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A horizontal gas engine running at $210 rpm$ has a bore of $220 mm$ and a stroke of $440 mm.$ the connecting rod is $924 mm$ long and the reciprocating parts weigh $20 kg.$
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When the crank has turned through an angle of $30°$ from the inner dead center, the gas pressure on the cover and the crank sides are $500 KN.m^2$ and $60 KN/m^2$ respectively. Diameter of the piston is $40 mm.$ Determine  Turning moment on the crankshaft.  Thrust on the bearings.  Acceleration of the flywheel which as a mass of $8 kg$ and radius of gyration of $600 mm$ while the power of engine is $22 kw.$

$N = 210\ rpm.$

$\therefore \omega = 21.99 rad/s$

$D = 220 mm$

Stroke $= L = 440 mm$

$\therefore$ $r \ = \ \frac {L}{2} = \frac{440}{2} = 220 \ mm$

$L = 924 mm$

$M_R = 20 kg$

$\theta$ = 30°

$P_1$ = 500 $KN/m^2$

$P_2$ = 60 $KN/m^2$

$d = 40 mm$

$\rightarrow$ $A_1 = \frac{\pi}{4} (0.220)^2 = 0.0380 \ m^2$

$\rightarrow$ $a = \frac{\pi}{4} (0.040)^2 = 1.2566 \times 10^{-3} \ m^2$

$\rightarrow$ $n = \frac{L}{r} = \frac{924}{220} = 4.2$

$\rightarrow$ $F_L = P_1 A_1 – P_2 (A_1 – a)$

$= (500 \times 0.0380) – \ 60 (0.0380 – 1.2566 \times 10^{-3})$

$F_L = 16.79 \ KN$

$\rightarrow$ $F_I = M_R. \ \omega^2 r \ [ cos \ \theta + \frac{cos \ 2 \ \theta}{n}]$

$= 20 \times \ (21.99)^2 \times (0.220) [cos \ 30 + \frac{cos \ 60}{4.2}]$

$F_I = 2.095 \times 10^3 \ N = 2.095 \ KN$

$\rightarrow$ $F_p = F_L – F_I$

$= 16.79 – 2.095$

$F_p = 14.695 KN.$

$\rightarrow$ $sin \ \phi = \frac{sin \ \theta}{n} = \frac{sin \ 30}{4.2} = 0.1190$

$\phi = 6.837°$

$\rightarrow$ $T = (\frac{F_p}{cos \ \phi}) . sin \ (\theta \ + \phi). r$

$= (\frac{14.695 \times 10^3}{cos \ 6.837}) sin \ (30 + 6.837). (0.220)$

$= 1.952 \times 10^3 \ N-m$ - - - - 1st Ans.

$\rightarrow$ Thrust on the bearings,

$F_B = F_Q . cos \ (\theta + \phi)$

$= \frac{F_p}{cos \ \phi} \ cos \ (\theta + \phi)$

$= \frac{14.695}{cos \ 6.837} . cos \ (30 + 6.877)$

$F_B = 11.84 \ N$ - - - - 2nd Ans.

$\rightarrow$ Acceleration of the flywheel,

Given: $P = 22 KW$

$M = 8 kg$

$K = 600 mm$

Mass moment of inertia of the flywheel,

$I = Mk^2 = (8) (0.6)^2$

$I = 2.88 \ kg.m^2$

$\therefore$ Accelerating Torque, $T_A = I . \alpha$

$= (2.88) \ \alpha$

Resisting Torque: $T_R = \frac{P \times 60}{2 \pi N}$

$= \frac{22 \times 10^3 \times 60}{2 \pi \times 210}$

$= 1 \times 10^3 \ N-m$

$\therefore$ $T_A \ = \ T \ - \ T_R$

$(2.88) \alpha \ = \ 1.952 \times 10^3 – 1 \times 10^3$

$\alpha = 330.55 \ rad/s^2$