# Step-by-Step Solution We are to **show that in steady, fully developed, turbulent flow in a pipe, the wall shear stress $\tau_{app}$ varies linearly with the radius $r$**. ## 1. **Understanding the Problem** - **Steady:** No change with time. - **Fully developed:** Velocity profile does not change along the pipe. - **Turbulent:** Type of flow, but the derivation for the shear stress distribution is similar for both laminar and turbulent. - **$\tau_{app}$:** This usually refers to the **apparent shear stress** at radius $r$ inside the pipe. ## 2. **Force Balance on Pipe Fluid** Consider a pipe of radius $R$ and length $L$. Take a cylindrical shell of fluid with radius $r$ and length $L$. ### **Free Body Diagram** - **Pressure force** acts at the two ends (left and right). - **Shear force** acts at the wall of the cylinder (at radius $r$). ### **Forces** - **Pressure force:** - At $x$: $P(x) \cdot \pi r^2$ - At $x+L$: $P(x+L) \cdot \pi r^2$ - Net: $\left[P(x) - P(x+L)\right]\pi r^2 = - \left(\frac{dP}{dx} L\right) \pi r^2$ (since $P$ decreases along the flow) - **Shear force:** - Shear stress at radius $r$ acts on surface area $2\pi r L$ - Total shear force: $\tau_{app}(r) \cdot 2\pi r L$ ### **Force Balance (Axial Direction)** \[ \text{Sum of forces} = \] \[ \left[P(x) - P(x+L)\right]\pi r^2 - \tau_{app}(r) \cdot 2\pi r L = \] \[ - \left(\frac{dP}{dx} L\right) \pi r^2 = \tau_{app}(r) \cdot 2\pi r L \] ## 3. **Solve for $\tau_{app}(r)$** \[ - \left(\frac{dP}{dx}\right) \pi r^2 = 2\pi r \tau_{app}(r) \] \[ - \left(\frac{dP}{dx}\right) r^2 = 2 r \tau_{app}(r) \] \[ \tau_{app}(r) = - \frac{1}{2} \frac{dP}{dx} r \] ## 4. **Interpretation** - $\frac{dP}{dx}$ is negative (pressure drops in the direction of flow), so $\tau_{app}(r)$ is positive. - $\tau_{app}(r)$ is **directly proportional to $r$**. ## 5. **Final Answer** \[ \boxed{ \tau_{app}(r) = - \frac{r}{2} \frac{dP}{dx} } \] - **The shear stress varies linearly with the radius $r$.** - This result is valid for both laminar and turbulent, steady, fully-developed pipe flow. --- **Summary:** For steady, fully-developed flow (laminar or turbulent) in a pipe, the shear stress at a distance $r$ from the center is proportional to $r$, i.e., varies linearly with radius.