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
"No soup for you!" - Yev Kassem (from Seinfeld)
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???[%-)]
-Don (Random stuff, mostly about trains - what else? http://blerfblog.blogspot.com/)
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
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
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
QUOTE: For our purposes WE measure horsepower in kilowatts from the main alternator/ generator.
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
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??? Reply Randy Stahl Member sinceJune 2004 From: roundhouse 2,747 posts 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. Randy Reply Anonymous Member sinceApril 2003 305,205 posts 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. Reply Edit ericsp Member sinceMay 2015 5,134 posts 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? "No soup for you!" - Yev Kassem (from Seinfeld) Reply Randy Stahl Member sinceJune 2004 From: roundhouse 2,747 posts 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 Reply ericsp Member sinceMay 2015 5,134 posts 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. "No soup for you!" - Yev Kassem (from Seinfeld) Reply GP40-2 Member sinceJuly 2004 803 posts 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 Reply GP40-2 Member sinceJuly 2004 803 posts 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] Reply 12 Join our Community! Our community is FREE to join. To participate you must either login or register for an account. 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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
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