Simple Mass/Acceleration/Power Question . . . 4

[AlbertG]

Good day, Folks.

I have a classically-simple question for the community today -- just checking my math and fishing for an additional equation ;o)

As a backgrounder, I'm attempting to estimate the power required to sustain a body in motion against the sum of all forces which are working to bring it to rest. To that end, a pair of proximal velocities are chosen for the body. The final speed of the body is reached by deactivation of the prime mover; yielding a “coast-down” scenario for the subsequent calculations. Simple enough.

OK.

Definitions:

U = Initial Velocity (m/s)
V = Final Velocity (m/s)
A = Acceleration (m/s squared)
S = Distance Traveled (m)
T = Time (sec)
F = Force (N)
P = Power (Nm/sec or Watts)
M = Mass (kg)


First, we need to determine the negative acceleration of the subject body from steady state to final speed; according to A = (V – U) / t.

Example calculation:

-3.67 = (235 – 246) / 3


Second, call in F = M * A.

Example calculation:

-77000 = 21000 * -3.67


Third, get the distance covered during the coast-down between U & V with S = (Vsquared – Usquared) / (2 * A).

Example calculation:

721 = (55225 – 60516) / -7.34


Finally, convert the assembled into an elucidation of the power represented during coast-down of the body using P = F * (S / T):

-18500 kW = -77000 * (721 / 3)


Now, assuming that I hashed this through correctly, at what velocity along this deceleration curve could I accurately place the above power figure as an equilibrium value sufficient to sustain the body in motion at a constant speed?

Thanks again.

 

[IRstuff]

You seem to calculating in circles. I don't see anything that quantifies the "sum of all forces which are working to bring it to rest."

Once you do that, the power is simply the force multiplied by the velocity.

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

Thank you for your reply.

"You seem to calculating in circles."

Alright. Fair enough.

In the above examples, we have a 21000 kg body which is coasting down from 246 m/s to 235 m/s. Drag coefficient and mechanical friction are unknown for the body. Force, acceleration, and distance traveled were computed to subsequently elucidate the power dissipated in the timeframe specified. In my humble opinion, I believe that all is linear so far.

"I don't see anything that quantifies the 'sum of all forces which are working to bring it to rest.'"

The figures yielded 18,000 kW as representative of the total power acting against the body to slow it down 11 m/s in 3 seconds' time. That amount of energy would appear to be working quite well to rapidly bring the body to rest.

Truly, if anything in the above-cited equations is in error, please tell me where my numbers have failed.

"...just checking my math..."

And finally,

"...assuming that I hashed this through correctly, at what velocity along this deceleration curve could I accurately place the above power figure as an equilibrium value sufficient to sustain the body in motion at a constant speed?"

Thank you again.

 

[desertfox]

Hi AlbertG

Your maths are correct and you have calculated the power
required to slow the mass down.
What you haven't said is what device your using to dissipate the energy.
The only velocity you can incorporate that power on the mass is when its 246m/s.
Then you need to release this braking power to allow the mass to continue, but without any other means keeping it going the mass will come to a stop due to friction and drag
forces.


regards
desertfox

 

[prex]

Your result is readily obtained as:
P=M(V²-U²)/(2[Δ]t)=18518500 W
Now your second question is not clear to me. If you continue to apply a constant braking power, the speed will go down faster and faster with the law:
v=[√](V²-2Pt/M)
and the time to stop is:
[Δ]t_o=MV²/(2P) (a little over 34 sec with your figures)

prex

http://www.xcalcs.com

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http://www.levitans.com

: Air bearing pads

 

[ivymike]

the acceleration you're using is also assumed to be constant if I read your equation correctly - that's probably not the case for aero drag

 

[Heckler]

Have you drawn a free body diagram? FBD helps to visualize what is happening to the system.

Heckler
Sr. Mechanical Engineer
SWx 2007 SP 4.0 & Pro/E 2001
o
_`\(,_
(_)/ (_)

This post contains no political overtones or undertones for that matter and in no way represents the poster's political agenda.

