Question 10.51: A vertical single cylinder engine has a cylinder diameter 30...

\text { Given: } D=300 mm , L=500 mm , \text { or } r=250 mm , n=\frac{l}{r}=4.5, N=180 rpm , M_{r}=280 kg .

r_{c}=\frac{V_{1}}{V_{2}}=14, V_{3}-V_{2}=0.1 V_{s}=0.1\left(V_{1}-V_{2}\right), p_{1}=0.1 N / mm ^{2}, \theta=45^{\circ} .

The p-V diagram for the diesel engine is shown in Fig.10.37.

\text { Angular speed of engine, } \omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 180}{60}=18.85 rad / s .

Process 1-2:    p_{1} V_{1}^{1.35}=p_{2} V_{2}^{1.35} .

p_{2}=p_{1}\left(\frac{V_{1}}{V_{2}}\right)^{1.35}=0.1(14)^{1.35}=3.526 N / mm ^{2} .

\text { Swept volume, } V_{s}=\left(\frac{\pi}{4}\right) D^{2} L=\frac{\pi}{4} \times 300^{2} \times 500 \times 10^{-9}=0.035343 m ^{3} .

Now          \frac{V_{1}}{V_{2}}=\frac{V_{2}+V_{s}}{V_{2}} .

14=\frac{1+0.035343}{V_{2}} .

Also    p_{2}=p_{3} \text { and } V_{3}=V_{2}+0.1 V_{s}=0.00272+0.0035343=0.00625 m ^{3} .

The displacement of the piston for 45° crank rotation from I.D.C. during expansion stroke is indicated by point 5 in Fig.10.37.

x=r\left[(1-\cos \theta)+\left\{n-\left(n^{2}-\sin ^{2} \theta\right)^{0.5}\right\}\right] .

=250\left[\left(1-\cos 45^{\circ}\right)+\left\{4.5-\left(4.5^{2}-\sin ^{2} 45^{\circ}\right)^{0.5}\right\}\right] .

= 87.2 mm or 0.0872 m.

V_{5}=V_{2}+\left(\frac{\pi}{4}\right) D^{2} x=0.00272+\left(\frac{\pi}{4}\right) \times 300^{2} \times 87.2 \times 10^{-9} .

=0.0089 m ^{3} .

Process 3-5:

p_{3} V_{3}^{1.35}=p_{5} V_{5}^{1.35} .

p_{5}=p_{3}\left(\frac{V_{3}}{V_{5}}\right)^{1.35}=3.526\left(\frac{0.00625}{0.0089}\right)^{1.35}=3.188 N / mm ^{2} .

\text { Difference of pressure on two sides of the piston, } \Delta p=p_{5}-p_{1}=3.188-0.1=3.088 N / mm ^{2} .

Force on the piston due to gas pressure during outstroke,

F_{p}=\left(\frac{\pi}{4}\right) D^{2} \times \Delta p=\left(\frac{\pi}{4}\right) \times 300^{2} \times 3.088=147592 N .

\text { Inertia force due to the reciprocating parts, } F_{i}=M_{r} \omega^{2} r\left(\cos \theta+\frac{\cos 2 \theta}{n}\right) .

=280 \times(18.85)^{2} \times 0.25 \times\left(\cos 45^{\circ}+\frac{\cos 90^{\circ}}{4.5}\right) .

=17587.6 N.

Piston effort during down stroke of a vertical four-stroke diesel engine.

F=F_{p}-F_{i}+M_{r} g=147592-17587.6+280 \times 9.81=132751.2 N .

(i) Crank pin effort.
Angle of inclination of the connecting rod with the line of stroke,

\sin \phi=\frac{\sin \theta}{n}=\frac{\sin 45^{\circ}}{4.5}=0.15713 .

\phi=9.04^{\circ} .

CPE , F_{t}=F \times \frac{\sin (\theta+\phi)}{\cos \phi} .

= 108803.9 N.

(ii) Thrust on bearings,

F_{r}=F \times \frac{\cos (\theta+\phi)}{\cos \phi} .

=132751.2 \times \frac{\left(45^{\circ}+9.04^{\circ}\right)}{\cos 0.94^{\circ}} .

= 78934.65 N.

(iii) Turning moment on the crankshaft or crank effort,

T=F_{t} \times r=108803.9 \times 0.25=27200.975 Nm .