Question

L50 DKI & DK2: 9.00 pm 10.30 pm QUESTION 2 (15 marks) (a) A twin cylinder V-engine has cylinders set at 120°, with both piston connected to a single crank as shown in Figure Q2(a). The stroke is 140 mm and the connecting rods are 210 mm long. The reciprocating mass per cylinder is 1.2 kg. Determine the maximum primary and secondary forces acting on the frame of the engine when running at 2,400 rpm. (b) Figure Q2(b) shows 4 cylinders radial engine with their connecting rods are coupled to a single crank. The stroke is 100 mm and the length of each connecting rod is 150 mm. The mass of the reciprocating parts per cylinder is 1.5 kg. Find the maximum primary and secondary forces acting on the frame of the engine when running at 1,000 rpm. Common Connecting rod crank 60° 609 Line of stroke Connecting rod 3 0 Uw JE Direction of rotation Line of stroke ܠܝܠܠܠܪ Common crank Direction of rotation Figure Q2(b) Figure Q2(a)

Answer 1

Solution:-

Given V-cylinder engine 22 = 120° crank radius gr = Stroke a 2 To 140/2 = 70 mm Connecting rod, d=210 mm mess per cylinder , mal.2 Kg O angle b/w & mid- line and crank Speed , N = 2,400 rpm Mr. 3 7 For menimum primary Force fp = 2m ² r / (Costa x Cos 0)² + ( sina x sings? W = 20N 60 251.2 rad /sec Putting values in Fr Ep = 10,501.04 fu cosoje+ te sinojen For 90 270° 2 Er is man i Femen 7.95 KN

F For monimum secondary force 2m * wer/(cosa cosa .Ces 201² wir V + (sina. Sinza Sin 2002 = 3,533.68 110.256320)+10.75 Sin2o n For o=45, 135 Es is manimum Esmane = 2.65 KN 4 Givepit 4 cylinder radial engine Crank radius, r = Stroke 12 r = 100/2 4 90 90 r = 50mm 90790 1412 connecting rod ed = 150 mm mass per cylinder m = 1.5kg Speed, N=1000 rpm W=2IN lou.67 rad /sec

For Manimum primary Force using primary direct and reverse crank position Let consider cylinder 4 as common crank 1. Since for cylinder 40=0, Both direct & reverse position Krank{o=c 2 cylinder 0 = 90° Primary direct crank > 90° anticlockwise reverse crank 90° clockwise 3 cylinder 2 direct crank - 180 anti clockwise reverse crank 180 rclockwise 4 cylinder 3 > o = 270 direct crank 270 anti cow. reverse crank 270°. c.w. 1, 2, 3, 4, 4x m O = 180 4 x 2 42 2xm 2 Direct primary crank 13 2xm Reverse primary crank . Erman Fomen - um w²r 2 1.643 KN

Similarly for Secondary, fonte m2 3 m/ m/h 2 2 m/ 2 m/ 2. cylinder 40=0 d.c.&R.CO=0 cylinder / o=90 doc. 180 acw&r.c.180 c.w cylinder 2 0 = 180 dec. 360 ac.w & ric360 c. w. cylinder 3 270 doc - 540 acw. & r.c-540c.w 3 Since both direct & reverse crank are balanced Esman- TO Please Rate ☆

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