Water is flowing in a pipe of diameter 6 cm with an average velocity `7.5 cms^(-1)` and its density is `10^(3) kg m^(-3)`. What is the nature of flow ? Given coefficient of viscosity of water is `10^(-3) kgm^(-1) s^(-1).`
A
Laminar
B
Streamline
C
Turbulent
D
Data is insufficient
Text Solution
Verified by Experts
The correct Answer is:
C
Reynolds number for the given situation is given as `R_(e )=(rho upsilon )/(eta)` Here `rho`, density `=10^(3) kgm^(-3)` Coefficient of visosity , `eta=10^(-3) kgm^(-1)s^(-1)` Average velocity of water, `upsilon =7.5 cms^(-1) = 0.075 ms^(-1)` Diameter of pipe, D=0.075=6 cm = 0.06 m Hence, `R_(e )=(10^(3)xx0.075xx0.06)/(10^(-3))` `=10^(6)xx0.0045=4500` `because " " R_(e ) gt 2000` Therefore, the flow is turbulent
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