In your replies, thanks in advance, please don’t get too technical, I’m not that knowledgable, but I’d really like to know how putting turbo on a GP38 to make it into a GP40 boosts the HP by 1000…
Also, why are SD38s preferred by steel mills over SD40s, when the SD40 is turbocharged. Wouldn’t that allow it to haul more? How much does turbocharging affect the tractive effort?
Thanks again,
Matthew
A turbo charger takes exhaust gasses from an engine and turns a ‘fan’ which in turn spins another fan that takes in combustion air to increase the pressure of the intake air. This denser air going into the cylinders allow more fuel to be burned and more horsepower to be generated. GE and Alco engines were always turbocharged and are 4 stroke engines. EMD uses a 2 stroke engine and needs a blower of some sort to force air through the cylinders each time the cylinder is on the down stroke. At first gear driven Roots blowers were used. To get more horsepower turbocharges have been used on EMD units but they are gear driven at lower throttle positions and are only exhaust driven at maximum throttle settings.
Turbochargers commonly have a failure rate of 100% after 5(?) years or so. Since horsepower is speed, iron ore roads opt for the simplification of no turbocharger since their train speeds are less than mainline speeds on other freight railroads. The turbocharged SD40 will not haul more tonnage than an SD38, it will just haul it faster while burning more fuel and suffering turbo failure after a time.
The turbocharger does not increase tractive effort. An SD38 and SD40 will both load to about 1200 amps per traction motor for a short time, load to 1100 amps while accelerating, hold 1000 amps while in the short time ratings and load 900 amps all day long.
In very simple terms, a turbocharger is nothing more than an exhaust driven air pump. The more air you get into the cylinders means more oxygen into the cylinders. The more oxygen you have means you can burn more fuel at a given time, thus more HP. Turbochargers also increase the thermodynamic efficiency of a diesel engine + give the engine the ability to hold its power output to a much higher altitude.
“The turbocharger does not increase tractive effort. An SD38 and SD40 will both load to about 1200 amps per traction motor for a short time, load to 1100 amps while accelerating, hold 1000 amps while in the short time ratings and load 900 amps all day long.”
That is not true. Horsepower is nothing more than tractive effort at a given speed. The more HP, the more tractive effort (assuming you have the adhesion to put the extra power to the rail).
Well, GP40-2, next time you are out on a mountain grade in Run 8, note the amps showing on the ammeter. Reduce the thottle to Run 7 and note the ammeter reading. It will probably not go down at all. The same may also hold true down to Run 6 if the grade is still enough and the train speed is slow enough. Amps = tractive effort.
In reality, an SD38 at 25 mph will carry the same amps as an SD40 at 25 mph. That is just the way it is. A set of SD38s with an 8,000 ton train on a grade will move the train at say 20 mph. The same number of SD40s with the same train on the same grade will move the train at a higher speed. If the size of the train behind the SD40s is increased the amount to reduce the train speed to 20mph on the same grade then the turbocharged units will have the same load and tractive effort as the nonturbocharged SD38 set.
Okay, since they fail at 5 years, where do the replacements come from? Do the manufacturers replace them n newer locomotives, and on older models are they scavenged from non working models?
Thanks again
I’ll try to make this as simple as possible. First of all, understand that the more air you can get into an engine (with the proper air/fuel ratio), the more power you can get out of that engine. This is where the turbo comes in. A turbo is mounted onto the engine’s exhaust. As the hot exhaust gases are rapidly pushed out of the the engine into the exhaust line, they flow over the turbo’s turbine (or “fins” to keep it in simple terms) and rapidly spin the turbo which in turn spins an air compressor. This air compressor adds boost (air), known as turbo boost, to the engine in turn giving it more power. The more RPMs the engine turns, the more boost the turbo puts out. This is because at higher RPMs there is more (and faster) exhaust gas flowing over the turbo, causing it to spin more rapidly and give more boost. That is the turbo explanation.
Now, to relate this more to locomotives, mostly all EMD turbo systems are different from regular turbos. In throttle knotches 1-6 the turbo is directly driven by the engine itself, actually making it a centrifugal type supercharger at these throttle settings. However, in throttle knotches 7 and 8 the turbo is a true “free wheeling” turbo that is directly driven by the exhaust gases. EMD does this to reduce turbo lag and because of the 2 cycle engine they use. GE is a different story on this whole deal. Let me know if you want me to explain this more.
