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A vertical double acting steam engine has a cylinder 30cm diameter and 450mm stroke and runs @ 200rpm. The reciprocating parts has a mass of 225kg and the piston rod is 50mm diameter.
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The connecting rod is 1.2m long. When the crank has turned through 125° from the TDC, The steam pressure above the piston is 30KN/m2 and below the piston is 1.5KN/m2, Calculate the effective turning moment on the crankshaft.

Given:

D=300mm

Stroke =L=450mm

Radius of crank = r = \frac{L}{2} = \frac{450}{2} = 225 \ mm

M_R = 225 kg

Piston rod dia. = d = 50 mm

l = 1.2 m

\theta = 125°

P_1 = 30 \ KN/m^2

P_2 = 1.5 \ KN/m^2

A_1 \ = \ \frac{\pi}{4} (0.300)^2 \ = \ 0.0706 \ m^2

a \ = \ \frac{\pi}{4} (0.050)^2 \ = \ 1.963 \times \ 10^{-3} \ m^2

N = 200 rpm

\omega = 20.94 rad/s

\rightarrow F_L = P_1 A_1 – P_2 [ A_1 – a]= 30 \times 0.0706 – 1.5 [0.0706 – 1.963 \times 10^{-3}] F_L = 2.015 \ KN \rightarrow n = \frac{l}{r} = \frac{1.2}{0.225} = 5.33

\rightarrow F_I \ = \ m_R. \ w^2.r. \ [cos \ \theta + \frac{cos \ 2 \ \theta}{n}]

= (225) . (20.94)^2 . \ (0.225) \ [ cos \ 125 + \frac{cos \ (2 \times 125)}{5.33}]

= -14.16 \times 10^3 \ N

\rightarrow Net force on piston or piston effort,

F_p = F_L – F_I + M_R . g \rightarrow As engine is vertical type therefore height of reciprocating is to be considered.

= 2.015 – (-14.16) + ( \frac{225 \times 981}{1000)})

= 18.38 KN

\rightarrow sin \ \phi \ = \frac{sin \ \theta}{n} = \frac{sin \ (125)}{5.33} = 0.1536

\phi = 8.84°

\rightarrow Effective turning moment on crank shaft.

T = F_T . r

= F_Q.sin \ (\theta \ + \ \phi) \ . r

= (\frac{FP}{cos \ \theta}). sin \ (\theta \ + \ \phi). r

= (\frac{18.38 \times 10^3}{cos \ 8.84}) . sin \ (125 + 8.84) (0.225)

= 3.0186 \times 10^3 \ N-m

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