The flow rate of water from a tap of diameter 1.25 cm is `0.48 L//mi n`. The coefficient of viscosity of water is `10^(-3) Pa-s`. After sometime, the flow rate is increased to `3L//mi n`. Characterise the flow for both the flow rates.

Explanation

Given, diameter D =1.25 cm
`=1.25 xx 10^(-2)m`
`eta=10^(-3) Pa s, p=10^(3) Kg//m^(3)`
Volume of water flowing out of tap per second is `Q=Av=(piD^(2))/4 xxv`
Speed of flow, `v=(4Q)/(pi D^2)`
Reynold.s number
`R_(e)=(pvD)/(eta)=(pD)/(eta). (4Q)/(piD^(2))=(4p Q)/(pi Dn)`
For, `Q_(1)=0.48 L/"min"`
`-(0.48 xx 10^(-3))/(60) m^(3) s^(-1)`
`=8 xx 10^(-6) m^(3) s^(-1)`
`R_(e)=(4 xx 10^(3) xx 8 xx 10^(-6))/(3.14 xx 1.25 xx 10^(-2) xx 10^(-3))=815`
Since `R_(e) lt 1000`, the flow is laminar
For `Q_(2)=3 L//"min"`
`=(3 xx 10^(-3))/(60) m^(3) s^(-1) =5 xx 10^(-5) m^(3) s^(-1)`
`R_(e)=(4 xx 10^(3) xx 5 xx 10^(-5))/(3.14 xx 1.15 xx 10^(-2) xx 10^(-3))`
`approx 5096`
Since `R_(e) gt 3000,` the flow is turbulent.