Energy Balance and Viscous Dissipation

Derivation

Energy Balance and Viscous Dissipation

D-327 Home PU-307 Threads energy · matter Depends on Navier-Stokes from the Newtonian Constitutive Law, first-law-of-thermodynamics

Statement

Contracting the Cauchy momentum equation with the velocity field yields the mechanical-energy equation for the specific kinetic energy \(k=\tfrac{1}{2}\,u_i u_i\). In it the stress-power term splits into a reversible pressure–dilatation part and an irreversible term equal to the viscous dissipation function \(\Phi=\tau_{ij}\,\partial u_i/\partial x_j\). For a Newtonian fluid \(\Phi=2\mu\,S_{ij}^{*}S_{ij}^{*}+\zeta(\nabla\!\cdot\!\mathbf{u})^{2}\ge 0\) is positive-definite, and appears in the thermal-energy equation with the opposite sign — it is the rate per unit volume at which mechanical energy is irreversibly converted to heat.

Why it matters

The mechanical-energy equation is not an independent conservation law: it is a scalar consequence of momentum balance. Its value is that it isolates exactly where kinetic energy goes — into transport, into reversible pressure work, and into an irreversible sink \(\Phi\). That sink is the microscopic origin of drag heating, of the temperature rise in lubricant films and re-entry boundary layers, and of the pumping power a pipe network dissipates.

Because \(\Phi\ge 0\) identically for a Newtonian fluid with non-negative viscosities, it is the fluid-mechanical face of the second law: mechanical energy degrades to internal energy but never spontaneously reverses. Splitting the stress power into \(-p\,\nabla\!\cdot\!\mathbf{u}\) (reversible) and \(\Phi\) (irreversible) is precisely what lets one write an entropy-production rate for a viscous flow.

Assumptions

Continuum, single-phase fluid obeying Cauchy momentum balance.Without a well-defined stress tensor field the starting point \(\rho\,Du_i/Dt=\rho g_i+\partial\sigma_{ij}/\partial x_j\) does not exist and no energy identity can be extracted.

Newtonian constitutive law \(\tau_{ij}=2\mu S_{ij}+\lambda(\nabla\!\cdot\!\mathbf{u})\delta_{ij}\) with \(\mu\ge 0\), bulk viscosity \(\zeta=\lambda+\tfrac{2}{3}\mu\ge 0\).If the fluid is non-Newtonian the algebraic form of \(\Phi\) changes and its positive-definiteness is no longer guaranteed by two scalar coefficients.

Symmetric stress tensor (no internal couple stresses / body torques).If \(\sigma_{ij}\ne\sigma_{ji}\) then \(\tau_{ij}\partial_j u_i\) keeps its antisymmetric part, the identification \(\Phi=\tau_{ij}S_{ij}\) fails, and angular-momentum balance must be carried separately.

Constant, isotropic viscosities over the material element considered.Strong \(\mu(T)\) variation couples \(\Phi\) back into the temperature field; the dissipation form still holds pointwise but \(\mu\) may no longer be pulled outside gradients.

Derivation

\[ \rho\,\frac{Du_i}{Dt}=\rho\,g_i+\frac{\partial \sigma_{ij}}{\partial x_j} \]

Cauchy momentum equation; \(D/Dt=\partial_t+u_k\partial_k\) is the material derivative and \(\sigma_{ij}\) the total stress. A

\[ \rho\,u_i\frac{Du_i}{Dt}=\rho\,u_i g_i+u_i\frac{\partial \sigma_{ij}}{\partial x_j} \]

Contract with the velocity \(u_i\) (form the scalar dot product). A scalar equation cannot add information to the vector one — it re-expresses it. A

\[ u_i\frac{Du_i}{Dt}=\frac{D}{Dt}\!\left(\tfrac{1}{2}u_i u_i\right)=\frac{Dk}{Dt} \]

Product rule for the material derivative, \(u_i\,Du_i/Dt=\tfrac12 D(u_iu_i)/Dt\), defines the specific kinetic energy \(k\). A

\[ u_i\frac{\partial \sigma_{ij}}{\partial x_j}=\frac{\partial (u_i\sigma_{ij})}{\partial x_j}-\sigma_{ij}\frac{\partial u_i}{\partial x_j} \]

