How to calculate force needed to push piston based on crankshaft value

[blackholefizix]

Homework Statement

How do you calculate the force needed to push against a piston based on the mass of the vehicle, the rpm of the crankshaft, radius of crankshaft (same distance as the center of the crankshaft to the attachment point of the connecting rod and the crankshaft), and the length of connecting rod. Although I am given these measurements, not all of them have to be used. Basically, I need to know how to calculate the piston force based on crankshaft measurements. I request that only people who are qualified or 100 percent sure about the solution answer this question. Also, it would be great if you could make up some hypothetical values in order to demonstrate the calculation process. Thanks!

Homework Equations

τ= I x α
τ= F x r
α= ω/t
τ- torque of crankshaft
α- angular acceleration of crankshaft
I- Moment of Inertia
t- time to reach max angular velocity (crankshaft)

The Attempt at a Solution

Crankshaft radius- .1524 m
Mass of vehicle- 204.12 kg
RPM of Crankshaft- 1260.45

1. 1260.45 rpm = 131.99 rad/s = ω

α= ω/t I interpreted "t" as the time needed to reach the goal rpm.

α= 131.99 rad/s / 10 sec = 13.20 rad/ sec^2

I = 1/2(204.12 kg)(.1524 m)^2 = 2.37 Kg*m^2

τ= I x α = (2.37 Kg*m^2)(13.20 rad/sec^2) = 31.29 N*M

I think everything is right to this point, (I could be wrong). But the calculations I am going to do now is what I am not sure about.

τ= F x r
31.29 N*M = F x .1524 m

F= 205.31 N

I thought that the value of "F" would be the linear force needed to push the piston in order to turn the crankshaft at the goal rpm.

 

[Simon Bridge]

I thought that the value of "F" would be the linear force needed to push the piston in order to turn the crankshaft at the goal rpm.

If you keep pushing on the piston with a constant force, won't it just just go to the bottom and stay there (or have some shm)?

I = 1/2(204.12 kg)(.1524 m)^2 = 2.37 Kg*m^2

You are expressing the moment of inertia of something (unspecified) in terms of the mass of the vehicle and the lever-arm of the crank? So you are using a non-inertial reference frame where the vehicle is rotating about the shaft?
The formula you used seems to have been ##I=\frac{1}{2}mr^2## which would be for a hoop, disk, or cylinder, being spun on it's axis. Is this a good model for the object in question?

α= ω/t I interpreted "t" as the time needed to reach the goal rpm.

That's fair enough - is the t=10s you used a figure you are given?

Anyway - that would tell you an average acceleration for the shaft.

I request that only people who are qualified or 100 percent sure about the solution answer this question. Also, it would be great if you could make up some hypothetical values in order to demonstrate the calculation process.

That's not how it works around here.
Your trouble comes, not so much from performing the needed calculation, but in thinking about the problem to start with.
This is much harder to help you with and a walkthrough won't help you in the long run.

The first step is usually to be clear about the model you are using to describe the problem.

I'd normally be thinking of a problem like this in terms of the geometry of the engine - the force to the piston is not always tangential to the rotation, for example, and the rotational motion of the crank is not always directly coupled to the mass that has to be moved (there's usually a gearbox at least).

But... if this is a theory problem that has been set for you, then it may be that you are expected to be more hand-wavey than that (especially considering how little information you are given).

Of course, if the idea is to work out the net force needed to maintain a constant rpm, then the answer is F=0.

From what you've written, the answer you want lies between these extremes.

 

[blackholefizix]

Hello Simon Bridge,

First of all, I am sorry for any disrespect when I said "people who know how to do this 100 percent". I am just facing a time constraint. This is actually not a book problem, but it is more of a design calculation. I can assign the measurements of the crankshaft to any specification that I choose to make it. But the problem I face is how I can calculate the force needed to push the piston for one full expansion, if I know information about the motion of the crankshaft. I had another idea for this possible calculation. Since we know the acceleration and velocity of the crankshaft, would it be possible to convert that rotational acceleration to a linear acceleration? And then using that linear acceleration (of the piston), use the mass of the vehicle, as well as the F=ma formula to find the force that is pushing on the piston to create that acceleration?

Also, you are right. The linear force from the piston is not tangential to the crankshaft. Another confusion I had though was that the torque will constantly be changing because the angle of the connecting rod and the center of the crankshaft will be changing. So how do I take into account the changing torque?

By the way, that will not be a constant force. The force on the piston will only occur until the piston has fully expanded.


Thank you very much for your first response.

 

[Simon Bridge]

The linear force from the piston is not tangential to the crankshaft. Another confusion I had though was that the torque will constantly be changing because the angle of the connecting rod and the center of the crankshaft will be changing. So how do I take into account the changing torque?

By the way, that will not be a constant force. The force on the piston will only occur until the piston has fully expanded.

Well it is probably even worse than that - because the applied force in this sort of arrangement is very seldom constant even during the down-stroke. The example that comes to me off the top of my head is where the piston is being driven by water from a large reservoir.

