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For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is – 10 Pa/m. The magnitude of axial velocity (in m/s) at a radial location of 0.2 cm is ________(up to two decimal places).

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Concept:

For a fully developed laminar flow, 

\(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}}\left( {{R^2} - {r^2}} \right)\)

where, R = radius of pipe, r = radial distance from centre (r = 0),  \(\frac{{\partial P}}{{\partial x}}\)= pressure gradient, μ is dynamic viscosity.

\(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\left( {1 - \frac{{\;{r^2}}}{{{R^2}}}} \right)\)

At, r = 0, \(\;u\; = \;{u_{max}} = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\)

\(u = {u_{max}}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)\)

Calculation:

Given,

 \(\frac{{\partial P}}{{\partial x}} = - 10\;Pa/m\) ,radius of pipe R = 5 cm = 0.05 m, μ = 0.001 Pa-s.

From,

  \(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\left( {1 - \frac{{\;{r^2}}}{{{R^2}}}} \right)\)

at r = 0.2 cm = 0.002 m,

 \(u = - \frac{1}{{4 \times 0.001}} \times \left( { - 10} \right) \times {0.05^2} \times \left( {1 - \frac{{{{0.002}^2}}}{{{{0.05}^2}}}} \right)\)

u = 6.24 m/s

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