1.

Water is flowing in a pipe of radius 1.5 cm with an average velocity 15 cm `s^(-1)`. What is the nature of flow? Given coefficient of viscosity of water is `10^(-3)` kg `m^(-1) s^(-1)` and its density is `10^(3)` kg `m^(-3)`

Answer» `R_(e) = (rho vD)/(eta)`
Here `rho`, density = `10^(3)` kg `m^(-3)`
Coeff. Of viscosity, `eta = 10^(-3)` kg `m^(-1) s^(-1)`
Average velocity of water, v = 15 cm `s^(-1)`
= 0.15 m `s^(-1)`
Diameter of pipe, D = 2 `xx` 1.5 cm
= 3 cm
= 0.03 m
Hence, `R_(e) = (10^(3) xx 0.15 xx 0.03)/(10^(-3))`
= `10^(6) xx 0.0045`
= 4500
`R_(e) gt 2000`
Therefore, the flow is turbulent.


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