As far as your 1000 HP boost question, turbos on big diesels like this can really boost the HP up a lot, but I really don’t know about 1000 HP. Maybe, maybe not. I don’t know why the steel industry prefers SD38s over SD40s. Turbo charging will allow the loco to haul faster, but not necessarily more. Turbocharging does not affect tracive effort (TE) at all, all it will do is allow the loco to haul faster. The only thing that affects TE is the weight of the loco, and the loco’s adhesion. HP is only speed, not TE. A 400,000 lb. SD40-2 with 10,000 HP would have the same amount of TE as a
trainboyh16-44’
The railroads buy remanufactured turbos from suppliers. They provide used turbos that have failed as a core and the supplier rebuilds them for the next customer. Just about what happens when you need a new alternator for your car.
Let me try some simple answers:
The way an engine makes power is by buring fuel. The more fuel you burn, the more energy you get out.
You burn fuel by combining it with oxygen in the air.
If you want to burn more fuel, you have to provide more air. The way you do this on an internal combustion engine is to blow more air into the cylinders under pressure.
Using an exhaust gas powered turbine to run a blower is called “turbocharging” and is a very cost effective way to do this.
An SD38 and an SD40 will provide the same maximum tractive effort, but the SD40 will be able to do it a 50% greater speed. They can haul the exact same amount, the SD40 will just haul it 50% faster.
You can calculate it from the equation:
HP= TE x speed/308
If an SD 38/40 can do a max of 80,000 lbf tractive effort, then:
2000 = 80,000 x speed/308 speed = 7.7 mph
3000 = 80,000 x speed/308 speed = 11.6 mph
When moving cars a couple of miles from one end of a steel mill to the other a few times a day, whether you can do it at 7 mph or 11matters little, so the steel mill will opt for the 2000 HP SD38. But, if you are trying to get at train from city A to city B which are separated by hundreds of miles, it can matter a great deal if you can move at 45 mph vs. 30 mph.
Here is a fairly straight forward explanation, with pictures
Wait a minute, if you increase the mass of the train behind the more powerful SD40s, you increase the resistance, thus you must increase the force (tractive effort) needed to pull the train at 20 MPH.
Your post makes no sense whatsoever.
An SD38 produces around 1750 DBHP or 32,800 lbs TE at 20 MPH.
An SD40 produces around 2700 DBHP or 50,600 lbs TE at 20 MPH.
The last time I looked 50,600 is much larger than 32,800.
While both locomotives (assuming they weigh the same and have the same adhesion properties) will have the same maximum STARTING tractive effort, the more powerful, turbocharged SD40 will have greater tractive effort at any speed above starting.
P.S. Kfleeman1, your little thesis that “HP is just speed, not TE” is totally wrong also.
Oltmannd: You 308 number is a little off for a SD40 era locomotive. The 308 is derived from the equation HP= (TE x SPEED)/375 assuming 82% efficiency (375 x 0.82 = 308) This is more accurate for a older first generation unit.
First of all, this is all assuming that the SD38 and SD40 weigh the same, have the same adhesion rating, have the same traction motors (TMs), and are pulling the same train.
GP40-2 is correct in that the SD40 will produce more TE than the SD38 at speeds above starting, but…
once you get into a grade and the speed of the train slows, there is a point where both the SD38 and SD40 will load their TMs up to their max and once again the SD38 will produce the same TE as the SD40. Given, that point will come a little quicker to the SD40, but the SD38 will get there too.
So all in all, the SD38 and SD40 will produce the same max (starting) TE, but at conditions or speeds above where the TMs are loaded to their max, the SD40 will produce more TE.
This brings us back to saying that the SD40 will not haul more than the SD38, it will just do it faster than the SD38.
Hope this clears things up.
hey go to howstuffworks.com this web site will tell you how everything works it is a really amazing
…and GP40-2, my “totally wrong little thesis” that “HP is just speed, not TE,” was assuming max TE. A 3000HP 400,000lb. loco has the same max TE as a 300HP 400,000lb. loco, but the 300HP loco won’t pull the load nearly as fast as the 3000HP loco. So the “totally wrong little thesis” that “HP is just speed, not TE,” isn’t totally wrong. Yes, it was my fault that I didn’t say max TE.
