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)