Sarthaks Test
0 votes
in Physics by (53.8k points)

A horizontal steam engine running at 240 r.p.m has a bore of 300 mm and stroke 600 mm. The connecting 1.25 m long and its mass of reciprocating parts is 60 kg. When the crank is 60o past its inner dead center, the steam pressure on the cover side of the piston is 1.125 N/mm2 while that on the crank side is 0.125 N/mm2 . Neglecting the area of the piston rod, determine: 

(a)  the force on the piston rod, and 

(b)  the turning moment on the crankshaft. 

1 Answer

+1 vote
by (53.5k points)
selected by
Best answer

ω = 2πN/60 = 2π(240)/60 = 25.13 m/s and n = I/r = 1.25/0.3 = 4.167

(a) Force on the piston rod, i.e., piston effort(Fp) :

we know that net load onthe piston,

FL = (P1 - P2) x (π/4) x D2

FL = (1.125 x 106 - 0.125 x 106) x (π/4) x 0.32 = 70685.83 N

Inertia force on reciprocating parts is given by

F1 = mω2r (cosθ + cos 2θ/n)

F1 = 60(25.13)2 x 0.3 x (cos 60 + cos2 x 60/4.167) = 4319.68 N

Therefore, Force on the piston rod or piston effort 

FP = FL - FI = 70685.83 - 4319.68 = 66.366 kN.

(b) Turning moment on the crankshaft (T): 

We know that turning moment on the crankshaft, 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.