.

To convert to a 15v 1500w dump load resistor :
Volts squared / resistance = Watts

15 ohms / 50" = .3 ohms per inch
.3 ohms x 5 = 5" for 1.5 ohms each.
5" sections x 10 = 1500w @ 15v
.15 ohms total in parallel

15v x 15v /.15 ohms = 1500 watts
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Old 24-04-2020, 07:10   #2
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Re: Dump load theory question

Yes, but I think you misplaced a decimal point.
Ohms = Volts˛ ÷ Watts
(230 x 230) ÷ 3500 = 15 Ohms
Watts = Volts˛ ÷ Ohms
(15 x 15) ÷ 15 = 15 Watts
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Old 24-04-2020, 07:11   #3
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Re: Dump load theory question

Yes indeed sir. I will be cutting this element into 10 equal lengths of 5" inches each and wiring them parallel.
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Old 24-04-2020, 07:22   #4
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Re: Dump load theory question

Coming at it from the other direction:

15V/1.5Ω=10A

10 Elements x 10A =100A

100A x 15V = 1500W

Just curious, is this a fixed dump load or are you using switching to control individual (or groups of) elements to give you more regulation of the amount of power being dumped?
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Old 24-04-2020, 07:25   #5
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Re: Dump load theory question

Greetings and welcome aboard the CF, Roger.

Power (Watts) is proportional to both Current (I) squared and/or Voltage (V or E) squared.

Handy wheel, describing Ohm's Law & iy's derivatives:

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Old 24-04-2020, 07:32   #6
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Re: Dump load theory question

It's going to be a fixed value dump load with a relay driven from a voltage monitor circuit board for dumping the load from a wind turbine generator around 14.7 volts and switching back to the batteries when battery voltage falls to a specific value. Both cut in to dump load and cut away from dump load voltages will be adjustable. I'm just getting started with this hobby at home. 🏡😁. I will be using 300w of solar panels with thier own charge controller and a thermodyne wind generator supposedly rated for 12vdc @ 1600w(I'll be happy to achieve 600w from it. As I know they are way hyped and overrated) 😁. Thank you for your time sir. So I can assume my math is correct?
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Old 24-04-2020, 07:43   #7
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Re: Dump load theory question

Quote:
Originally Posted by Roger Clement View Post
So I can assume my math is correct?
Don't know about that, but we both came up with the same answer

I get 0.15Ω for your equivalent parallel resistance as well, and it all follows from there.

Have fun with the project.
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Old 24-04-2020, 07:54   #8
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Re: Dump load theory question

Why not just connect the dump load to the existing water heater?
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Old 24-04-2020, 07:59   #9
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Re: Dump load theory question

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Originally Posted by GordMay View Post
Yes, but I think you misplaced a decimal point.
Ohms = Volts˛ ÷ Watts
(230 x 230) ÷ 3500 = 15 Ohms
Watts = Volts˛ ÷ Ohms
(15 x 15) ÷ 15 = 15 Watts
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Yes indeed sir. I will be cutting this element into 10 equal lengths of 5" inches each and wiring them parallel.
How hot do you hope this oven to get?
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Old 24-04-2020, 08:03   #10
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Re: Dump load theory question

Quote:
Originally Posted by Roger Clement View Post
Yes indeed sir. I will be cutting this element into 10 equal lengths of 5" inches each and wiring them parallel.
I missed that.
Why would you bother cutting them up, to reduce the total effective resistance?

For resistors in parallel, the overall resistance (R total) is the reciprocal, of the sum of reciprocals, of the individual resistors.
1/Rt = 1/R1 + 1/R2 + 1/R3 +...
Hence:
1/Rt = (1/0.15) x 5 = 1/33.3
Rt = 1 ÷ 33.3 = 0.030 Ohms
Rt = R total
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Old 24-04-2020, 08:11   #11
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Re: Dump load theory question

Gord,

they're 1.5Ω per section, and there are 10 sections.

1/1.5 * 10 = 6.67, 1/6.67 = 0.15Ω.

He's cutting up a 15Ω resistor to get 10 resistors at 1.5Ω each, then paralleling those 10 gets 0.15Ω equivalent resistance.
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Old 24-04-2020, 08:18   #12
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Re: Dump load theory question

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Originally Posted by Dsanduril View Post
Gord,
they're 1.5Ω per section, and there are 10 sections.
1/1.5 * 10 = 6.67, 1/6.67 = 0.15Ω.
He's cutting up a 15Ω resistor to get 10 resistors at 1.5Ω each, then paralleling those 10 gets 0.15Ω equivalent resistance.
My arithmetic, and reading comprehension deficiencies, at work, again.
Thanks!
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Old 24-04-2020, 09:33   #13
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Re: Dump load theory question

if your dump resistor is .15 ohms it will take 100amps to create 15v across the resistor.
V= Rxi
15volts = .15ohms x 100amps

would it make more sense to size the resistor to see 15v across it when the source output is approaching it's maximum?
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Old 24-04-2020, 10:11   #14
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Re: Dump load theory question

Question to you guys. If I have 640 watts of solar will I have at 12 v 53.3 amps theoretically speaking without clouds or shading.
Thanx Ernie on the Mary Jane
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Old 24-04-2020, 10:27   #15
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Re: Dump load theory question

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Originally Posted by sailon46 View Post
Question to you guys. If I have 640 watts of solar will I have at 12 v 53.3 amps theoretically speaking without clouds or shading.
Thanx Ernie on the Mary Jane
Theoretically yes. And 44.4A at 14.4V charging voltage.

All at standard test conditions (STC) which are rarely achieved on a boat. But, in theory at least, yes.
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