# Calculating rev-gain based on engine specs?

1. Jun 18, 2010

### ISX

Is there any way of calculating rev-gain based on engine specs, like bore and stroke? What other variables are involved? Maybe I am thinking of rev-gain in a wrong sense.

If the former isn't capable, is it capable to find out the rotating mass of an engine, like the force the engine has to overcome just to turn itself at a certain RPM, how would I calculate something like that?

2. Jun 19, 2010

### jack action

What do you mean by "rev-gain"?

The force the engine has to overcome just to turn itself at a certain RPM is based on friction alone. The friction power (kW) of a typical 4-stroke automobile engine can be estimated by:

Friction power = (100 + 10.5 * (rpm * Stroke / 30 000) ) * volume flow rate

The stroke is in mm and the volume flow rate in m³/s.

Example: What is the friction power of a 2 L engine at 4000 rpm which has a 90 mm stroke?

volume flow rate = 2 L * 4000 rev/min / 2 / (1000 L / m³) / (60 s/min) = 0.067 m³/s

Friction power = (100 + 10.5 * (4000 * 90 / 30 000) ) * 0.067 = 15 kW

As for the rotating mass of an engine, it will only affect the acceleration of the engine, i.e. the time taken to go from one rpm to the next.

3. Jun 19, 2010

### ISX

Rev gain is precisely the last part you stated. It is the rate at which the engine accelerates. I think you have to have a load factored in too, like other than engine frictional load.

When you say frictional power, do you mean the 2L needs 15KW just to turn 4000 rpm with no load? Meaning if there was no momentum force, laying my finger on the flywheel would stop it?

4. Jun 19, 2010

### jack action

When you accelerate a crankshaft or a piston, it is just like any other mass where a = F / m for linear motion or $$\alpha$$ = T / I in rotation. If the engine is connected to a clutch, transmission, axle and wheels, you must accelerate those too. This rotational effect can be related the mass of the vehicle itself by a factor $$\gamma$$ such that a = F / $$\gamma$$m. Values for $$\gamma$$ for a typical car can be estimated by:

$$\gamma = 1.04 +0.0025G^2$$

Where G is the overall gear ratio between the engine and the wheels.

Check this http://stephenmason.com/cars/rotationalinertia.html" [Broken] for more info.

Frictional power is the power required to fight the friction between the pistons and the cylinders and from the different bearings. It usually also includes the power needed from the essential components such as the oil and water pumps, alternator, etc.

So if there was no combustion (but all internal parts would keep the same temperature such that all tolerances would be the same), just to spin the 2L engine at 4000 rpm you would need a 15 kW electric motor.

Last edited by a moderator: May 4, 2017
5. Jun 20, 2010

### ISX

Alright so now your going over my head a bit. On the frictional power part, how would you factor in pistons? I mean a 2L 4 cylinder would take more force to turn than a 2L single cylinder..

6. Jun 21, 2010

### jack action

Nope! Let me go through the equation I gave you:

Friction power = (100 + 10.5 * (rpm * Stroke / 30 000) ) * volume flow rate

The part in parenthesis have the unit of pressure. It represents the equivalent average pressure necessary to fight friction. We call it the friction mean effective pressure.

The part within the inner parenthesis is actually what we call the mean piston speed. When the piston travel through the cylinder, starting at the top, it has its speed = to zero. Then it accelerates going down, reaching a maximum speed around mid-stroke. It then decelerates until it reaches the bottom of the cylinder where its speed will be again = to zero. It then "backs up", accelerating until mid-stroke, decelerating again, reaching its initial position where its speed is again zero. What is the average speed of the piston (vp)?

For 1 rev, the total distance traveled is twice the stroke (S) (once going down, once going up) and that was done at a given rpm. The time (t) taken to travel 1 rev should be equal to the time taken to travel 2S, so:

t = 1 rev / rpm = 2S / vp

Therefore: vp = 2S * rpm

If S is in mm and you want vp in m/s, you will need to add a conversion factor of 60 000 such that:

vp = 2S * rpm / 60 000 = S * rpm / 30 000

So the friction MEP (in kPa) can be rewritten simply:

friction MEP = 100 + 10.5 * vp

So the friction MEP depends only on the piston speed. This is of course an empirical relationship, but it has been found to be true for any engine. Only the constant 100 and 10.5 can slightly vary from one engine to another (for example diesel usually have tighter tolerances to withstand the higher pressure and therefore experience slightly more friction or small industrial engines will have roller bearings which reduces the friction).

Now let's look at the volume flow rate (m³/s):

volume flow rate = engine displacement * rpm / 2 / 60 000
volume flow rate = (Ap * S / 1 000 000) * (vp / S * 30 000) / 2 / 60 000
volume flow rate = Ap * vp / 4 000 000

Where Ap = number of piston * pi / 4 * bore² and is the equivalent area based on the bore of every cylinder of the engine (mm²). So the friction power becomes:

friction power = (100 + 10.5 * vp) * (Ap * vp / 4 000 000)

This is the more "pure" definition of friction power. You can see how important is the mean piston speed. The bore and number of piston are included in Ap.

