Viscous Polymer Pipeline Flow (Chemical & Petroleum Engineering) Solution

A chemical engineer monitors the laminar flow of a highly viscous polymer solution through a rectangular processing channel in a refinery plant. Due to no-slip boundary conditions at the solid duct walls, fluid velocity drops to zero at the channel perimeter and peaks along the central core line.

The axial velocity $v(x, y)$ (in meters per second, $\text{m/s}$) across the rectangular cross-section $0 \le x \le 4\text{ cm}$ and $0 \le y \le 2\text{ cm}$ is modeled by the separable polynomial profile:

1. Explain how to compute $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ by treating terms involving the other coordinate as constant scaling factors.
2. Determine the analytical expressions for the transverse velocity gradients $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$.
3. In Newtonian fluid mechanics, shear stress across fluid layers is proportional to the velocity gradient:
   $$\tau_{zx} = \mu \frac{\partial v}{\partial x}, \quad \tau_{zy} = \mu \frac{\partial v}{\partial y}$$
   Evaluate $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ at the cross-sectional evaluation point $(x, y) = (1, 0.5)\text{ cm}$. State their units and identify which coordinate axis experiences greater internal fluid shear stress.

Step-by-Step Solution:

1. State the Constant Scaling Principle:

   • When differentiating with respect to $x$, the entire vertical term $(2y - y^2)$ behaves as a constant multiplier and can be factored outside the derivative operator:
     $$\frac{\partial v}{\partial x} = 1.5(2y - y^2) \frac{\partial}{\partial x}(4x - x^2)$$
   • Similarly, when differentiating with respect to $y$, $(4x - x^2)$ behaves as a constant multiplier:
     $$\frac{\partial v}{\partial y} = 1.5(4x - x^2) \frac{\partial}{\partial y}(2y - y^2)$$
     (Defined in Partial Derivatives.)

2. Compute Analytical Expressions:

   • Differentiate the polynomial factors using the single-variable power rule:
     $$\frac{\partial}{\partial x}(4x - x^2) = 4 - 2x = 2(2 - x)$$
     $$\frac{\partial}{\partial y}(2y - y^2) = 2 - 2y = 2(1 - y)$$
   • Substitute back into the partial derivatives:
     $$\frac{\partial v}{\partial x} = 1.5(2y - y^2)(4 - 2x) = 3(2 - x)(2y - y^2)$$
     $$\frac{\partial v}{\partial y} = 1.5(4x - x^2)(2 - 2y) = 3(1 - y)(4x - x^2)$$

3. Evaluate at $(x, y) = (1, 0.5)\text{ cm}$:

   • Compute coordinate factors at $(1, 0.5)$:
     $$4x - x^2 = 4(1) - (1)^2 = 3$$
     $$2 - x = 2 - 1 = 1$$
     $$2y - y^2 = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$$
     $$1 - y = 1 - 0.5 = 0.5$$
   • Evaluate $\frac{\partial v}{\partial x}$:
     $$\left.\frac{\partial v}{\partial x}\right|_{(1, 0.5)} = 3(1)(0.75) = 2.25\text{ (m/s)/cm} = 225\text{ s}^{-1}$$
   • Evaluate $\frac{\partial v}{\partial y}$:
     $$\left.\frac{\partial v}{\partial y}\right|_{(1, 0.5)} = 3(0.5)(3) = 4.5\text{ (m/s)/cm} = 450\text{ s}^{-1}$$
   • Units and Interpretation: The physical units are meters per second per centimeter ($(\text{m/s})/\text{cm}$, or $100\text{ s}^{-1}$ in SI base units).
   • Internal Shear Comparison: Because $\left|\frac{\partial v}{\partial y}\right| = 4.5 > \left|\frac{\partial v}{\partial x}\right| = 2.25$, the velocity gradient in the vertical direction is twice as large as the horizontal gradient. Since shear stress $\tau = \mu \frac{\partial v}{\partial s}$, the fluid experiences significantly greater internal shear stress in the vertical direction ($\tau_{zy} = 2\tau_{zx}$) at this location.