Well I’ve got to the point where I understand about Tractive Effort, and the I get Continuous Tractive Effort thrown in my face. Looking at data for some popular models GP38-2’s and GP40-2’s are rated at 54,700# CTE at 11.1mph. From what I understand this means that at a minimum sustainable speed of 11.1mph there is 54,700 lbs of tractive force available. But what about higher speeds? 11mph would take over 8 hours to travel 100 miles. Is there a way to figure out the CTE at different speeds for a locomotive? Or maybe even charts? I got my data from thedieselshop.
My understanding is that tractive effort is the tonnage that can be pulled disregarding speed, which is largely the function of weight on the driving wheels. However, traction motors on locomotives will burn up if operated at maximum power below a certain speed. That “certain speed” is the continuous tractive effort (maximum load at the minimum, undamaging speed).
Mark
Yes. How about an equation? In general, HP = TE * speed/308 (TE in pounds force, Speed in mph) The limit on the low end is usually the thermal capacity of the traction motors. Try to push more amps through them on a continuous basis and the insulation in the windings go away. The limit on the high end is the maximum voltage output of the main generator. This is usually somewhere above 60 mph. There are lots of exceptions, caveats and asterisks here, and they have be hashed and rehashed ad nauseum elsewhere on these forums. But, this general equation will get you close. If you plug in 54,700# TE and 11.1 mph, you get 1970 HP - right on the money for your GP38. 2000 HP at 40 mph would give you 15,400# TE. (A GP40 that shows 11.1 mph for the continuous TE speed has a special feature that derates the HP as the locomotive slows down below about 20 mph. This feature allow it to run behind a GP38 or SD40 without having to worry about damaging the traction motors on the GP40. As long as the SD40’s load meter is in the green, the one in the trailing GP40 will be, too. Without it, 3000 = 54,700 * speed/308, solve for speed = 16.7 mph. So, if you were in notch 8 at 12 mph, the SD40 would be OK and the GP40 would be in trouble.)
Say a GP38-2 is pulling a 2000-ton train on level track at 20 mph. They move onto a 1% upgrade and speed naturally drops; on paper it shouldn’t drop below 11 mph, required TE will be less than 50000 lb, required current thru each traction motor will be less than 1050 amps, and the blowers can cope for however long it takes-- the motor heats up, but the blower can keep it from heating beyond its maximum allowed temperature.
But if they have 3000 tons, speed will drop below 10 mph, TE will continue to climb, amps will climb to 1500 (if adhesion allows) and the traction motor blower can no longer keep the motor cool enough. The motor can stand 1500 amps for a few minutes, but then the engineer needs to shut off. In reality, speed may drop to zero before they even use up the allowed few minutes-- 3000 tons is way too much for a GP on 1%.
Minor clarification: In Don’s formula above, the ‘308’ has been reduced a little bit - down to about 82 % - from the theoretical number of ‘375’ to allow for friction and other mechanical losses and non-traction uses in the locomotive’s machinery. If you derive the formula from scratch, it’s as follows:
1 HP = 550 ft.-lbs./ second = 550 lbs. x 1 ft./ sec.
Since 1 MPH = 5,280 ft. / (60 mins. x 60 secs.) = 1.47 ft. /sec.,
Divide by the 1.47 ft. / sec. per 1 MPH = 550/ 1.47 = 374.15 - call it 375.
Also note that the 308 factor needs to be changed to 345 when discussing modern AC motored locomotives. This accounts for the improvements in mechanical and electrical efficiency of today’s locomotives compared to the old GP38s and SD40s which were first designed nearly 50 years ago.
Paul,
The 375 figure is exact.
Original definition of 1 HP = 33,000 ft-lbs/minute (equals 550 ft-lb/sec)
1 MPH = 5280 ft per mile/ 60 minutes per hour = 88 feet per minute
33,000 / 88 = 375
- Erik
Erik -
Thanks ! [tup] I thought it should be, and was still mildly curious as to why it wasn’t. In my math, I was using the short-cut of 1 MPH = 1.47 ft./ sec., and so that’s why the slight discrepancy. Moral: When in doubt, go back to the original derivation.
Altogether, an enlightening little diversion. Thanks again.
- Paul North.
As always, it’s a puzzle just what percentage of their rated power tocomotives actually produce at the wheelrim (as distinct from what they’re supposed to produce). An additional puzzle is how slow the engine can be moving and still maintain its hoped-for percentage. How often do ES44s actually produce 101,200 lb of TE at 15 mph? The only way we fans can hope to find out is to go out and clock them, hoping that we know the train’s actual tonnage.
If a 7250-ton eastward train with four ES44s could climb Donner without dropping below 15 mph, that would be a good indication they were living up to the 345 factor.
Actually, what’s a puzzle to me is why anybody would bring up the performance of am outdated locomotive such as a GP38 or SD40 in 2009. What’s the short time rating of a SD40-2? Who cares anymore. Might as well be discussing the performance of a Model T. AC traction is the future of railroading. Railfans should be spending their time researching the latest motive power technology, not some inefficient 40 year old pile of junk.
Who peed in your cornflakes?
Okay, okay…you admit to being puzzled. Why not just enjoy the show unfold while they discuss a diesel about the same age as your own namesake?
