3. Boundaries and interfaces

A differential equation needs its surroundings

The same conductor can remain cold in a bath, warm indefinitely in insulation, or approach an intermediate temperature in moving air. The material equation alone cannot distinguish these experiments. Boundary conditions state how the body exchanges energy with its surroundings.

This chapter teaches you to select those conditions, track normal directions, and explain how prescribed values and fluxes enter a weak formulation.

Three common boundary descriptions

A prescribed temperature, $T=T_b$, idealizes a reservoir that can supply or remove whatever power is needed to maintain its temperature. It fixes a value and leaves the heat flux to be determined. This is also called a Dirichlet or essential condition.

A prescribed outward flux, $\boldsymbol q\cdot\boldsymbol n=g$, fixes the power per area leaving the body. Insulation is the special case $g=0$. An inward heater has $g<0$ with this convention. Such a condition is called a Neumann condition.

An exchange condition relates the two:

$$\boldsymbol q\cdot\boldsymbol n=h(T-T_\infty).$$

Here $h$ has units W/(m² K), and $T_\infty$ is the surrounding fluid temperature. It is a Robin condition. If the surface is warmer than the surroundings, heat leaves; if cooler, heat enters. The coefficient describes an assumed exchange regime. It does not resolve the fluid velocity and temperature fields. This surface-exchange idealization is discussed in Glicksman’s convection notes.

For a slab exchanging through one face, compare its internal resistance $L/(kA)$ with the boundary resistance $1/(hA)$. Their ratio is $hL/k$. A large ratio allows a substantial internal temperature difference; a small ratio suggests internal conduction can keep the body nearly uniform. A chosen length must accompany any reported Biot number.

A natural boundary term has a sign

Write the steady equation as $-\nabla\cdot(k\nabla T)=s$. Multiply by a test function $w$ and integrate by parts:

$$\int_\Omega k\nabla w\cdot\nabla T\,\mathrm dV +\int_{\partial\Omega}wq_n\,\mathrm dA =\int_\Omega ws\,\mathrm dV, \qquad q_n=-k\nabla T\cdot\boldsymbol n.$$

On a prescribed-temperature boundary, admissible changes in $T$ vanish, so choose $w=0$ there. On the remaining boundary, a prescribed outward flux $g$ contributes $-\int wg$ on the right side. That minus sign is physical: outward power reduces the energy available to heat the interior.

For an exchange boundary, substitute $q_n=h(T-T_\infty)$:

$$\int_\Omega k\nabla w\cdot\nabla T\,\mathrm dV +\int_{\Gamma_h}whT\,\mathrm dA =\int_\Omega ws\,\mathrm dV +\int_{\Gamma_h}whT_\infty\,\mathrm dA.$$

This derivation explains why flux conditions are often called natural: they appear directly in the boundary integral. Omitting that term on a free boundary imposes zero flux. It does not leave the exterior unspecified. In the heated square we use next, temperature is prescribed everywhere, so $w$ vanishes on the entire exterior and only the volume integrals remain.

An interface has two outward normals

Consider two slabs in perfect thermal contact with no interfacial storage or source. Temperature is continuous and flux is conserved:

$$T_1=T_2,\qquad \boldsymbol q_1\cdot\boldsymbol n_1+ \boldsymbol q_2\cdot\boldsymbol n_2=0.$$

text

          slab 1                 slab 2
     ─────────────────│─────────────────────→ x
            qx →      │      qx →
               n1 →   │   ← n2
                      interface

Original orientation diagram. The same rightward physical heat flux is outward from slab 1 and inward to slab 2. Equal flux vectors give opposite outward normal components.

Integrate the energy account across a vanishingly thin pillbox around the interface. With no surface source or storage, whatever enters one side must leave the other. Temperature continuity is a separate perfect-contact assumption. A contact resistance would permit a temperature jump while preserving energy transfer.

For a common area $A$, no volumetric source, end temperatures $T_L,T_R$, and resistances $R_1=L_1/(k_1A)$ and $R_2=L_2/(k_2A)$,

$$P=\frac{T_L-T_R}{R_1+R_2},\qquad T_I=T_L-PR_1.$$

This hand calculation connects the slab model to the algebraic network chapter. Take $R_1=1$ K/W, $R_2=3$ K/W, $T_L=340$ K, and $T_R=300$ K. Then $P=10$ W and $T_I=330$ K. The larger resistance receives the larger temperature drop. We use this analytic example to reason about interfaces; the computational square in the next chapter is a single homogeneous body.

Two useful failure experiments

First, prescribe zero outward flux everywhere while retaining positive uniform generation in a steady problem. Integrating gives $0=sV$, a contradiction. A solver difficulty here originates in the model.

Second, make a fully insulated, source-free steady problem. Any uniform absolute temperature solves it. The missing temperature reference creates a family of solutions; changing linear solver tolerance cannot select the physical reference for you.

Exercises

1. For a heat flux directed rightward, list the outward signs at the two slab faces. Explain why adding coordinate components is the wrong balance.
2. At a perfectly conducting contact, should you equate the two outward fluxes or their negatives?
3. Add a contact resistance $R_c$ in K/W to the two-slab calculation. Derive the power and temperature jump.
4. Derive the weak equation with a prescribed inward flux of magnitude $p>0$.

Solution hints

1. Left negative, right positive. A normal component uses the local normal.
2. Their negatives; the outward fluxes sum to zero.
3. $P=(T_L-T_R)/(R_1+R_c+R_2)$ and the left-to-right temperature drop at contact is $PR_c$.
4. Set $g=-p$, producing a positive $\int wp$ load on that boundary.

References

• Leon R. Glicksman, Heat Transfer, MIT 4.42J (2010), §2 and the introduction to convection.
• Chris Schuh, Materials Processing, MIT 3.044 (2013), Lecture 2, boundary conditions.

Previous: Steady conduction · Next: A heated body in Eqiora