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Kinetic Energy -> Electrical Energy

  1. Mar 14, 2008 #1
    Okay, so the situation is a hydro-electric plant, assumed to have a vertical inlet pipe, of (s) metres high, and a opening to the turbine of (d) metres in diameter. The water is assumed to be stationary before it begins falling down the pipe (s) metres. So, I have the speed of the water when it reaches the bottom, beng the potential verses the kinetic gained, hence;
    v = [tex]\sqrt{}(2gh)[/tex] (h being s, sorry lol..)

    Anyways I alos have the mass of water entering the pipe in 1 second, from which I assume I can work out the power output of the generato, assuming that the turbine is direclty under the inlet opening etc. Now, a watt is a joule/sec, hence finding the mass entering the turbine in one second should help me obatin P, (= W/t). So;

    the mass of water in 1 second = 250[tex]\pi[/tex]d^2 ([tex]\sqrt{}(2gh))[/tex] [in kilograms].
    The question I pose, is it legitimae, or, rather, correct, to try and find the Power output by first using F=ma, to find the force exerted by the mass of water, then use W = fs, for the work that can be done by this, and then to use P = w/t? Or is the a way to directly convert from Ke, to electrical enegery, or power, in watts?

  2. jcsd
  3. Mar 15, 2008 #2


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    I'm not really clear on the quesiton. "Directly" converting is what you were describing. With metric units, the units are easy to convert. Watts are joules per second. In everyday situations, how you do the calculation depends on the situation.
  4. Mar 15, 2008 #3
    K, basically, have botho those values, teh velocity, and the mass. Now I am asked to work out the expression for the generating capacity of the turbine in watts.

    so I have, for a direct vertical fall of h, into a opening d metres wide,

    v = [tex]\sqrt{}(2gh)[/tex]

    and m of the water falling through the opening in 1 sec, m = (250*pi*d^2)* [tex]\sqrt{}(2gh)[/tex]

    how do i find this expression?
  5. Mar 15, 2008 #4


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    The turbines run mostly at a constant speed, so the only acceleration involved is the vertical decleration of water, and the angular acceleration of that same water as it passes through valves and later the turbine.

    It's probably easier to calculate the power of the water based on the distance it falls until past the turbine blades, and then use the actual electrical output to determine the efficiency of the turbine / generator combination.

    Wiki has a few diagrams in their articles:



    Hoover dam uses Francis turbines (as mentioned in above Wiki article):

    Last edited: Mar 15, 2008
  6. Mar 15, 2008 #5
    (a) a hydroelectric power plant has water falling a vertical heigh of h through an inlet pipe of diameter d to a turbine. Assume that there are no loses due to friction, turbulence, or similar effects. Use g = 9.8 ms^-2, be careful with units.

    (i) if the water is motionaless before it falls, and the reduction in potential enegry of the water as it fallsi is equal to the increase in its kinetic energy, find the velocity of teh water at the bottom of its fall (before it passes through the turbine)

    (ii) Find the expression for the mass of water that moves through the turbine in 1 sec (hint, you need to consider both velocity of the water amd the cross sectional area f the pipe)

    (iii) Assuming the turbines are 100% efficient at converting kinetic energy to electrical energy, find an expression for the generating capacity of the turbine in watts.

    Number (iii) is the one I am having trouble with. I need to know I can use the sequence F=ma, then W=fs, and then P=W/t to find the capcity in watts... Sorry...
  7. Mar 15, 2008 #6


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    All you need to know is the kinetic energy produced per second by the water after it falls from a height h in a pipe with diameter d.

    Note that 1 Newton = 1 kg m / s^2, where m = meter and s = second

    so 1 Joule = 1 N m = 1 kg (m/s)^2, where (m/s) is meters per second,

    and 1 Watt = 1 Joule / s = (1 kg (m/s)^2) / s

    Assuming a temperature of 4 C, the mass of water is 1000kg / m^3.

    v = sqrt(2 g h)

    Assuming v is expressed in (m/s), the mass of water flow for each second in the pipe is (where s is 1 second):

    m/s = v(m/s) s (d(m)/2)^2 pi (1000kg / m^3) = 250 pi v(m/s) s d(m)^2 kg / m^3

    at this point, the energy produced per second will be 1/2 m v^2 / s and will produce units of (kg (m/s)^2) / s or Watts.
    Last edited: Mar 15, 2008
  8. Mar 15, 2008 #7
    Haha, thanks. Your unit analysis makes sense. I understand where I am trying to get to now.

