Question 10.50: The turning moment diagram for a four-stroke gas engine may ...

\text { Given: } N_{\max }=202 rpm , N_{\min }=198 rpm , R=1.2 m .

N_{m}=\left(N_{\max }+N_{\min }\right) / 2=(202+198) / 2=200 rpm .

\text { Coefficient of fluctuation of speed, } C_{s}=\left(N_{\max }-N_{\min }\right) / N_{m} .

= (202 – 198)/200=4/200=0.02.

\omega=2 \pi \times 200 / 60=20.944 rad / s .

The turning moment diagram is shown in Fig.10.36.

\text { Net area, } A=A_{3}-\left(A_{1}+A_{2}+A_{4}\right)=\left[6.8-(0.45+1.7+0.65] \times 10^{-3}\right. .

=4 \times 10^{-3} m ^{2} .

\text { Net work done }=A \times \text { Scale }=4 \times 10^{-3} \times 3 \times 10^{6}=12,000 Nm .

\text { Work done per cycle }=T_{\text {mean }} \times 4 \pi Nm .

T_{\text {mean }}=\frac{12,000}{4 \pi}=954.93 Nm .

\text { Work done during expansion stroke }=A_{3} \times \text { Scale }=6.8 \times 10^{-3} \times 3 \times 10^{6}=20.400 Nm .

= Area ΔABC.

=\left(\frac{1}{2}\right) \times A C \times B F=\left(\frac{1}{2}\right) \times \pi \times B F .

B F=\frac{20400 \times 2}{\pi}=12987 N m .

\text { Excess torque, } B G=B F-F G=12987-954.93=12033.07 Nm .

Now Δs ABC and DBE are similar. Thus

\frac{D E}{A C}=\frac{B G}{B F} .

D E=\left(\frac{12033.07}{12987}\right) \pi=3.91 \text { radian } .

\text { Maximum fluctuation of energy, } E_{f}=\text { Area } \Delta B D E=\left(\frac{1}{2}\right) D E \times B G .

=\left(\frac{1}{2}\right) \times 3.91 \times 12033.07=17506.7 Nm .

Mass of flywheel,    m=\frac{E_{f}}{R^{2} \omega^{2} C_{s}} .

=\frac{17506.7}{(1.2)^{2} \times(20.944)^{2} \times 0.02}=1385.8 kg .