6. Conservation laws

Does the slab lose energy or merely pass it along?

A steady slab conducts heat from a hot reservoir to a cold one without changing its own stored energy. That does not mean zero heat flux. It means the total outward power is zero: power entering one face leaves the other.

text

hot side          slab of area A, thickness L          cold side
           jx →   ┌─────────────────────────┐   jx →
                  │       stored energy     │
                  └─────────────────────────┘
             n = −ex                         n = +ex

Conceptual control volume. Heat travels to the right at both faces, while the outward normals point in opposite directions. Side faces are insulated.

If $j_x=20\,\mathrm{W/m^2}$ and $A=0.1\,\mathrm{m^2}$, the left face’s outward power is $-2\,\mathrm W$ and the right face’s is $+2\,\mathrm W$. Their sum is zero. If the right-face flux increases to $30\,\mathrm{W/m^2}$ while the left remains unchanged, stored energy must decrease at $1\,\mathrm W$ in the absence of internal generation. This is our sign check before calculus.

Account for what can change

A conservation law is an accounting identity. For an extensive quantity $Q$ inside a fixed region $\Omega$,

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_\Omega q\,\mathrm{d}V + \int_{\partial\Omega}\boldsymbol{j}\!\cdot\boldsymbol{n}\,\mathrm{d}A = \int_\Omega s\,\mathrm{d}V.$$

The three terms are storage, outward flux, and production. This sign convention makes positive $\boldsymbol{j}\!\cdot\boldsymbol{n}$ a loss from the region. Changing the orientation changes the flux sign, not the physics.

Learning outcomes

After this chapter, you should be able to:

• identify storage, flux, and source terms in a balance;
• use the divergence theorem to relate integral and local forms;
• distinguish conservation from a constitutive law that supplies a flux; and
• catch a sign error by checking a simple control-volume limit.

From a control volume to a local relation

For a sufficiently smooth flux, the divergence theorem gives

$$\int_{\partial\Omega}\boldsymbol{j}\!\cdot\boldsymbol{n}\,\mathrm{d}A = \int_\Omega \nabla\!\cdot\boldsymbol{j}\,\mathrm{d}V.$$

If the balance holds for every suitable subregion, its local form is

$$\partial_t q + \nabla\!\cdot\boldsymbol{j} - s = 0.$$

The integral form states the accounting principle directly. The local form is often convenient for authoring field relations. Moving between them requires regularity and a clear orientation; it is not a change of numerical method.

Derive the one-dimensional heat equation

Let $u_v$ be internal energy per volume, $j_x$ heat flux, and $s_v$ internal heating per volume. Integrating a short slab from $x$ to $x+\Delta x$ gives

$$\frac{\mathrm d}{\mathrm dt}\int_x^{x+\Delta x} u_v A\,\mathrm dx +A\bigl[j_x(x+\Delta x)-j_x(x)\bigr] =\int_x^{x+\Delta x}s_v A\,\mathrm dx.$$

Divide by $A\Delta x$ and take the smooth limit. With stationary material and constant volumetric heat capacity $c_v=\rho c$, where $c$ is the appropriate specific heat under this solid approximation, storage becomes $\partial_tu_v=c_v\partial_tT$. Thus

$$c_v\partial_tT+\partial_xj_x=s_v.$$

The units are watts per cubic metre in all three terms. This equation does not yet tell us $j_x$. If we additionally choose Fourier conduction $j_x=-\kappa\partial_xT$, it becomes

$$c_v\partial_tT-\partial_x(\kappa\partial_xT)=s_v.$$

For constant $\kappa$, steady state, and no source, it reduces to $T''=0$. Two integrations give $T=a+bx$; the two prescribed end temperatures then recover the linear profile from the fields chapter. The straight line comes from the whole combination of balance, material law, and boundary data.

For constant positive $s_v$, steady balance instead gives $\partial_xj_x=s_v$: more power exits than enters because energy is added inside. With equal fixed temperatures at both ends, heat flows outward on both faces. The heat-transfer path explores these profiles.

Recover the small model by integration

If the bead’s temperature is uniform, integrate its volumetric storage over the whole volume to obtain $C\dot T$. Replacing the total outward surface power with $G(T-T_\infty)$ recovers the first chapter’s ODE. The lumped equation and spatial balance therefore share the same energy accounting; their difference is the assumed spatial detail.

A moving fluid needs care: a fixed control volume can receive energy or mass by advection as well as conduction. A balance written only for conductive flux must not be reused as the complete transport law for moving material.

Two steady fluid balances

For constant-density incompressible flow, the corresponding mass statement is $\nabla\cdot\boldsymbol u=0$. It says there is no net volumetric outflow from a small fixed volume; it does not determine velocity by itself. Momentum balance adds the force accounting. In the fluid component, find incompressibility and momentum:

steady balance component

public component SteadyNewtonianBalance2d(
  support body: volume(ambient_dimension = 2),
  variable velocity: vector<m / s, 2> on body,
  variable pressure: kg / (m * s ^ 2) on body,
  variable force_potential: kg / (m * s ^ 2) on body,
  parameter dynamic_viscosity: kg / (m * s)
) {
  relation momentum on body {
    -div(
      2 * dynamic_viscosity * symmetric_part(grad(velocity))
      - isotropic_lift(pressure)
    ) - grad(force_potential) = 0;
  }
  relation incompressibility on body {
    div(velocity) = 0;
  }
}

The divergence in momentum acts on stress. Its viscous term supplies a material relation, while pressure enforces the incompressibility constraint. A force potential supplies the body-force density through its gradient. The next chapter separates those material assumptions from conservation itself.

Read the standard fluid components.

Deliberate failure: reverse only one flux sign

Imagine a source-free, steady one-dimensional region with more flux leaving the right face than entering the left. Its contents cannot remain steady. A residual that declares this imbalance zero after reversing only the right-face sign has lost the chosen outward-normal convention.

Test the signs on this two-face control volume before manipulating the full field equation. A compiler can verify compatible types and dimensions, but it cannot infer which orientation you intended.

Exercises

1. Set $s=0$ and storage to zero. What must the total outward flux be?
2. Derive the local balance from the integral statement using the divergence theorem.
3. Integrate div(velocity) = 0 over a region and interpret the boundary integral.
4. In the momentum excerpt, identify the conserved balance and the constitutive ingredients separately.
5. Draw a one-dimensional control volume and label the outward normal at both ends. Use it to check the flux signs.

Exercise guidance

In exercise 1, zero total outward flux permits a nonzero flux through individual faces. For exercise 5, write the boundary sum as $A[j_x(L)-j_x(0)]$ before assigning numerical values. The minus sign at the left face comes from its outward normal, not from reversing the physical heat-flow direction.

Generation check. A slab with volume $AL$ produces heat uniformly at $s_v\,\mathrm{W/m^3}$. Show that steady outward power must be $s_vAL$. If symmetry sends equal power through both ends, find each end’s outward flux. This integrated check can expose a sign or source error even when a temperature plot looks smooth.

References

• Gilbert Strang and Edwin “Jed” Herman, Calculus, volume 3 (OpenStax, 2016), section 6.8, The divergence theorem, for converting an outward surface integral to a volume integral.

Continue with heat transfer for energy balances or fluid mechanics for mass and momentum balances.

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