Numerical : 6

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Numerical : 6

written 5.3 years ago by

bhaveshkumarn.pasi • 510

dynamics of machinery

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written 5.3 years ago by

bhaveshkumarn.pasi • 510

A six cylinder four stroke vertical in-line engine has a firing order 1-4-5-2-3-6. Firing takes place with equal angular interval. The mass of reciprocating parts per cylinder is 2 kg, Stroke 100 mm and connecting rod length 200 mm, The cylinder centre lines are spaced at 800 mm apart. The crankshaft speed is 300 rpm. Examine the engine for the balance of the primary and secondary forces and couples.

Four stroke cycle,

$\therefore $ Angle between each crank = $\frac {720}{6}$ = $120 ^\circ$

Firing order : $1-4-5-2-3-6$

$m = 2 kg$

$r \ = \ \frac {100}{2} = 50 \ mm = 50 \times 10^{-3} m $

$L = 200 \ mm$

$\frac{\text { Ratio of length of } \mathrm{CR}}{\text { crank radius }}= n = \frac{L}{r} = 4$

$w=\frac{2 \pi \times 300}{60}=31.416$ rad/s

enter image description here

Plane	Mass(m) kg	Radius(r) m	cent. $ \div \ \omega^2$ (m.r) $kg \cdot m$	Dist from R.P.(l)	couple $\div \ \omega^2$ (m.r.l) $kg \cdot m^2$
1	2	0.05	0.1	-0.75	-0.075
2	2	0.05	0.1	-0.45	-0.045
3	2	0.05	0.1	-0.15	-0.015
4	2	0.05	0.1	0.15	0.015
5	2	0.05	0.1	0.45	0.045
6	2	0.05	0.1	0.75	0.075

Primary force polygon : 1 cm = 0.05 kgm

enter image description here

Primary couple polygon : 1 cm = 0.015 $kg \cdot m^2$

enter image description here

$K_p = mrw^2$

$= 2 \times 0.05 \times (31.416)^2$

$= 98.696$

UPC \ = $\ K_p \ \times \ 0.212$ $= 20.923 \mathrm \ {kg}-\mathrm{m}^{2} \quad(\mathrm{or} \ N-m)$

Secondary force polygon :

enter image description here

Secondary couple polygon :

enter image description here

$Ks = \frac {Kp}{h}$ $ = \frac {98.696}{4}$ $ = 24.674$

USC = $Ks \times 0.212$ $ = 5.23 \mathrm{kg} \cdot \mathrm{m}^{2}(N-m)$

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