-Crandell
This discussion began with an inquiry about the definition of continuous tractive effort; and the point that I took from “GP40-2’s” posting is that the concept of CTE has changed significantly over time.
Locomotive manfacturers still cite “continuous” TE and “continuous” speed figures; however because of thermal protection circuitry in DC-traction units and the fact that AC-traction motors are much more heat-resistant than DC-traction motors, the “continuous” concept is applied differently today than it was three or more decades ago. The most extreme example of this is AC-traction technology. For AC-traction purposes, CTE is defined entirely differently than it is for DC-traction purposes. It doesn’t relate to the maximum amount of tractive effort that a locomotive is capable of producing “continuously” without risk of overheating its traction motors but, rather, to the amount of tractive effort that the locomotive will produce at the level of adhesion that it can be relied upon “continuously” to maintain.
So I think that the issue here is whether the original question related to the concept of CTE as it existed 30-some years ago or to the concept as it exists today. “GP40-2” seemed to assume that the question related to the latter; and that assumption seems reasonable to me.
Point me to a Class 1 railroad operating officer with responsibility for moving trains over any sizeable territory that doesn’t need to know about or care about the minimum continuous speed of an SD40-2.
RWM
Good question and I don’t have a quantitative answer. I’ve seen plenty of data on motor efficiency versus speed and tractive effort for DC electric locomotives (e.g. from CERA’s Electrification by GE), but nothing for recent AC inverter driven traction motors. There have been a few comments on this forum asserting that the transmission efficiency of an AC diesel-electric locomotive is on the order of 94%. What was not said was what ranges of speed and tractive effort for which the 94% figure was valid. The figure is certainly not valid for max tractive effort at speeds much less than 10 MPH due to slip inherent with induction motors (will be different with synchronous motors).
I have seen the efficiency map for the UQM permanent magnet motor intended for electric vehicles and it implies that the efficiency for a synchronous traction motor will fall off significantly with speed due to windage, hysteresis and eddy current losses. An induction motor will probably show a slower fall-off with increasing speed and decreasing load as the magnetic field can be reduced (lowering hysteresis losses) in proportion to to tractive effort.
- Erik
For how much longer? SD40-2s are being taken out of mainline service about as fast as railroads can speed dial the sales department at GE for new motive power. A trend I see accelerating once we are out of this recession OR the cost of diesel fuel increases, whichever comes first. Like I said, you might as well be discussing the performance of a Model T…
Reality check: The railroads have to work with the euquipment they have, not the equipment they may one day have. I’d say that any railroader who didn’t concern himself with the equipment he has now, won’t have a job when the the new stuff arrives.
I genuinely wish you could be on my next 0500 conference call to express your opinions.
RWM
Sounds like you’re saying an AC unit can produce, say, 180,000 lb for eight continuous hours, if adhesion is that good for that long. In the past, didn’t GM say an SD70ACe was good for 137,000 lb or 145,000 lb (or whatever it was) “continuous tractive effort”? If that depends on adhesion and nothing else, where did that figure come from?
Seems like a “CTE” figure that depended on adhesion and nothing else would have to be based on probability-- the unit has an X% chance of maintaining Y pounds for an indefinite period of time. Next question: does that percentage assume dry rail? If so, what’s the percentage for wet rail? How wet? Does it apply to the lead unit? Or just trailing units? Does it assume the possibility of curve greasers?
Yes, I think that the CTE concept, particularly as applied to AC-traction units, is “based on probability”. And as the term “continuous” implies, variables such as consist position and rail conditions apparently have been factored into whatever the assigned probabilities are. For example, EMD estimated (calculated, determined, or whatever) that an SD50 could be relied upon to maintain 24% adhesion 99% of the time and 31% adhesion 95% of the time and that an SD70MAC could be relied upon to maintain 33% adhesion 99% of the time and 39% adhesion 95% of the time.
Timz asked where figures like those come from; and my answer is that I really don’t know. And I presume that if railroads viewed them as absolutes, there would be no need for the railroads actually to test locomotive models before acquiring them.
So my personal for-whatever-it’s-worth view of the CTE concept is that CTE is a parameter that is more useful for comparing the estimated performance of one locomotive model against another locomotive model than it is for estimating the actual performance of any given locomotive model.
Evidently at least one Class I doesn’t see it that way, and is quite happy with the SD40 platform - so much so that they are now essentially building their own.
Norfolk Southern has now rebuilt about 20 SD50’s to ‘SD40-E’ specs for helper service on the Altoona-Cresson-Pittsburgh grades. See -
Juniata Diesel Conversions
SD50/SD50S to SD40-E
at http://www.altoonaworks.info/rebuilds/ns_sd40e.html
and especially the details/ scope of the work at the bottom of the page.
I also see them in many lash-ups on the mainline runs, and even taking turns on the local freights. They all look well-maintained, and have probably outlived several other models of both EMD and GE parentage. I wouldn’t call them Model T’s - more like 1960’s Chevy pick-up trucks. Not glamorous and and the most efficient and not state-of-the-art - but easy to understand, repair, and basically dependable power - as close to ‘bullet-proof’ as we’re ever likely to see. At least they can be fixed with standard wrenches - don’t need to re-boot a computer for most functions . . . [;)]
- Paul North.