written 5.8 years ago by |
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