A Fluid of viscosity 0.2 Pa.s and specific gravity 0.8 flows thro

SOLUTION

Concept:

The pressure drop per meter length of the pipe is given by

\(\frac{{{\rm{\Delta }}P}}{{{\rm{\Delta }}L}} = \frac{{32\;\mu {V_{avg}}}}{{{D^2}}}\)

Volumetric flow rate is given by

\(Q = AV = \frac{{\pi {D^2}}}{4}{V_{avg}}\)

Power required to overcome resistance is given by

P = Q × ΔP

The shear stress distribution in laminar flow in a pipe is given by

\(\tau = - \frac{{\partial P}}{{\partial x}}\left( {\frac{r}{2}} \right)\)

The velocity distribution is given by

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

Calculation:

Given μ = 0.2 Pa.s, ρ = 0.8 × 1000 = 800 kg/m³, d = 30 mm = 0.03 m, ΔP/ΔL = 10 kPa;

\(10 \times {10^3} = \frac{{32 \times 0.2 \times {V_{avg}}}}{{{{0.03}^2}}} \Rightarrow {V_{avg}} = 2.8\;m/s\)

Reynolds number will be

\(Re = \frac{{800 \times 2.8 \times 0.03}}{{0.2}} = 337.5\;\left( {laminar} \right)\)

\(Q = \frac{{\pi \times {{0.03}^2}}}{4} \times 2.8 = 1.98 \times {10^{ - 3}}\;{m^3}/s\)

Power required will be

P = 1.98 × 10^-3 × 10 × 10³ = 19.8 W

The shear stress at wall is

\(\tau = - \frac{{\partial P}}{{\partial x}}\left( {\frac{R}{2}} \right) = 10 \times {10^3} \times \frac{{0.015}}{2} = 75\;Pa\)

We know V_avg = 0.5 V_max,

V_max occurs at r = 0 ⇒ \({V_{max}} = - \frac{1}{{4\mu }}\left( {\frac{{\partial {\rm{P}}}}{{\partial x}}} \right){R^2}\)

\( \Rightarrow {V_{avg}} = - \frac{1}{{8\mu }}\left( {\frac{{\partial {\rm{P}}}}{{\partial x}}} \right){R^2}\)

Substituting in velocity distribution,

\(- \frac{1}{{8\mu }}\left( {\frac{{\partial {\rm{P}}}}{{\partial x}}} \right){R^2} = - \frac{1}{{4\mu }}\left( {\frac{{\partial {\rm{P}}}}{{\partial x}}} \right)\left( {{R^2} - {r^2}} \right) \Rightarrow r = \frac{R}{{\sqrt 2 }}\)