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Tony's Intertwingled Memex • Reference Manual

Viscous Polymer Pipeline Flow (Chemical & Petroleum Engineering)

Formulas & Definitions

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)v(x, y)v(x,y) (in meters per second, m/s\text{m/s}m/s) across the rectangular cross-section 0≤x≤4 cm0 \le x \le 4\text{ cm}0≤x≤4 cm and 0≤y≤2 cm0 \le y \le 2\text{ cm}0≤y≤2 cm is modeled by the separable polynomial profile:

v(x,y)=1.5(4x−x2)(2y−y2)v(x, y) = 1.5(4x - x^2)(2y - y^2)v(x,y)=1.5(4x−x2)(2y−y2)
where xxx is horizontal distance from the left wall (in cm\text{cm}cm) and yyy is vertical distance from the duct floor (in cm\text{cm}cm).

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