## Temperature distribution in Fuel Cladding

**Cladding** is the outer layer of the fuel rods, standing between the **reactor coolant** and the **nuclear fuel** (i.e. **fuel pellets**). It is made of a corrosion-resistant material with low absorption cross section for thermal neutrons, usually **zirconium alloy**. **Cladding** prevents radioactive fission products from escaping the fuel matrix into the reactor coolant and contaminating it. Cladding constitute one of barriers in ‘**defence-in-depth**‘ approach.

Consider the fuel cladding of inner radius **r _{Zr,2} = 0.408 cm** and outer radius

**r**. In comparison to fuel pellet, there is almost no heat generation in the fuel cladding (cladding is slightly heated by radiation). All heat generated in the fuel must be transferred via

_{Zr,1}= 0.465 cm**conduction**through the cladding and therefore the inner surface is hotter than the outer surface.

To find the temperature distribution through the cladding we must solve the **heat conduction equation**. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. In this example, we will assume that there is strictly no heat generation within the cladding. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C_{1} and C_{2} are the constants of integration.

1)

Calculate the temperature distribution, T(r), in this fuel cladding, if:

- the temperature at the inner surface of the cladding is T
_{Zr,2}= 360°C - the temperature of reactor coolant at this axial coordinate is T
_{bulk}= 300°C - the heat transfer coefficient (convection; turbulent flow) is h = 41 kW/m
^{2}.K. - the averaged material’s conductivity is k = 18 W/m.K
- the linear heat rate of the fuel is q
_{L}= 300 W/cm and thus the volumetric heat rate is q_{V}= 597 x 10^{6}W/m^{3}

From the basic relationship for heat transfer by convection we can calculate the outer surface of the cladding as:

As can be seen, also in this case we have given surface temperatures T_{Zr,1} and T_{Zr,2}. This corresponds to the Dirichlet boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

Solving for C_{1} and C_{2} and substituting into the general solution, we then obtain:

**∆T – cladding surface – coolant**

Detailed knowledge of geometry, outer radius of cladding, linear heat rate, convective heat transfer coefficient and the coolant temperature determines **∆T **between the coolant (T_{bulk}) and the cladding surface (T_{Zr,1}). Therefore we can calculate the cladding surface temperature (T_{Zr,1}) simply using the **Newton’s Law**:

**∆T in fuel cladding**

Detailed knowledge of geometry, outer and inner radius of cladding, linear heat rate, and the cladding surface temperature (T_{Zr,1}) determines **∆T **between outer and inner surfaces of cladding. Therefore we can calculate the inner cladding surface temperature (T_{Zr,2}) simply using the **Fourier’s Law**:

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