More on vp and Ap

IMHO, mean piston speed is a engine characteristic that is far more important than engine rpm. Ever wonder how some engines can go up to 20 000 rpm and others can't go over 5000 rpm? The answer lies in the mean piston speed. Except for the valvetrain, rpm is not a limiting factor for tearing apart an engine: mean piston speed is. Here are typical maximum mean piston speeds:

typical automotive engine: 13-18 m/s
race car engine: 20-26 m/s
absolute max (dragster engine): 30 m/s

These values are mainly based on material strength versus piston acceleration. No matter what, it will be very difficult to go over 30 m/s as, at this speed, the piston can go faster than the flame propagation of the combustion (which leads to power loss).

So let's compare two engines: A 4-cyl 2L engine with a stroke of 80 mm @ 5000 rpm or a 1-cyl 2L engine with a 130 mm @ 4000 rpm?

4-cyl: vp = 80 * 5000 / 30 000 = 13.3 m/s
1-cyl: vp = 130 * 4000 / 30 000 = 17.3 m/s

So even if the 1-cyl has a slower rpm, its piston actually travels faster in the cylinder than the ones in the 4-cyl. This means that the 1-cyl engine needs stronger components and that the friction losses are greater too. To compare fairly the two engine's potentials, the 4-cyl needs to rev at 6500 rpm.

This is why smaller engines have higher rpm: they have smaller stroke. The same thing goes for Formula 1 engine: compare with production car engine, they have really smaller stroke leading to smaller displacement but giving room for higher rpm to respect the maximum mean piston speed before blowing up.

We can also relate the power of the 4-stroke engine to mean piston speed instead of the rpm (everything in SI units):

Power = Torque * rpm
Power = (BMEP * Engine displacement / (4 pi) ) * (vp / S * pi)
Power = (BMEP * (Ap * S) / (4 pi) ) * (vp / S * pi)
Power = BMEP * Ap * vp / 4

Where BMEP is the brake mean effective pressure (which depends mainly on the thermodynamic cycle and the FMEP). Note that the stroke has no influence on the power output of an engine. Why? Let's see an example:

A 4-cyl with bore and stroke = 90 mm and a 1-cyl with a bore and stroke = 180 mm. Compare the engines, assuming BMEP = 10 bar and vp = 15 m/s.

displacement = N * pi/4 * D² * S / 1 000 000
4-cyl ---> 2.3 L
1-cyl ---> 4.6 L

rpm = vp / S * 30 000
4-cyl ----> 5000 rpm
1-cyl ----> 2500 rpm

Ap = N * pi/4 * D²
4-cyl ---> 25447 mm²
1-cyl ---> 25447 mm²

power = BMEP * Ap * vp / 4 /10000
4-cyl ---> 95 kW
1-cyl ---> 95 kW

Both engines have the same power output even though the 1-cyl displacement is twice as large as the 4-cyl! Why? Because, due to the shorter stroke, the smaller 4-cyl can rev up twice as fast as the larger 1-cyl. Hence both of them can displace the same amount of air with the same mean piston speed. Note that Ap is the same for both engine and that is the reason they give the same power output, if we assume similar construction. The true size of an engine is not measured with its displacement, it is measured with its total piston area.

So to resume how to evaluate the potential of an engine:

- BMEP represents the efficiency of the engine to convert combustion energy into mechanical energy;

- Ap represents the size of the engine;

- vp represents the speed of the engine.

Torque, rpm and displacement are dependent of these parameters, but say very little about the engine's true potential.

Last small remark about power (always in SI units):

Power = BMEP * Ap * vp / 4
Power = BMEP * volume flow rate

If 2 engines draw the same amount of air, assuming similar efficiencies, they will produce the same amount of power, no matter their displacement or rpm.

Damn! I think I went overboard with my answer!

7. Jun 21, 2010

### ISX

Internet has been down so I couldn't reply, hitting off strand of signal from afar. Anyhow, that is one hell of a reply! I will kick myself if I can't find any answers to my questions within that. Thanks for your unparalleled work! Just incredible.

I'm basically trying to figure out the differences in inline 6 engines and V8 in the diesel world. I have found some things that give the I6 the edge but am trying to find more, I mean the V8 has 2 more pistons which have advantages and disadvantages, just trying to get some calculations showing it. That's why I wanted to know about factoring in number of pistons but it seems it doesn't matter. The thing I have found is that if you add up the area of the pistons that are on their power stroke in an I6 and V8 of equal displacement, the I6 always has more area. Things like that are things I am trying to calculate, things that show why an I6 is more efficient and has more torque than a V8.

Thanks again!

8. Jun 22, 2010

what if the stroke was in inches?

9. Jun 22, 2010

### ISX

I have just been converting it to mm by multiplying inches by 25.4.

So a 3" stroke would be 76.2mm.