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