Product rule on the stress term splits it into a divergence (energy transport by surface stresses) and a local stress-power density. B

\[ \rho\,\frac{Dk}{Dt}=\rho\,u_i g_i+\frac{\partial (u_i\sigma_{ij})}{\partial x_j}-\sigma_{ij}\frac{\partial u_i}{\partial x_j} \]

Assemble steps 2–4. This is the mechanical-energy equation in raw form; the last term is where the physics lives. B

\[ \sigma_{ij}\frac{\partial u_i}{\partial x_j}=\bigl(-p\,\delta_{ij}+\tau_{ij}\bigr)\frac{\partial u_i}{\partial x_j}=-p\,\frac{\partial u_i}{\partial x_i}+\tau_{ij}\frac{\partial u_i}{\partial x_j} \]

Substitute the stress decomposition \(\sigma_{ij}=-p\,\delta_{ij}+\tau_{ij}\); the isotropic part gives \(-p(\nabla\!\cdot\!\mathbf{u})\), reversible compression work. B

\[ \tau_{ij}\frac{\partial u_i}{\partial x_j}=\tau_{ij}\,S_{ij}\equiv\Phi,\qquad S_{ij}=\tfrac12\!\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) \]

Because \(\tau_{ij}=\tau_{ji}\) is symmetric, its contraction with \(\partial_j u_i\) annihilates the antisymmetric (rotation) part, leaving only the strain rate \(S_{ij}\). Rigid rotation does no work against viscous stress. B

\[ \Phi=\bigl(2\mu S_{ij}+\lambda(\nabla\!\cdot\!\mathbf{u})\delta_{ij}\bigr)S_{ij}=2\mu\,S_{ij}S_{ij}+\lambda\,(\nabla\!\cdot\!\mathbf{u})^{2} \]

Insert the Newtonian constitutive law and use \(\delta_{ij}S_{ij}=S_{kk}=\nabla\!\cdot\!\mathbf{u}\). B

\[ S_{ij}=S_{ij}^{*}+\tfrac13(\nabla\!\cdot\!\mathbf{u})\delta_{ij},\quad S_{ij}^{*}S_{ij}^{*}\ge 0 \;\Rightarrow\; \Phi=2\mu\,S_{ij}^{*}S_{ij}^{*}+\zeta\,(\nabla\!\cdot\!\mathbf{u})^{2} \]

Split the strain rate into its trace-free (deviatoric) part \(S_{ij}^{*}\) and its isotropic part; regrouping with \(\zeta=\lambda+\tfrac23\mu\) gives a manifest sum of squares. With \(\mu\ge 0,\ \zeta\ge 0\) this proves \(\Phi\ge 0\). C

\[ \underbrace{\rho\frac{De}{Dt}}_{\text{thermal}}=\underbrace{\bigl(\rho\tfrac{DE}{Dt}\bigr)}_{\text{first law: total}}-\underbrace{\rho\frac{Dk}{Dt}}_{\text{mechanical}}=-p\,(\nabla\!\cdot\!\mathbf{u})+\Phi-\nabla\!\cdot\!\mathbf{q} \]

Subtract the mechanical-energy equation from the total-energy (first-law) balance, \(E=e+k\). The transport divergence and gravity cancel; \(\Phi\) reappears with a plus sign — mechanical energy lost is internal energy gained. Since \(\Phi\ge0\), this term only ever heats. C

Result

\[ \rho\,\frac{Dk}{Dt}=\rho\,\mathbf{u}\!\cdot\!\mathbf{g}+\nabla\!\cdot\!(\boldsymbol{\sigma}\!\cdot\!\mathbf{u})+p\,(\nabla\!\cdot\!\mathbf{u})-\Phi,\qquad \Phi=2\mu\,S_{ij}^{*}S_{ij}^{*}+\zeta\,(\nabla\!\cdot\!\mathbf{u})^{2}\ge 0 \]

Reading. The kinetic energy of a material element changes by the power of gravity, the net working of surface stresses (a divergence, so a pure transport that integrates to a boundary flux), a reversible exchange with internal energy through \(p(\nabla\!\cdot\!\mathbf{u})\), and a strictly non-negative loss \(\Phi\). Only \(\Phi\) is one-way: it is the local rate per unit volume at which ordered kinetic energy is degraded into disordered thermal motion. For incompressible flow \(\nabla\!\cdot\!\mathbf{u}=0\) and \(\Phi=2\mu\,S_{ij}S_{ij}\).

