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
"No soup for you!" - Yev Kassem (from Seinfeld)
-Don (Random stuff, mostly about trains - what else? http://blerfblog.blogspot.com/)
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???[%-)]
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
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