Correct Answer - Option 4 : 84.8%

__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 a certain amount of energy, which can be found by finding the work done against friction without air vessel and with air vessel. The difference between 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 the 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 single-acting reciprocating pump

Loss of head due to friction for single-acting

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

∴ Work lost due to friction per stroke

W2 = Area of the rectangle

\({{\rm{W}}_2} = {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{{\rm{A}}}}{{\rm{a}}} \times \frac{{{\rm{\omega r}}}}{{\rm{\pi }}}} \right)^2} = \frac{1}{{{{\rm{\pi }}^2}}} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)^2}\)

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

\({\rm{S}} = \frac{{\frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)}^2} - \frac{1}{{{{\rm{\pi }}^2}}} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}}{\rm{\omega }}{{\rm{r}}^2}} \right)}^2}}}{{\frac{2}{3} \times {\rm{L}} \times \frac{{4{\rm{fl}}}}{{{\rm{d}} \times 2{\rm{g}}}} \times {{\left( {\frac{{\rm{A}}}{{\rm{a}}} \times {\rm{\omega r}}} \right)}^2}}} = \frac{{\left( {\frac{2}{3}} \right) - \left( {\frac{1}{{{{\rm{\pi }}^2}}}} \right)}}{{\left( {\frac{2}{3}} \right)}} = 0.848 = 84.8{\rm{\% }}\)