Question

Consider the steady, uni-directional, fully-developed, pressure-driven laminar flow of an incompressible Newtonian fluid through a circular pipe of inner radius \( 5.0 \, \text{cm} \). The magnitude of shear stress at the inner wall of the pipe is \( 0.1 \, \text{N} \, \text{m}^{-2} \). At a radial distance of \( 1.0 \, \text{cm} \) from the pipe axis, the magnitude of the shear stress, in \( \text{N} \, \text{m}^{-2} \), rounded off to 3 decimal places, is \_\_\_\_\_\_\_.

Hint:

For fully-developed laminar flow in a pipe, shear stress decreases linearly with radial distance from the wall to the pipe axis.

Answer 1

Step 1: Shear stress variation in a circular pipe. For steady, fully-developed, laminar flow in a circular pipe, the shear stress varies linearly with the radial position \( r \): \[ \tau(r) = \tau_w \left( 1 - \frac{r}{R} \right), \] where: - \( \tau_w \) is the shear stress at the inner wall, - \( R \) is the inner radius of the pipe, - \( r \) is the radial distance from the pipe axis. Step 2: Substitute the given values. Given: \[ \tau_w = 0.1 \, \text{N} \, \text{m}^{-2}, \quad R = 5.0 \, \text{cm} = 0.05 \, \text{m}, \quad r = 1.0 \, \text{cm} = 0.01 \, \text{m}. \] Substitute into the equation: \[ \tau(r) = 0.1 \left( 1 - \frac{0.01}{0.05} \right). \] Step 3: Simplify the expression. \[ \tau(r) = 0.1 \left( 1 - 0.2 \right) = 0.1 \cdot 0.8 = 0.02 \, \text{N} \, \text{m}^{-2}. \] Step 4: Conclusion. The magnitude of the shear stress at \( r = 1.0 \, \text{cm} \) is \( 0.020 \, \text{N} \, \text{m}^{-2} \).