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Mathematics of fluid dynamics University of Bath 25-26, MA32051/MA52129

 

Section 7.6 Viscous Dissipation of Energy

Non-examinable.
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In this section we derive an expression for the rate of dissipation of kinetic energy in terms of the rate-of-strain tensor.
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We consider a volume \(V(t)\) of fluid moving with the fluid. The total kinetic energy in \(V\) is
\begin{equation*} T=\frac12\int_V\rho\left|\bu\right|^2\de V =\sum_{i=1}^3\frac12\int_V\rho u_i^2\de V. \end{equation*}
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Proof.

We need derive an expression for the rate of change:
\begin{equation*} \dd{T}{t}=\sum_{i=1}^3\int_V\rho u_i\DD{u_i}{t}\de V \end{equation*}
Using the Cauchy equation (7.5.2):
\begin{align*} \dd{T}{t}=&\sum_{i=1}^3\int_Vu_i\left(\sum_{j=1}^3\pd{\sigma_{ij}}{x_j}+\rho g_i\right)\de V\\ =&\sum_{i=1}^3\int_V\left(\sum_{j=1}^3\pd{}{x_j}\left(u_i\sigma_{ij}\right)-\sum_{j=1}^3\sigma_{ij}\pd{u_i}{x_j}+\rho g_iu_i\right)\de V\\ =&\sum_{i=1}^3\left(\int_S\sum_{j=1}^3u_i\sigma_{ij}n_j\de S+\int_V\left(-\sum_{j=1}^3\sigma_{ij}\pd{u_i}{x_j}+\rho g_iu_i\right)\de V\right)\\ =&\sum_{i=1}^3\left(\int_Su_i\tau_i\de S+\int_V\left(-\sum_{j=1}^3\sigma_{ij}\pd{u_i}{x_j}+\rho g_iu_i\right)\de V\right),\\ =&\int_S\btau\cdot\bu\,\de S-\int_V\sum_{i=1}^3\sum_{j=1}^3\sigma_{ij}\pd{u_i}{x_j}\,\de V+\int_V\rho\bg\cdot\bu\,\de V, \end{align*}
where, in the above calculation, we used the divergence theorem to obtain the penultimate line and the definition of stress to obtain the fourth line. We need to simplify
\begin{align*} \sum_{i=1}^3\sum_{j=1}^3\sigma_{ij}\pd{u_i}{x_j} =&\frac12\sum_{i=1}^3\sum_{j=1}^3\left(\sigma_{ij}+\sigma_{ji}\right)\pd{u_i}{x_j}\\ =&\frac12\sum_{i=1}^3\sum_{j=1}^3\sigma_{ij}\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)\\ =&\frac12\sum_{i=1}^3\sum_{j=1}^3\left(-p\delta_{ij}+\mu\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)\right)\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)\\ =&-\frac{p}2\sum_{i=1}^3\pd{u_i}{x_i}+\frac{\mu}2\sum_{i=1}^3\sum_{j=1}^3\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)\\ =&2\mu\sum_{i=1}^3\sum_{j=1}^3e_{ij}^2 \end{align*}
where in the first line of the above we used the fact that \(\sigma_{ij}\) is symmetric, in the second line we swapped the \(i\) and \(j\) indices on the second term, in the third line we used the constitutive relationship for a Newtonian fluid (7.4.1), and in the final line we used the fact that the divergence of \(\bu\) is zero due to mass conservaion.
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Hence,
\begin{equation*} \dd{T}{t}=\int_S\btau\cdot\bu\,\de S-\Phi+\int_V\rho\bg\cdot\bu\,\de V, \end{equation*}
and the result follows.
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Remark 7.6.2. Interpretation of the terms in the viscous dissipation.

The terms on the right-hand side of (7.6.1) have physical interpretations as follows:
  • The first term \(\int_V\rho\bu\cdot\bg\,\de V\) is the decrease in gravitational potential energy associated with the flow. This energy is converted into kinetic energy.
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  • The second term \(\int_S\btau\cdot\bu\,\de S\) equals the rate at which the surrounding fluid is doing work on the fluid in \(V\) via the surface stresses \(\btau\text{.}\)
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  • Throughout the volume of fluid, the viscous stresses \(\Phi=2\mu\sum_{i=1}^3\sum_{j=1}^3e_{ij}^2\) act to reduce the energy, so this term is integrated over the volume.
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Remark 7.6.3. Viscous stresses.

Recall from Exercise 7.9.9 that any incompressible flow may locally be written as a sum of a translation, a rotation and a straining component. The viscous stresses \(2\mu\sum_{i=1}^3\sum_{j=1}^3e_{ij}^2\) are zero for the translational and rotational components of the flow and non-zero for the straining component.
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Since the rate-of-strain tensor is symmetric, and writing
\begin{equation*} \be=PDP^{-1} \end{equation*}
where \(D\) is symmetric, we have
\begin{align*} 2\mu\sum_{i=1}^3\sum_{j=1}^3e_{ij}^2 =&2\mu\,\mathrm{tr}\left(\be^2\right)\\ =&2\mu\,\mathrm{tr}\left(PDP^{-1}PDP^{-1}\right)\\ =&2\mu\,\mathrm{tr}\left(PD^2P^{-1}\right)\\ =&2\mu\,\mathrm{tr}\left(D^2\right)\\ =&2\mu\,\left(\lambda_1^2+\lambda_2^2+\lambda_3^2\right), \end{align*}
where \(\lambda_i\text{,}\) \(i=1,2,3\text{,}\) are the eigenvalues of \(\be\text{.}\)
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Remark 7.6.4. The energy equation.

We can also use the Reynolds Transport Theorem to perform conservation of energy in the control volume instead of mass. The energy is the sum of the internal energy, kinetic energy and the gravitational potential energy
\begin{equation*} e=\hat{u}+\frac12\left|{\bu}\right|^2-{\bg}\cdot{\bx}, \end{equation*}
where \(\hat{u}\) is the internal energy per unit mass (which is often expressed as \(d\hat{u}\approx c_v\,dT\)) and \({\bg}\) is the acceleration due to gravity.
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In the special case that the fluid is incompressible and Newtonian and that the specific heat \(c_v\) is constant and the fluid is homogeneous and isotropic, and, upon letting the volume of the control volume tend to zero, we obtain
\begin{equation*} \rho c_v\left(\pd{T}{t}+{\bu}\cdot\bnabla T\right)=k\nabla^2 T+\Phi+\dot{S}, \end{equation*}
where
  • \(c_v\) is the specific heat at constant volume, which is the rate of change of internal energy of the fluid with respect to temperature,
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  • \(T\) is the temperature of the fluid,
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  • \(k\) is the coefficient of thermal conductivity, which is the heat flux per unit area per unit temperature gradient (by Fourier’s law, which is \({\bq}=-k\bnabla T\text{,}\) where \({\bq}\) is the flux of heat per unit area – cf Fick’s law), and
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  • \(\Phi\) is the rate of heating due to viscous stresses.
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  • \(\dot{S}\) is the energy production per unit volume per unit time.
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This is the diffusion equation with an extra term, \(\Phi\text{,}\) representing the heating due to viscous stresses in the fluid. It is the energy loss due to the viscous forces per unit volume of fluid per unit time. Often this term is neglected to give the usual advection--diffusion equation. In a non-Newtonian fluid the equation is significantly more complicated.
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