Question

Consider fluid flow over a surface with a velocity profile given as $$ u(y)=100\left(y+2 y^{2}-0.5 y^{3}\right) \mathrm{m} / \mathrm{s} $$ Determine the shear stress at the wall surface, if the fluid is (a) air at $1 \mathrm{~atm}\( and (b) liquid water, both at \)20^{\circ} \mathrm{C}$. Also calculate the wall shear stress ratio for the two fluids, and interpret the result.

Short answer

Answer: The wall shear stress ratio for liquid water and air at 20°C is approximately 55.56. This means that the wall shear stress is 55.56 times greater for liquid water than for air, indicating that the fluid flow exerts a significantly higher force along the wall surface for water than for air. This difference in wall shear stress can have a significant impact on erosion and friction between the fluid and the wall surface.

Step-by-step solution

Calculate the velocity gradient at the wall surface

Taking the derivative of the velocity profile with respect to y, we get: $$ \frac{du}{dy} = \frac{d}{dy}(100(y + 2y^2 - 0.5y^3)) $$ $$ \frac{du}{dy} = 100(1 + 4y - 1.5y^2) $$ The wall surface corresponds to y = 0, so we can find the velocity gradient at the wall surface by plugging y = 0 into the expression above: $$ \left(\frac{du}{dy}\right)_{y=0} = 100(1 + 4(0) - 1.5(0)^2) = 100\,\mathrm{s}^{-1} $$

Find the shear stress at the wall surface for both fluids

The shear stress at the wall surface can be calculated using the formula: $$ \tau = \mu \frac{du}{dy} $$ Where τ is the shear stress and μ is the dynamic viscosity of the fluid. For air at 1 atm and \(20^{\circ} \mathrm{C}\), the dynamic viscosity can be found as \(\mu_{air} = 1.8 \times 10^{-5}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}\). So, the shear stress at the wall surface for the air is: $$ \tau_{air} = \mu_{air}\frac{du}{dy} = (1.8 \times 10^{-5}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1})(100\,\mathrm{s}^{-1}) = 1.8 \times 10^{-3}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2} $$ For liquid water at \(20^{\circ} \mathrm{C}\), the dynamic viscosity can be found as \(\mu_{water} = 1.0 \times 10^{-3}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}\). So, the shear stress at the wall surface for the water is: $$ \tau_{water} = \mu_{water} \frac{du}{dy} = (1.0 \times 10^{-3}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1})(100\,\mathrm{s}^{-1}) = 1.0 \times 10^{-1}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2} $$

Calculate the wall shear stress ratio and interpret the result

The wall shear stress ratio can be calculated as the ratio of the shear stress for water to the shear stress for air: $$ \text{Wall Shear Stress Ratio} = \frac{\tau_{water}}{\tau_{air}} = \frac{1.0 \times 10^{-1}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2}}{1.8 \times 10^{-3}\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2}} \approx 55.56 $$ This means that the wall shear stress is approximately 55.56 times greater for liquid water than for air. This result indicates that the fluid flow exerts a significantly higher force along the wall surface for water than for air due to the higher viscosity and density of water compared to air. This difference in wall shear stress can have a significant impact on the erosion and friction between the fluid and the wall surface.