Question: Determine .
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A horizontal gas engine running at 210 rpm has a bore of 220 mm and a stroke of 440 mm. The connecting rod is 924 mm long and the reciprocating parts weight 20 kg. when the crank has turned through an angle of 300 from the inner dead center, the gas pressures on the Cover and the crank sides are 500 kN/m2 and 60 kN/m2 resp. Diameter of the piston is 40 mm. Determine i) turning moment on the crank shaft. ii) thrust on bearings iii) acceleration of the flywheel which has a mass of 8 kg and radius of gyration of 600 mm while the power of engine is 22 kW.

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

Marks: 10 Marks

Year: Dec 2015

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modified 2.8 years ago  • written 2.8 years ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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BA=270mm

AO=60mm

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A20◦ r=.44/2=.22m l= 0.924m N=210rpm M=20kg η=l/r=0.924/.22=4.2 ω=2ΠN/60=2Π210/60=22rad/s sinβ=sin 30◦/η=sin 30◦/4.2=.119 or, β=6.837◦

$Fp=(p_1A_1-p_2A_2) \\ =500*10-3*Π/4*(.22)2-60*103*Π/4(.222 – 0.042) \\ =19007-2206 \\ Fp=16801N$

Inertia force $Fb=mf=mrω2[\cosθ+\cos 2θ/η] \\ =20*0.22*222[\cos 30◦+\cos 60◦/4.2] \\ =2098N$

Piston effort=$F=F_p-F_b \\ =16801-2098 \\ =14703N$

  1. Turning moment $T=F/cosβ sin(θ+β)*r =14703/cos 6.837 sin(30+6.387)*.22 \\ =1953Nm$

  2. Thrust on the bearings $Fr= F/ \cosβ \cos(θ+β) \\ =14703/cos 6.837* sin(30+6.837) = 11852N$

  3. Acceleration Torque= Turning moment-resisting torque

    Resisting torque=$P=T*ω \\ 22*103=T*22 \\ T=1000 Nm$

    Acceleration torque= 1953-1000=953Nm

    $Iα=mk_2.α=953 \\ 8*0.62*α=953 \\ α=330.9 rad/sec$

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written 2.8 years ago by gravatar for Sayali Bagwe Sayali Bagwe2.2k
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