Physical model of the heat transfer problem

The geometry of the heat transfer problem pertinent to most commercial nuclear reactors is shown in Figure 11.1. As depicted we assume cylindrical fuel rods cooled by ordinary water with uranium dioxide, U ${\mbox{O}}_2$ , as fuel and zircaloy-2, Zr-2, as clad material.

$\begin{figure} % latex2html id marker 37936\rule{\textwidth}{0.2mm} \rule{0cm}... ...{{\sl Figure~\thefigure}\hspace{1em}Heat transfer problem geometry.}\end{figure}$

Heat is generated in the fuel by a volumetric source, q'''(r). Since the ceramic U ${\mbox{O}}_2$ fuel material consists of a sintered powder which has a relatively low conductivity, large temperature gradients exists in the radial direction during operation. Together with generation of gaseous fission products the fuel material will after short time exhibit a complicated pattern of cracks which depends on factors such as power histories and burn-up.

When we want to describe heat transport through a material with a structure as complicated as the ceramic fuel material one can pursue two fundamental different ways.

One way utilizes a microscopic treatment of the cracks by assuming some distribution of cracks through the material. As a consequence, every crack is considered.

Another model uses a macroscopic point of view and uses a effective thermal conductivity which in some average way describes the heat transmission properties of the cracked fuel material.

We will use the macroscopic model to describe the heat transmission in the fuel because it is the simplest to use and in practice quite adequate. This model choice implies that heat in the fuel material is transported solely by conduction and therefore governed by the heat diffusion equation (see, for instance, [35, p. 58])

$\begin{displaymath} \nabla \cdot [ k_f \nabla T ] + q''' = \rho_f c_{p,f} \frac{\partial T}{\partial t} \end{displaymath}$

(11.1)

effective conductivity of fuel material [W/(m $\cdot$ K)].
l: temperature [ ${}^\circ\mbox{C}$ ].
l: heat generation per unit volume [W/ ${\mbox{m}}^3$ ].
l: density of fuel material [kg/ ${\mbox{m}}^3$ ].
l: specific heat of fuel material [J/(kg $\cdot$ K)].
l: time [s].

Since the temperature gradient in the radial direction is many times larger than in the axial z-direction we neglect heat conduction in the axial direction, ie we make the following approximation

$\begin{displaymath} \frac{\partial T}{\partial z} \simeq 0 \end{displaymath}$

(11.2)

Furthermore, in the steady-state case with angular symmetry Eq. (11.1) in a cylindrical (r,z) coordinate system simplifies to

$\begin{displaymath} \frac{1}{r} \frac{d}{dr} \left( k_f r \frac{d T}{dr} \right) + q''' = 0 \end{displaymath}$

(11.3)

This ordinary differential equation can be solved either by analytical or numerical means. A discretization of (11.3), for instance, by finite differences would yield the most flexible and general applicable solution algorithm.

However, if the functions k_f=k(T) and q'''=q'''(r) are known analytically integrable functions it is very simple and affordable to obtain an analytical expression for the temperature distribution T = T(r). We will use the analytical approach in the subsequent analysis.

Rearranging (11.3) and integrating with respect to r yields

$\begin{displaymath} \left[ k_f r \frac{d T}{dr} \right]_0^r = - \int\limits_0^r q''' r dr \end{displaymath}$

(11.4)

Let us write the volumetric source term, q''', as

$\begin{displaymath} q'''(r) \;\hbox{$=$\kern-0.68em\raise1.1ex \hbox{$\scriptscriptstyle\triangle$}}\;\overline{q'''}{} F(r) \end{displaymath}$

(11.5)

where F(r) is a normalized power distribution within the fuel material in the rod. We assume that

$\begin{displaymath} \int\limits_0^{R_f} F(r) r dr = \frac{R_f^2}{2} \end{displaymath}$

(11.6)

where R_f is the radius of the fuel material (see Figure 11.1) such that $\overline{q'''}{}$ is the average volumetric source [W/ ${\mbox{m}}^3$ ].

