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 Message Boards » » CE 382 Hydraulics: Question about shaft work Page [1]  
BigBadWulf07
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In one of my problems it says that the pump in the pipe delivers 16 kW. What do I do with this 16 kW so that I can use it in my energy equation? I'm not sure what Ws stands for and how I get Ws from this 16 kW. Thanks

11/5/2005 9:12:11 PM

Dumbass
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kW = 1000*W

just like kJ or kN

11/5/2005 11:22:35 PM

BigBadWulf07
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I'm using the equation

Pout/p + V^2(out)/2 + gZ (out)= Pin/p + V^2(in)/2 +gZ(in) + Ws - loss

I am pluging in the 16000 watts in for Ws directly. I'm not sure if I need to convert the 16000 watts to something else first though.

11/6/2005 8:47:32 AM

Nerdchick
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LOL shaft work!

11/6/2005 11:23:25 AM

jwb9984
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UNITS

make sure they're all the same

[Edited on November 6, 2005 at 2:30 PM. Reason : v]

11/6/2005 2:29:54 PM

 Message Boards » Study Hall » CE 382 Hydraulics: Question about shaft work Page [1]  
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