damper energy dissipation

[MJW]

Hi,
I hope you can help. I need to know what the energy dissipated by a damper in a single stroke is? It seems a simple, and obvious question but I am tying myself in knots with it! Please put me out of my misery.
F=c*velocity,
E=integral(0 to Length)F.dx?
F=c*Length*acceleration?
Aaargh!
Thanks, Mike

 

[GregLocock]

F(t)=c*v()
W=int F(t)*[int(v(t))dt] dt

over one cycle. Acceleration does not appear, and you can't integrate v outside of the other equation.

Cheers

Greg Locock

 

[electricpete]

p(t)=v(t)*f(t)
where
p(t) is instantaneous power
v(t) is instantaneous velocity
f(t) is instantaneous force

Assume the system is sinusoidally excited (force, displacement, velocity are all sinusoidal in steady state).

The most general sinusoidal case will be give average power
<p(t)>=Vmax*Fmax*cos(theta) where theta is the angle between the voltage and force.

I assume the system can be modeled as simple linear single degree of freedom mass spring system. theta varies based upon relationship between damping c and spring k.

IF there is only damping (no spring constant), then the angle is 90 degrees. Further more in this case Vmax=Fmax/C and therefore <p(t)>=Fmax^2 / (C)

The more general case can be solved by laplace/fourier analysis of sdof mass spring system
X(s)/F(s) = 1/(m*s^2+c*s+k)
substitue s=jw to convert from laplace to fourier
multiply by (jw) to convert from X(jw) to V(jw)

 

[MJW]

Thanks for your help. It isnt actually a vibrating system; it's more an energy dissipation mechanism before an end stop; a buffer if you like.

Energy dissipated is therefore simply 1/2 cvL, where v=initial velocity, L=length of cushion and c is the damping coefficient. This assumes a zero final velocity and linear decelleration.
But am I right to assume a linear decelleration? Force is proportional to the velocity, so as the velocity drops so will the force proportionally? Therefore linear deceleration?

 

[electricpete]

F=m* d^2/dt^2 (x) = -c*dx/dt
d/dt(v) = -c*v
Solution:
v = v0*exp(-c*t/m)+v0
x = int(v)dt = ....

Gotta run. Note if you have assumed vf=0, you have assumed all kinetic energy is dissipated.

 

[electricpete]

sorry
v = v0*exp(-c*t/m)

F = Int(F)dx = Int(Fv)dt
= Int Int(-cv^2)dt
substitute above expression for v(t) and you will have a difficult to evaluate expression. Maybe we need some more info/assumptions to clarify what you have and what you're after.

Once again ,if you are assuming that the final velocity is zero, then the damper absorbs exactly an amount of energy equal to the initial kinetic energy.

 

[electricpete]

Sorry last two equations should have been
W = Int(F)dx = Int(Fv)dt
= Int(-cv^2)dt
= Int (-v0^2*exp(-2c*t/m)dt

 

[GregLocock]

While electricpete sorts his algebra out (it usually comes out right in the end), no you can't use 1/2cvL, if you want an accurate answer.

This is because you don't really know L until you've got the equation right, obviously if c is larger then L will be shorter.

The force applied is proportional to the instantaneous velocity, whereas for linear acceleration the force must be constant.





Cheers

Greg Locock

 

[electricpete]

Once again, I don't think there can be any meaningful solution unless you assign an m. i.e. if we assume initial velocity v0, we shouldn't expect the energy absorbed by the damper over any fixed time interval or distance to be the same for a pea as for a bowling ball. (as shown below v(t) depends on m).

IF we assume the final condition is zero velocity, the answer is simple… the amount of kinetic energy absorbed by damped is equal to initial kinetic energy 0.5*m*v0^2.

More interesting is to stop at some intermediate time or distance.

F=m* d^2/dt^2 (x) = -c*dx/dt
d/dt(v) = -c*v

Solution:
v = v0*exp(-c*t/m)
x(t) = v0*m/c*[1-exp(-c/m*t)]


Energy E=W = Int(F)dx = Int(Fv)dt
= Int(-cv^2)dt
= Int (-v0^2*exp(-2c*t/m)dt
> E:=int(-c*v(t)^2,t=0..tfinal);
E := 1/2 m*V0^2 * [exp(-2*tfinal*c/m) -1]
(yes, work is negative since force is acting opposite velocity)

If you don't like having tfinal left in your expression, you can convert based upon the above relation for x(t) as follows:.

> solve(subs(t=tfinal,x(t))=L,tfinal);

tfinal = -ln(-(-m*V0+L*c)/m/V0)*m/c

No guarantees that I haven't made an algebra blunder.