Question

In a single cylinder piston machine, stroke 20 cm, connecting rod length 30 cm, connecting rod mass 3 kg, piston mass 2 kg, piston 20 cm, crankshaft speed 3000 rpm, cylinder piston piston friction coefficient 0.5 and connecting rod mass crank The length of the journal is 10 cm. What is the pressure of the piston, where the crank passes the TURN 37° and the torque coming to the crankshaft will make 425 mN in clockwise? (Rotational masses are balanced.) 1 sin 20 V=-ro sin + 2n 9). a=-re”[c cos 20 cos + 72 O=ot, n=- 7 0= tanu

Answer 1

2 10 0811:57 For vertical engine Solution Univen, coading of crearkko - Shakem olm 0=34, Simo – Sime - 317370 Ø = 11.870 Torque (T) = 425 Nm we know, Torque (U) = FXD = Fp sin(@+8)x0 0 425=Ep sin(37+11°53) X0.), > F=1553:26N Fp53:46N, Tp Piston effort Piston effort (Fp) = FL - FI + mg. Loheve inertia poole (FI) = mo-(6988+ (220) m=total al mass of reciprocating prot conmeeting rool maes=3 kg Angulars velocity of casuank (a) = 2AN 20 X 3000 p=314 wad)s Promeonation FI = 5X(314)2x0.xx{20837+ 68(2x37) 43900 6N Ang FL=6483 P., p=pressure on piston FL-1 (02)2xp = 0.031up m=3+2=5kg, piston mass=2kg 60

from equation fp = FL - Fiting 35553.26–0-0314 P-43900.6+5X9-8) :: P=1.573X106 Pa = 1573 kPa Piston pressure (P) =1573kPa