Units check. \([\mu]=\mathrm{Pa\cdot s}=\mathrm{kg\,m^{-1}s^{-1}}\) and \([S_{ij}]=\mathrm{s^{-1}}\), so \([\Phi]=\mathrm{kg\,m^{-1}s^{-1}}\cdot\mathrm{s^{-2}}=\mathrm{kg\,m^{-1}s^{-3}}=\mathrm{W\,m^{-3}}\) — a power per unit volume, matching \([\rho\,Dk/Dt]=\mathrm{kg\,m^{-3}}\cdot\mathrm{m^{2}s^{-2}}\cdot\mathrm{s^{-1}}=\mathrm{W\,m^{-3}}\).

Limiting cases

Incompressible flow (\(\nabla\!\cdot\!\mathbf{u}=0\)): pressure–dilatation and bulk terms vanish, \(\Phi=2\mu S_{ij}S_{ij}\) — dissipation is purely deviatoric (shear).
Inviscid limit (\(\mu,\zeta\to 0\)): \(\Phi\to 0\) and mechanical energy is conserved along particle paths save for reversible pressure work — recovers the Bernoulli/Euler energy budget.
Rigid rotation (\(\mathbf{u}=\boldsymbol{\Omega}\times\mathbf{r}\)): \(S_{ij}=0\) so \(\Phi=0\) — a fluid in solid-body motion dissipates nothing, as it must.
Pure dilatation (\(\mathbf{u}=\alpha\mathbf{r}\), \(S_{ij}^{*}=0\)): \(\Phi=\zeta(3\alpha)^{2}\) — only bulk viscosity acts; vanishes under the Stokes hypothesis \(\zeta=0\).
Simple shear \(u_x=\dot\gamma\,y\): \(\Phi=\mu\dot\gamma^{2}\) — the canonical rheometric heating rate.

Breaks when

Non-Newtonian / viscoelastic fluids. The linear stress–strain-rate law fails; stress depends on strain history, elastic energy can be stored and returned, and \(\tau_{ij}S_{ij}\) is no longer a guaranteed-positive sum of squares. Part of the "dissipation" may be recoverable elastic storage.
Rarefied gas / high Knudsen number. When the mean free path is comparable to the flow scale the continuum stress tensor breaks down; the Navier–Stokes closure and hence this \(\Phi\) lose meaning, and one must go to kinetic (Boltzmann) theory.
Negative or ill-defined viscosities. Turbulence models employing an effective eddy viscosity can produce local backscatter (\(\mu_{\text{eff}}<0\)), so the modelled \(\Phi\) is no longer sign-definite even though the true molecular dissipation is.
Non-symmetric (polar / micropolar) stress. With internal spin and couple stresses the antisymmetric part of \(\partial_j u_i\) contributes to the stress power, and \(\Phi=\tau_{ij}S_{ij}\) is incomplete.

Failure modes

Contracting \(\tau_{ij}\) with the full velocity gradient and keeping the rotation part. Writing \(\Phi=\tau_{ij}\partial_j u_i\) is correct, but students then forget the symmetry argument (step 7) and wrongly believe vorticity dissipates energy. It does not — only strain does.
Calling \(-p(\nabla\!\cdot\!\mathbf{u})\) a dissipation. Pressure–dilatation is reversible; it shuttles energy between kinetic and internal reversibly. Only \(\Phi\) is irreversible. Lumping them together violates the second-law bookkeeping.
Using \(\Phi=\mu(\partial u_i/\partial x_j)^2\) instead of the symmetric-strain form. This over-counts and gives the wrong coefficient; the correct incompressible form is \(2\mu S_{ij}S_{ij}=\mu(\partial_j u_i+\partial_i u_j)\partial_j u_i/\,\)… only after symmetrisation.
Dropping the bulk-viscosity term for compressible flow. Setting \(\zeta=0\) (Stokes hypothesis) is a modelling choice, not an identity; for acoustic absorption and shock structure it is quantitatively wrong.
Double counting \(\Phi\) in a total-energy budget. \(\Phi\) is internal transfer: it is a sink in the mechanical equation and an equal source in the thermal equation. It must cancel in the total-energy equation, appearing there not at all.

