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Re: [microhydro] prony brake design equation

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  • Hugh Piggott
    ... If the torque is in Newton -metres than the second version is correct. Except that it has too many brackets. P(kW)= T(Nm)* ((2*3.14)/60)* N(rpm) If you
    Message 1 of 4 , Oct 1, 2003
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      At 8:19 AM +0000 29/9/03, blastronics wrote:
      >
      >one formula gives :
      >
      >P(kW) = (T(lb-ft) * rpm)/ 5252
      >
      >P(kW)= T* omega
      >
      >or P(kW)= T(N)* ((2*3.14)/60))* N(rpm)
      >
      >which should i use?

      If the torque is in Newton -metres than the second version is
      correct. Except that it has too many brackets.

      P(kW)= T(Nm)* ((2*3.14)/60)* N(rpm)

      If you must use pounds and feet then you will need a constant, and I
      presume that 1/5252 is correct there - I have not checked it. A
      Newton is about 100 grams-weight force (1000g/9.81). A metre is 3.28
      ft.
      --
      Hugh

      http://www.scoraigwind.co.uk/
    • Radu Babau - VARSPEED Hydro
      Hello Tony, I am a motor guy myself, and have experienced variable speed operated turbines, driven by bi-directional power flow converters with asynchronous
      Message 2 of 4 , Oct 1, 2003
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        Hello Tony,

        I am a motor guy myself, and have experienced variable speed operated
        turbines, driven by bi-directional power flow converters with asynchronous
        generators. These electronic gadgets (such as ACS611 from ABB) incorporate a
        motor control philosophy called DTC (Direct Torque Control). It can show you
        eveything about the motor (U,I,P), inclusing shaft torque and speed (T, n).
        On Kaplan and Francis turbines, we have noticed the same decrease of torque
        with speed. It is also a throretical curve shown on the Layman Handbook,
        available on http://europa.eu.int/comm/energy/library/hydro/layman2.pdf , on
        page 180, fig. 6.30.

        So, what you have noticed is correct, don't feel confused, it comes with the
        turbine. Regarding the formula, here it is what I use since my university
        years:
        P [kW]=T[Nm]*N[rpm]/9550

        If the torque-speed characteristic is the one you have noticed, it means
        that the power-speed is a hill, also shown in the same figure. You can see
        that if the generator (turbine) is left out of load, then the speed will go
        higher, up to a maximum speed (around 2xrated speed), where the generated
        power is zero. Well, the happy thing is that the turbine will not go higher
        in speed (it would become a motor then), which als means that will not
        succeed making a turbo-plane engine which will blow sky-high if you loose
        the output (self-protect feature).

        Why the torque is going down with speed ? Not beeing a hydraulic engineer, I
        can only imagine that it is connected on what angle does the water hits the
        blades. If the turbine speed is zero, than the blades are deflecting
        practically all the water, thus the torque is maximum. When you start moving
        more and more water will become tangent to the blades, and that means
        decreasing torque. You will then reach a optimum ratio between the tangent
        and the axial forces (corresponding to the maximum power and efficiency, top
        of the hill), and then finally the zero torque and power (here the blade is
        almost running faster than water tangent speed). But, perhaps the hydraulic
        guru's in the group can explain you better this turbine behaviour.

        Have fun.

        Radu

        [Non-text portions of this message have been removed]
      • Lancie1@aol.com
        Tony, Generally the faster an output shaft is turning, the less reserve power is left in the motor, turbine, or machine, to turn it even faster. So you can
        Message 3 of 4 , Oct 4, 2003
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          Tony,

          Generally the faster an output shaft is turning, the less reserve power is
          left in the motor, turbine, or machine, to turn it even faster. So you can say
          that the torque (twisting force) decreases as the rpm's increase.
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