4.1.14. Conservation of Energy¶

4.1.14.1. What is the conservation of total energy in integral form?¶

• Rate of change of total energy:

${\partial \over {\partial t}} \int_V \rho E dV$
• Flux of total energy:

$\int_S \rho E (\vec{u} \cdot \vec{n}) dS$
• Work done by fluid by surface forces (work is done in the positive direction of $$\vec{n}$$):

$\int_S \overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{u} \cdot \vec{n} dS$
• Heat added to fluid by surface forces (heat is added in the negative direction of $$\vec{n}$$):

$-\int_S \vec{q} \cdot \vec{n} dS$
• Conservation of energy (neglecting gravity):

${\partial \over {\partial t}} \int_V \rho E dV + \int_S \rho E (\vec{u} \cdot \vec{n}) dS = - \int_S \vec{q} \cdot \vec{n} dS + \int_S \overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{u} \cdot \vec{n} dS$

4.1.14.2. What is the conservation of total energy in differential form?¶

• Gauss divergence theorem:

${\partial \over {\partial t}} \int_V \rho E dV + \int_V \nabla \cdot (\rho E \vec{u}) dV = - \int_V \nabla \cdot \vec{q} dV + \int_V \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) dV$
• Volume fixed and arbitary gives:

${\partial \over {\partial t}} (\rho E) + \nabla \cdot (\rho E \vec{u}) = - \nabla \cdot \vec{q} + \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n})$
• Hence:

${D \over {D t}} (\rho E) = - \nabla \cdot \vec{q} + \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n})$

4.1.14.3. What is the kinetic energy equation in integral form?¶

• Rate of change of kinetic energy:

${\partial \over {\partial t}} \int_V \rho E_k dV$
• Flux of kinetic energy:

$\int_S \rho E_k (\vec{u} \cdot \vec{n}) dS$
• Work done by fluid by surface forces (work is done in the positive direction of $$\vec{n}$$):

$\int_S \overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{u} \cdot \vec{n} dS$
• Work done by fluid by volumetric forces (work is done in the positive direction of $$\vec{n}$$):

$\int_V p \nabla \cdot \vec{u} dV$
• Work done on fluid by friction (work is done in the negative direction of $$\vec{n}$$):

$- \int_V \Phi dV$
• Kinetic energy equation:

${\partial \over {\partial t}} \int_V \rho E_k dV + \int_S \rho E_k (\vec{u} \cdot \vec{n}) dS = \int_S \overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{u} \cdot \vec{n} dS + \int_V p \nabla \cdot \vec{u} dV - \int_V \Phi dV$

4.1.14.4. What is the kinetic energy equation in differential form?¶

• Gauss divergence theorem:

${\partial \over {\partial t}} \int_V \rho E_k dV + \int_V \nabla \cdot (\rho E_k \vec{u}) dV = \int_V \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) dV + \int_V p \nabla \cdot \vec{u} dV - \int_V \Phi dV$
• Volume fixed and arbitary gives:

${\partial \over {\partial t}} (\rho E_k) + \nabla \cdot (\rho E_k \vec{u}) = \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) + p \nabla \cdot \vec{u} - \Phi$
• Hence:

${D \over {D t}} (\rho E_k) = \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) + p \nabla \cdot \vec{u} - \Phi$

4.1.14.5. How is the internal energy equation derived from the total energy equation and the kinetic energy equation?¶

• By definition:

${{D} \over {Dt}} (\rho e) = {{D} \over {Dt}} (\rho E) - {{D } \over {Dt}} (\rho E_k)$
• By substitution:

${{\partial } \over {\partial t}} (\rho e) + \nabla \cdot (\rho e \vec{u}) = - \nabla \cdot \vec{q} + \nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) - (\nabla \cdot (\overset{\underset{\mathrm{\rightrightarrows}}{}}{\sigma} \cdot \vec{n}) + p \nabla \cdot \vec{u} - \Phi)$
• By cancellation of shear stress:

${{\partial } \over {\partial t}} (\rho e) + \nabla \cdot (\rho e \vec{u}) = - \nabla \cdot \vec{q} + - p \nabla \cdot \vec{u} + \Phi$
• where: $$\Phi$$ = viscous dissipation function

4.1.14.6. What is the viscous dissipation function?¶

• This is the viscous work put into fluid element deformation and is irreversible.

