The piston of a single acting reciprocating pump moves in simple harmonic motion. Show that ratio of its work done against friction with air vessel and without air vessel is 3/2π².
We have already showed that work done by the pump per stroke against friction without air vessel is given by Eq. (15.29) as:
\mathrm{WD}_{1}=\mathrm{Area~of~the~parabola}={\frac{2}{3}}\times\mathrm{Base}\times\mathrm{Height}or \mathrm{WD}_{1}={\frac{2}{3}}\times L\times h_{f\operatorname*{max}}
={\frac{2}{3}}L\times{\frac{f l}{2g d}}\left\lgroup{\frac{A}{a}}r\omega\right\rgroup ^{2} (15.29)
and work done with rectangular air vessel is
\mathrm{WD}_{2}=\ { L}\times\frac{f \mathit{l}}{2g\ d}\left\lgroup\frac{A}{a}\times\frac{r\omega}{\pi}\right\rgroup ^{2}or \mathrm{WD}_{2}={\frac{1}{\pi^{2}}}\times L\times{\frac{f l}{2g d}}\left\lgroup{\frac{A}{a}}r\omega\right\rgroup ^{2} (15.30)
Hence, ratio of Eq. (15.30) and Eq. (15.29) gives
\frac{\mathrm{WD}_{2}}{\mathrm{WD}_{1}}=\frac{\left[\frac{1}{\pi^{2}}L\frac{f l}{2g d}\left\lgroup\frac{A}{a}r\omega\right\rgroup ^{2}\right]}{\frac{2}{3} L\times\frac{f l}{2g d}\left\lgroup\frac{A}{a}r\omega\right\rgroup ^{2}}or {\mathrm{Raito}}={\frac{3}{2\pi^{2}}}