Question
Download Solution PDFA flywheel is fitted to the crank shaft of an engine having W amount of indicated work per revolution. Permissible limits of coefficient of fluctuation of energy and speed are CE and CS respectively. The kinetic energy of the flywheel is given by
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Coefficient of fluctuation of energy:
\({C_E} = \frac{{Maximum\;fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)
\({C_E} = \frac{\Delta E}{W} \Rightarrow \Delta E=C_E.W\)
Coefficient of fluctuation of speed:
\(\begin{array}{l} {C_S} = \frac{{Maximum\;fluctuation\;of\;speed}}{{Mean\;speed}}\\ {C_S} = \frac{{{ω _1} - {ω _2}}}{ω } = \frac{{2\left( {{ω _1} - {ω _2}} \right)}}{{\left( {{ω _1} + {ω _2}} \right)}} \end{array}\)
ω = mean angular speed during the cycle (rad/s)
\(ω = \frac{{{ω _1} + {ω _2}}}{2}\)
Maximum fluctuation of energy: ΔE = CE. W
\(\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}Iω _1^2 - \frac{1}{2}Iω _2^2 = \frac{1}{2}I\left( {{ω _1} + {ω _2}} \right)\left( {{ω _1} - {ω _2}} \right)\\ {\rm{\Delta }}E = Iω \left( {{ω _1} - {ω _2}} \right) = I{ω ^2}\frac{{\left( {{ω _1} - {ω _2}} \right)}}{ω } = I{ω ^2}{C_S}\\ K.E = \frac{1}{2}I{ω ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{C_S}}} = \frac{{{C_E}W}}{{2{C_S}}} \end{array}\)
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