**1 Answer**

written 5.1 years ago by |

**The work done by the gases during the expansion stroke is 3 times the work done on gases during the compression stroke, The work done during the suction and exhaust strokes being negligible. If the total fluctuation of speed is not to exceed 2% of the mean speed and the TMD during compression and expansion is assumed to be triangular in shape. Find the MOI of flywheel.**

**Given:** $w_1 – w_2 = 4%w$

$\frac{w_1 – w_2}{w} = 0.04$

$= C_s$

$\rightarrow$ $W_E = 2W_c$

$\rightarrow$ $W.D/cycle = \frac{p \times 60}{n} = \frac{20 \times 10^3 \times 60}{150} = 8000 \ N-m$

$n = \frac{N}{2} = \frac{300}{2} = 150$

$\rightarrow$ $W.D/Cycle = W_E – W_c$

$8000 = W_E - \frac{W_E}{3}$

$\therefore$ $W_E = 12000 \ N-m$

$\rightarrow$ $W_E \ = \ Area \ of \ triangle \ ABC \ = \ \frac{1}{2} \ \times \ BC \times \ AF$

$12000 = \frac{1}{2} \times \pi \times \ AF$

$AF = 7.63 \times 10^3 \ N – m = \ T_{max}$

$\rightarrow$ $P = \frac{2 \pi N Tmean}{60}$

$Tmean = \frac{20 \times 10^3\times 60}{2 \pi \times 300} = 636.61 \ N – m = \ Gf$

$\therefore$ $AG = AF – GF = 7.63 \times 10^3 – 636.61$

$AG = 6.99 \times 10^3 \ N - m$

$\rightarrow$ $\frac{DE}{AG} = \frac{BC}{AF}$

$\therefore$ $DE = 6.99 \times 10^3 \times \frac{\pi}{7.63 \times 10^3} = 2.879 \ rad$

$\rightarrow$ $\triangle E = Area \ of \ \triangle \ ADE = \frac{1}{2} \times DE \times AG$

$= \frac{1}{2} \times 2.879 \times 6.99 \times 10^3$

$= 10.06 \times 10^3 \ N - m$

Also, $\triangle E = I. (\frac{\pi^2}{900}) N^2. \ C_s$

$10.06 \times 10^3 =I. [ \frac{\pi^2}{900}] (300)^2 \times 0.04$

$I = 254.3 \ Kgm^2$