2. Steady conduction

Begin with a slab

Take a slab occupying $0\le x\le L$, with cross-sectional area $A$ and constant conductivity $k$. Assume temperature depends only on $x$: side faces are insulated, end temperatures are uniform, and lateral variations are negligible. The steady equation from the previous chapter becomes

$$-kT''(x)=s.$$

We will integrate this equation twice and use two boundary conditions. Before doing so, predict the curvature: a positive source makes $T''$ negative. An internally heated slab should bulge upward above the line joining its end temperatures.

No internal generation: a straight line

With $s=0$, $T''=0$, so $T=a+bx$. Impose $T(0)=T_L$ and $T(L)=T_R$:

$$T(x)=T_L+\frac{T_R-T_L}{L}x, \qquad q_x=k\frac{T_L-T_R}{L}.$$

The heat rate is $P=Aq_x$. Rearranging gives a useful circuit-like relation:

$$T_L-T_R=PR,\qquad R=\frac{L}{kA}.$$

Thermal resistance has units K/W. The area-normalized resistance $L/k$ has units m² K/W; specify which quantity you use. The physical slab and circuit analogy are developed in Glicksman, §2.

For an original arithmetic example, choose $L=0.2$ m, $A=0.5$ m², $k=2$ W/(m K), $T_L=320$ K, and $T_R=300$ K. Then $R=0.2$ K/W and $P=100$ W. Doubling thickness halves this rate; doubling area doubles it. These are consequences of the assumed one-dimensional geometry.

Uniform generation: a parabola

Now take $s$ constant and both ends at $T_b$. Integrating gives

$$T=-\frac{s}{2k}x^2+ax+b.$$

The left condition gives $b=T_b$, and the right gives $a=sL/(2k)$:

$$T(x)=T_b+\frac{s}{2k}x(L-x).$$

Differentiate to find the maximum at $x=L/2$:

$$T_{\max}-T_b=\frac{sL^2}{8k},\qquad q_x(x)=s\left(x-\frac L2\right).$$

text

       x = 0                  x = L/2                 x = L
       T = Tb                 maximum                T = Tb
         │                       •                      │
         │                   .       .                  │
         │               .               .              │
         •                                               •
    outward ←                qx = 0                  → outward
      sL/2                                              sL/2

Conceptual profile, not numerical output. With equal end temperatures and uniform generation, heat leaves through both ends. The arrows label outward flux in W/m², while the curve indicates temperature in K.

At the left boundary $q_x=-sL/2$, but its outward normal is $-1$: $q_n=sL/2>0$. The right outward flux is also $sL/2$. Thus total outward power is $A(sL/2+sL/2)=sAL$, exactly the generated power.

For $s=1000$ W/m³, $L=0.2$ m, and $k=2$ W/(m K), the maximum rise is 2.5 K. Each end carries 100 W/m² outward. With $A=0.5$ m², each end removes 50 W and the whole slab generates 100 W. These numbers come from the equations, before any computer calculation.

Which parameter controls what?

The profile shows three useful sensitivities:

• Double $s$: double the temperature rise and the outward power.
• Double $k$: halve the temperature rise; outward power is unchanged because steady conservation still requires removal of the same generated power.
• Double $L$ at fixed $s,k,A$: quadruple the temperature rise and double the total generated power.

A larger conductivity reduces the gradient needed to carry the same power. This distinction between temperature response and energy throughput will be valuable when interpreting the two-dimensional example.

Define $\xi=x/L$ and $\Theta=k(T-T_b)/(sL^2)$. Then $\Theta=\xi(1-\xi)/2$. All these slabs share one dimensionless curve. Such a collapse is a useful model prediction: a change in a constant parameter should rescale the curve, not change its shape.

A square is a different boundary-value problem

Our Eqiora investigation cools all four sides of a square. A slab profile is constant in the transverse direction, so its warm interior would also touch the transverse boundary. It cannot satisfy a cool temperature on all four sides. Do not use the slab’s center value as an exact answer for that square.

The square has additional routes for heat loss. We will retain the slab’s qualitative predictions—positive interior rise, linear scaling with $s$, inverse scaling with $k$—and derive the square’s small finite element system separately.

Exercises

1. Derive the profile when uniform generation is present but $T_L\ne T_R$.
2. Find its stationary point. Under what condition does it lie inside the slab?
3. An engineer doubles $k$ and reports double the generated steady outward power with fixed $s,L,A$. What has gone wrong in the interpretation?
4. Show that the mean temperature rise with equal ends is $sL^2/(12k)$.

Solution hints

1. Add the source-free linear solution to $sx(L-x)/(2k)$.
2. $x_*=L/2+k(T_R-T_L)/(sL)$ for $s>0$; check $0<x_*<L$.
3. Conductivity changes the required gradient. Integrating the source still gives $sAL$, so the report must involve changed inputs, a flux convention, or an erroneous result.
4. Integrate the parabola over $0\le x\le L$ and divide by $L$.

References

• Leon R. Glicksman, Heat Transfer, MIT 4.42J (2010), §2, Conduction Heat Transfer.
• Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2 (OpenStax, 2016), §1.6, conduction and thermal resistance.

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