Question

Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should water?

Answer 1

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

$Re_{5}$ = $Re_{12}$

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$= $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

$Re_{5}$ = Reynold number of water pipe

$Re_{12}$ = Reynold number of oil pipe

$V_{5}$ = Velocity of water 5 diameter pipe = ?

$V_{12}$ = Velocity of oil 12 diameter pipe = 2.30

$v_{5}$ = Kinetic Viscosity of water = 1 x $10^{-6}$ $m^{2}$/s

$v_{12}$ = Kinetic Viscosity of oil =  4 x $10^{-6}$ $m^{2}$/s

$D_{5}$ = Diameter of pipe used for water = 0.05 m

$D_{12}$ = Diameter of pipe used for oil = 0.12 m

Use the formula

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$= $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

By Removing square rots on both sides

${\frac{V_{5}XD_{5} }{v_{5}}}$= ${\frac{V_{12}XD_{12} }{v_{12}}}$

${V_{5}$= ${\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}$x$v_{5}$

${V_{5}$= [ (0.23 x 0.12m ) / (4 x $10^{-6}$ $m^{2}$/s) x 0.05 ] 1 x $10^{-6}$ $m^{2}$/s

${V_{5}$ = 1.38 m/s