• Tractive Effort/Throttle/While Slip Question

  • General discussion about locomotives, rolling stock, and equipment
General discussion about locomotives, rolling stock, and equipment

Moderator: John_Perkowski

  by Denver Dude
 
I am in a discussion with an class 1 RR engineer who claims that he run in run 8 even down to 1 MPH. My answer is that that's impossible - the power HAS to be reduced or there would be major wheel slip. I contend that the throttle may be at 8, but the actual power to the rails is far less.
I know that maximum tractive effort is the weight of the locomotive times the adhesion percentage - 180,600 lbs. for a 432,000 lb. unit with an adhesion of 43%. I know that tractive effort decreases with speed, but how is that calculated? (I also know that the more powerful the prime mover, the higher the minimum speed at full power). I can't find that anywhere.
I am looking for a logical and accurate way to respond.
  by Pneudyne
 
I think that there are two key issues here:

1. The relationship between power, tractive effort and speed.

2. What happens to these parameters if a given locomotive is left in notch 8 down to zero speed, assuming that it does not slip or otherwise damage itself in the process.

Ignoring the constants involved, power is the product of tractive effort (a force) and speed (in the direction of that force). So at a given power level, tractive effort will decrease with increasing speed. That is shown in this old (late 1950s) GE chart:

from DRT 195807 p.285.jpg

The exact relationship will vary by locomotive according to transmission, load control accuracy and so on. For a given locomotive, it will also vary over the speed range. For example, transmission efficiency varies over the range, tending to be lower at higher main generator/alternator currents. Roughly though, on the full power curve, doubling the speed results in halving of the tractive effort. The constant power curve is proximate to a rectangular hyperbola.

Now to the case where the locomotive stays in Run 8 down to a standstill, without slipping. Another old GE chart will suffice here.

from DRT 195807 p.286.png

The central, concave part of the notch 8 curve is the constant power section, with 1600 hp drawn from the diesel engine. The electrical power is around 1100 kW or so, varying a little over the curve. Since 1600 hp is roundly 1195 kW, the difference represents heating (I²R) losses in the generator and traction motor windings. As may be seen, at the bottom (right-hand) end, the notch 8 curve departs from constant power, following unmodified generator curve and drops down to the standstill line. At the standstill line, the current is around 3100 amps and the voltage is 100 volts. So the electrical power is 310 kW, roundly 415 hp. The demand on the engine will be more than that, perhaps around 500 hp, as even with zero electrical load, the generator requires some power to be turned at speed. (Including power to windage and its internal fam.) The engine is clearly underloaded. And all of the electrical power is produced is going into heating the generator and traction motor windings, not into propulsion, as zero speed means zero power at the drawbar. It may also be seen that engine power output is progressively reduced from 1600 hp as one progresses from the bottom end of the constant power section down to the standstill line. The engine power output at 1 mile/h, near the bottom of the curve, will be much closer to 500 hp than to 1600 hp.

Thus your contention that at 1 mile/h, the throttle may be at 8, but the actual power to the rails is far less is shown to be correct.

This chart gives a picture of power distribution over the speed range. One may assume that for the locomotive in question, getting down to zero power at full throttle was not possible, either because of self-protection or self-destruction.

Locomotive Power Curves.gif


Cheers,
You do not have the required permissions to view the files attached to this post.