What expression gives the inertia force of the reciprocating parts at crank angle theta (with obliquity ratio n)?

m r w^2 (cos theta + cos 2theta / n)

m r w^2 (sin theta + sin 2theta / 2n)

m r w (cos theta + cos 2theta / n)

During the out (expansion) stroke when the piston is accelerating, how is the net force on the piston found?

Gas force only; inertia is ignored

Practice · 8 marks

Piston effort and crank-pin turning moment

Short answer

At 360 rpm, ω=37.70 rad/s, gas force 31416 N. Inertia force = 20708 N, net piston effort = 10708 N, connecting rod thrust = 10849 N, side thrust = 1743 N, turning moment = 1233 N·m.

The crank radius of a horizontal single-cylinder engine is 150 mm and the connecting rod is 600 mm long. The cylinder bore is 200 mm and the engine runs at 360 rpm. The mass of the reciprocating parts is 120 kg. When the crank is $40^\circ$ from the inner dead centre on the expansion (out) stroke, the gas pressure on the piston is 1.0 MPa (gauge, back pressure neglected). Neglecting friction and the weight of the parts, determine: (i) the inertia force, (ii) the net force on the piston (piston effort), (iii) the thrust in the connecting rod, (iv) the side thrust on the cylinder wall, and (v) the turning moment on the crankshaft.

Data: $r = 0.15$ m, $l = 0.6$ m, $n = l/r = 4$ , $N = 360$ rpm, $m = 120$ kg, $\theta = 40^\circ$ , $p = 1$ MPa, $d = 0.2$ m.

Angular speed and gas force

\omega = \frac{2\pi N}{60} = \frac{2\pi \times 360}{60} = 37.70\ \text{rad/s}

F_P = p\,\frac{\pi}{4}d^2 = 1\times10^{6}\times\frac{\pi}{4}(0.2)^2 = 31416\ \text{N}

(i) Inertia force

\begin{aligned} F_I &= m r\omega^2\left(\cos\theta + \frac{\cos 2\theta}{n}\right) \\ &= 120 \times 0.15 \times (37.70)^2 \left(\cos 40^\circ + \frac{\cos 80^\circ}{4}\right) \\ &= 25582.0 \times 0.8095 = 20708\ \text{N} \end{aligned}

(ii) Net force on piston

During the out stroke the piston is accelerating, so the inertia force opposes the gas force:

F = F_P - F_I = 31416 - 20708 = 10708\ \text{N}

(iii) Thrust in the connecting rod

\sin\phi = \frac{\sin\theta}{n} = \frac{\sin 40^\circ}{4} = 0.1607 \Rightarrow \phi = 9.25^\circ

F_Q = \frac{F}{\cos\phi} = \frac{10708}{0.9870} = 10849\ \text{N}

(iv) Side thrust on the cylinder wall

F_N = F\tan\phi = 10708 \times 0.1628 = 1743\ \text{N}

(v) Turning moment on the crankshaft

T = F_Q \cdot r\sin(\theta+\phi) = 10849 \times 0.15 \times \sin(49.25^\circ) = 1233\ \text{N·m}

Answer: Inertia force = 20708 N; piston effort = 10708 N; connecting-rod thrust = 10849 N; side thrust = 1743 N; turning moment = 1233 N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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