Question

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. 

Answer 1

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(F_p) :

we know that net load onthe piston,

F_L = (P₁ - P₂) x (π/4) x D²

F_L = (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 = m_R ω²r (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 

F_P = F_L - F_I = 70685.83 - 4319.68 = 66.366 kN.

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

We know that turning moment on the crankshaft,