Question

the length of a connecting rod of an engine is 500 mm measured between the centres and its mass is 18 kg. the centre of gravity is 125 mm from the crank pincentre and crank radius is 100 mm

Answer 1

Similar question:

The length of the connecting rod of a gas engine is 500 mm and its center of gravity lies at 165mm from the crankpin center. The rod has a mass of 80 Kg and a radius of gyration of 182 mm about an axis through the center of mass. The stroke of the piston is 225 mm and the crank speed is 300 rpm. Determine the inertia force on the crankshaft when the crank has turned 30 ° from the inner dead center.

Answer: Inertia force in

case 1: fi = 26.17 * 108.661

= 2876.25N

case 2: fi = 26.17 * 78.512

= 2078.212

Inertia:

When an object experiences inertia, it keeps moving in the same direction or at the same pace until another force changes it. The phrase "the principle of inertia" as it is used in Newton's first rule of motion is correctly understood as the term "inertia."

Given that:

The connecting rod's length = 500nm

Center of the gravity,  CG = 165mm from crack pin

A connecting rod's weight $M_{2}$ = 80Kg

The radius of gyration , 182mm

The stroke, K = 225mm

Speed of the pistons, N = 300rpm

To find: Inertia force

Calculations,

Stroke , K = 2 * radius of crank

crank radius, r = K/2 = 225/2 = 112.5mm

Angular velocity,

$w = 2\pi N/60\\ = 2*\pi * 300/60\\ =10\pi rad/sec$

connected to the crank by the connecting rod's mass ,    

Lr = 500 - 165 = 335mm

So, Mcr = 500 - (335/500) * 80

Now , Inertia force, fi = Mcr * acceleration of the piston(ap)

So, the acceleration of the piston at

30 degree  crank angle from inner dead center

$ap30 = w^{2}r(costeta +cos\frac{2teta}{n} )\\n = \frac{1}{r} = \frac{500}{112.5} = 1.11\\ap30 = 10\pi ^{2}(112.5) (cos30 + cos\frac{60}{1.11})\\= 108661.163 mm/s^{2}$

135° crank angle from the inner dead center

$ap135 = 10\pi ^{2}(112.5) (cos135 + cos\frac{270}{4.44})\\\\=-78512.22mm/s^{2} \\=-78.512.22m/s^{2} = 78.512.22m/s^{2}$

(-ve) retardation

Inertia force in

case 1: fi = 26.17 * 108.661

= 2876.25N

case 2: fi = 26.17 * 78.512

= 2078.212

More questions and answers

https://brainly.in/question/13041483

https://brainly.in/question/36312538

#SPJ1