Boundary layer shear force

Homework Help
1

The x component of velocity in a laminar boundary layer in water is approximated as u=U sin(πy/2δ), whereU=3 m/s and δ=2 mm. The y component of velocity is much smaller than u. Obtain an expression for the net shearforce per unit volume in the x direction on a fluid element. Calculate its maximum value for this flow.

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This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. ENGINEERING CONTEXT:
    We are analyzing the viscous shear forces acting on a fluid element in a laminar boundary layer over a flat surface, which is essential for predicting viscous dissipation effects in devices such as aerodynamic surfaces.

  2. PHYSICS:
    The dominant physical effect is the transport of momentum by viscosity, with the net force per unit volume arising from the gradient of the shear stress.

  3. ASSUMPTIONS:
    We assume that the flow is unidirectional (the y-velocity is negligible) and the water’s viscosity is constant; these assumptions are justified within a laminar boundary layer of small thickness where the velocity variation is primarily in the y-direction.

  4. SOLUTION TECHNIQUE:
    The net shear force (a.k.a. viscous force) per unit volume in the x direction is given by

f_x = \mu\frac{d^2u}{dy^2}\,.

Given the velocity profile

u(y)= U\sin\!\left(\frac{\pi y}{2\delta}\right),

the first derivative is

\frac{du}{dy} = U\,\frac{\pi}{2\delta}\cos\!\left(\frac{\pi y}{2\delta}\right)\,,

and the second derivative is

\frac{d^2u}{dy^2} = -U\,\left(\frac{\pi}{2\delta}\right)^2\sin\!\left(\frac{\pi y}{2\delta}\right)\,.

Thus,

f_x = -\mu U\,\left(\frac{\pi}{2\delta}\right)^2\sin\!\left(\frac{\pi y}{2\delta}\right)\,.

The maximum magnitude of f_x occurs when \left|\sin\!\left(\frac{\pi y}{2\delta}\right)\right| = 1, so

\max\{|f_x|\} = \mu U\,\left(\frac{\pi}{2\delta}\right)^2\,.

Substituting U=3~\mathrm{m/s}, \delta=2~\mathrm{mm}=0.002~\mathrm{m}, and \mu= 1\times10^{-3}~\mathrm{Pa\ s} (for water),

\max\{|f_x|\} = 1\times10^{-3}\times 3\left(\frac{\pi}{2\times 0.002}\right)^2\,.

Evaluating,

\frac{\pi}{0.004} \approx 785.4~\mathrm{m^{-1}}\,,

so

\left(785.4\right)^2 \approx 6.17\times10^5~\mathrm{m^{-2}}\,.

Therefore,

\max\{|f_x|\} \approx 3\times10^{-3}\times6.17\times10^5\approx 1.85\times10^3~\mathrm{N/m^3}\,.
  1. REFLECTION:
    The derived expression shows that the net shear force per unit volume varies sinusoidally with y and reaches a maximum value of approximately 1.85\times10^3~\mathrm{N/m^3}. This force points in the negative x (i.e. upstream) direction. To get perspective on this, it is interesting to recast this in terms of G’s of acceleration:
a = \frac{f}{\rho} = 1.85 \frac{\mathrm{m}}{\mathrm{s}^2}

which is about 0.2 G’s. That acceleration is comparable to a normal person accelerating or braking in a car during a normal drive around town in Ithaca, NY.

In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 10.3: Boundary Layers: Overview Part 3 Part 1 (HIGHLY RELEVANT)
  2. Chapter 10.3: Boundary Layers: Overview Part 1 (HIGHLY RELEVANT)
  3. Chapter 10 Engineering Models: Boundary-Layer Equations (HIGHLY RELEVANT)
  4. Chapter 10.3: Boundary Layers: Overview Part 2 (HIGHLY RELEVANT)
  5. Chapter 10.4: Boundary Layers: Details (HIGHLY RELEVANT)

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