The equation (11.6) implies that the total linear heat generation, q', can be written as

$\begin{displaymath} q' = \int\limits_0^{R_f} q''' dA = \overline{q'''}{} \int\limits_0^{R_f} F(r) 2\pi r dr = \pi R_f^2 \overline{q'''}{} \end{displaymath}$

(11.7)

Noting that the lower limit on the left hand side of (11.4) vanishes (due to a symmetry condition (dT/dr)|_r=0 = 0), introducing (11.5) and integrating reveals

$\begin{displaymath} \int\limits_{T(R_f)}^{T(r)} k_f(T) dT = - \overline{q'''}{}... ...left[ \frac{1}{r''} \int\limits_0^{r''} F(r')r'dr'\right]dr'' \end{displaymath}$

(11.8)

In (11.8) we integrate from the fuel surface r=R_f and inwards, since it is the surface temperature which is specified by a boundary condition.

For a detailed investigation the power distribution F(r) is calculated by neutron transport calculations but in general a parabolic profile is a good approximation, ie we assume

$\begin{displaymath} F(r) = 1 + \left[ \left( \frac{r}{R_f} \right)^2 - \frac{1}{2} \right] f \end{displaymath}$

(11.9)

where the parameter f determines the amount of flux depression, ie the decrease in the thermal neutron flux in the center of the rod compared to the surface flux. For f=0 we have an uniform internal heat generation and hence no flux depression. The flux depression factor, $\cal F$ , is defined by

$\begin{displaymath} {\cal F} \;\hbox{$=$\kern-0.68em\raise1.1ex \hbox{$\script... ... _{r=R_f}} \simeq \frac{q'''\vert _{r=0}}{q'''\vert _{r=R_f}} \end{displaymath}$

(11.10)

where $\phi_{\mbox{\protect\scriptsize th}}$ is the thermal neutron flux [neutrons/ $({\mbox{s}}\cdot{\mbox{cm}}^2)$ ]. The last expression for $\cal F$ assumes that heat is generated solely by thermal fission--an acceptable assumption valid in most thermal power reactors.

With F(r) given by (11.9) we have

$\begin{displaymath} {\cal F} = \frac{1 - 0.5 f}{1 + 0.5 f} \end{displaymath}$

(11.11)

$\begin{displaymath} f = \frac{2(1-{\cal F})}{1-{\cal F}} \end{displaymath}$

(11.12)

The flux depression factor can be estimated by diffusion theory [11, p. 163] or accurately calculated by the analytical ABH-method^11.1. In general, $\cal F$ lies around 0.95-1.0 (which corresponds to $f \in [0.0;0.05]$ ) for light water reactors with fairly low enriched fuel (< 5 %).

With the parabolic power distribution, F(r), given by (11.9) the double integral on the RHS of (11.8) can be written as

$\begin{displaymath} -\int\limits_{R_f}^r \left[ \frac{1}{r''} \int\limits_0^{r'... ...1 - \left(\frac{r}{R_f}\right)^2 \right) \right]^2 f \right\} \end{displaymath}$

(11.13)

Before we can calculate the result of the LHS integral of (11.8) we have to assume a functional form of k_f(T). According to Lahey and Moody [18] the equivalent thermal conductivity of U ${\mbox{O}}_2$ , $k_{{\mbox{\protect\scriptsize UO}}_2}$ [Btu/(h $\cdot$ ft $\cdot$ ${}^\circ\mbox{F}$ )], can be approximated by the function

$\begin{displaymath} k_{{\mbox{\protect\scriptsize UO}}_2} = \frac{3978.1}{692.61 + T} + 6.02366\cdot 10^{-12} (T + 460)^3 \end{displaymath}$

(11.14)

where

thermal conductivity in US units, ie [Btu/(h $\cdot$ ft $\cdot$ ${}^\circ\mbox{F}$ )].
l: temperature in US units, ie [ ${}^\circ\mbox{F}$ ].

Note that (11.14) is a dimensional equation which implies that it is of vital importance to use the US units as described.

The integral on the LHS of (11.8) may now be evaluated in terms of the function

$\begin{displaymath} {\cal N}(T) \;\hbox{$=$\kern-0.68em\raise1.1ex \hbox{$\scr... ...66\cdot 10^{-12}}{4} \left[ (T+460)^4 - (T_0 + 460)^4 \right] \end{displaymath}$

(11.15)

where T₀ is a reference temperature, eg 32 ${}^\circ\mbox{F}$ (= 0 ${}^\circ\mbox{C}$ ) and $\cal N$ has the US unit [Btu/(h $\cdot$ ft)].

