Exercises

1. Equations (1.66), (1.75), and (1.87) can be combined to give the following energy conservation equation for a non-ideal compressible fluid:
   $$\rho\,\frac{D{\cal E}}{Dt} - \frac{p}{\rho}\,\frac{D\rho}{Dt} = \chi - \nabla\cdot{\bf q},$$
   where $\rho$ is the mass density, $p$ the pressure, ${\cal E}$ the internal energy per unit mass, $\chi$ the viscous energy dissipation rate per unit volume, and ${\bf q}$ the heat flux density. We also have

   $$\frac{D\rho}{Dt} = -\rho\,\nabla\cdot{\bf v},$$

   $${\bf q} =-\kappa\,\nabla T,$$

   
   
   where ${\bf v}$ is the fluid velocity, $T$ the temperature, and $\kappa$ the thermal conductivity. According to a standard theorem in thermodynamics (Reif 1965),
   $$T\,d{\cal S} = d{\cal E}- \frac{p}{\rho^{\,2}}\,d\rho,$$
   where ${\cal S}$ is the entropy per unit mass. Moreover, the entropy flux density at a given point in the fluid is (Hazeltine and Waelbroeck 2004)
   $${\bf s} = \rho\,{\cal S}\,{\bf v} + \frac{{\bf q}}{T},$$
   where the first term on the right-hand side is due to direct entropy convection by the fluid, and the second is the entropy flux density associated with heat conduction.

   Derive an entropy conservation equation of the form

   $$\frac{dS}{dt} + {\mit\Phi}_S = {\mit\Theta}_S,$$
   where $S$ is the net amount of entropy contained in some fixed volume $V$ , ${\mit\Phi}_S$ the entropy flux out of $V$ , and ${\mit\Theta}_S$ the net rate of entropy creation within $V$ . Give expressions for $S$ , ${\mit\Phi}_S$ , and ${\mit\Theta}_S$ . Demonstrate that the entropy creation rate per unit volume is
   $$\theta = \frac{\chi}{T} + \frac{{\bf q}\cdot{\bf q}}{\kappa\,T^2}.$$
   Finally, show that $\theta\geq 0$ , in accordance with the second law of thermodynamics.

2. The Navier-Stokes equation for an incompressible fluid of uniform mass density $\rho$ takes the form
   $$\frac{D{\bf v}}{Dt}= - \frac{\nabla p}{\rho}-\nabla{\mit\Psi} + \nu\,\nabla^{\,2} {\bf v},$$
   where ${\bf v}$ is the fluid velocity, $p$ the pressure, ${\mit\Psi}$ the potential energy per unit mass, and $\nu$ the (uniform) kinematic viscosity. The incompressibility constraint requires that
   $$\nabla\cdot{\bf v} = 0.$$
   Finally, the quantity
      $\omega$ $$\equiv \nabla\times {\bf v}$$
   is generally referred to as the fluid vorticity.

   Derive the following vorticity evolution equation from the Navier-Stokes equation:

   $$\frac{D\mbox{\boldmath $\omega$}}{Dt} = (\mbox{\boldmath $\omega$}\cdot\nabla)\,{\bf v} + \nu\,\nabla^{\,2}\mbox{\boldmath $\omega$}.$$

3. Consider two-dimensional incompressible fluid flow. Let the velocity field take the form
   $${\bf v} = v_x(x,y,t)\,{\bf e}_x + v_y(x,y,t)\,{\bf e}_y.$$
   Demonstrate that the equations of incompressible fluid flow (see Exercise 2) can be satisfied by writing

   $$v_x =-\frac{\partial \psi}{\partial y},$$

   $$v_y =\frac{\partial\psi}{\partial x},$$

   
   
   where
   $$\frac{\partial\omega}{\partial t}+[\psi,\,\omega] = \nu\,\nabla^{\,2}\omega,$$
   and
   $$\omega = \nabla^{\,2}\psi.$$
   Here, $[A,\,B]\equiv {\bf e}_z\cdot\nabla A\times \nabla B$ , and $\nabla^{\,2}= \partial^{\,2}/\partial x^{\,2}+ \partial^{\,2}/\partial y^{\,2}$ . Furthermore, the quantity $\psi$ is termed a stream function, because ${\bf v}\cdot\nabla\psi = 0$ . In other words, the fluid flow is everywhere parallel to contours of $\psi$ .

