The crank and connecting rod of vertical single cylinder gas engine running at 1800 rpm are 60 mm and 240 mm resp. The distance of the piston is 80 mm, and mass of reciprocating part is 1.2 kg. At a point during power stroke when piston has moved 20 mm from top dead center position, the pressure on the piston is $800 kN/m^2$. Determine

1. Net force on the piston
2. Net load on gudgeon pin.
3. the thrust on cylinder walls
4. the speed at which the gudgeon pin load is reverse in direction.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: May 2016

1. N=1800 rpm
2. r=60mm=0.06m
3. l=240mm=.24m
4. m=1.2 kg
5. d=0.08m
6. θ=20◦ - ( correction in the paper, it should be ◦ and not mm)

$η =l/r =0.24/0.06=4 \\ ω=2πN/60=2π*1800/60=188.5 rad/ \sec \\ \cos β=1/η \sqrt{η^2 – \sin^2⁡θ} = 1/4 \sqrt{4^2 – \sin^2⁡20} \\ \cos β= 0.9962 \\ β=4.94◦$

force due to pressure Fp= A*pressure

$=π/4d2*p \\ =π/4 (0.08)2*800*103 \\ =4021 N.$

Inertia force $Fb=mrω^2[ \cos θ + \cos 2θ/η]$

$=1.2*0.06*(188.5)^2[\cos 20◦ + \cos 40/4] \\ =2893.98 N$

1. Net effective force on the piston

   $F= Fp-Fb+mg \\ =4021-2893.98+1.2*9.81 \\ =1139.772N$

2. Net load on the cylinder = force exerted by the connecting rod = F/cos β = 1139.772/0.9962=1144.11 N

3. Thrust on the wall cylinder = F tanβ = 1139.772*tan 4.94 = 98.51N

4. Speed at which the cylinder load is in the reverse direction

   $F=Fp-mrω^2[ \cos θ+\cos 2θ/η]+mg \\ 0=1139.772-1.2*(0.06)ω^2 [\cos 20 + \cos 40/4]+1.2*9.81 \\ 1151.544/1.2*0.06=ω^2[ \cos 20 + \cos 40/4] \\ 15993.66/[ \cos 20 + \cos 40/4] =ω^2 \\ 14138.62=ω^2 \\ ω=118.90 \\ ω=2πN/60 \\ 118.90=2πN/60 \\ N=1135$.