# Mathematical modelling of the crank in a two-stroke (diesel) engine

I need to use a mathematical model of the crank in a two-stroke (diesel) engine. Since a real two-stroke engine endures anharmonic motion, I need to implement higher order harmonics - preferably into a second order ODE, e.g.:

$$\ddot x +2\zeta\omega_n\dot x+\omega_n^2x=\mu r\omega^2\cos (\omega t+\phi)+\Gamma\omega_n^2\cos(\omega t)$$

How do one model the anharmonics that a two-stroke diesel engine suffers from by using higher order harmonics? Is it adequate to implement e.g.:

$$\sum _{i=1}^m\kappa_i\omega_{n,m}^2\cos(i\omega t+\phi_i)$$

• Welcome to Engineering! The second part of your question, where you ask for specific articles, falls under the category of resource-hunting questions. Such questions are prone to becoming out-dated, and are therefore considered off-topic for this site. I'd say the first part is fine, though. – Wasabi Sep 1 '16 at 10:27
• Removed resource-hunting part from the question: "Could you maybe refer to scientific articles discussing these matters (I've searched and searched...) Articles which experimentally, analytically and/or numerically discuss these matters are preferred." – peterh Sep 4 '16 at 17:36

## 2 Answers

You don't need a differential equation.

Write the kinematics for the piston as

\begin{aligned} x & = r \sin(\varphi) - \ell \sin(\beta) =0 \\ y & = r \cos(\varphi) + \ell \cos(\beta) \end{aligned}

Where $r$ is the crank radius, $\ell$ the conrod length, $\varphi$ the crank angle and $\beta$ the conrod angle. Piston is constrained at $x=0$ and the vertical position is $y$

The above is solved for the conrod angle $\beta=\arcsin\left(\frac{r}{\ell} \sin \varphi \right)$ and the piston position $y$.

Now start differentiating the above with the chain rule $\frac{{\rm d}\square{}}{{\rm d}t} = \frac{\partial \square{}}{\partial \varphi} \dot{\varphi} + \frac{\partial \square{}}{\partial \beta} \dot{\beta}$

\begin{aligned} 0 & = r \dot{\varphi} \cos\varphi - \ell \dot{\beta} \cos \beta \\ \dot{y} &= -r \dot{\varphi} \sin\varphi -\ell \dot{\beta} \sin\beta \end{aligned}

which is solved for the conrot rot. speed $$\boxed{\dot{\beta} =\dot{\varphi} \frac{r \cos\varphi}{\ell \cos\beta}}$$ and the piston speed $$\boxed{\dot{y}=-r (\dot\varphi+\dot\beta)\sin\varphi}$$. Rinse and repeat

\begin{aligned} 0 & = -\ell \ddot{\beta} \cos \beta + \ell \dot{\beta}^2 \sin\beta + r \ddot{\varphi} \cos\varphi - r \dot{\varphi}^2 \\ \ddot{y} & = -\ell \dot{\beta}^2 \cos\beta -\ell \ddot{\beta} \sin\beta -r \dot{\varphi}^2 \cos\varphi -r \ddot{\varphi} \sin \varphi \end{aligned}

to get the conrod rot. acceleration

$$\boxed{\ddot{\beta} = -\frac{\dot{\beta} (\dot{\varphi}^2-\dot{\beta}^2)\tan \varphi}{\dot\varphi}}$$

and piston acceleration (<= what you are looking for)

$$\boxed{\ddot{y} = -r ( \ddot\varphi + \ddot\beta) \sin\varphi - r \dot\varphi (\dot\varphi+\dot\beta)\cos\varphi }$$

Example If the crank speed was constant at $\dot\varphi = \Omega$ the result can be expanded out in terms of $\varphi$ only as:

$$\frac{\ddot{y}}{\Omega^2 r} = -\cos \varphi - r \frac{r^2 \cos^4\varphi + 2 (\ell^2-r^2)\cos^2\varphi-\ell^2+r^2}{(\ell^2-r^2 \sin^2\varphi)^\tfrac{3}{2}}$$

Now that you know the piston acceleration, you can do sum of forces on the piston (including gas pressure) to find the wristpin reaction load and the piston side load. Typically the conrod is split into rotating mass and reciprocating mass and added to the piston mass in this step.

If you want to be more elaborate you will need a system of equations considering the mass moment of inertia of the conrod. The unknowns are the 6 pin reaction force components and 1 piston side load and 1 crank torque. The equations are 2 for the piston, 3 for the connecting rod and 3 for the crank. A total of 8 equations and 8 unknowns.

This is completely solvable fast as I have done so in a simulation at work.

• As stated in OP, I need higher harmonics in the equation of motion. The reason for this is that $\varphi$ in my case, and in all real cases, is non-constant (varying) [the simplification $\dot{\varphi}=\Omega$ is not valid in my case]. I need to model this anharmonic (and more complex) motion. How would you approach that from your second last equation? – User4536124 Sep 5 '16 at 6:43
• The equation for piston acceleration $\ddot{y}$ contains crank acceleration terms $\ddot{\varphi} \neq 0$. In the end you will need to Fourier transform the acceleration (or forces) in order to get all the harmonics. – John Alexiou Sep 5 '16 at 14:27
• @Stefan are you talking about torsional vibrations, or like a balance shaft for the entire engine? – John Alexiou Sep 5 '16 at 14:36
• @Stefan you should probably consider investing into getting a multibody simulation software. While you can write multibody dynamics by hand they have a tendency to be really tedious and require special solvers. Also from modeling point of view you might need to build several models to account for several facts like maybe the internals of your bearings etc. in which case you could end up with hundreds of equations to write. Also you may need to do a flexible multibody simulation. – joojaa Sep 5 '16 at 20:28
• @joojaa I was about to suggest the same thing. We use RecurDyn for this sort of scenario. Even if you develop equations by hand, we sue MBD software to validate the results before they are used by designers. – John Alexiou Sep 6 '16 at 15:16

Try searching for the term "reciprocating mass balancing", google returns several useful-looking results.

In case they aren't detailed enough, I would suggest using the oldschool paper-based literature. My favourite in this topic is "Kraefte, Momente und deren Ausgleich in der Verbrennungskraftmaschine", Springer. Looking at your name the language may be ok for you.

The whole topic is way too broad for a post, but if you need specific equations I can look them up for you.

• The question is very specific, and the right researchers will most likely know which scientific articles to redirect me towards. But thank you for your time anyhow :) – User4536124 Sep 1 '16 at 17:18