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Horsepower: Where do we measure it?

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Posted by Anonymous on Saturday, August 7, 2004 12:38 PM
Thats one heck of a toaster!!!!!!!!!!!!!!!!!!!!!!!!!
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Posted by Randy Stahl on Friday, August 6, 2004 10:08 PM
really really hot!!!!!!! Calculate the watts and compare to a household appliance.
Randy
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Posted by ericsp on Friday, August 6, 2004 9:57 PM
QUOTE: Originally posted by Randy Stahl

ericsp had the right idea i commend him for the thought on the matter. Each grid section is .43 ohms each grig has two .43 sections for a total of .86 ohms each locomotive (SD40-2) has 2 grids in series for a total of 1.72 ohms , in self load the grids are arranged in series parellell in pairs. That means voltage will be set at 1204 and current will be 700 amps per grid or 4200 amps plenty of capacity to test a 5057 horse power locomotive. On an SD45 the voltage limit settings are 900 volts and 4000 amps total 3600 horsepower.
Randy



Do you know the about what temperature the grids usually reach when doing the self-load test?

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Posted by Randy Stahl on Friday, August 6, 2004 9:47 PM
ericsp had the right idea i commend him for the thought on the matter. Each grid section is .43 ohms each grig has two .43 sections for a total of .86 ohms each locomotive (SD40-2) has 2 grids in series for a total of 1.72 ohms , in self load the grids are arranged in series parellell in pairs. That means voltage will be set at 1204 and current will be 700 amps per grid or 4200 amps plenty of capacity to test a 5057 horse power locomotive. On an SD45 the voltage limit settings are 900 volts and 4000 amps total 3600 horsepower.
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Posted by ericsp on Friday, August 6, 2004 9:35 PM
QUOTE: Originally posted by drailed1999

Ok Randy. Since i'm a machinist and not a " sparky ", when I would put a SD40-2 in self-load test, I was only using 2 of the 6 availible DB grids???[%-)]


If you use all 6, it just lowers the current.

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Posted by Anonymous on Friday, August 6, 2004 9:09 PM
Thats why i'll stick to pulling wrenches and not wire. Me no savvy. Besides that stuff bites hard!!!!!!!!!!!!!!!![bow]
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Posted by Randy Stahl on Friday, August 6, 2004 9:02 PM
You use all six grids.
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Posted by Anonymous on Friday, August 6, 2004 8:56 PM
Ok Randy. Since i'm a machinist and not a " sparky ", when I would put a SD40-2 in self-load test, I was only using 2 of the 6 availible DB grids???[%-)]
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Posted by oltmannd on Friday, August 6, 2004 7:33 AM
Back to the original question, the "Traction HP" for a locomotive is the shaft horsepower going into the main generator (or alternator) that will be converted to electrical HP for traction. It is to be measured at AAR std conditions, 28.86" Hg (1000 ft elevation) 60 deg F ambient air and fuel temp, 7.043 #/gal fuel density and 19350 BTU/# HHV.

It is engine brake HP less external auxillaries (such as TM blower, cooling fans, air compressor). The brake HP does not include internal auxillaries such as water pumps and lube oil pumps.

In practice, you measure the electrical output from the main generator and adjust for generator efficiency (that is why Randy divides by 700 instead of 743), calculate the auxillaries, add them up, then adjust results back to standard conditions.

-Don (Random stuff, mostly about trains - what else? http://blerfblog.blogspot.com/

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Posted by ericsp on Friday, August 6, 2004 2:45 AM
QUOTE: Originally posted by ericsp
QUOTE: Originally posted by Randy Stahl

Thats the answer I was looking for...good job!!
Randy


Using 1 HP = 745.7 W, 3000 HP = 2237100 W
Assume the voltage is 1250 V
Use W=(V^2)/R to get R=W/(V^2)
R=2237100/(1250^2)=0.6984 Ohms
To make sure that the current does not exceed the current for 3000HP and 1250V the resistance cannot drop below the above value.
Therefore 2 grids are used.


QUOTE: Originally posted by Randy Stahl

Thats the answer I was looking for...good job!!
Randy


It seems like Sierra Railroad has done that up at Oakdale, CA.

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Posted by Randy Stahl on Friday, August 6, 2004 12:21 AM
Thats the answer I was looking for...good job!!
Randy
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Posted by ericsp on Thursday, August 5, 2004 12:44 AM
QUOTE: Originally posted by Randy Stahl

Here's a simple question for you guys: You are building a load test cell out of locomotive dynamic brake grids, each grid is .43 ohms, how many do you need to test 3000 hp locomotives?
Randy


Using 1 HP = 745.7 W, 3000 HP = 2237100 W
Assume the voltage is 1250 V
Use W=(V^2)/R to get R=W/(V^2)
R=2237100/(1250^2)=0.6984 Ohms
To make sure that the current does not exceed the current for 3000HP and 1250V the resistance cannot drop below the above value.
Therefore 2 grids are used.

