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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

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

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

$K_p = mrw^2$

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

$= 98.696$

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

**Secondary force polygon :**

**Secondary couple polygon :**

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

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