Solved Problem on Heat Transfer

A wall consists of alternating sheets made of thickness d of two different materials. The heat transfer coefficients of the sheets are equal to k₁ and k₂. The cross-section areas of the sheets are the same, and the number of sheets of each material is the same, the temperatures of the external surfaces of the walls are equal to T₁ and T₂ (T₁ > T₂) and remain constant. Determine the heat transfer coefficient of the wall.

Problem data:

Heat transfer conductivity of sheet 1: k₁;
External temperature of plate 1: T₁;
Heat transfer conductivity of sheet 2: k₂;
External temperature of plate 2: T₂.

Solution

a) The heat flux is given by

\[ \bbox[#99CCFF,10px] {\phi =kA\frac{\left(t_{1}-t_{2}\right)}{e}} \]

The heat passes from the higher temperature medium T₁ to the lower temperature medium T₂, where n is the number of sheets of each material, and the total area of each will be nA. The flux through plate 1 is given by

\[ \begin{gather} \phi_{1}=k_{1}nA\frac{\left(T_{1}-T_{2}\right)}{d} \tag{I} \end{gather} \]

the flux through plate 2 is given by

\[ \begin{gather} \phi_{2}=k_{2}nA\frac{\left(T_{1}-T_{2}\right)}{d} \tag{II} \end{gather} \]

Figure 1

Since the surfaces are kept at constant temperatures, the heat flux is steady, so the total flux through the two plates will be the sum of the flux through each plate (Figure 1-A), adding expressions (I) and ( II)

\[ \begin{gather} \phi _{T}=\phi _{1}+\phi _{2}\\ \phi_{T}=k_{1}nA\frac{\left(T_{1}-T_{2}\right)}{d}+k_{2}nA\frac{\left(T_{1}-T_{2}\right)}{d}\\ \phi_{T}=\left(k_{1}+k_{2}\right)nA\frac{\left(T_{1}-T_{2}\right)}{d} \tag{III} \end{gather} \]

The area of two consecutive sheets 1 and 2 will be 2A (Figure 1-B) as there are n sheets of each material the total area of the wall will be 2nA, and the total flux through the wall will be

\[ \begin{gather} \phi_{T}=k2nA\frac{\left(T_{1}-T_{2}\right)}{d} \tag{IV} \end{gather} \]

equating the expressions (III) and (IV), we obtain the heat transfer coefficient

\[ \begin{gather} 2k\cancel{n}\cancel{A}\frac{\cancel{\left(T_{1}-T_{2}\right)}}{\cancel{d}}=\left(k_{1}+k_{2}\right)\cancel{n}\cancel{A}\frac{\cancel{\left(T_{1}-T_{2}\right)}}{\cancel{d}}\\ 2k=k_{1}+k_{2} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {k=\frac{k_{1}+k_{2}}{2}} \]

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