## Example – Convection – Cladding Surface Temperature

**Example – Convection – Problem with Solution **

**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, therefore its **coolability** is one of key safety aspects.

Consider the fuel cladding of inner radius **r**_{Zr,2}** = 0.408 cm** and outer radius **r**_{Zr,1}** = 0.465 cm**. 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 **conduction** through the cladding and therefore the inner surface is hotter than the outer surface.

Assume that:

- the outer diameter of the cladding is:
**d = 2 x r**_{Zr,1}= 9,3 mm - the pitch of fuel pins is:
**p = 13 mm** - the thermal conductivity of saturated water at 300°C is:
**k**_{H2O}**= 0.545 W/m.K** - the dynamic viscosity of saturated water at 300°C is:
**μ = 0.0000859 N.s/m**^{2} - the fluid density is:
**ρ = 714 kg/m**^{3} - the
**specific heat**is:**c**_{p}**= 5.65 kJ/kg.K** - the core flow velocity is constant and equal to
**V**_{core}**= 5 m/s** - the temperature of reactor coolant at this axial coordinate is:
**T**_{bulk}**= 296°C** - the linear heat rate of the fuel is
**q**_{L}**= 300 W/cm**(F_{Q}≈ 2.0) and thus the volumetric heat rate is q_{V}= 597 x 10^{6}W/m^{3}

Calculate the Prandtl, Reynolds and Nusselt number for this flow regime (internal forced turbulent flow) inside the rectangular fuel lattice (fuel channel), then calculate the **heat transfer coefficient** and finally the **cladding surface temperature**, **T _{Zr,1}**.

To calculate the **cladding surface temperature**, we have to calculate the **Prandtl**, **Reynolds** and **Nusselt number**, because the heat transfer for this flow regime can be described by the **Dittus-Boelter equation**, which is:

## Calculation of the Prandtl number

To calculate the Prandtl number, we have to know:

- the thermal conductivity of saturated water at 300°C is:
**k**_{H2O}**= 0.545 W/m.K** - the dynamic viscosity of saturated water at 300°C is:
**μ = 0.0000859 N.s/m**^{2} - the
**specific heat**is:**c**_{p}**= 5.65 kJ/kg.K**

Note that, all these parameters significantly differs for water at 300°C from those at 20°C. Prandtl number for water at 20°C is around **6.91. **Prandtl number for reactor coolant at 300°C is then:

## Calculation of the Reynolds number

To calculate the Reynolds number, we have to know:

- the outer diameter of the cladding is:
**d = 2 x r**(to calculate the hydraulic diameter)_{Zr,1}= 9,3 mm - the pitch of fuel pins is:
**p = 13 mm**(to calculate the hydraulic diameter) - the dynamic viscosity of saturated water at 300°C is:
**μ = 0.0000859 N.s/m**^{2} - the fluid density is:
**ρ = 714 kg/m**^{3}

**The hydraulic diameter, D**** _{h}**, is a commonly used term when handling flow in

**non-circular tubes and channels**. The

**hydraulic diameter of the fuel channel**,

*D*

*, is equal to 13,85 mm.*

_{h}See also: Hydraulic Diameter

The **Reynolds number **inside the fuel channel is then equal to:

This fully satisfies the **turbulent conditions**.

## Calculation of the Nusselt number using Dittus-Boelter equation

For fully developed (hydrodynamically and thermally) turbulent flow in a smooth circular tube, the local **Nusselt number** may be obtained from the well-known **DittusBoelter equation**.

To calculate the **Nusselt number**, we have to know:

- the Reynolds number, which is
**Re**_{Dh}= 575600 - the Prandtl number, which is
**Pr = 0.89**

The **Nusselt number **for the forced convection inside the fuel channel is then equal to:

## Calculation of the heat transfer coefficient and the cladding surface temperature, T_{Zr,1}

Detailed knowledge of geometry, fluid parameters, outer radius of cladding, linear heat rate, convective heat transfer coefficient allows us to calculate the temperature difference **∆T **between the coolant (T_{bulk}) and the cladding surface (T_{Zr,1}).

To calculate the the cladding surface temperature, we have to know:

- the outer diameter of the cladding is: d = 2 x
**r**_{Zr,1}**= 9,3 mm** - the Nusselt number, which is
**Nu**_{Dh}**= 890** - the hydraulic diameter of the fuel channel is:
*D*_{h}**= 13,85 mm** - the thermal conductivity of reactor coolant (300°C) is:
**k**_{H2O}**= 0.545 W/m.K** - the bulk temperature of reactor coolant at this axial coordinate is:
**T**_{bulk}**= 296°C** - the linear heat rate of the fuel is:
**q**_{L}**= 300 W/cm**(F_{Q}≈ 2.0)

The convective heat transfer coefficient, **h**, is given directly by the definition of Nusselt number:

Finally, we can calculate the cladding surface temperature (T_{Zr,1}) simply using the **Newton’s Law of Cooling**:

For PWRs at normal operation, there is a compressed liquid water inside the reactor core, loops and steam generators. The pressure is maintained at approximately **16MPa**. At this pressure water boils at approximately **350°C**(662°F). As can be seen, the surface temperature T_{Zr,1} = 325°C ensures, that even subcooled boiling does not occur. Note that, subcooled boiling requires T_{Zr,1} = T_{sat}. Since the inlet temperatures of the water are usually about** 290°C** (554°F), it is obvious this example corresponds to the lower part of the core. At higher elevations of the core the bulk temperature may reach up to 330°C. The temperature difference of 29°C causes the subcooled boiling may occur (330°C + 29°C > 350°C). On the other hand, **nucleate boiling** at the surface effectively disrupts the stagnant layer and therefore nucleate boiling significantly increases the ability of a surface to transfer thermal energy to bulk fluid. As a result, the convective heat transfer coefficient significantly increases and therefore at higher elevations, the temperature difference (T_{Zr,1} – T_{bulk}) significantly decreases.