A horizontal gas engine running 210 rpm has a bore of 220 mm and a stroke of 440mm. The connecting rod is 924 mm long and the reciprocating parts weights 20 kg, when the crank has turned through an angle of 33 degrees from the inner dead center, gas pressure on the cover and the crack sides are 500KN/m and 60 KN/ m respectively. Diameter of the piston rod is 40mm. - Determine - 1. turning moment on crank shaft. - 2. thrust on the bearing. - 3. acceleration of the flywheel which has a mass of 8 kg and radius of gyration of 600mm while the power of engine is 22Kw. -

Given:

D=0.22m; Ls= 0.44m

Θ= 30; N = 210rpm

Lc= 0.924m

M= 20kg

P1 = 500kPa; P2= 60kPa

Dp= 0.04m

Mf = 8kg; r= 0.6m; P= 22kw

To find:

Moment

Thrust

acceleration

First we need to determine all angles.

Now by applying sine rule, we can get the angle of the connecting rod. Let it be φ

Lc/sinθ = r/sinφ

0.924/sin30 = 0.6/sinφ

$Φ = 18.950^0$

$. α= 180 – 30 – 18.95 = 1310^0$

Force on piston due to Gas pressure

Fg $= Pressure \times area$ $= (p2-p1) \times Ap = (p2-p1) \times πDp^2/4 \\ = (500 – 60) \times 1000 \times π \times 0.04^2/4 = 553 N$

Force due to reciprocating parts

Fr = m x a

But, $a = rw^2(\cosθ + (\cos2θ)/n)$

Where n = Lc/r = 0.924/0.6 = 1.54

And w = 2πN/60 = 22 rad/s

So, $a = 0.6 \times 22^2 \times (\cos30 + (\cos60)/1.54) = 345.8 N$

So total force on piston

Fp = Fg + Fr = 898.8 N

Force in the connecting rod can be found by applying ∑Fx=0 at the piston pin

Fcosφ – Fp = 0

F cos 18.95 = 898.8

F = 950N

At the flywheel, the Force F will resolve into two components, Ft and Fn.

Ft will produce turning moment.

Ft = F sin α = 950 x sin 131 = 717 N

Turning Moment = T = Ft x r = 717 x 0.6 = 430 Nm

Thrust = Fn = Fcos 131 = 623N $\hspace{3cm}$ (-ve sign is ignored)

P = I α w

$= mr^2 \times α \times w$

$22000 = 8 \times 0.6^2 \times α \times 22$

α = 346 rad/s2

$at = rα = 0.6 \times 346 = 208.5 m/s^2$

Moment = 430 Nm

Thrust = 623 N

$\text{Acceleration} = 208.5 m/s^2$