With the definition (11.15) the conductivity integral on the LHS of (11.8) may be state as

$\begin{displaymath} \int\limits_{T(R_f)}^{T(r)} k_{{\mbox{\protect\scriptsize UO}}_2} (T') dT' = {\cal N}(T(r)) - {\cal N}(T(R_f)) \end{displaymath}$

(11.16)

Inserting (11.16) and (11.13) into (11.8) gives us the final expression for the temperature distribution

$\begin{displaymath} {\cal N}(T(r)) = \overline{q'''}{} R_f^2 \left\{ \frac{1}{4... ...R_f}\right)^2 \right) \right]^2 f \right\} + {\cal N}(T(R_f)) \end{displaymath}$

(11.17)

which means that the temperature distribution can be written as

$\begin{displaymath} T(r) = {\cal N}^{-1} \left[ \overline{q'''}{} R_f^2 \left\{... ...ht)^2 \right) \right]^2 f \right\} + {\cal N}(T(R_f)) \right] \end{displaymath}$

(11.18)

where ${\cal N}^{-1}$ is the inverse function of $\cal N$ .

In order to calculate the fuel temperature distribution we have to specify the following quantities

average volumetric heat source in the fuel [W/ ${\mbox{m}}^3$ ] (see alternatively (11.7) for a relation to q'.).
l: radius of the fuel pellets (see Figure 11.1) [m].
l: factor which determines the flux depression (see (11.12)) [--].
l: temperature on the outer surface of the fuel [ ${}^\circ\mbox{C}$ ].

Note that since $\cal N$ is a transcendental function it cannot be solved by analytical means. However, for our purposes it suffices to construct a look-up table for the values of ${\cal N}(T)$ and perform a linear interpolation in order to obtain $T({\cal N})$ . Alternatively, we could solve (11.17) numerically with some specified accuracy requirement.

We notice that in order to calculate the temperature distribution in the fuel we naturally have to specify the fuel surface temperature T(R_f). The fuel surface temperature is calculated by utilizing the specified (calculated by the hydraulics code) bulk liquid temperature, $T_\ell$ , the linear heat generation rate (calculated by the neutronics code), q', and the equivalent thermal circuit depicted in Figure 11.2 in which the concentric compositions' thermal conductivities are modeled by thermal resistors.

$\begin{figure} % latex2html id marker 38131\rule{\textwidth}{0.2mm} \rule{0cm}... ..._fig} (note that the fuel material is not included in the circuit).}\end{figure}$

The thermal resistances per unit (axial) length, R' [(m $\cdot$ K)/W], is defined by

$\begin{displaymath} R' \;\hbox{$=$\kern-0.68em\raise1.1ex \hbox{$\scriptscriptstyle\triangle$}}\;\frac{\Delta T}{q'} \end{displaymath}$

(11.19)

The temperature drop across the fuel-cladding interface is normally expressed by Newton's law of cooling, ie

$\begin{displaymath} q''(R_{\mbox{\protect\scriptsize cl,i}}) = h_{\mbox{\protec... ...ct\scriptsize gap}} \Delta T_{\mbox{\protect\scriptsize gap}} \end{displaymath}$

(11.20)

where $h_{\mbox{\protect\scriptsize gap}}$ [W/( ${\mbox{m}}^2\cdot$ K)] is the gap convection coefficient and the temperature drop across the fuel-cladding interface, $\Delta T_{\mbox{\protect\scriptsize gap}}$ [ ${}^\circ\mbox{C}$ ], is defined by

$\begin{displaymath} \Delta T_{\mbox{\protect\scriptsize gap}} \;\hbox{$=$\kern-... ...tyle\triangle$}}\;T_{\mbox{\protect\scriptsize cl,i}} - T(R_f) \end{displaymath}$

(11.21)