Discussion

The deepest point is that the mechanical-energy equation carries no new physics: it is the momentum equation dotted into \(\mathbf{u}\). What it buys is a clean partition of the stress power \(\sigma_{ij}\partial_j u_i\) into a reversible piece \(-p(\nabla\!\cdot\!\mathbf{u})\) and an irreversible piece \(\Phi\). That partition is exactly the split needed to write the first and second laws for a moving continuum. Subtracting the mechanical from the total (first-law) balance leaves the thermal-energy equation, where \(\Phi\) reappears as a heat source of definite sign.

Positive-definiteness is not an accident of algebra; it is a thermodynamic constraint. The requirement that entropy production be non-negative for every admissible flow forces \(\mu\ge 0\) and \(\zeta\ge 0\), which is precisely what makes \(\Phi\) a sum of squares. Dividing by temperature, \(\Phi/T\) is the viscous contribution to the local entropy-production rate, and it vanishes only where the strain rate vanishes — i.e. only for locally rigid motion.

Physically, \(\Phi\) is the macroscopic shadow of microscopic irreversibility: velocity gradients drive momentum diffusion between fluid layers, and the associated work against intermolecular friction thermalises into random molecular motion. This is why viscous heating is always a warming, never a cooling, and why it scales as the square of the strain rate — the same quadratic that makes drag heating explode in high-speed boundary layers and lubricant films.

The scaling \(\Phi\sim\mu U^{2}/L^{2}\) versus a convective enthalpy flux \(\rho c_p U\,\Delta T/L\) defines the Brinkman number \(\mathrm{Br}=\mu U^{2}/(k\,\Delta T)\) (equivalently \(\mathrm{Br}=\mathrm{Pr}\cdot\mathrm{Ec}\) via the Eckert number). When \(\mathrm{Br}\gtrsim 1\) self-heating feeds back on \(\mu(T)\) and the isothermal Navier–Stokes closure is no longer self-consistent — the regime of polymer-melt extrusion, journal bearings, and hypersonic aeroheating, where the dissipation term ceases to be a passive rider on the flow and becomes a governing effect.

Common misconceptions. "Vorticity dissipates energy" — false; only strain does, and a purely rotating fluid dissipates nothing. "Pressure work is a loss" — false; it is reversible. "Dissipation appears in the total-energy equation" — false; it is an internal transfer that cancels there and shows up only in the mechanical and thermal equations, with opposite signs.

Worked examples

Example 1 — Plane Couette flow: local dissipation and its energy balance.

\[ u_x(y)=U\,\frac{y}{h},\quad u_y=u_z=0 \]

Incompressible flow between plates gap \(h\), top plate speed \(U\). Only non-zero gradient is \(\partial u_x/\partial y=U/h\). A

\[ S_{xy}=S_{yx}=\tfrac12\frac{\partial u_x}{\partial y}=\frac{U}{2h},\qquad \Phi=2\mu\,S_{ij}S_{ij}=2\mu\,(S_{xy}^{2}+S_{yx}^{2})=\mu\left(\frac{U}{h}\right)^{2} \]

Incompressible dissipation form with the only two equal off-diagonal strain components. A

\[ \mu=1.0~\mathrm{Pa\,s},\ U=2~\mathrm{m/s},\ h=1\times10^{-3}~\mathrm{m}\;\Rightarrow\;\Phi=1.0\left(\frac{2}{10^{-3}}\right)^{2}=4\times10^{6}~\mathrm{W/m^{3}} \]

