Inertia force and turning moment in an engine

Inertia force of reciprocating parts

Let $r$ = crank radius, $l$ = connecting rod length, $n = l/r$, $\theta$ = crank angle from the inner dead centre (IDC), $\omega$ = crank speed (constant), $m$ = reciprocating mass.

Displacement of the piston from IDC:

$$x = r(1 - \cos\theta) + l(1 - \cos\phi), \qquad \sin\phi = \frac{\sin\theta}{n}$$

Using $\cos\phi = \sqrt{1 - \sin^2\theta/n^2} \approx 1 - \dfrac{\sin^2\theta}{2n^2}$ and $\sin^2\theta = \dfrac{1 - \cos2\theta}{2}$, the displacement from the crank centre is

$$x_c \approx r\cos\theta + \frac{r}{4n}\cos2\theta + \text{constant}$$

Differentiating twice with $\theta = \omega t$, acceleration of the piston:

$$a = -\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right)$$

The inertia force is equal and opposite to $ma$ (D'Alembert's principle). Taking the working-stroke direction (towards the crank) as positive, its magnitude is

$$F_I = m\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right)$$

A positive value means the inertia force opposes the gas force (as near IDC); it is negative (aids motion) in the later part of the stroke. The net force on the piston is $F = F_P - F_I$.

Numerical

Data: $D = 100$ mm, $r = 0.1$ m, $l = 0.4$ m so $n = 4$, $N = 600$ rpm, $m = 1.5$ kg, $\theta = 30^\circ$, $p = 0.8$ MPa.

$$\omega = \frac{2\pi \times 600}{60} = 62.832\ \text{rad/s}, \quad \omega^2 = 3947.8$$

Gas force on piston:

$$F_P = p\times\frac{\pi}{4}D^2 = 0.8\times10^6 \times 0.00785 = 6283.2\ \text{N}$$

Inertia force:

$$F_I = 1.5 \times 3947.8 \times 0.1\,(\cos 30^\circ + \tfrac{\cos 60^\circ}{4}) = 1.5 \times 3947.8 \times 0.1\,(0.8660 + 0.1250) = 586.9\ \text{N}$$

(a) Net force on piston (along the line of stroke)

$$F = F_P - F_I = 6283.2 - 586.9 = 5696.3\ \text{N}$$

(b) Thrust on cylinder walls

$$\sin\phi = \frac{\sin30^\circ}{4} = 0.125 \Rightarrow \phi = 7.18^\circ$$ $$F_N = F\tan\phi = 5696.3 \times \tan7.18^\circ = 717.7\ \text{N}$$

(c) Force in connecting rod

$$F_Q = \frac{F}{\cos\phi} = \frac{5696.3}{\cos 7.18^\circ} = 5741.4\ \text{N}$$

(d) Turning moment

Tangential component of the rod force on the crank pin: $F_T = F_Q\sin(\theta + \phi)$, with $\theta + \phi = 37.18^\circ$.

$$F_T = 5741.4 \times \sin 37.18^\circ = 3469.7\ \text{N}$$ $$T = F_T\times r = 3469.7 \times 0.1 = 347.0\ \text{N m}$$

The radial component on the crank pin is $F_R = F_Q\cos(\theta + \phi) = 4574.3$ N, which loads the shaft bearings.

Answer: $F = 5696.3$ N, $F_N = 717.7$ N, $F_Q = 5741.4$ N, $T = 347.0$ N m.