Forces on a Fluid Package

Linear Deformation

• If $u_{in}=u_{out}$ volume does not change
• Otherwise, volume will increase or decrease according to the rate of $u$ in and out
  

If we want to change volume we need to change our velocity

Changes in Volume

For a volume after some change in time…
$V\left(t_o+\Delta t\right)=\left(L_x+\Delta x\right)\left(L_y+\Delta y\right)\left(L_z+\Delta z\right)$

$$\begin{gathered} \Delta V=V(t_o+\Delta t)-V(t_o) \\ =L_y{L}_z\Delta x+L_x{L}_z\Delta y+L_x{L}_y\Delta z+\text{higher order terms} \end{gathered}$$

Looking Specifically at a Fluid Parcel

$\frac{\partial }{\partial x}\cong \frac{\Delta u}{Lx}$ where you use $\Delta u=\frac{\Delta x}{\Delta t}$

• $\Delta x=\frac{\partial u}{\partial x}{L}_x\Delta t$
• $\Delta y=\frac{\partial v}{\partial y}{L}_y\Delta t$

Estimation of Volumetric Strain Rate
$\frac{\Delta V}{\Delta t}=\left(\frac{du}{dx}+\frac{dv}{dy}+\frac{dw}{dz}\right)L_x{L}_y{L}_z$

Where $L_x{L}_y{L}_z=\forall$

If the volume of fluid does not chance, the flow is incompressible which means $\left(\frac{du}{dx}+\frac{dv}{dy}+\frac{dw}{dz}\right)=0$ just like in Thermodynamics (see Specific Heat and Entropy)

This is the same as saying that the velocity field is divergence free or $\nabla V=0$

Boundary Layer

• A thin layer of fluid where the presence of a solid wall leads to high shear stress

Mixing of Momentum and Energy Loss

2 parcels of fluids, both of mass m
The kinetic energy component is here

Shear Strain

Assuming a small angle… $\Delta \alpha \cong \frac{\Delta y}{L_x},\tan \theta \cong \theta$

$\Delta y=\Delta v\Delta t=\left(\frac{\partial V}{\partial x}{L}_x\right)\Delta t$ Since…
$\Delta \alpha =\frac{\Delta y}{L_x}=\frac{\partial v}{\partial x}\Delta t$
Then…
$\frac{d\alpha }{dt}\cong \frac{\Delta \alpha }{\Delta t}=\frac{\partial V}{\partial x}$

Summary of Shear Strain Rates

• ${ϵ}_{xy}^{\prime }=\frac{\partial V}{\partial x}+\frac{\partial U}{\partial y}$
• ${ϵ}_{yz}^{\prime }=\frac{\partial W}{\partial x}+\frac{\partial U}{\partial z}$
• ${ϵ}_{xz}^{\prime }=\frac{\partial W}{\partial x}+\frac{\partial U}{\partial z}$

Shear Stress Relationship (Couette Flow Questions)

You can recognize these because they are “linear velocity profile” or velocity questions…
Shear stress is proportional to strain rate and for a newtonian fluid the relationship is linear.

If $u=f\left(x\right)$ only then…
${\tau }_{xy}=\mu \left|\frac{du}{dx}\right|$
Which usually is just $\frac{du}{dx}=\frac{V}{h}$ since the velocity profile is linear…
This force is produces… $\left|F\right|=\tau A$

No Slip Condition

A fluid parcel is contact with a boundary will have the same velocity as the boundary

• Linear velocity profile
• Couette flow

Viscometers

Unit for Viscosity = $Pa\ast s$

Viscometer Questions

You usually need to solve for some expression, just see my Goodnotes

Newtons Law on a Fluid Parcel

$\frac{dV}{dt}=F_{pressure}+f_g+f_{friction}$