Formulas For Engine Cylinder Perssure

[Jason Louison]

does anyone know what unit of measure would the bottom equation be in?

((P–Patm)*A*r)÷(sin(a+p))/cos(p))

where P = Cylinder Pressure (psi)
Patm = Atmospheric Pressure (psi)
A = Piston Crown Area (mm^2)
r = Crankshaft Radius (mm)
a = Crankshaft Angle (deg)
p = Piston-Rod Angle (deg)

It plots the pressure contribution of the cylinder.

 

[xxChrisxx]

Hmm, I'm not comprehending something here.

Are you asking what the unit of the equation you wrote are in?

 

[Jason Louison]

Hmm, I'm not comprehending something here.

Are you asking what the unit of the equation you wrote are in?

Yes. I'm confused because the pressure volume diagram plots cylinder pressure as Newtons/meter/meter, Which I think is megaPascals, and now somehow the pressure contribution, Inertia contribution, and Turning Moment are all in Newton Meters.

 

[xxChrisxx]

I think ultimately you need to be much more careful with your units as Nidum said. Note pressure as N/m^2 rather than N/m/m (which is confusing).
The answer is all in the equation. Ignore units for now and conceptually break down the equation into steps.

What is happening at each step:
1. (P-Patm)
2. (P-Patm)*A
3. ((P-Patm)*A)*r

What do the cos and sin terms do to this?

Or try doing this from first principles to gain a conceptual understanding. We are plotting torque yes?
T=F*d
So how to F and d change as we go through the engine cycle? Start with d first.

 

[Jason Louison]

I think ultimately you need to be much more careful with your units as Nidum said. Note pressure as N/m^2 rather than N/m/m (which is confusing).
The answer is all in the equation. Ignore units for now and conceptually break down the equation into steps.

What is happening at each step:
1. (P-Patm)
2. (P-Patm)*A
3. ((P-Patm)*A)*r

What do the cos and sin terms do to this?

Or try doing this from first principles to gain a conceptual understanding. We are plotting torque yes?
T=F*d
So how to F and d change as we go through the engine cycle? Start with d first.

So far, this is what iv'e got:
(P-Patm)
Where
P= Cylinder Pressure (atm)
P= Atmospheric Pressure (atm)

atm–atm=atm

atm*A

Where A= Piston Cross Sectional Area (mm^2)

atm*mm^2= atm*mm^2

atm*mm^2*r

Where r= Crankshaft Radius (mm)

atm*mm^2*mm= atm*mm^3

Now this is were things start to get looney

atm*mm^3*(sin(ac+acr)/cosine(acr))

Where

ac= crank angle
acr = angle between bore centerline and connecting rod

 

[xxChrisxx]

You need to abstract away from the units. They are actually irrelevant to what's going on.

1. Is net (gauge) cylinder pressure. Can be Psi, Bar, atm, MPa, whatever.
2. Pressure * area = ?
3. ^_^ * distance = ?

Draw the crank throw and conrod vertical. If you apply pressure here, what rotating torque do you have?

 

[Dr.D]

The proper phasing depends upon the geometry of the engine, specifically (1) the relative crank angles and (2) the relative cylinder inclination angles (in-line, V, opposed, etc). All of this goes into the kinematics that relate one reference crank angle to the effective crank angle for each other cylinder.

Depending on the purpose for the analysis, it may, or may not, be appropriate to simple add the torques from all cylinders.
(1) If the crank is to be considered rigid, then all of the torques are additive.
(2) If the crank is to be considered as flexible, as is necessary for a torsional vibration analysis of the machine, then each cylinder torque should be considered as a separate excitation, with appropriate phasing.

 

[Jason Louison]

You need to abstract away from the units. They are actually irrelevant to what's going on.

1. Is net (gauge) cylinder pressure. Can be Psi, Bar, atm, MPa, whatever.
2. Pressure * area = ?
3. ^_^ * distance = ?

Draw the crank throw and conrod vertical. If you apply pressure here, what rotating torque do you have?

at 360°, where pressure=atm, area=mm^2, and distance= mm, I have 85864 atm's*mm^3

 

[xxChrisxx]

You're missing the point of what I'm asking and seemingly blindly applying an equation you have found. To understand properly you need to know how to derive the equation for torque.

Draw(or google) the picture of a crank slider with the piston at TDC.

 

[Jason Louison]

Ignore the values and the figure on the right please.

