Friday

August 29, 2014

August 29, 2014

Posted by **Slice** on Saturday, November 6, 2010 at 3:10pm.

a) Calculate the power in this flow. Note that the velocity and height of the water are not changed appreciably by the generators, and the pressure at the exit point is at atmospheric level. Take the density of water to be 1000 kg/m3.

answer in MW

b) What is the speed of the water at the exit point?

answer in m/s

c) What is the radius of the exit hole in the dam?

answer in m

- Physics -
**Damon**, Saturday, November 6, 2010 at 3:26pmDensity of water = 1000 kg/m^3

pressure difference = density *g * h =10^3*9.81*110 = 1.08*10^6 Newtons/m^2

power = pressure*Q = 1.08*10^6 * 740

= 8*10^8 Joules meters/second or Watts

or 800 megawatts

b) (1/2) rho v^2 = rho g h

v = sqrt (2 g h)

c) pi r^2 v = 740

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