The Heat Equation — Question 4

Question 4 – Solution

Strategy. Add the two layer resistances and the contact resistance, retaining the interface temperature drop.

Step 1: Apply the interface balance. No interfacial storage or source means incoming flux equals outgoing flux. Thus the same $q$ satisfies $u_x=-q/k_1$ in the first layer and $u_x=-q/k_2$ in the second. The boundary and contact relations are $$u(0)=T_h,\quad u(L_1+L_2)=T_c,\quad u_--u_+=Rq.$$ Temperature continuity is a special case of zero resistance, not the general interface law.

Step 2: Add all temperature drops. Writing $\mathcal R=L_1/k_1+R+L_2/k_2$ gives $$\boxed{q=\frac{T_h-T_c}{\mathcal R},\quad u_-=T_h-\frac{qL_1}{k_1},\quad u_+=T_c+\frac{qL_2}{k_2}.}$$ The profiles are $T_h-qx/k_1$ on the first layer and $T_c+q(L_1+L_2-x)/k_2$ on the second. Their traces differ by $Rq$.

Step 3: Verify the numerical example. Here $\mathcal R=1+1/2+1/2=2$, so $$\boxed{q=50,\quad u_-=50,\quad u_+=25,\qquad u(x)=\begin{cases}100-50x,&0\leq x<1,\\ 50-25x,&1<x\leq 2.\end{cases}}$$ The flux is $50$ on both sides, although the slopes differ. The temperature jump is $25=Rq$. Open circles show separate traces, not a single assigned temperature at an idealized zero-thickness contact.

Step 4: Take the resistance limits. As $R\to 0$, the flux tends to $(T_h-T_c)/(L_1/k_1+L_2/k_2)$ and the traces meet. As $R\to\infty$, $q\to 0$, each layer becomes nearly constant at its own outer temperature, and $Rq\to T_h-T_c$. A large resistance can therefore sustain a finite temperature jump while the flux vanishes. For finite $R>0$, forcing $u_-=u_+$ would give $q=0$, contradicting $T_h>T_c$ and the finite total resistance.