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Figure 1.18 A hollow cylinder.

(1.107)

      Based on Fourier's law, the rate at which heat is transferred by conduction across the cylindrical surface in the solid is expressed as

      (1.108)

      where A = 2𝜋rL is the area normal to the direction of heat transfer.

To determine the temperature distribution in the cylinder, it is necessary to solve Eq. (1.107) under appropriate boundary conditions, by assuming that k is constant. By integrating Eq. (1.107) twice, the following heat transfer equation is obtained:

      (1.109)

      If we now consider a composite hollow cylinder, the heat transfer equation is found to be as follows, where interfacial contact resistances are neglected:

      (1.110)

      where R_t,t = (1/2πr₁Lh₁) + (ln(r₂/r₁)/2πk₁L) + (ln(r₃/r₂)/2πk₂L) + ⋯(1/2πr_nLh_n)

      1.6.7 The Sphere

The case of heat transfer through a sphere is not as common as the cylinder problem. Consider a hollow sphere of internal radius r₁ and external radius r₂ (Figure 1.19). Also, consider the inside and outside temperatures to be T₁ and T₂, respectively, and constant thermal conductivity with no heat generation. We can express the heat conduction across the sphere wall in the form of Fourier's law:

(1.111)

Figure 1.19 Heat conduction in a hollow sphere.

where, A = 4πr² is the area normal to the direction of heat transfer.

After integrating Eq. (1.111), we obtain the following expression:

      (1.112)

      If we now consider a composite hollow sphere, the heat transfer equation is determined to be as follows, neglecting interfacial contact resistances:

      (1.113)

      where,

      1.6.8 Conduction with Heat Generation

      (a) The Plane Wall

Consider a plane wall, as shown in Figure 1.20a, in which there is uniform heat generation per unit volume. The heat conduction equation becomes

(1.114)

By integrating Eq. (1.114) with the prescribed boundary conditions, T(−L) = T₁ and T(L) = T₂. The temperature distribution can be obtained as

(1.115)

The heat flux at any point in the wall can be found, depending on x, by using Eq. (1.115) with Fourier's law.

      If T₁ = T₂ ≡ T_s, the temperature distribution is symmetrical about the midplane (Figure 1.20b). Then,

(1.116)

Figure 1.20 Heat conduction in a slab with uniform heat generation: (a) asymmetrical boundary conditions, (b) symmetrical boundary conditions.

At the plane of symmetry dT/dx = 0, and the maximum temperature at the midplane is

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