Differential equation for pressure and heat release in combustion engine

Question

I have a differential equation on the following form, and I am interested in finding $p(\theta)$

$\frac{dp}{d \theta}=\frac{\gamma-1}{V(\theta)}\frac{dQ_{HR}}{d \theta} - \gamma \frac{p}{V(\theta)} \frac{dV}{d \theta}$

and I know the following,

$\frac{dQ_{HR}}{d \theta} = \frac{dm_{burnt}}{d \theta}H_u$

$\frac{dm_{burnt}}{d \theta} = k_1 \sin\left( \pi \frac{\theta-\theta_0}{\Delta \theta_c}\right)$

and,

$\frac{dV}{d \theta} = k_2\left(\sin(\theta)+k_3\sin(2\theta) \right)$

I have tried integration by parts, but this only seems to dig my hole deeper. What are your suggestions to tackling a problem like this?

Answer 1

The best I can get is the following:

Firstly, integrate the last equation gives $$V(\theta) = k_2 \left(-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right)$$ where $C_1$ is some constant to be determined later.

Now substitute everything into the first equation, we get $$\frac{\mathrm{d}p}{\mathrm{d}\theta} = \frac{\gamma - 1}{k_2 \left(-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right)} \cdot k_1 \sin \left(\pi \frac{\theta - \theta_0}{\Delta \theta_c} \right) H_u - \gamma \cdot \frac{p}{k_2 \left(-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right)} \cdot k_2 (\sin \theta + k_3 \sin 2 \theta)$$

Notice it is a $1^{\text{st}}$ order ODE. Re-writing it into the standard form $y' + P(x) y = Q(x)$ gives

$$\frac{\mathrm{d}p}{\mathrm{d}\theta} + \gamma \frac{\sin \theta + k_3 \sin 2 \theta}{-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1} p = \frac{k_1}{k_2}(\gamma - 1)H_u \frac{\sin \left(\pi \frac{\theta - \theta_0}{\Delta \theta_c} \right)}{-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1}$$

Next, we compute $$\begin{align} \int \gamma \frac{\sin \theta + k_3 \sin 2 \theta}{-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1} \mathrm{d}\theta &= \gamma \int \frac{\mathrm{d} (-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1)}{-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1} \\ &= \gamma \log \left|-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right| \\ &= \log \left|-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right| ^ \gamma \end{align}$$

So the integrating factor $\mu(\theta) = \left|-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right| ^ \gamma $

Finally, we need to compute $$\int \frac{k_1}{k_2}(\gamma - 1)H_u \frac{\sin \left(\pi \frac{\theta - \theta_0}{\Delta \theta_c} \right)}{-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1} \cdot \left|-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right| ^ \gamma \mathrm{d} \theta \\ = \frac{k_1}{k_2}(\gamma - 1)H_u \int \sin \left(\pi \frac{\theta - \theta_0}{\Delta \theta_c} \right) \cdot \operatorname{sgn} \left(-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right) \cdot \left|-\cos \theta - \frac{k_3}{2} \cos 2 \theta + C_1 \right| ^ {\gamma - 1} \mathrm{d} \theta$$ but this is how far I've got.

Answer 2

I ended up using a forward Euler numerical integration for this. I don't think the analytical solution is worth anybody's time for this.