4. Consider incompressible irrotational flow: that is, flow that satisfies
   $$\nabla\times {\bf v} = {\bf0},$$
   as well as

   $$\frac{D{\bf v}}{Dt} = -\frac{\nabla p}{\rho} -\nabla{\mit\Psi} +\nu\,\nabla^{\,2}{\bf v},$$

   $$\nabla\cdot{\bf v} =0.$$

   
   
   Here, ${\bf v}$ is the fluid velocity, $\rho$ the uniform mass density, $p$ the pressure, ${\mit\Psi}$ the potential energy per unit mass, and $\nu$ the (uniform) kinematic viscosity.

   Demonstrate that the previous equations can be satisfied by writing

   $${\bf v}= -\nabla\phi,$$
   where
   $$\nabla^{\,2}\phi = 0,$$
   and
   $$-\frac{\partial\phi}{\partial t} + \frac{1}{2}\,v^{\,2}+ \frac{p}{\rho} + {\mit\Psi} ={\cal C}(t).$$
   Here, ${\cal C}(t)$ is a spatial constant. This type of flow is known as potential flow, because the velocity field is derived from a scalar potential.

5. The equations of inviscid adiabatic ideal gas flow are

   $$\frac{D\rho}{Dt} = -\rho\,\nabla\cdot{\bf v},$$

   $$\frac{D{\bf v}}{Dt} = - \frac{\nabla p}{\rho} - \nabla{\mit\Psi},$$

   $$\frac{D}{Dt}\!\left(\frac{p}{\rho^{\,\gamma}}\right) =0.$$

   
   
   Here, $\rho$ is the mass density, ${\bf v}$ the flow velocity, $p$ the pressure, ${\mit\Psi}$ the potential energy per unit mass, and $\gamma$ the (uniform) ratio of specific heats. Suppose that the pressure and potential energy are both time independent: that is, $\partial p/\partial t= \partial{\mit\Psi}/\partial t=0$ .

   Demonstrate that

   $$H = \frac{1}{2}\,v^{\,2} + \frac{\gamma}{\gamma-1}\,\frac{p}{\rho} + {\mit\Psi}$$
   is a constant of the motion. In other words, $DH/Dt = 0$ . This result is known as Bernoulli's theorem.

6. The equations of inviscid adiabatic non-ideal gas flow are

   $$\frac{D\rho}{Dt} = -\rho\,\nabla\cdot{\bf v},$$

   $$\frac{D{\bf v}}{Dt} = - \frac{\nabla p}{\rho} - \nabla{\mit\Psi},$$

   $$\frac{D{\cal E}}{Dt} - \frac{p}{\rho^{\,2}}\,\frac {D\rho}{Dt} =0.$$

   
   
   Here, $\rho$ is the mass density, ${\bf v}$ the flow velocity, $p$ the pressure, ${\mit\Psi}$ the potential energy per unit mass, and ${\cal E}$ the internal energy per unit mass. Suppose that the pressure and potential energy are both time independent: that is, $\partial p/\partial t= \partial{\mit\Psi}/\partial t=0$ . Demonstrate that
   $$H = \frac{1}{2}\,v^{\,2} + {\cal E}+ \frac{p}{\rho} + {\mit\Psi}$$
   is a constant of the motion. In other words, $DH/Dt = 0$ . This result is a more general form of Bernoulli's theorem.

7. Demonstrate that Bernoulli's theorem for incompressible, inviscid fluid flow takes the form $DH/Dt = 0$ , where
   $$H = \frac{1}{2}\,v^{\,2} + \frac{p}{\rho} + {\mit\Psi}.$$