**1 Answer**

written 3.9 years ago by |

**Find the mass of the rim of a flywheel required to keep the speed between $202 \ and \ 198 \ rpm.$ The mean radius of the rim is 1.2m.**

Suction stroke = $0.45 \times 10^{-3} \ m^2$

Suction stroke = $0.45 \times 10^{-3} \ m^2$

Compression stroke = $1.7 \times 10^{-3} \ m^2$

Expansion stroke = $6.8 \times 10^{-3} \ m^2$

Exhaust stroke = $0.65 \times 10^{-3} \ m^2$

Each “$m^2$” of area represents 3MNm of energy assuming the resisting torque to be uniform, find the mass of the rim of a flywheel required to keep the speed between 202 and 198 rpm. The mean radius of the rim is 1.2 m.

**Given:** $N_1 = 202 \ RPM, \ N_2 = 198 \ rpm$

$C_s = \frac{ (202 – 198)}{(\frac{200 + 198}{2})} = 0.02$

$a_1 = 0.45 \times 10^{-3} \ m^2$

$a_2 = 1.7 \times 10^{-3} \ m^2$

$a_3 = 6.8 \times 10^{-3} \ m^2$

$a_4 = 0.65 \times 10^{-3} \ m^2$

$\rightarrow$ Net area = Net work done.

$= a_3 – (a_1 + a_2 + a_4)$

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

$\rightarrow$ scale = $1m^2 = 3MN – m = 3 \times 10^6 \ N - m$

$\therefore$ $Net \ work done/cycle \ = \ T_{mean} \ \times \ \theta$

$12000 \ = \ T_{mean} \ \times 4 \pi$

$T_{mean} \ = \ 954.92 \ N - m$ - G.f.

$\rightarrow$ Work Done during expansion stroke,

$W_E = a_3 \ \times \ scale$

$= 6.8 \times 10^{-3} \times 3 \times 10^6$

$= 20.4 \times 10^3 \ N - m$

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

$20.4 \times 10^3 = \frac{1}{2} \times (\pi) \times AF$

$AF = 12.987 \times 10^3 N - m \ = \ T_{max}$

$\rightarrow$ $AG = AF – GF = 12.98 \times 10^3 – 954.92 = 12.03 \times 10^3 \ N - m$

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

$= \frac{1}{2} \times 2.911 \times 12.03 \times 10^3$

$\triangle E = 17.5 \times 10^3 \ N - m$

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

$DE = 12.03 \times 10^3 \times \pi$

$12.487 \times 10^3$

$DE = 2.911\ rad$

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

$17.5 \times 10^3 = I \ [ \frac{\pi^2}{900}] (200)^2 . (0.02)$

$I = 1994.76 \ kg – m^2 \ = \ m k^2$

$\frac{1994.76}{(1.2)^2} \ = \ m$

$m = 1385\ kg.$