Advantage of many-V engines over V-twin engines

[ManiacChicken]

Main Question or Discussion Point

What is the advantage of a big V engine over a V-twin engine? Why not just use extra-large V-twin engines instead of using V-12 engines and such?

 

[jack action]

That is because the power you can get out of an engine is proportional to the surface area of the cylinder bore (and not to the displacement, as most people think)

So if you have 2 engines with N cylinders of similar proportions (cylinder bore D, stroke S and displacement V) you have:

V₁ = PI / 4 * N₁ * D₁² * S₁

V₂ = PI / 4 * N₂ * D₂² * S₂

To have the same power output:

N₁ * D₁² = N₂ * D₂²

If the cylinders are proportional then:

S₁ / S₂ = D₁ / D₂

So:

V₂ / V₁ = ( N₁ / N₂ )^½

Hence, the displacement of a 12-cylinder engine would be equivalent of 40,8% of the one of a 2-cylinder engine to get the same power output, given the designs are the same. So, to get the same power of a V-12 of 5,0 L, the V-2 would need to be 12,3 L in size!

 

[Mech_Engineer]

Also, power output is proportional to engine speed, and one huge v-twin engine would not be capable of very high rpms. However it is true that for very large engine applications like cruise ships, very large displacement diesel engines are utilized rather than lots of smaller ones.

 

[ManiacChicken]

Thanks guys, that makes a lot of sense!

 

[jack action]

Also, power output is proportional to engine speed, and one huge v-twin engine would not be capable of very high rpms. However it is true that for very large engine applications like cruise ships, very large displacement diesel engines are utilized rather than lots of smaller ones.

Actually no. Engine power is not proportional to RPM but to piston speed:

P = BMEP * A_p * V_p / 29 840

where P is the engine power (hp), BMEP is the brake mean effective pressure (bar), V_p is the mean piston speed (m/s) and A_p is the total cylinder bore area (mm²).

And the relation between the RPM and V_p is:

RPM = 30 000 * V_p / S

where S is the stroke (mm).

Piston speed is proportional to piston acceleration and this is what determines the inertial forces that the piston, connecting rod and crankshaft will be submitted to, hence the stress limit they can endure with today's material strength. But, for very short stroke, it is possible that the valvetrain doesn't support the high RPM.

The greater the stroke, the slower the RPM for a same given V_p. If Formula 1 racing car have such high RPM, it is mostly because they have shorter stroke than production engine (hence smaller displacement --> that's why engine displacement is not a good variable to determine power output).

Here are typical values for mean piston speed for engine and BMEP:

production: 13-18 m/s; 13-18 bar (max combined = 270 bar.m/s)
racing: 22-26 m/s; 15 bar
max: 30 m/s; 22-30 bar

 

[Cyrus]

Interesting posts, Jack!

 

[Ranger Mike]

That is because the power you can get out of an engine is proportional to the surface area of the cylinder bore (and not to the displacement, as most people think)

Forgive me but is not the displacement of one cylinder the total volume of that bore? and is not volume, the surface area of that bore?? Or did you mean the piston top surface area? Or does surface area of the cylinder bore include the combustion chamber? Would this include the volume of the cylinder head gasket?

 

[Mech_Engineer]

Actually no. Engine power is not proportional to RPM but to piston speed:

I understand you know a great deal about how engines generate power, but piston speed IS related directly to the engine's rotational speed.

 

[jack action]

That is because the power you can get out of an engine is proportional to the surface area of the cylinder bore (and not to the displacement, as most people think)

Forgive me but is not the displacement of one cylinder the total volume of that bore? and is not volume, the surface area of that bore?? Or did you mean the piston top surface area? Or does surface area of the cylinder bore include the combustion chamber? Would this include the volume of the cylinder head gasket?

I'm sorry if I did not express myself clearly, but I didn't say piston top surface area as most of the time these are not flat. The area I'm talking about is really the "imaginary" one based on the bore of the cylinder, i.e. A = pi / 4 * D². The displacement (or "imaginary" volume) of that cylinder is this area multiply by the stroke of the piston. It corresponds to the theoretical amount of air displaced by the piston's motion (In practice, it can be lower or greater).

I understand you know a great deal about how engines generate power, but piston speed IS related directly to the engine's rotational speed.

Yes, but if you read well my post, you'll notice that piston speed IS ALSO related directly to piston stroke, meaning that it is possible for an engine to have a very low RPM and still deliver great amount of power if it has a very long stroke.

The important thing with engines is that, compared to piston speed, it is almost limitless in how fast they can rotate or how long their stroke can be. But, in most cases, the piston speed is the very clear, lower limit for all engines. No matter if it's a R/C model engine or a huge cruise ship engine.

Let's just play with the equations from my previous posts.:

P = BMEP * A_p * V_p / 29 840
P = BMEP * A_p * ( S * RPM / 30 000 ) / 29 840

P = BMEP * Vol_cyl * RPM / 895 200 000

Here we have the power proportional to displacement (Vol_cyl) and RPM. Just like everybody likes to look at it. Let's say you have a 4-cyl, 2 L engine with a 85 mm stroke that produces 300 hp @ 8800 rpm. Based on the last equation, you might say " let's double that stroke to 170 mm, my displacement will double to 4 L and I'll double my power to 600 hp." You would be wrong, because you would also double your mean piston speed, which was already at a high value of 25 m/s. The new engine would destroy itself long before you would reach 8800 rpm. In fact, your camshaft would also be mismatch and the pistons would even go faster than your flame propagation! So the maximum RPM of that new engine would need to be halved to 4400 rpm to respect to 25 m/s mean piston limit. Hence you double your engine displacement, but you have to cut in half the RPM, which means you get the same power output (but at a lower RPM, which may be an advantage).

And this is why basing potential power output on displacement and RPM is misleading: when you change the stroke, one goes up and the other goes down. It is better to do it with the area and mean piston speed, which are completely independent variables.

 

[Ranger Mike]

that was a lot better...especially your last sentence. You finally got there, amigo.
this forum will pin you down to exact concise statements and specif facts...universal statements will make us bonkers...
in general i agree with your opinion regarding estimation of power based upon piston speed and area...as long as the total picture includes all the related parts..
welcome!!
rm