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Posted by jchnhtfd on Wednesday, August 4, 2004 10:06 AM
QUOTE: Originally posted by Randy Stahl

Fly wheel horse power has it's application in EMD marine engines. There is not an application on the RR.
Randy

and Karn -- in both cases what you want to know -- and are finding out -- is the useful output of the prime mover. The losses in a railroad alternator are very small (relatively speaking) and the two values would be quite close. There are, incidentally, some marine applications where one looks at the KW from the alternators rather than 'flywheel' horsepower -- a surprising number of ships are diesel electric drive, even some very big ones (e.g. Queen Mary 2, the biggest passenger liner in the world![:D])
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Posted by Randy Stahl on Wednesday, August 4, 2004 10:00 AM
Fly wheel horse power has it's application in EMD marine engines. There is not an application on the RR.
Randy
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Posted by Anonymous on Wednesday, August 4, 2004 9:53 AM
QUOTE: For our purposes WE measure horsepower in kilowatts from the main alternator/ generator.


seems to be conflict with

QUOTE: If you're also asking how the horsepower is rated, as in a 3,000-hp SD40-2, it's the horsepower that's supposed to be available at the flywheel of the engine at maximum rpm, after deduction for parasitic loads


Karn[:)]
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Posted by Randy Stahl on Tuesday, August 3, 2004 11:30 PM
If you don't care about damaging the prime mover, don't worry about it, on a locomotive with a horse power problem on the high end I usually shut them down when I blow off a loadbox cable or if some thing else blows to hell. You can't fix the problem if you can't load em up and put your meter to it.
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Posted by jrw249 on Tuesday, August 3, 2004 11:24 PM
Oh, so you can exceed the rated hp. Back to my original question, how do you know when to stop loading the engine for the load test?
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Posted by Randy Stahl on Tuesday, August 3, 2004 11:16 PM
J ruppert is right ,, you should not be able to exeed the rated out put of the machine, however I have seen 4400 horse power SD45s, of course there is something wrong with the locomotive and needs an electrician to straighten it out before the prime mover is ruined. Here's a simple question for you guys: You are building a load test cell out of locomotive dynamic brake grids, each grid is .43 ohms, how many do you need to test 3000 hp locomotives?
Randy
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Posted by jrw249 on Tuesday, August 3, 2004 11:14 PM
So the testing is just basically verifing the the engine can put out the hp the manufacturer says it can and not really testing to see the maximum hp it can put out.
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Posted by ericsp on Tuesday, August 3, 2004 10:37 PM
All of the components of the locomotive and the test equipment have been designed to handle so much current, heat, stress, etc. If any of this equipment is operated beyond what the manufacturer says it can operate at there is a risk of serious damaging or destroying the equipment or injuring or killing the operator. Unless as Mr. Ruppert points out it is designed (most likely the computer program the controls the equipment) to not allow operation beyond safe limits.

There is usually a factor of safety built into the design, however, most people will not know what it is and therefore should not exceed the operating limits.

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Posted by Anonymous on Tuesday, August 3, 2004 4:49 PM
"rated" output cannot be exceeded, my experience is with gen sets not locomotives, but the answer is the same, rated output cannot be exceeded. If load increases beyond rated output, exitation is lowered or stopped depending on design. This is called "dropping the load". It is necessary to design a unit this way to protect itself from overload damage.

An engine's rpm can drop as the unit is being loaded, but that is a function of governor design. A "droop" governor falls from "high idle" - maximum no load speed, a predetermined amount to "rated speed" - full load rpm. A governor without droop is called an isochronus governor.

So, when you are load testing, you are looking to achieve a certain number, if you get it, your job is done. If you exceed that number, something is wrong, just the same as if you couldn't get the number.
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Posted by jrw249 on Tuesday, August 3, 2004 2:09 PM
Originally posted by Randy Stahl

RKARN ,
If you want to find the horsepower of your locomotive connect the generator to a load box, put the engine in throttle 8 and calculate volts times amps divided by 700.
Randy
Assuming that you could keep adding load current, how do you know when you reach full load on the engine? When the manufacturer's full load rpm starts to drop???
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Posted by Randy Stahl on Tuesday, August 3, 2004 9:38 AM
700 is the industry standard, you are correct that one HP=745.7 watts, how ever we have to compensate for fuel temp, air density, ambient air temp,water inlet temp. 700 is simply the generator efficiency factor.
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Posted by Anonymous on Tuesday, August 3, 2004 6:21 AM
Randy is right on, because a generator's resistance to rotation when its being loaded is no different than a dynanometer's resistance to turning. In fact, an electrical dyno (vs. a water brake, or friction brake type) is the exact same thing.