 

[IRstuff]

OK, as stated, then, you can only apply that power to that velocity, since power is the product of force times velocity, and you only know the force in that velocity regime..

You've calculated the power extracted from the moving object through drag and friction. Drag is proportional to V², while friction could be relatively constant.

I don't know if you have the actual velocity curve as a function of time, but if you did, you could hypothetically curve fit to separate the drag component from the friction and extrapolate to other velocities.

TTFN

FAQ731-376

 

[zekeman]

Most of your equations are wrong since they make the assumption of constant force along the path. In fact since the force is from drag, it is some funtion of instaneous velocity and the power absorbed at that moment is that force times the velocity.
So your answer is that the force necessary to sustain the motion at any time is equal to the drag force but opposite so that the net force on the object becomes zero. Since power equals force times velocity, the power required is numerically the power absorbed by the drag.
The equations are not linear and the solution is not linear.
Start from basics and don't make any assumptions, only
1)F=MdV/dt
2)F=function of V. (you must get this from empirical data)
These two equations together with the initial conditions define your problem, period andf their solution yields the motion. If at any time, T you chose to make the system move at constant velocity, then you can write a third equation
3 F1=-F(T) which then renders equation 1
F-F1=0Mdv/dt for t>T.

 

[AlbertG]

@Everyone:

Thank you for your input. I truly appreciate the discussion which we're having over this issue.

As has been keenly pointed out, it is not possible using my math to neatly establish at what place along this deceleration segment we can accurately assign the power figure which was yielded. I am sorry for having caused any confusion here; as the aero and mech drag curves must be analyzed in order to make this determination.

That being said, I'd like to respond to some of the points which have now been elucidated.


@desertfox:

“Your maths are correct and you have calculated the power required to slow the mass down.”

Thanks for the confirmation here. I was fairly certain of my arithmetic; but it never hurts to get a second opinion...


@prex:

“Your result is readily obtained as: P=M(V2-U2)/(2?t)=18518500 W.”

Thank you, prex, for the cross-confirmation of my figures. I'll tuck this approach away for future reference.


@IRstuff:

Good points. Thanks again --


@zekeman:

“So your answer is that the force necessary to sustain the motion at any time is equal to the drag force but opposite so that the net force on the object becomes zero.”

That was the heart of the approach which I was seeking to explore here.

“Since power equals force times velocity, the power required is numerically the power absorbed by the drag.”

Good.

“The equations are not linear and the solution is not linear.”

True, indeed.

However, if we bring U and V more closely together (oh, say a mere 2 m/s apart), we should be able to safely engage a reasonably linear segment of the composite deceleration curve for the body.

Therefore, the simple linear math which was originally chosen should yield a reasonably accurate figure for the power required to sustain the body at a constant speed (defined somewhere between U and V) working against those discretely-unknowable forces.

Again, if those points are moved closer together, I believe a conservative assumption can be made that by applying the derived power figure (P) to the final (lowest) velocity point (V), we should be able to safely consider P an equilibrium value sufficient to sustain the body in motion at V.

Is this an accurate deduction, or am I missing something here?

Thanks again.

 

[IRstuff]

Sounds plausible. The two velocities you have should only result in about 10% difference in power, so two values that are closer should give you a decent answer.

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

Thanks again for the input.

Great go-round, everyone!

'Til next time ;o)

 

[JamesBarlow]

This would be much easier to model if you simply used the original equations for motion.

I mean, it's easier if you remember how to solve second order differential equations.

 

[hydtools]

Use a work = change in energy approach.

Assume only change in kinetic energy.

work = 0.5*mass*(U^2-V^2) = 55 555 500 N-m or 15.43 kw-hrs

work/unit time = power = 15.43*3600/3 = 18518.5 kw

Ted

 

[AlbertG]

@hydtools:

Thanks, Ted, for the tip!

This'll save a little finger skin ;o)