## Re: [microhydro] prony brake design equation

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• ... 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
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/
• 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
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.