When a brand new gen-set is finished, it is evaluated with a load bank, which is what dynamic brakes are. Kw/hr, Hp, it's all the same.

Larger engines like locomotive engines are not dyno tested because it's not practical to build a dyno for such amounts of power, so other methods of determining performance have to be used, like comparing actual fuel consumption to a BSFC chart, or fuel consumption vs. power generated. Locomotive engines are also far from the biggest engines, some marine engines develope 40,000 Hp, engines of this size are typicaly rated in Kw.

Rating output at the flywheel, makes sense to me because while an engine's output can be pretty constant, the power that gets "to the ground" can vary a lot, and this is true for any vehicle or application. In the case of locomotives, it is convenient to measure electrical output even though we know the the engine as a x,000 Hp unit.

SD-40-2, right on with the formula.
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Posted by ericsp on Tuesday, August 3, 2004 12:11 AM
QUOTE: Originally posted by Randy Stahl

RKARN ,
If you want to find the horsepower of your locomotive connect the generator to a load box, put the engine in throttle 8 and calculate volts times amps divided by 700.
Randy


Randy:

With all of the posts, you should have started with "To make a short story long..."

Also, why do you use 700? According to the conversion table on page 646 Fundamentals of Thermodynamics, 5th Edition by Sonntag, Borgnakke, and Van Wylen one horsepower is approximately 745.7 Watts. With the power of locomotives, couldn't this result in a large error?

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Posted by Randy Stahl on Monday, August 2, 2004 11:30 PM
RKARN ,
If you want to find the horsepower of your locomotive connect the generator to a load box, put the engine in throttle 8 and calculate volts times amps divided by 700.
Randy
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Posted by ericsp on Monday, August 2, 2004 9:08 PM
I’d like to say some things about the formulas used in this thread.

Never pluralize the units when they are abbreviated, it could be mistaken for seconds.

The English units for torque are lb-ft, not lb/ft. Lb/ft is pounds per foot..

A moment (torque) equals force multiplied by the lever arm (distance). Moment and force are not interchangeable or comparable.

Other than this the posts appear to be fairly accurate.

Here is how the constant in the HP formula is derive. 1HP=550lb-ft/s. If we know the torque and the speed of the engine we can multiple to gets its power. However, if RPM is being used the units are wrong. We have to convert the revolutions per minute into radians/second (rad/s). There are 2pi radians in one revolution, therefore 1 RPM = 2pi rad/min. To convert to rad/s we multiple by 1min/60sec, 1RPM = (2/60)*pi rad/s which is approximately 0.1047 rad/s. An engine producing 1000 lb-ft of torque and turning at 600 RPM is producing 62831.85 lb-ft/s of power. To convert this to HP divide by 550 HP/(lb-ft/s), this gives 114.24 HP. The 5252 comes from (2*pi/60)/550 which is approximately 1/5252.

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Posted by GP40-2 on Monday, August 2, 2004 7:35 PM
As you can see HP is nothing more than torque (a twisting force created by the downward motion of the engine's pistons that the crankshaft converted into a rotational motion) multiplied by speed (the engine RPM) divided by a derived mathamatical constant (5252).

This same formula can calculate the TE of a locomotive if you know the traction motors HP and the speed of wheel rotation. You will also need to know the adhesion of the locomotive, as this is the limiting factor for maximum TE.

This formula also explains why an engine of small displacement (such as an Indy car engine) and relatively low maximum torque (torque is related to engine displacement) can produce a crap load of HP. The engine may be small and low torque, but it spins at such high RPM (10,000+) that its HP output is very high (1000+ HP)

jruppert: Top Fuel engines produce 6000 HP at 8000 RPM!!! Of couse, this is only for a few seconds. At the end of a run, engines must be totally rebuilt. Many of the internal parts are melted together!! To find out the torque:

Torque = (6000)(5252)/8000 = 3,939 lbs*ft@8000 RPM


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Posted by GP40-2 on Monday, August 2, 2004 7:17 PM
The formula that relates HP to torque is:

HP = (torque) X (rpm)/ 5252

The GE FDL produces a maximum of 4700 HP at the crankshaft at 1050 RMP, therefore:

torque = (HP) X (5252)/RMP = 23,508 lbs*ft torque@1050 rpm

Of the 4700 hp, a minimum of 4400 hp is reserved for traction, with the remainder reserved for auxillary use.

That 23,508 lbs*ft is a tad more than your truck! [:D]

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