    Thanks, w_b
  9. Mar 15, 2008 #8
    So does that mean that my Ek value is the power output ? in other words, I use the 2 equations I have, put them into 1/2mv^2, and I have my generating capacity, in watts?

    I must still not understand - that sequence of units brings me to use the F=ma, then W = fs, and then P= W/t, exactly....
    Last edited: Mar 15, 2008
  10. Mar 15, 2008 #9


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    Yes, since you used seconds for the time period to determine the mass of water per time period. I updated my previous post. The confusing thing here are the time factors. In order to calculate the mass of water flow per time period, you have to multiply velocity by the time period to calculate the height of a cylinder of water, in order to calculate the volume, then the mass of water per time period. Throughout this series of equations, you need to leave the time period in the denominator on the left hand side of the equations. In this case since the goal is to calculate Watts, which is also based on a time period, when done, you only need to convert (1/time period) to (1 / second). Since your time period was a second, no conversion was required.

    More correctly stated:

    v = sqrt(2 g h)

    volume / t = v t pi (d/2)^2, where v = velocity, t = time period, and d = diameter of pipe.
    mass / t = (volume / t) (m^3) (1000 kg / m^3)

    energy / t = 1/2 (mass / t) v^2
    power = (energy / t) (t / s), where (t/s) is how to convert 1/time into 1/second;
    for example if t = 1 min, then (t/s) is (1 min / 60 sec).
    Last edited: Mar 15, 2008
  11. Mar 15, 2008 #10
    Okay...so let me test this...that would mean for an inlet pipe of h = 150m, and a diameter of d = 2m, and an efficiency of 80%, the power output would be 200322020 watts...wow...

    that doesn't seem right...hmm. I understand what you've said, but have I implemented it properly..?
  12. Mar 15, 2008 #11


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    Using a calculator and not rounding off, I get

    200323020 watts, while you got

    200322020 watts, maybe a typo on the 6th digit where you show a 2 and I show a 3?

    You're talking about 170342.7 kg of water at 54.22 m/s speed every second, that's 250403775 watts of input power.

    For comparison, Hoover dam has a maximum effective water height of 590 feet, and an average water height of 510 to 530 feet (from the link above). There are 4 main water penstocks (pipes) with 30 foot diameters. 4 penstocks with 13 foot diameters are tapped from each of the main water penstocks. Each of the 13 foot diameter penstocks flow into a circular shaped inlet with wicket gates into a turbine. The wicket gates are similar to blinds and control the water flow. I'm not sure of the turbine diameters, so I'm not sure how much water actually passes through the main turbines. There are 15 main turbines rated at 178,000 hp, or 132,700,000 watts, while actual output from 13 of the generators is rated at 130,000,000 watts. I doubt that efficiency is 98%, so I assume these quoted numbers are rounded off. Back to the main 30 foot diamter penstocks, the water flows past the 4 penstocks, to other pipes and to the river below with yet another gate to control the flow of water not going through the turbines. In adddition, there are two gated side tunnels in the canyon walls that bypass the dam and turbines for flood control. Also, regardless of the number of taps into the main penstocks, the pressure of the water is strictly a function of effective water height, assuming pipe diameter is large enough that friction and viscosity aren't significant issues.
    Last edited: Mar 15, 2008
  13. Mar 15, 2008 #12
    Yeah, I asked my dad, who is an electrical engineer, and he says that 200 megawatts is totally normal - in fact, one in our region generates exactly that. By the way, thankyou for your help. I will take more care is paying attention to the base units next time!

  14. Mar 15, 2008 #13


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    You did show a 2 in the 6th digit of your answer, where I got a 3, was this a typo?
    Last edited: Mar 15, 2008
  15. Mar 15, 2008 #14
    Okay, now I have a follow on question;

    'A small wildlife unit etc etc requires a constant 1kW of electricity to operate. Water is diverted etc etc. to power the monitering station, assume that the water falls 10 m.

    (i) If the hydroelectric plant is 70% efficient, what volume of water r second is required to power the monitering unit?'

    Okay, so since we don't have the inlet diamter, i am assuming i just use 1/2mv^2, and find m that way, and then multiply by 1000 to get the volume per second?

  16. Mar 15, 2008 #15


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    Water density is 1000kg / m^3 at 4 C. If you know the mass in kg, you need to divide by 1000 to get the volume in m^3. You'll also need to divide the 1000 watts of power by .7 to get the actual power of the falling water = 1428.57 watts. Since the height is given as 10m, then v is known = sqrt(2 g 10) = 14 m/s.
    Last edited: Mar 15, 2008
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