Correct Answer - Option 1 : 39.2 percent

__Concept:__

By fitting air vessel the head loss due to friction in suction and delivery pipe is reduced. This reduction in the head loss saves certain amount of energy, which can be found by finding the work done against friction without air vessel and with air vessel. The difference of the two gives the saving in work done.

i) Work done against friction without air vessels

Consider a single-acting reciprocating pump without any air vessels on the pipes. The velocity of flow through the pipes is given by

\({\rm{v}} = \frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega }} \times {\rm{r}}\sin {\rm{\theta }}\)

And loss of head due to friction is given by

\({{\rm{h}}_{\rm{f}}} = \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left[ {\frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega r}}\sin {\rm{\theta }}} \right]^2}\)

The variation of hf with θ is parabolic in nature and hence indicator diagram for the loss of head due to friction in pipes will be a parabola. The work done by pump against friction per stroke is equal to the area of the indicator diagram due to friction.

∴ Work done by pump per stroke against friction.

W1 = Area of the parabola \(= \frac{2}{3} \times {\rm{Base}} \times {\rm{Height}}\)

\({{\rm{W}}_1} = \frac{2}{3} \times {\rm{L}} \times \left[ {\frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{a}}}{{\rm{a}}}{\rm{\omega r}}} \right)}^2}} \right]\) (∵ Height = ff at θ = 90°)

\({{\rm{W}}_1} = \frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{ft}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega r}}} \right)^2}\)

ii) Work saved in a double-acting reciprocating pump

The work lost in friction per stroke in case of double-acting reciprocating pump without air vessel is the same as given in case of single-acting reciprocating pump.

\({{\rm{W}}_1} = \frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{ft}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega r}}} \right)^2}\)

When the air vessel is fitted to the pipe near the cylinder, the mean velocity of flow, V̅ for double-acting is given by

\({\rm{\bar V}} = \frac{{{\rm{Discharge}}}}{{{\rm{Area\;of\;pipe}}}} = \frac{{\rm{Q}}}{{\rm{a}}} = \frac{{2{\rm{ALN}}}}{{60{\rm{a}}}}\)

\({\rm{V}} = \frac{{2{\rm{A}} \times 2{\rm{r}}}}{{60{\rm{a}}}} \times \frac{{60{\rm{\omega }}}}{{2{\rm{\pi }}}}\) (∵ L = 2r and \(\;{\rm{N}} = \frac{{60{\rm{\omega }}}}{{2{\rm{\pi }}}}\))

∴ Loss of head due to friction for double-acting

\({{\rm{h}}_{\rm{f}}} = \frac{{4{\rm{fl}} \times \overline {{{\rm{V}}^2}} }}{{{\rm{d}} \times 2{\rm{g}}}} = \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{2{\rm{A}}}}{{\rm{a}}} \times \frac{{{\rm{\omega r}}}}{{\rm{\pi }}}} \right)^2}\)

∴ Work lost due to friction per stroke

W2 = Area of the rectangle

\({{\rm{W}}_2} = {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{2{\rm{A}}}}{{\rm{a}}} \times \frac{{{\rm{\omega r}}}}{{\rm{\pi }}}} \right)^2} = \frac{4}{{{{\rm{\pi }}^2}}} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)^2}\)

∴ Saving in work done per stroke \(\left( {\rm{S}} \right) = \frac{{{{\rm{W}}_1} - {{\rm{W}}_2}}}{{{{\rm{W}}_1}}}\)

\({\rm{S}} = \frac{{\frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)}^2} - \frac{4}{{{{\rm{\pi }}^2}}} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega }}{{\rm{r}}^2}} \right)}^2}}}{{\frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)}^2}}} = \frac{{\left( {\frac{2}{3}} \right) - \left( {\frac{4}{{{{\rm{\pi }}^2}}}} \right)}}{{\left( {\frac{2}{3}} \right)}} = 0.392 = 39.2{\rm{\% }}\)