I think the 308 factor is good for everything through the Dash 2’s. The only real eff. difference between the 1st gen and Dash 2s is the DC main gen vs. AC traction alt., but that’s not worth that many eff. pts.
CR SD40-2 perfomrance numbers per EMD:
389,000#
83,000# TE
3000 THP
MCS 11.1 mph
3000=83,000 x 11.1/ factor
factor = 307.1
Oltmannd, ok I see where you are coming from.
The conditions you listed are when the locomotive is producing maximum amps. The electrical efficiency (and thus factor) of any locomotive is not a static number, it will vary depending on the operating conditions. Diesel-Electrics are least efficient when producing their power via maximum amps. Efficiency is lost by increased electrical backpressure and heat losses. The same SD40 will be more electrically efficient when it is producing its power via higher voltage and lower amps (higher speed)
According to EMD engineering data, the typical overall electrical efficiency of a SD/GP40-2 is in the 87%-88% range.
Another example is that we have found that the C44ACs are running just under 90% efficient when producing their maximum pull (180,000 lbs continious TE @ 7.4 MPH), yet the same unit will record an efficiency over 94% when used in higher speed general frieght / intermodal service.
[quote]
Originally posted by kfleeman1
This air compressor adds boost (air), known as turbo boost, to the engine in turn giving it more power. The more RPMs the engine turns, the more boost the turbo puts out. This is because at higher RPMs there is more (and faster) exhaust gas flowing over the turbo, causing it to spin more rapidly and give more boost. That is the turbo explanation.
[/quote
Sorry to have to disagree here about engine RPM = turbo boost pressure. The turbocharger works off the exhaust pressure which varies in accordanced with the load put on the engine. Turbo boost is, of course, lowest at engine idle and no load. As the engine is asked to work, fuel is increased, causeing the exhaust pressure to in crease, causing the turbocharger to spin faster, thus increasing the intake pressure to rise. A diesel engine turning 900 RPM under a heavy load will have more turbo boost then it would running at 1800 RPM and no load on it. Now a gear driven blower as used on 2 cycle engines is RPM related. A 2 cycle engine depends on the blower not only to supply air to the combustion chamber but it also pushes the exhaust gases out of the cylinder. Ken
2 cycle diesels are scavenged, as in when the piston drops to the bottom of the cylinder it opens a set of ports and the air blows the exhaust gases out thereby also supplying air for the compression stroke, has only an exhaust valve.
4 stroke diesels have 2 or more valves, and operate like your car motor, the turbo only supplies more air to burn more fuel, but puts much higher internal loads on the motor. It is also a free source of power, as it is driven by the velocity, and heat load of the exhaust gas, whereby a blower is driven from the crank by a belt or gear train. EMD’s are gear driven till the clutch disengages when they reach a certain output…
Turbos are usefull for emissions, high RPM operation, and altitude compensation. At low speeds, ie 15 to 20 MPH, they really are a maintenance headache, and for switching, will be a very large unnecessary expense…
[quote]
QUOTE: Originally posted by Eriediamond
Boost goes up with RPMs, and falls as RPM’s decrease…
…Lets use a Ford Powerstroke as an example,
at 2800 RPM on a grade pulling my 5500 pound jetboat, I will run about 19 to 20 pounds of boost, but if I am only turning 2400 RPM, the boost rapidly falls off while exhaust gas temps skyrocket as there is no longer enough air to burn the fual being injected. As RPM’s decrease, boost will too, thats why wastegates exist, they will compensate for this by having a smaller turbine wheel, exhaust side, which will spin faster under low rpm conditions. The drawback to this is at high RPM conditions, it becomes a chokeing point and will raise the exhaust gas temps thereby limiting power,
[quote]
QUOTE: Originally posted by m1ashooter
[quote]
QUOTE: Originally posted by Eriediamond
Boost goes up with RPMs, and falls as RPM’s decrease…
…Lets use a Ford Powerstroke as an example,
at 2800 RPM on a grade pulling my 5500 pound jetboat, I will run about 19 to 20 pounds of boost, but if I am only turning 2400 RPM, the boost rapidly falls off while exhaust gas temps skyrocket as there is no longer enough air to burn the fual being injected. As RPM’s decrease, boost will too, thats why wastegates exist, they will compensate for this by having a smaller turbine wheel, exhaust side, which will spin faster under low rpm conditions. The drawback to this is at high RPM conditions, it becomes a chokeing point and will raise the exhaust ga