Insert numbers for an oil film. A

\[ \text{per unit plate area: } \Phi\,h=4\times10^{6}\times10^{-3}=4000~\mathrm{W/m^{2}};\quad \tau_w U=\mu\frac{U}{h}\,U=(2000~\mathrm{Pa})(2~\mathrm{m/s})=4000~\mathrm{W/m^{2}} \]

Cross-check: integrated dissipation equals the mechanical power the plate feeds in through wall shear \(\tau_w=\mu U/h\). They match — the mechanical-energy balance closes. B

\[ \boxed{\ \Phi=4\times10^{6}~\mathrm{W/m^{3}},\qquad \int\Phi\,dV/A=\tau_w U=4000~\mathrm{W/m^{2}}\ } \]

Reading. Every cubic metre of the film dissipates 4 MW; the plate does exactly that much work, all of it thermalised. In steady state with adiabatic walls this heat leaves by conduction, setting the film temperature.

Units check. \(\mathrm{Pa\,s}\cdot(\mathrm{m/s}/\mathrm{m})^{2}=\mathrm{Pa\,s\cdot s^{-2}}=\mathrm{Pa/s}=\mathrm{W/m^{3}}\). ✓

Example 2 — Laminar pipe flow: total dissipation, pumping power, and temperature rise.

\[ \dot W_{\text{diss}}=\int_V \Phi\,dV = Q\,\Delta p = 8\pi\mu L\,\bar U^{2} \]

For Hagen–Poiseuille flow the total dissipation equals the pumping power \(Q\Delta p\) with \(\Delta p=8\mu L\bar U/R^{2}\) and \(Q=\pi R^{2}\bar U\); substitute to eliminate \(R\). B

\[ \mu=0.1~\mathrm{Pa\,s},\ \rho=900~\mathrm{kg/m^3},\ R=0.01~\mathrm{m},\ L=10~\mathrm{m},\ \bar U=1~\mathrm{m/s} \]

Oil; check laminarity: \(\mathrm{Re}=\rho\bar U(2R)/\mu=900(1)(0.02)/0.1=180\ll 2300\). ✓ A

\[ \dot W_{\text{diss}}=8\pi(0.1)(10)(1)^2=8\pi\approx 25.1~\mathrm{W} \]

Evaluate. Cross-check: \(\Delta p=8(0.1)(10)(1)/(10^{-4})=8\times10^{4}~\mathrm{Pa}\), \(Q=\pi(10^{-4})(1)=3.14\times10^{-4}~\mathrm{m^3/s}\), \(Q\Delta p=25.1~\mathrm{W}\). ✓ B

\[ \dot m=\rho Q=900(3.14\times10^{-4})=0.283~\mathrm{kg/s},\quad \Delta T=\frac{\dot W_{\text{diss}}}{\dot m\,c}=\frac{25.1}{0.283\times 2000} \]

Adiabatic steady flow: all dissipation raises the bulk temperature; take \(c=2000~\mathrm{J\,kg^{-1}K^{-1}}\). B

\[ \boxed{\ \dot W_{\text{diss}}=Q\Delta p\approx 25.1~\mathrm{W},\qquad \Delta T\approx 0.044~\mathrm{K}\ } \]

Reading. The 25 W of pumping power is not "lost" — the mechanical-energy equation shows it becomes internal energy via \(\Phi\), warming the oil by \(0.044~\mathrm K\) over the pipe length. Viscous heating is genuine but usually small at ordinary scales; it becomes dominant only at high \(\mathrm{Br}\).

Units check. \(\mathrm{W}/(\mathrm{kg\,s^{-1}}\cdot\mathrm{J\,kg^{-1}K^{-1}})=\mathrm{W\,s\,K/J}=\mathrm{K}\). ✓

