View Single Post
Clausius2 is offline
Sep28-04, 06:49 AM
Sci Advisor
PF Gold
Clausius2's Avatar
P: 1,481
Quote Quote by derekbeau

P = Mgh
P = 1710000 * 9.81 * 100

and I got 1677510000 (watts?)

Well that wasnt the right answer, so i tried converting to kilowatt-hours

1677510000 W / 3600000 J = 465.975 kWh

Still not right.

I think my problem is because either the height of the water is changing, or i am using an incorrect height. I dont know. But any help would be great. (it is due tomorrow)

The link does not work.

Anyway, the power obtained in the turbine is:

[tex]W=Q*\Delta P_o[/tex]; where [Q]=[m^3/s] is the volumetric flow; [P_o]=[Pa] is the total pressure in both sides of the turbine.

In your case: [tex]P_{oentrance}-P_{oexit}=P_a+\rho g H[/tex]-P_a[/tex]

So that: [tex]W=\rho g H Q[/tex] where the units are:

[tex] [W]=\frac{Kg}{m^3} * \frac{m}{s^2}* m *\frac{m^3}{s}=\frac{J}{s}=Watt[/tex]

You said it doesn't work. Although the height of the reservoir is changing, in my opinion that change is cuasi-steady, so that the turbine power also changes with time W=W(H(t)). I need to view the drawing of your link to answer you better.