But you are specifying a model where the pressure at the piston-head is a constant for the entire down-stroke. That's OK.

The piston is connected to the crank by some mechanism - you use that mechanism to figure out the relationship between force (F=PA) at the piston-head and the torque at the crank. It will depend on the angle the crank has already turned.

The mass of the piston will be important, as will the mass of the crank and other moving parts. Friction is usually significant too.

I would normally be approaching this sort of thing in terms of conservation of energy though ...

 

[blackholefizix]

I appreciate the feedback that you gave me and I understand that my original questions was a little unclear. However, this is the real concept behind my question. The connecting rod actually connects the piston directly to the wheel of the vehicle, not the crankshaft. So, as the piston is expanding and contracting, the wheel is rotating. I said crankshaft because it works like a piston-wheel relationship. The force is not constant. I am trying to calculate how much force will be needed to push the piston based on my initial values. So, for example, I can say that the maximum velocity of the vehicle is 60 mph, the mass is 100 kg, the radius of the wheel is 1 ft, and so on. Based on this, since the piston is directly connected to a connecting rod that connects to the wheel, what force is needed to push on the piston, for the vehicle to move at a max velocity of 60 mph. I hope this clears my question up a little bit. Thanks.

 

[Simon Bridge]

Piston driving the wheel is not a problem - you need the mass of the wheel.
Not all of the mass of the vehicle is rotating so it does not all factor into the moment of inertia (unless you mean that the whole vehicle is one big wheel?)

The radius of the wheel and the distance from the center the piston-rod attaches tell you the coupling of the piston's energy to the road. The length of the rod, and the distance to the attachment, tells you how the torque would vary with each cycle. Sketch it out.

I still think this is best as conservation of energy - the piston falls under gravity + some force (or mount the piston horizontally) which pushes the piston rod. Depending on the angle the wheel has turned, there will be a radial and tangential component to the force applied to the wheel. You can work out the unbalanced force needed to accelerate the wheel to the speed you want by looking carefully at what happens in one single stroke.

 

[blackholefizix]

Yes, the piston is actually mounted horizontally. In one single stroke, the wheel will rotate 180 degrees. However, throughout that 180 degree rotation, the torque will be changing at every moment. I need one uniform torque to work with in order to convert it into a linear force which would be the force of the piston. So would i just use the maximum torque at the tangential angle as the uniform torque?

In addition, are you saying that I should use the mass of the wheel only when I am calculating the torque, and then separately find the force of friction and gravity using the vehicle mass. Then I can find the find the force that the whole vehicle needs to move in one direction because I know its mass and acceleration, F=ma. So at this point, wouldn't I be able to find the force needed to push against the piston because it will have to balance out all of the forces of resistance (torque, gravity,friction), and then even surpass those forces to match the forward force needed?

Thanks

 

[Simon Bridge]

If you postulate a constant torque, then the force at the piston head will vary continuously.
You use calculus to deal with continuously changing values - could you be biting off more than you can chew here?

You may be able to get away with average values for a back-of-envelope calculation:
Since the actual torque will vary from 0 (at the top and bottom) and some maximum when the piston-rod is 90deg to the lever arm, you may prefer to use rms rather the straight averages.

Or you can just say that ##\tau_{ave}=(\omega_{bottom}-\omega_{top})/T_{single stroke}##

The way you handle the different masses depends on how the vehicle is constructed.
What I'm saying is that unless the wheel is the whole vehicle, you should not use the whole vehicle mass to calculate the moment of inertia.

You have to account for all the places energy can go ... some goes into turning the wheels, some into linear motion of the vehicle and some to friction, noise etc.

 

[blackholefizix]

I see. Since the piston arrangement for this vehicle is horizontal, would it be best to first calculate all of the energy that goes to various places, like you had said about the turning of the wheels and moving of the car in a linear motion, and then the sum of all of that energy will be the amount of energy that will be needed to push the piston. I will then have to take into account the extra energy to push the piston head and connecting rod.

But, like you said, as torque changes with the position of the piston, the force needed to push the piston will also change.

SO, do you think it would be possible to add up all of these sources of energy loss, except the torque force. This would be a constant energy expenditure. Then a graph of the addition of the changing torque, with the constant energy expenditure, based on various instantaneous time and piston position intervals, would mean that the force being pushed on the piston to maintain that constant velocity will fluctuate. And that curved fluctuation of force will be the force that this piston will need to be pushed at based on the various intervals.

Thanks a lot for your help.

 

[Simon Bridge]

It is the sort of thing that steam-engine engineers had to do all the time so, yes, it can be done. Usually it is not easy and you usually need calculus.

Their machines were similar to yours - horizontal pistons directly coupled to the wheels.
There would be an additional complication that the pressure to the piston-head was not a nice function of time.
They developed rules of thumb to help them with their designs.

It may be easier for you to start out assuming a simple force to the piston-head and work out the equation that relates that to torque on the wheel ... or, the other way, from the static friction on the wheel you get the max acceleration of the vehicle, and so the max angular acceleration of the wheels, which gives a max torque ... the max torque is the applied force times the lever arm.