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  • Newbie question about engine torque.

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General discussion about locomotives, rolling stock, and equipment

Moderator: John_Perkowski

 #961860  by chapel
 
I build cars... so I'm used to engine power ratings in horsepower and in torque.
Now when I look up specs on the engines that power the locomotives, they're generally rated in horsepower and 'tractive effort'...

so, my question is, how does that translate into ft/lbs of torque?
in my experience, diesels make massive torque, while making significantly less horsepower due to their low operating speeds.
When I see an diesel developing 3000+bhp, I can only imagine what kind of torque that's producing.

Could someone possibly elaborate on that?

now, from my basic understanding of these locomotives, the engine itself does not directly drive the locomotive, but simply turns an electric generator, which in turn propels the locomotive?
That's the only explanation I can see as to why torque is not really spoken of in the literature...

side question: how much torque does the electric motor generate? again, from my experience with cars, I know that electric motors generate epic amounts of torque at initial spinover (or what we refer to, perhaps incorrectly, as zero RPM)

Torque is sort of a thing I'm obsessed with when building cars. My own personal car is an Audi S6 with a turbocharger matched for low end torque (and it produced about 375ft lbs of torque and only 350bhp). I also have a friend who builds two stroke diesel busses, and I'm floored by the sheer volume of torque they produce (not to mention how the 2 stroke diesel cycle works)
 #961900  by Allen Hazen
 
Hi, welcome aboard!

I'll let you do the torque calculations. First, the engine. The GE and EMD diesel engines used in current locomotives have nominal top speeds of 1050 rpm and 900 rpm respectively (about: I may be a few percent off in memory). At this speed the 12 cylinder(four-stroke) GE engine and the 16 cylinder (two stroke) EMD engine both produce around 4400 or 4300 hp as input to the main generator: brake horsepower would be a couple of hundred more.

The electric motors that finally tun the wheels. (At this sort of power rating the electric transmission -- wildly complicated and round-about as it may intuitively seem -- seems to be the most efficient available: locomotive-style diesel engines and electric tranmissions are also used in very large off-road dump trucks.) The motors are geared to the axles, so the motor rotates at roughly four times the rpm of the wheel: think around 2300 rpm as a good top speed for the traction motor. With six powered axles (standard on contemporary North American freight locomotives, each motor soaks up about one sixth of the locomotive's power: since not all the motors draw the same current at the same time (typically the motors at the front, where the rail might be wet and slippery, operate at lower power than the motors on the rear axles at the same speed), assume that the motor is capable of taking rather more: maybe a quarter of the locomotive's total power. Full power would spin the wheels at low speeds, but modern control systems allow the locomotive to produce tractive force (= pull) equivalent to about a third of its total weight: say 25,000 pounds of tractive force per axle. A locomotive pulling a heavy coal train up a hill will be doing this at, say, 10 m.p.h. The wheels are typically about 42 inches in diameter these days.

I'm sorry: never having worked with cars or motorcycles, and not being a professional engineer, I don't think in terms of torque. But I think those figures (ballpark estimates!) ought to be enough to calculate some very rough torque estimates for engine and motor.
 #961959  by chapel
 
Allen Hazen wrote:Hi, welcome aboard!

I'll let you do the torque calculations. First, the engine. The GE and EMD diesel engines used in current locomotives have nominal top speeds of 1050 rpm and 900 rpm respectively (about: I may be a few percent off in memory). At this speed the 12 cylinder(four-stroke) GE engine and the 16 cylinder (two stroke) EMD engine both produce around 4400 or 4300 hp as input to the main generator: brake horsepower would be a couple of hundred more.
Oh great, that's all I need is horsepower @ rpm to determine engine torque in ft/lbs

Ok... so horsepower is TxN/5252 (where T is torque and N is RPM)
so...

GE 12 cylinder four stroke
4400 = (T*1050)/5252...
*grabs an algebra calculator because he's totally forgotten how to solve for X in this instance...
22,008 ftlbs of torque at 1050rpms

EMD V16 Two stroke
4300 = (T*900)/5252
25,093 ftlbs of torque at 900rpms

dang.