Problems

(A) Simple shear. A fluid of viscosity \(\mu=0.5~\mathrm{Pa\,s}\) undergoes simple shear at rate \(\dot\gamma=200~\mathrm{s^{-1}}\). Find the volumetric dissipation.
Solution
Simple shear \(u_x=\dot\gamma y\) gives \(\Phi=\mu\dot\gamma^2=0.5\times(200)^2=0.5\times4\times10^4=2\times10^4~\mathrm{W/m^3}=20~\mathrm{kW/m^3}\).
(A) Rigid rotation check. Show that a fluid in solid-body rotation \(\mathbf{u}=\boldsymbol\Omega\times\mathbf r\) has \(\Phi=0\).
Solution
With \(u_i=\epsilon_{ijk}\Omega_j x_k\), \(\partial u_i/\partial x_l=\epsilon_{ijl}\Omega_j\) is antisymmetric in \(i,l\). Hence the symmetric strain rate \(S_{il}=\tfrac12(\partial_l u_i+\partial_i u_l)=\tfrac12(\epsilon_{ijl}+\epsilon_{lji})\Omega_j=0\) since \(\epsilon_{ijl}=-\epsilon_{lji}\). Therefore \(\Phi=2\mu S_{ij}S_{ij}=0\): rotation stores no strain and dissipates nothing.
(B) Couette temperature rise. For Example 1's oil film (\(\Phi=4\times10^{6}~\mathrm{W/m^3}\), \(h=1~\mathrm{mm}\)), the plates are held at temperature \(T_0\) and the fluid has conductivity \(\kappa=0.15~\mathrm{W\,m^{-1}K^{-1}}\). Find the peak steady temperature rise at mid-gap.
Solution
Steady 1-D conduction with uniform source: \(\kappa\,d^2T/dy^2+\Phi=0\), \(T(0)=T(h)=T_0\). Solution \(T-T_0=\frac{\Phi}{2\kappa}y(h-y)\), peak at \(y=h/2\): \(\Delta T_{\max}=\frac{\Phi h^2}{8\kappa}=\frac{4\times10^6\times(10^{-3})^2}{8\times0.15}=\frac{4}{1.2}\approx 3.3~\mathrm{K}\).
(B) Compressible dissipation. A gas has a purely radial expansion \(\mathbf u=\alpha\mathbf r\) with \(\alpha=50~\mathrm{s^{-1}}\), shear viscosity \(\mu=2\times10^{-5}~\mathrm{Pa\,s}\), bulk viscosity \(\zeta=1\times10^{-5}~\mathrm{Pa\,s}\). Compute \(\Phi\).
Solution
Here \(S_{ij}=\alpha\delta_{ij}\) so \(\nabla\!\cdot\!\mathbf u=3\alpha\) and the deviatoric part \(S_{ij}^*=S_{ij}-\alpha\delta_{ij}=0\). Thus only bulk viscosity acts: \(\Phi=\zeta(\nabla\!\cdot\!\mathbf u)^2=\zeta(3\alpha)^2=10^{-5}\times(150)^2=10^{-5}\times2.25\times10^4=0.225~\mathrm{W/m^3}\). (Under the Stokes hypothesis \(\zeta=0\) this would vanish.)
(C) Brinkman number and self-consistency. Estimate the Brinkman number for Example 2's pipe (\(\mu=0.1~\mathrm{Pa\,s}\), \(\bar U=1~\mathrm{m/s}\), \(\kappa=0.15~\mathrm{W\,m^{-1}K^{-1}}\)) taking the wall–centre temperature difference from viscous heating, and comment on whether isothermal Navier–Stokes is self-consistent.
Solution
The characteristic self-heating scale is \(\Delta T\sim\mu\bar U^2/\kappa=0.1\times1/0.15\approx 0.67~\mathrm K\), giving \(\mathrm{Br}=\mu\bar U^2/(\kappa\,\Delta T)\sim 1\) by construction — the definition is \(\mathrm{Br}=\mu\bar U^2/(\kappa\Delta T)\) so the self-generated \(\Delta T\) yields \(\mathrm{Br}=O(1)\). But the absolute rise (\(\sim 0.7~\mathrm K\)) is tiny compared with the ambient temperature, so \(\mu(T)\) barely changes and the isothermal closure is self-consistent here. Self-heating only governs the flow when the temperature rise is large enough to swing \(\mu(T)\) significantly — e.g. concentrated-polymer extrusion or high-speed bearings where \(\bar U\) and \(\mu\) are orders of magnitude larger.