5. Describing Flow CH EN 374: Fluid Mechanics

Theory and Derivations I want to know: What are your thoughts on theory and derivations?

Where We’ve Been Definition of a fluid Fluid properties Density Viscosity, shear rate, shear stress Types of flow Pressure and pressure forces

Where We’re Going Fluid Kinematics (motion) Property balances: Today: Mass Momentum Energy Today: Different ways of viewing fluid flow Property balance math

Frames of Reference We can describe fluid motion in two ways: Eulerian: a control volume is defined, and fluid flows in and out. We define field variables. Field variables are functions of space or time. Lagrangian: The Fluid is divided into parcels of fluid (“fluid particles). We keep track of each fluid particle’s position vector and velocity vector.

Eulerian Description Pressure Field Velocity Field

Lagrangian Description Fluid particles ≠ “real” particles Deform—can be hard to keep track of. https://www.youtube.com/watch?v=DhNt_A3k4B4 https://www.youtube.com/watch?v=RuZQpWo9Qhs

Rates of Change Consider a leaf on a river.

Rates of Change How does the velocity of the river change? Is that a Eulerian or Lagrangian way of looking at it?

Rates of Change How does the velocity of the leaf change? Is that a Eulerian or Lagrangian way of looking at it?

𝑑𝑓 𝑑𝑡 Eulerian: Lagrangian: 𝑓=𝑓(𝑥,𝑦,𝑧, 𝑡) 𝑓 is any property 𝑓 field changes with time 𝑑𝑓 𝑑𝑡 Lagrangian: How will 𝑓 change from the point of view of the leaf? (Or a fluid particle) Will change as the field changes and as we move through the field

The Material Derivative 𝐷𝑓 𝐷𝑡 = 𝜕𝑓 𝜕𝑡 + 𝑣 ∙𝛻𝑓 Notice the capital D’s—this is the material derivative This relates the Lagrangian and Eulerian points of view. The material derivative can be applied to any property. It represents how that property changes as we follow a fluid particle.

Math Reminder: 𝛻𝑓= 𝜕𝑓 𝜕𝑥 𝒆 𝒙 + 𝜕𝑓 𝜕𝑦 𝒆 𝒚 + 𝜕𝑓 𝜕𝑧 𝒆 𝒛 The del operator: 𝛻𝑓= 𝜕𝑓 𝜕𝑥 𝒆 𝒙 + 𝜕𝑓 𝜕𝑦 𝒆 𝒚 + 𝜕𝑓 𝜕𝑧 𝒆 𝒛 The velocity vector: 𝑣 =𝑢 𝒆 𝒙 +𝑣 𝒆 𝒚 +𝑤 𝒆 𝒛 So what’s 𝑣 ∙𝛻𝑓? 𝑣 ∙𝛻𝑓=𝑢 𝜕𝑓 𝜕𝑥 +𝑣 𝜕𝑓 𝜕𝑦 +𝑤 𝜕𝑓 𝜕𝑧

Example: Acceleration Consider a steady, incompressible, two-dimensional velocity field given by: 𝑣 = 𝑢,𝑣 = 0.5+0.8𝑥 𝒆 𝒙 +(1.5−0.8𝑦) 𝒆 𝒚 where 𝑥 and 𝑦 are in m and velocities are in m/s. Calculate the material acceleration (the acceleration of a parcel of fluid) at the point (𝑥=2, y=3).

𝐷 𝑣 𝐷𝑡 = 𝑣 ∙𝛻 𝑣 =𝑢 𝜕 𝑣 𝜕𝑥 +𝑣 𝜕 𝑣 𝜕𝑦 +𝑤 𝜕 𝑣 𝜕𝑧 General form of material derivative: 𝐷𝑓 𝐷𝑡 = 𝜕𝑓 𝜕𝑡 + 𝑣 ∙𝛻𝑓 Newtonian relation for acceleration: 𝑎= 𝑑 𝑣 dt so 𝑓 is 𝑣 𝐷 𝑣 𝐷𝑡 = 𝜕 𝑣 𝜕𝑡 + 𝑣 ∙𝛻 𝑣 Material acceleration: 𝐷 𝑣 𝐷𝑡 = 𝜕 𝑣 𝜕𝑡 + 𝑣 ∙𝛻 𝑣 This flow is steady: 𝐷 𝑣 𝐷𝑡 = 𝑣 ∙𝛻 𝑣 =𝑢 𝜕 𝑣 𝜕𝑥 +𝑣 𝜕 𝑣 𝜕𝑦 +𝑤 𝜕 𝑣 𝜕𝑧

𝐷 𝑣 𝐷𝑡 = 𝑣 ∙𝛻 𝑣 =𝑢 𝜕 𝑣 𝜕𝑥 +𝑣 𝜕 𝑣 𝜕𝑦 𝑣 = 𝑢,𝑣 = 0.5+0.8𝑥 𝒆 𝒙 +(1.5−0.8𝑦) 𝒆 𝒚 𝑢=0.5+0.8𝑥 𝑣=0.5−0.8𝑦 𝑤=0 𝑑 𝑣 𝑑𝑥 =0.8 𝒆 𝒙 𝑑 𝑣 𝑑𝑦 =−0.8 𝒆 𝒚 𝐷 𝑣 𝐷𝑡 = 0.5+0.8𝑥 0.8 𝒆 𝒙 +(1.5−0.8𝑦)(−0.8) 𝒆 𝒚

𝐷 𝑣 𝐷𝑡 = 0.4+0.64𝑥 𝒆 𝒙 +(−1.2+0.64𝑦) 𝒆 𝒚 So @ x = 2, y = 3: 𝐷 𝑣 𝐷𝑡 = 0.4+0.64𝑥 𝒆 𝒙 +(−1.2+0.64𝑦) 𝒆 𝒚 So @ x = 2, y = 3: 𝑎 =1.68 𝒆 𝒙 +0.72 𝒆 𝒚 A different way of expressing this would be a magnitude and direction: 𝑎=1.82 at 66.8°