I knew that two stroke was going to end up making more torque than the four stroke despite the horsepower rating.
The electric motors that finally tun the wheels. (At this sort of power rating the electric transmission -- wildly complicated and round-about as it may intuitively seem -- seems to be the most efficient available: locomotive-style diesel engines and electric tranmissions are also used in very large off-road dump trucks.) The motors are geared to the axles, so the motor rotates at roughly four times the rpm of the wheel: think around 2300 rpm as a good top speed for the traction motor. With six powered axles (standard on contemporary North American freight locomotives, each motor soaks up about one sixth of the locomotive's power: since not all the motors draw the same current at the same time (typically the motors at the front, where the rail might be wet and slippery, operate at lower power than the motors on the rear axles at the same speed), assume that the motor is capable of taking rather more: maybe a quarter of the locomotive's total power. Full power would spin the wheels at low speeds, but modern control systems allow the locomotive to produce tractive force (= pull) equivalent to about a third of its total weight: say 25,000 pounds of tractive force per axle. A locomotive pulling a heavy coal train up a hill will be doing this at, say, 10 m.p.h. The wheels are typically about 42 inches in diameter these days.

I'm sorry: never having worked with cars or motorcycles, and not being a professional engineer, I don't think in terms of torque. But I think those figures (ballpark estimates!) ought to be enough to calculate some very rough torque estimates for engine and motor.
In terms of a wheeled vehicle, this is what we see as ultimate torque
the Detroit Diesel 24V-71
http://youtu.be/v6KWsIFqoso
(skip to the 1:00 mark to see it in action and about 4:00 to see it run up)
27.9 litres, 3424bhp
this is without turbocharging, only with superchargers (which are required due to the design of a two stroke diesel) and nitrous... though it does have TWELVE superchargers (8-71 blowers)...
there's some really cool pictures afterwards with them mocking up the blower gearbox

here's his 24V-71 with turbos:
http://www.youtube.com/watch?v=Jp-qcZr_ ... re=related
makes "only" 2800hp

Now, in these engines, the -71 is the displacement per cylinder in cubic inches and the first number is the amount of cylinders

So to make 3424bhp from 27.9 liters is 'relatively' impressive, considering the EMD 645 is 645 cubic inches per cylinder... and on a GP40, they're 16 cylinders... that's a nearly 170 LITRE ENGINE! and it only makes 3000bhp.

Though I don't know how what RPM Mike Harrah's motor made it's 3400bhp... There's not a lot of info on the 24Vs, but the 16Vs have an operating speed of 1900rpms
So toss that into the equation and it's making just about 9500ft lbs of torque.

So, despite having horsepower ratings that aren't that far apart, the big locomotive engines make almost 3 times as much torque... though they carry 6 times the displacement.

So, I guess there really is no replacement for displacement :)
 #961961  by FarmallBob
 
Torque is easily calculated if you know the horsepower and RPM. The equation:

Torque = Horsepower x 5252 / RPM

----

Example -

For a GE diesel delivering 4,400 HP at the flywheel at 1,050 RPM. Torque is:

4,400 x 5,252 / 1,050 = 22,008 ft-lb

----

Similarly the torque delivered by a locomotive's traction motors can be calculated. However the calculation is a bit more complex. You need to know the locomotive speed, drawbar pull, the number of driven axles, the rail wheel diameter and motor to axle gear ratio. And as you suspect for traction motors the maximum torque is delivered at 0 RPM ("locked rotor" condition).

Does this help?
 #961964  by chapel
 
FarmallBob wrote:Torque is easily calculated if you know the horsepower and RPM. The equation:

Torque = Horsepower x 5252 / RPM

----

Example -

For a GE diesel delivering 4,400 HP at the flywheel at 1,050 RPM. Torque is:

4,400 x 5,252 / 1,050 = 22,008 ft-lb

----

Similarly the torque delivered by a locomotive's traction motors can be calculated. However the calculation is a bit more complex. You need to know the locomotive speed, drawbar pull, the number of driven axles, the rail wheel diameter and motor to axle gear ratio. And as you suspect for traction motors the maximum torque is delivered at 0 RPM ("locked rotor" condition).

Does this help?
oh right, I always forget the obverse calculation for torque..

*facepalm*

that could have saved me like 5 minutes...