Wall shear stress from a pipe force balance: 16-Civ-A5 Hydraulic Engineering, December 2018, Q4

Question 4: Wall shear stress from a pipe force balance (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Steady, fully-developed flow in a horizontal circular pipe; pressure drop $\Delta P=15\ \text{kPa}$ over length $L=2\ \text{m}$; diameter $D=0.150\ \text{m}$ ($R=0.075\ \text{m}$).

Given data
Quantity	Value
Pressure difference $\Delta P$	15 kPa = 15,000 Pa
Length $L$	2 m
Diameter $D$ (radius $R$)	0.150 m (0.075 m)

Find. A regime-independent relation between wall shear stress $\tau_w$ and average velocity $V$, and the numerical $\tau_w$ for the stated data.

Figure 4. Free body of the fluid cylinder: pressure force drives flow; wall shear resists it.

Approach. Apply Newton’s second law to the cylindrical fluid core of radius $R$ and length $L$. In steady, fully-developed flow the momentum flux in equals out, so the net pressure force balances the wall shear force — a statement independent of whether the flow is laminar or turbulent. Then fold in the Darcy–Weisbach definition to express the same $\tau_w$ through the average velocity.

Force balance on the fluid cylinder. Pressure acts on the end area $\pi R^{2}$; shear acts on the lateral area $2\pi R L$: $$(P_1-P_2)\,\pi R^{2}=\tau_w\,(2\pi R L).$$
Solve for wall shear. Cancelling and using $R=D/2$, $$\boxed{\tau_w=\frac{\Delta P\,R}{2L}=\frac{\Delta P\,D}{4L}.}$$ This holds for any regime because it is pure statics — no constitutive (viscous) law was invoked.
Relate to average velocity. Darcy–Weisbach writes the same drop as $\Delta P=f\dfrac{L}{D}\dfrac{\rho V^{2}}{2}$. Substituting into $\tau_w=\Delta P\,D/4L$ gives the regime-independent form $$\tau_w=\frac{f}{8}\,\rho V^{2},$$ with $f$ from the laminar law $64/Re$ or the Colebrook/Moody chart for turbulent flow.
Numerical value. With $\Delta P=15{,}000\ \text{Pa}$, $D=0.150\ \text{m}$, $L=2\ \text{m}$, $$\tau_w=\frac{15000(0.150)}{4(2)}=\frac{2250}{8}\;\Rightarrow\;\boxed{\tau_w\approx 281\ \text{Pa}.}$$

Question 4 — results
Quantity	Value
Closed-form relation	$\tau_w=\Delta P\,D/4L=(f/8)\rho V^{2}$
Wall shear stress $\tau_w$	281.25 Pa

Topic. Momentum (force) balance for fully-developed pipe flow; wall shear and the friction factor.

Key relations. $\tau_w=\Delta P\,D/4L$ (statics); $\Delta P=f\tfrac{L}{D}\tfrac{\rho V^{2}}{2}$ (Darcy–Weisbach) $\Rightarrow \tau_w=(f/8)\rho V^{2}$.

Why this works. In steady, fully-developed flow the velocity profile does not change along the pipe, so there is no net momentum change and the control-volume force balance is exact: the pressure force on the ends must equal the shear force on the wall. Because this argument never uses a viscosity model, the resulting $\tau_w=\Delta P\,D/4L$ is valid for laminar and turbulent flow alike. The Darcy–Weisbach substitution simply repackages the pressure gradient into a velocity-based friction factor.

Common pitfall. Using radius where diameter belongs (the factor of 4 vs 2) or assuming the laminar $\tau_w=8\mu V/D$ is the general answer — it is only the laminar special case of $(f/8)\rho V^{2}$.

Source. Munson, Young & Okiishi, Fundamentals of Fluid Mechanics, viscous pipe-flow chapter (wall shear & friction factor).

Question 4 of 6: Wall shear stress from a pipe force balance

Question 4: Wall shear stress from a pipe force balance (20 marks)