• It represents the irreversible conversion of mechanical energy to thermal energy through the action of viscosity.

$\Phi = 2 \mu \left( e_{ij} - {1 \over 3} \delta_{ij} {{\partial u_j} \over {\partial x_j}} \right)^2$
• By expansion:

$\Phi = 2 \mu \left[ \left( {{\partial u_1} \over {\partial x_1}} \right)^2 + \left( {{\partial u_2} \over {\partial x_2}} \right)^2 + \left( {{\partial u_3} \over {\partial x_3}} \right)^2 + {1 \over 2} \left( {{\partial u_2} \over {\partial x_1}} + {{\partial u_1} \over {\partial x_2}} \right)^2 + {1 \over 2} \left( {{\partial u_3} \over {\partial x_2}} + {{\partial u_2} \over {\partial x_3}} \right)^2 + {1 \over 2} \left( {{\partial u_1} \over {\partial x_3}} + {{\partial u_3} \over {\partial x_1}} \right)^2 \right]$
• where:

$$\Phi$$ = viscous dissipation function = $${1 \over 2} \mu \gamma^2$$

$$e_{ij}$$ = strain rate tensor = $${1 \over 2} \left( {{\partial u_j} \over {\partial x_i}} + {{\partial u_i} \over {\partial x_j}} \right)$$

$${1 \over 3} \delta_{ij} {{\partial u_j} \over {\partial x_j}}$$ = compressibility term

4.1.14.7. What is Fourier’s Law?¶

$\vec{q} = -k \nabla T$

where:

$$q$$ = heat flux per unit area $$(W/m^2)$$

$$k$$ = thermal conductivity $$(W/mK)$$

$$\nabla T$$ = temperature gradient $$K/m$$

• Used in the Navier-Stokes equations to reduce the vector $$vec{q}$$ of three unknowns to one unknown, temperature $$T$$

• This applies to fluids and solids

4.1.14.8. What is thermal conductivity?¶

• Thermal conductivity is the property of a material to conduct heat

• For water it’s a fifth order polynomial

4.1.14.9. How is the heat transport equation derived from the internal energy equation for an inviscid incompressible flow?¶

• From the internal energy equation:

${{\partial } \over {\partial t}} (\rho e) + \nabla \cdot (\rho e \vec{u}) = - \nabla \cdot \vec{q} + - p \nabla \cdot \vec{u} + \Phi$

where:

$e = C_v T$

and

$\vec{q} = -k \nabla T$
• For an incompressible, inviscid flow: $$\nabla \vec{u} = 0$$, $$\Phi = 0$$

${D \over {Dt}} (T) = {k \over {\rho C_v}} \nabla^2 T = \alpha \nabla^2 T$
• Thermal diffusivity $$\alpha$$ plays the role of viscosity for temperature:

$\alpha = {k \over {\rho C_v}}$

or:

$\alpha = {k \over {\rho C_p}}$

(as $$C_v = C_p$$ for incompressible flow)

4.1.14.10. How is the entropy production equation derived from the internal energy equation for incompressible flow?¶

• From the internal energy equation:

${{\partial } \over {\partial t}} (\rho e) + \nabla \cdot (\rho e \vec{u}) = - \nabla \cdot \vec{q} + - p \nabla \cdot \vec{u} + \Phi$
• For a reversible process:

$T ds = de + pdv$
• For incompressible flow:

$dv = 0$
$\rho = \text{constant}$
$\nabla \cdot \vec{u} = 0$
• Hence:

$\rho {{De} \over {Dt}} = \rho T {{Ds} \over {Dt}} = -\nabla \cdot \vec{q} + \Phi$
• Hence:

$\rho {{Ds} \over {Dt}} = - {{\nabla \cdot \vec{q}} \over T} + {\Phi \over T}$

4.1.14.11. What is the cause of entropy production?¶

• Heat conduction (heat added to the fluid)

• Viscous dissipation (work done by the fluid)