- At 8:19 AM +0000 29/9/03, blastronics wrote:
>

If the torque is in Newton -metres than the second version is

>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?

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

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] - 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.