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The crank pin circle radius of a horizontal engine is $300 mm.$ the mass of the reciprocating parts is $250 kg.$ When the crank has traveled $60^\circ$ from IDC,
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The difference between the driving and back pressure is $0.35 N/mm^2$. The connecting rod length between centers is $1.2 m$ and the cylinder bore is $0.5 m.$ If the engine runs at $250 rpm$ and if the effect of piston rod diameter is neglected, Calculate:

1] Pressure on slide bars.

2] Thrust in the connecting rod.

3] Tangential force on the crank pin.

4] Turning moment on the crankshaft.

Given: $r = 300mm$

$M_R = 250 kg$

$\theta = 60°$

$\triangle P \ = \ P \ = \ 0.35 \ N/mm^2$

$L = 1.2m$

$D = 0.5m$

$N = 250\ rpm$

$\therefore \omega = \frac{2 \pi (250)}{60} = 26.18 \ rad/s.$

$\rightarrow$ $n = \frac{L}{r} = \frac{1.2}{0.3} = 4$

$\rightarrow$ Net load on piston = $F_L = 0.35 \times \frac{\pi }{4} (500)^2$

$F_L = 68.72 \times 10^3 N$

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

$= (250) (26.18)^2 (0.3) [ cos \ 60 + \frac{cos \ (2 \times 60)}{4}]$

$F_I = 19.28 \times 10^3 \ N$

$\rightarrow$ Piston Effort = $F_P \ = \ F_L \ – \ F_I$

$=(68.72 – 19.28) \times 10^3$

$F_P = 49.44 \times 10^3 N$

$\rightarrow$ $sin \ \phi = \frac{sin \ 60}{4} = 0.2165$

$\phi \ = \ 12.50°$

$\rightarrow$ Pressure on slide bars:

$F_N \ = \ F_Q . \ sin \ \phi$

$= (\frac{F_p}{cos \ \phi}) sin \ \phi$

$= f_p \ tan \ \phi$

$= (49.44 \times 10^3) tan \ (12.50)$

$F_N = 10.96 \times 10^3 \ N$

$\rightarrow$ Thrust in the connecting rod:

$F_Q = \frac{F_p}{cos \ \phi} = \frac{49.44 \times 10^3}{cos \ (12.5)}$

$F_Q = 50.64 \times 10^3 \ N$

$\rightarrow$ Tangential force on crank pin:

$F_T = F_Q . sin \ (\theta \ + \ \phi)$

$= (50.64 \times 10^3) sin \ (60 + 12.5)$

$F_T = 48.29 \times 10^3 \ N$

$\rightarrow$ Turning moment on crankshaft:

$T \ = \ F_T . \ r$

$= (48.29 \times 10^3) \ (0.3)$

$= 14.48 \times 10^3 \ N-m$