48V in a 36v system

[cadstarsucks]

16 AWG wire is rated at 10 amps at 120 volts and can carry 1,200 watts. Reduce the voltage to 1 volt and 16 AWG can carry 1,200 amps because the derived wattage rating is not exceeded.

Post a reference... If you have one other than your own faulty logic then you should share this gem with all the wire manufacturers, none of whom appear to know this.

Dan

[myelectricbike]

See:  Reply #73 on: Today at 11:10:37 AM

Manufactures, especially, of products not intended for use with line voltage, may use the intended voltage to compute the current rating of their product.

[cadstarsucks]

See:  Reply #73 on: Today at 11:10:37 AM

Ok so you have a reference to ohms law and watts law, that still has nothing to do with your erroneous claims. 

P=I²R, ie motor current squared times copper resistance - VOLTAGE HAS NOTHING TO DO WITH IT!

Dan

[myelectricbike]

Except that I²R = V²/R and R=V/I.

P then equals (V²)/(V/I) and I=P/V.

[cadstarsucks]

Except that I²R = V²/R and R=V/I.

P then equals (V^2)/(V/I) and I=P/V.

Not true.  The back EMF of the motor shows up as a battery between one terminal and the resistance of the winding.  If you care to on a BDC, sit it on a rack, spin it up to 10MPH and take a voltmeter to it.  Subtract that number from the rotor voltage to get the "winding" voltage. 

As I said before, the power that moves the bike has to come from somewhere.  What you are saying is that you have a perpetual motion machine.  700W in, 500W mechanical out, and 700W of heat out for a total of 1900W out for 700W in.  Last I knew, and what you implied earlier, was that your physics class said that was impossible.

Dan

[myelectricbike]

I've never said anything like that. In fact even when Counter EMF is captured and used to charge the batteries, the load of charging the batteries requires the applied power to be even more. Thus Counter EMF capture or regeneration is limited to situations in which power is not being applied, such as in braking or going downhill or pedaling above the point of freewheel.

However, that is off topic and is a new topic you can start if you wish to find others willing to discuss.

Hijacking topics is frowned upon here.

[cadstarsucks]

Not hijacking at all.  BEMF is there all the time and it is directly proportional to the motor speed in a PM motor.  If you take the 500W Golden for example you can easily find the voltage on the "winding" from the ratio of the no load speed to the current speed at a given voltage.

You really should stop trying to guess what is happening electrically when discussing with an electronic engineer.  The motor would be modeled, overly simplistically, by a series resistor representing the losses in series with a perfect motor.  More realistically you would throw in some caps and inductors.  The ideal motor ALWAYS has BEMF, that is the ONLY thing that prevents the current from continuing to go up till it burns out.  Just because you can not see or use it does not mean that it is not there.

Dan

[myelectricbike]

This is the last time I will respond to any comments here not related to "48V in a 36V system." If you want to discuss CEMF then start a new topic.

Potential power is of course present and you can measure the voltage which indicates so but the potential power is not being tapped to power a load. If you did tap it then the speed would drop until applied power was sufficient to make up for the CEMF that was tapped to power the load ,otherwise you are saying that BLDC motors are capable of operating as perpetual motion machines, which they are not and never can be. 

Now start a new topic or my responses to comments unrelated to the designated topic will come to an end.

[pdonahue]

16 AWG wire is rated at 10 amps at 120 volts and can carry 1,200 watts. Reduce the voltage to 1 volt and 16 AWG can carry 1,200 amps because the derived wattage rating is not exceeded.

I'm not sure if you're joking here or not...  I will assume that you are, however, just in case there is someone in the crowd who doesn't understand:

The only knowledge required is:
POWER  = I^2 * R

I will use a 3 foot peice of 16 AWG wire since that is what I have on my bike.  The resistance of a 3 foot peice of 16 AWG wire is approximately 0.0014 ohms.  This is roughly a constant until the wire gets VERY hot.

When there are 10 amps running through it, the power lost as heat in the wire is:
 I^2 R = 10 * 10 * 0.0014 = 1.4 watts.  (Note that the power dissapated does not depend on the applied voltage)
At 1.4 Watts, the wire is easily able to pass that heat on to the surrounding air, and will not feel hot to the touch.

When there are 1200 amps running through it, the power lost as heat in the wire is:
 I^2 R = 1200 * 1200 * 0.0014 = 20520 watts. 
At 20520 Watts (approximately the same heating capacity as two typical home furnaces), the wire very quickly reaches the point where the insulation and/or the metal melts.

Pete




       

[myelectricbike]

Show me the calculation then when voltage is zero.

I've cut a 3 ft length of 16AWG and its sitting right here on my desk. Its not red hot or melting, in fact it even feels cold.

Wonder how can that be when your calculations clearly show it should be burning a hole through my desk.

[pdonahue]

Show me the calculation then when voltage is zero.

I've cut a 3 ft length of 16AWG and its sitting right here on my desk. Its not red hot or melting, in fact it even feels cold.

Wonder how can that be when your calculations clearly show it should be burning a hole through my desk.

When the voltage is zero, the current through the wire is:
I = V/R = 0/0.0014 = 0, so using the same formula as in my post, the heat dissapated in the wire is:
 POWER = I^2 * R = 0 * 0 * 0.0014 = 0 Watts
 0 Watts is enough power to do absolutely nothing to your wire.  I assume that is why it is not red hot or melting.

Pete

[myelectricbike]

So what voltage then would have to be applied to each end of my wire get 20,520 watts?

[Dave]

Show me the calculation then when voltage is zero.

I've cut a 3 ft length of 16AWG and its sitting right here on my desk. Its not red hot or melting, in fact it even feels cold.

Wonder how can that be when your calculations clearly show it should be burning a hole through my desk.

When the voltage is zero, the current through the wire is:
I = V/R = 0/0.0014 = 0, so using the same formula as in my post, the heat dissapated in the wire is:
 POWER = I^2 * R = 0 * 0 * 0.0014 = 0 Watts
 0 Watts is enough power to do absolutely nothing to your wire.  I assume that is why it is not red hot or melting.

Pete

Exactly, Pete. 0 times anything equals what? Zero.

[pdonahue]

So what voltage then would have to be applied to each end of my wire get 20,520 watts?

That would require that you apply a voltage of:
V=IR = 1200 * 0.0014 = 1.68V.

If you have a 1200A power supply around you can try it out for yourself!!!  However, you can verify the effect using a 100A power supply (I have an Agilent 100A, 20V supply and I have verified (by accident) that this works.  If you hook up a single piece of wire and apply 0.15V across it ( I = V/R = 0.15 / 0.0014 = 107A) you will find that it heats up quite significantly.

[myelectricbike]

What equation did you use to obtain a resistance of .0014 per 3 ft of 16AWG and what were the values for voltage, current and power?

R=P/I²
R=V/I
R=V²/P