question_answer

A pump motor is used to deliver water at a certain rate from a given pipe. To obtain 'n' times water from the same pipe in the same time by what amount (a) the force and (b) power of the motor should be increased.

Answer:

                Let a = area of cross section of pipe, v = velocity of flow of water, p = density of water. \[\therefore \] Mass of water flowing out/sec = \[\frac{dm}{dt}=\text{av}\rho \]. To get n times water in the same time, \[(\text{dm}/\text{dt})'=\text{n}\left( \text{dm}/\text{dt} \right)\]       \[\therefore \]             \[\text{a}'\text{v}'\rho '\text{ }=\text{n }(\text{av}\rho )\] But \[\rho '=\rho \] and a' = a (pipe is same) \[\therefore \]        \[~\upsilon '=\text{n}\upsilon \]                      ...(i) As \[\text{F}=\upsilon \frac{dm}{dt}\]  \[\therefore \]   \[\frac{F'}{F}=\frac{\upsilon '(dm/dt)'}{\upsilon (dm/dt)}=\frac{n\upsilon \left( n\frac{dm}{dt} \right)}{\upsilon \,dm/dt}={{n}^{2}}\]    \[\therefore \]\[F'={{n}^{2}}F\] Also, \[\frac{P'}{P}=\frac{F'\times \upsilon '}{F\times \upsilon }=\frac{{{n}^{2}}F\times n\upsilon }{F\times \upsilon }={{n}^{3}}\] Therefore, \[\text{P}'={{\text{n}}^{\text{3}}}\text{P}\]