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Chapter 6: Problem 38

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

Expert verified
Answer: The wall shear stress ratio between air and liquid water at 20°C is 0.018, which means the wall shear stress on the surface with air flowing over it is 1.8% of the wall shear stress with water flowing over it.

Step by step solution

01

Differentiation of velocity profile

$$ \frac{du}{dy} = \frac{d}{dy}\left(100(y + 2y^{2} - 0.5y^{3})\right) $$ $$ \frac{du}{dy} = 100(1 + 4y - 1.5y^{2}) $$ #Step 2: Calculate the dynamic viscosity for air and water# The given fluid temperatures are 20°C. We need to find the dynamic viscosity at 20°C for both air and liquid water.
02

Dynamic viscosity of air at 20°C

μ_air = 1.8*10^{-5} kg/(m s)
03

Dynamic viscosity of water at 20°C

μ_water = 1.002*10^{-3} kg/(m s) #Step 3: Calculate the wall shear stress for air and water# To calculate the wall shear stress, we will use the wall condition which is y = 0
04

Shear stress of air at the wall

$$ τ_{air} = μ_{air} * \left(\frac{du}{dy}\right)_{y=0} = 1.8*10^{-5} * (1 + 4(0) - 1.5(0)^2) = 1.8*10^{-5} \,\text{kg/(m s)} $$
05

Shear stress of water at the wall

$$ τ_{water} = μ_{water} * \left(\frac{du}{dy}\right)_{y=0} = 1.002*10^{-3} * (1 + 4(0) - 1.5(0)^2) = 1.002*10^{-3} \,\text{kg/(m s)} $$ #Step 4: Calculate the wall shear stress ratio and interpret the result# The wall shear stress ratio is the ratio of the wall shear stress of air to the wall shear stress of water.
06

Wall shear stress ratio calculation

$$ \text{Wall Shear Stress Ratio} = \frac{τ_{air}}{τ_{water}} = \frac{1.8*10^{-5}}{1.002*10^{-3}} = 0.018 $$ This result means that the wall shear stress on the surface with air flowing over it is 1.8% of the wall shear stress with water flowing over it. Since the wall shear stress is a major factor in drag and frictional force, air flowing over the surface produces significantly less drag and friction compared to water.

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