OCPD tripping on 24V DC circuit

[hurk27]

I would say that this in in the location of a Train locomotive engine, I would say the wiring is of a high temp design, and might allow a higher rating that the building wiring we are use to. we see this even in cars and trucks, many times will have conductors protected by a fuse or breaker larger then would be allowed to by the NEC, Since almost all locomotives are manufactured by GE, I would contact them as to how this should be solved.

 

[philly]

091029-1715 EST

philly:

Your new schematic is whole different story.

For my estimates of lamp current I will double the values in my table. Now the intersection point is approximately 10 V for the lamp voltage with 2 ohms in series with a pair of lamps.




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Can you explain this part how you come up with the values in the table and how this works with plotting the intersection point?

As a _rough_ approximation, the power consumed by an incandescent lamp scales as the V^(1.5) Since power is simply voltage * current, this approximation can be restated as the current consumed by an incandescent lamp scales as V^(0.5) You can calculate the resistance from the 're-rated' voltage and current if you wish.

Your lamps are rated as 200W at 30V, or 6.67A at 30V. At 24V the approximation suggests 5.96A.

If you want to solve the series circuit of a resistor and a lamp, I'd suggest writing the equation in terms of voltage and current rather (in a resistor I scales as V^1 rather than converting lamp voltage/current into resistance.

-Jon4

So you are recomending solving the circuit in terms of voltage and current so that you dont have to deal with the changing resistance? The one problem I had with solving these circuits was dealing with the changing resistance in the equations due to the changing temperature.

 

[gar]

091030-0743 EST

philly:

The table of currents I created were based on measurements of current vs voltage of a 25 W 120 V tungsten bulb. Tungsten whether in a small bulb or large will have the same relative change in resistance vs temperature. There may be some problems in scaling from the gases and other factors in different bulb construction affecting radiation and conduction, but as a first order approximation the results of scaling should be useful.

It would be best if you ran a I - V curve for the actual lamp you are using. Knowing that your bulb was rated at 200 W at 30 V the calculated current at this operating point is 6.67 amperes. I have to assume the color temperature, temperature of the filament, of your bulb is the same as the 25 W lamp I used to create the basic curve. My current at 120 V for the 25 W bulb was 0.207 A and this calculates to 24.8 W. Good correlation. Then I measured the current at 120, 110, 100, .... 0 V. Next I took the ratio of 6.67/0.207 = 32.21 and applied this to each of my current readings. Since my 120 V corresponds to your 30 V I divided each of my voltage values by 4 to obtain the corresponding voltage relating to your bulb.

Assuming the color temperatures of both bulbs are the same at their rated voltages, then operating the 30 V bulb at 24 V makes very good sense because it greatly increases the bulb life.

Once I have an I-V curve for a lamp, then with a constant series resistance I can determine the operating point by plotting a straight line on the lamp I-V that represents the constant resistance. Where the two curves intersect is the steady state operating point.

The resistance curve is plotted by picking two points. One point is with zero current thru the resistor, and that is placed at the value of the source voltage. In your case 24 V. So on the graph put a point at zero current and 24 V. Next calculate the current thru the resistor if the full source voltage is applied to the resistor. For 2 ohms and a source of 24 V this is 12 A. So at 0 V on the plot put a point at 12 A. Draw a straight line between the two points. For other source voltages you simply draw lines parallel to the line for 24 V.

Also note for two lamps in parallel the current at a given voltage is double that of a single bulb. Next plot both the single bulb and two bulb plots and observe how the intersection point with the fixed resistor varies.

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[hurk27]

Philly

just for a note, because you probably have already checked, those lamps also come in a 250 watt version, so make sure someone hasn't installed the wrong lamp, I still have 3 of them in my garage.