 

[Nidum]

Recommended reading

 

[xxChrisxx]

So in that left hand image, imagine we apply maximum cylinder pressure (as happens at TDC on an ideal otto cycle). At that instant, how much torque does it make?

 

[Jason Louison]

So in that left hand image, imagine we apply maximum cylinder pressure (as happens at TDC on an ideal otto cycle). At that instant, how much torque does it make?

I can't specify...

 

[xxChrisxx]

You can, its one of the cases where you don't need to see any units or values.

If torque = force * distance?

Force acts vertically downwards, what is th length of the moment arm?

 

[Jason Louison]

You can, its one of the cases where you don't need to see any units or values.

If torque = force * distance?

Force acts vertically downwards, what is th length of the moment arm?

One meter.

 

[xxChrisxx]

I'm afraid not. The answer is zero; no moment arm, no torque. Can you see why?

 

[Jason Louison]

I'm afraid not. The answer is zero; no moment arm, no torque. Can you see why?

does it have to do with the fact that the crank angle is zero?

 

[xxChrisxx]

It does indeed. You are still applying the force on the piston though.

If it's not making torque what is it doing?

 

[Jason Louison]

well I know this:

When you generate torque on a static (non-moving/no velocity of any kind) Object, You are still indeed applying torque, but you are not generating any WORK. Correct?

 

[xxChrisxx]

The force is bending the crank. The thing to take away from this is that for every crank angle. The force acting vertically down resolved into a bending component and a turning component.

Much in the same way that gravity on an inclined plane resolves into a normal and perpendicular force.

 

[Dr.D]

There are inertial terms arising from both the piston acceleration and the connecting rod (linear and angular) acceleration. These are not constants, and are phased to be associated with each cylinder in turn.

This difficulty with the slider-crank machine is largely due to the kinematics. If you have a good device to handle the kinematics, then the correct equation for every cylinder is easily written using energy methods. This will have every cylinder moving in its proper phase, with the associated terms due to piston and connecting rod motion. To try to cobble together torque functions, etc. on an ad hoc basis is just asking for trouble.

 

[Jason Louison]

Yeah, the equations support this. Now, what I'm TRYING to figure out is how to get from (atm*mm^3+kg*degree/mm) to Newton Meters.
I'm sorry I'm probably frustrating you a lot.


*Note: the above graph is an approximation.

 

[Work Hard Play Hard]

Jason,

Thanks for the msg and I'll be answering soon but Chris and others will be helping you from here on.
Remember Newton's 3 laws of motion. You're dealing with rotational or angular analogs but they still apply. Body at rest, body in motion stuff. What happens at TDC?
It's also time to take off your "physics" hat and put on your "ME" hat with this project of yours. In physics, torque and moment are interchangeable terms. They are not to an ME. Wikipedia even covers this because it's so often misused.
Torque = (moment of inertia) x (angular acceleration)
Newton gave us this.
Furthermore remember equilibrium or here, "Rotational Equilibrium." At a steady velocity, steady can be zero mph or 100 mph, the net external forces of torque equals zero or the sum of all external torque's equals zero. You are going to have to start breaking things down and putting what you are looking for into context to properly organize and then understand it. But that's alright. It's what makes all this fun and you're doing well. As it was once put also, Chris and others are doing you a solid.
By the way, the compression ratio of your engine is 10:1 or 10.5:1 with VV.
----not everything was here when I wrote this----

 

[Randy Beikmann]

View attachment 114636

Yeah, the equations support this. Now, what I'm TRYING to figure out is how to get from (atm*mm^3+kg*degree/mm) to Newton Meters.
I'm sorry I'm probably frustrating you a lot.


*Note: the above graph is an approximation.

Before you get too far trying to convert this result, check the dimensions of your terms. The first part has atm*mm^3, or pressure x volume. This correctly gives you work units.

However, the second part, in kg*degree/mm, is in mass/distance, which does not make physical sense here.

I would go back and redo the problem, putting everything into SI units, and track the units every step of the way.

 

[Jason Louison]

Hi guys, Iv'e been SUPER busy the past few weeks so I didn't think about posting for a while. The spreadsheet has gone through a series of MASSIVE updates, But I won't put in multi-cylinder engine support until a few roadblocks are cleared up. For now, I will post this screenshot and upload the spreadsheet for you guys to check out!

 

[Jason Louison]