Question

An oil of kinematic viscosity 0.5bstokes flow through a pipe of diameter 7.5 cm. The flow is critical at a velocity of

Answer 1

Given that,

Kinematic viscosity = 0.5 stokes

Diameter = 7.5 cm

Suppose, The flow is laminar

We need to calculate the velocity

Using formula of  viscosity

$R=\dfrac{\rho d v}{\eta}$

$v=\dfrac{R\eta}{\rho d}$

Where, R = reynold number

$\rho$ = density of oil

$\eta$ = viscosity

d = diameter

Put the value into the formula

$v=\dfrac{2100\times0.5\times10^{-4}}{950\times7.5\times10^{-2}}$

$v=0.00147\ m/s$

$v=1.47\times10^{-3}\ m/s$

Hence, The velocity is $1.47\times10^{-3}\ m/s$

Answer 2

Given that,

Kinematic viscosity = 0.5 stokes

Diameter = 7.5 cm

Suppose, The flow is laminar

We need to calculate the velocity

Using formula of  viscosity

$R=\dfrac{\rho d v}{\eta}$

$v=\dfrac{R\eta}{\rho d}$

Where, R = reynold number

$\rho$ = density of oil

$\eta$ = viscosity

d = diameter

Put the value into the formula

$v=\dfrac{2100\times0.5\times10^{-4}}{950\times7.5\times10^{-2}}$

$v=0.00147\ m/s$

$v=1.47\times10^{-3}\ m/s$

Hence, The velocity is $1.47\times10^{-3}\ m/s$

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