In terms of the linear heat generation rate equation (11.20) is given by

$\begin{displaymath} q' = h_{\mbox{\protect\scriptsize gap}} 2 \pi R_{\mbox{\protect\scriptsize cl,i}} \Delta T_{\mbox{\protect\scriptsize gap}} \end{displaymath}$

(11.22)

This means that the thermal gap resistance per unit length, $R_{\mbox{\protect\scriptsize gap}}'$ [(m $\cdot$ K)/W], (see (11.19)) is given by

$\begin{displaymath} R_{\mbox{\protect\scriptsize gap}}' = \frac{1}{h_{\mbox{\protect\scriptsize gap}} 2 \pi R_{\mbox{\protect\scriptsize cl,i}}} \end{displaymath}$

(11.23)

The gap convection coefficient, $h_{\mbox{\protect\scriptsize gap}}$ , is extremely difficult to calculate accurately since it depends on phenomena which at present time are not well understood. Despite these difficulties there exist several models for estimating $h_{\mbox{\protect\scriptsize gap}}$ ([46] and [18, p. 271]) but these models require a coupling to a structural analysis of the fuel and cladding (in order to estimate, for instance, the contact pressure) which is beyond the scope of this text.

As a consequence we will assume a user specified constant value of $h_{\mbox{\protect\scriptsize gap}}$ . According to Lahey and Moody [18, p. 254] and [37, p. 524] we may assume $h_{\mbox{\protect\scriptsize gap}} \simeq 6000 {\mbox{ (W/(m}}^2\cdot{\mbox{K}}))$ which is a value on the rather conservative side.

In the case with the cladding we can with good accuracy assume a constant conductivity. This assumption is reasonable if we consider that the cladding is a thin layer of high conductivity material. Furthermore, assuming negligible internal heat generation (q''' = 0) the heat diffusion equation (11.3) can be solved to yield a thermal resistance per unit length of the clad, $R_{\mbox{\protect\scriptsize cl}}'$ [(m $\cdot$ K)/W] [35, p. 97-8]

$\begin{displaymath} R_{\mbox{\protect\scriptsize cl}}' = \frac{1}{2 \pi k_{\mbo... ...scriptsize cl,o}}}{R_{\mbox{\protect\scriptsize cl,i}}}\right] \end{displaymath}$

(11.24)

The thermal conductivity for the Zr-2 cladding material, $k_{\mbox{\protect\scriptsize Zr-2}}$ [Btu/(h $\cdot$ ft $\cdot$ ${}^\circ\mbox{F}$ )], can be calculated from [18, p. 252]

$\begin{displaymath} k_{\mbox{\protect\scriptsize Zr-2}} = 7.151 + (2.472 \cdot 10^{-3})T + (1.674 \cdot 10^{-6})T^2 - (3.334 \cdot 10^{-10})T^3 \end{displaymath}$

(11.25)

where

thermal conductivity in US units, ie [Btu/(h $\cdot$ ft $\cdot$ ${}^\circ\mbox{F}$ )].
l: temperature in US units, ie [ ${}^\circ\mbox{F}$ ].

Note that (11.25) is a dimensional equation which implies that it is of vital importance to use the US units as described.

At this point a discussion of the temperature drop across the convective film only remains. This subject is treated in depth in the next section. For now we simply state that according to Newton's law of cooling we have

$\begin{displaymath} q' = \overline{h}{}(z) 2 \pi R_{\mbox{\protect\scriptsize cl,o}} \Delta T_{\mbox{\protect\scriptsize film}} \end{displaymath}$

(11.26)

where

$\begin{displaymath} \Delta T_{\mbox{\protect\scriptsize film}} \;\hbox{$=$\kern... ...}\;R(R_{\mbox{\protect\scriptsize cl,o}}) - T_m = T_w - T_\ell \end{displaymath}$

(11.27)

and $\overline{h}{}$ is the average convection coefficient [W/( ${\mbox{m}}^2\cdot$ K)] and depends on z since it varies with the flow state.

According to (11.19) we may define

$\begin{displaymath} R_{\mbox{\protect\scriptsize film}}' \;\hbox{$=$\kern-0.68e... ...{ \overline{h}{}(z) 2 \pi R_{\mbox{\protect\scriptsize cl,o}}} \end{displaymath}$

(11.28)

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