current flow efficiencies - proper terminology

[motolectric]

Hi,

I'm writing some documentation for some low voltage (11-14 V) circuits where the steel and brass terminals have been replaced by copper terminals.

I just want to be sure I am stating things properly. The data I have found says that brass flows only 26% of the current that copper flows and steel only 10%.

Am I correct in claiming that the rest of the energy was *all* converted to heat? Or is there any other action that is taking place that is using the energy or degrading the flow of the energy?

If a consumer asks for 10 amps and an all copper circuit gives the consumer the 10 amps, would the consumer only get 2.6 amps if the circuit was all brass?

Thanks for any help/tips/advice.

M./

 

[KeepItSimpleStupid]

OK, here's the deal. There is a material properly called resistivity (p=rho) has units of (ohm-cm) which is independent of the amount of materal. It's sort of like the parameter density.

There is a prarameter Resistance, which is dependent on the cross-sectional area and the length Thus the resistance R = pL/A. Resistances usually mean heating losses in terms of I^2*R where I is the current flowing in amps and R is in ohms.

Thus to carry the same amount of current safely using aluminum wire, the wire mad out of aluminum must be thicker. This is appropriate for power transmission.

Dissimilar metals causes other effects and so does temperature differences. These are negligible in most instances.

 

[motolectric]

A small clarification and expansion

Thanks for the reply.

But let me be more clear.

The circuit I'm talking about is the starter circuit on motorcycles and so the current draw runs between about 60 and 90 amps depending on the particular bike. When the starter is run for 5-10-15 seconds (because of the losses the bike is not starting properly and they keep cranking the motor) on some mid-90s bikes the wiring actually heats up (like very hot).

So I'm trying to give laypeople some concepts they can understand easily. I'd like to give them real data on what portion of the 60-90 amps is turning into heat. I have done clamp on current readings with the original circuits and of course the losses vary widely bike to bike. After I upgrade the circuit to all copper (and new 250% larger wire) we find that the current flow is exactly what would be expected for the wattage of 700 watts (as used on the Ducati motorcycles for example).

So my goal is to give these bike owners (across various brands, which means various makes of starter motor and various current draws) some numbers which are generic to the types of terminals they have on their bikes and not have to have a lot of different figures here and there that might confuse them.

All the bikes used plated brass and some use plated steel terminals (the Ducatis) in their starter circuit, (none use copper terminals, not even plated copper). So they all experience large losses quite quickly and when they go to their MC brand dealer they get sold a new battery for $75-$150 and for 2 months it can overcome the losses and start the bikes. But after a few months they are back where they started. This means they are eating through batteries on a 12-18 month clip and that contributes to the hazardous waste problem that affects us all (unless you think motorcycle guys are stringent in their battery disposal routines - hint: they're not).

These losses are also there on cars but because cars have batteries that are 2-3 times the capability of a motorcycle battery (usually as small as they think they can get away with for weight and cost reasons) the cars don't really have the problems that motorcycles do.

So any axiomatic type explanations would help me the most.

Things like "yes, if the terminals are the same size and one is brass then it will only flow 26% of the current the copper terminal would flow and all the rest is turned into heat".

But if a portion of the 74% does not turn into heat, where does it go?

I just want to contribute real data and not bogus BS to the dissemination of this information. In the motorcycle field I sound like I am wearing a tin hat because many of these guys have spent anywhere from $100-$1000 at their (highly regarded in the industry) dealer trying to troubleshoot why their bikes don't start and I come along and say it's due to brass/steel terminals and barely within spec wire. (as you can imagine) I get a lot of disbelievers. But every guy who buys one of my all copper kits marvels at the the fact that his expensive (or in many cases just old) motorcycle now starts in a third to a half of a second. Which to my way of thinking defines the way the battery-wire-starter circuit was supposed to work since the consumer is fixed at its stated wattage.

Again, thanks for your reply, any and all commentary is appreciated.

M./

 

[KeepItSimpleStupid]

Ignore terminals for the time being

Here is a voltage drop calculator for various wire gages: Voltage Drop Calculator

So you can use the following. Since P=VI and I is constant, then the voltage drop in % is the amount converted into heat.

You can easily create a table that uses that looks like:

Assume a 2' length or whatever is standard; 100 Amps;

Starter current; OEM size wire (##AWG); % voltage drop at 12V;Upgraded wire size (##AWG); % voltage drop at 12V

Note power lost in % is the same as voltage drop in % because we assume constant current.
There are effects based on temperature and conduit.

You might be able to show the temperature rise of the wire based on say a 30 second crank and say cooler wires last longer.

 

[MikeMl]

Bottom line is that replacing a brass terminal with a copper wont make a noticeable difference.

Look at the circuit below. Notice what happens to the power lost at the terminal as the terminal resistance goes from ~0.2m to 0.4mΩ, such as what might happen if you replace brass with copper.

Note that the power in the starter motor (red trace) is ~700W, with the power lost in the wire (purple trace) is ~7W, while the power lost in the terminal is ~1W, a tiny fraction of the total. If you reduce the power loss in the terminal from 1 to 0.5W, BFD!!!!

Selection of terminal metallurgy is usually driven by compatibility with the wire...

 

[crutschow]

I just want to be sure I am stating things properly. The data I have found says that brass flows only 26% of the current that copper flows and steel only 10%.

Am I correct in claiming that the rest of the energy was *all* converted to heat? Or is there any other action that is taking place that is using the energy or degrading the flow of the energy?

If a consumer asks for 10 amps and an all copper circuit gives the consumer the 10 amps, would the consumer only get 2.6 amps if the circuit was all brass?
/

It's true that the resistance of brass and steel terminals are higher than copper but the terminal resistance is only a small part of the total resistance that determines the current flow. This includes the battery resistance, the wire resistance, the motor equivalent resistance when cranking, and the frame resistance (the current return path).

So even though the steel terminal may have significantly more resistance than a copper terminal, it's likely still a small portion of the overall resistance and thus only has a small effect on the current flow. Changing the wire to a larger gauge probably has more effect in reducing circuit resistance then changing the terminal material.

It's true than any resistance in the circuit will cause heat dissipation.

 

[#12]

That statement, "steel only flows some percentage of what copper flows" is the problem. Throw it out. That "specification" is worthless.
Yes, all losses turn into heat.

Other people here have explained very well how to add up the losses of each part and how to do it correctly. Trust them.