water flow?

Question

water flows (v=0) over a dam at the rate of 650kg/s and falls vertically 81 m before striking the turbine blades.
Calculate (A) the speed of the water just before striking the turbine blades (ignore air resistance)
and (B) the rate at which mechanical energy is transferred to the turbine blades, assuming 58% efficiency.

Answer 1

Given data:

w = 650 Kg/s
g = 9.81 m/s²
h = 81 m

Part A: Velocity before striking the blades:

v = sqrt(v0² + 2gh)
v = sqrt( 0 + 2(9.81 m/s²)(81m)]
v = 39.86 m/s

Part B: Rate of Energy transfered to the turbine

E = mgh + (1/2)mv² .......sum of potential energy and kinetic energy
E = m (gh + (1/2)v²)
E = (650 Kg/s)((9.81 m/s) (81m) + 0.5(39.86 m/s)²)
E = 1.0328 x 10^6 Joules/s
E = 1.0328 x 10^6 watts
E = 1032.8 kW

Assuming 58% of efficiency:

E = (0.58)(1032.8 kW) = 599.02 kW

That´s it!

Good luck!

Answer 2

The vertical velocity can be determined using v^2 = v_0^2 + 2gy, where v is final velocity and v_0 is initial velocity. Here, v_0 is zero, so it's just v = sqrt(2gy) = sqrt(2*9.8*81) = 39.8 m/s. I guess you're meant to assume the horizontal velocity is zero (v = 0), because you can't calculate it without knowing the cross-sectional area of the water flow.

Given the speed, you can find the kinetic energy of the water hitting the turbine each second. It's 0.5*m*v^2 = 0.5*650*39.8^2 = 515970 J. (Take note that the gravitational potential energy of the water has been converted into kinetic energy. The potential energy the water had at the top of the falls should not be added back into the equation at any point.) If the efficiency is only 58%, the actual energy transferred each second is 0.58*515970 = 299263 J, meaning the energy transfer rate is 299 kW.

Answer 3

V^2 = v'^2 + 2gh
Lets take initial velocity as 0
hence V^2 = 2gh
or V = sqrt(2gh)
Plug in the values
we get V = sqrt(2x9.8x81)
A) V = 39.8m/s

The falling water possesses Kinetic Energy = 1/2mV^2
K.E. = 1/2* 650kg*(39.8)^2 = 515kJ
assuming 58% efficiency
Energy transferred = 0.58x515kJ =298kJ

Answer 4

regardless of how much water is falling, it will accelerate at a rate of 9.8 m/sec^2. distance = (at^2)/2. ie: 81 = (9.8 t^2) /2
81 = 4.8 t^2. (solve for t)
v = a* t so speed of water = 9.8 * t
the rate at which KINETIC energy is transferred is .58 (58%)

ps
when you mention "v=0", what are you talking about?

Answer 5

velocity=sqrt2*g*h,
where,
g=gravitational constant=9.8m/sec^2
h=head=81 meter.
velocity=sqrt2*9.8*81
velocity=39.845 meter/second.

free flowing water is having kinetic energy=1/2*m*v^2
=.5*650*39.845^2
=515977.81 joule
now, efficiency=58%,so,
the energy transferred
=.58*515977.81
=299267.129 joule/second.

Answer 6

A]v^2-u^2=2gs, u=o
v^2=2*9.81m/sec^2*81m=m^2/sec^2
v=sqrt[2*9.81*81]m/sec
=39.86m/sec
B]energy transfer.
The potential energy/sec
=m*g*h
=850kg/sec*9.81m/sec^2*81m
=675418.5 kg*m/sec^2*m* 1/sec
=675418.5 Joules/sec
PE becomes Kinetic energy
efficiency=.58
Energy transferred
=.58*675418.5 J/sec or Watts
=391742.73 Watt
=391.743 KW

Answer 7

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Answer 8

kinetic energy = gravitational potential enrgy, do some rearanging to get what v equals.
and the rest is simple.
enjoy