What is the moment of inertia and bearing friction of a rotating turbine?

AI Thread Summary
The discussion revolves around calculating the moment of inertia and bearing friction for a turbine that accelerates to 60 RPM and then continues to rotate after the force is removed. The known parameters include a torque of 260 Nm, an acceleration of 0.35 rad/s², and a deceleration of 0.16 rad/s². Participants suggest using the formula τ = Iα to find the moment of inertia, with one user indicating that the expected answer is 509.8 kgm². There's a concern about the non-constant acceleration and deceleration due to friction, which complicates the calculations. The discussion highlights the challenges in applying the equations correctly to solve for the required values.
confused2006
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Moment of inertia problem...please help !

Hi,
I have been sat in front of this computer all day trying to work this problem out.

A turbine rota accelerates to 60rpm from rest in 9 revs then the force is removed and it continues to rotate for a further 20 revolutions.

Basically I know the torque is 260Nm
I know the acceleration is 0.35 rads/s^2
I know the deceleration is 0.16 rads/s^2

I need to find the moment of inertia in kgm^2

And i need to find the bearing friction in Nm

Please help, think my head is going to explode
 
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Did you try just use this formula

\tau=I\alpha

where \tau is torque and \alpha is angular acceleration.
 
What i didnt explain is I know the answer is 509.8Kgm^2 and I can't get anywere near it with any equation
 
I don't know,but I also cannot get the answer.

Here's the my solution.

\tau_f=I\frac{d\omega}{dt}=I\frac{d\omega}{dt}\frac{d\theta}{d\theta}

\int_{0}^{20\times 2\pi}d\theta\tau_f=I\int_{0}^{1 rad/s}\omega d\omega

40\pi \tau_f=I\frac{\omega^2}{2}

when we have \tau_0=260 Nm

\tau_0-\tau_f=I\frac{d\omega}{dt}\frac{d\theta}{d\theta}

d\theta(\tau_0-\tau_f)=I\omega d\omega

9\times 2\pi\left(\tau_0-\frac{\omega^2}{80\pi}\right)=I\frac{\omega^2}{2}

I=\frac{18\pi\tau_0}{\frac{\omega^2}{2}+\frac{9\omega^2}{40}}
 
ermm at a glance
torque=I*alpha + friction*omega

if it max's out at 60rpm (2pi rads/s) then alpha is 0
so 260nm=friction* 2pi

friction= 260/2pi nm/rad/s ? roughly 40nm/rad/s ??
dunno if that's any help?
 
also when you say that

Basically I know the torque is 260Nm
I know the acceleration is 0.35 rads/s^2
I know the deceleration is 0.16 rads/s^2

i'm pretty sure that the acc/deelertaion is NOT constant if you have friction which is usually proportional to velocity
 
also, i don't know how to do that fancy maths typing!
but

using o for theta
w for omega

T=I*Alpha +D*w

T=I*(d2o/dt2)+D(do/dt)

in laplace(if you know it?)

T=Ios^2 + Dos

lol i should really have put this allin one post, i'll look at it again if I've time later!

also its
i have been sitting,
and a turbine rotar
 
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