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Test cases--description

In order to facilitate the solution of the analytical equations presented in section 9.1.1 the author has implemented a set of Matlab functions (m-scripts) which calculates the solution, ie for instance $\Delta p_{\mbox{\protect\scriptsize tot}}$, $<\!{\alpha}\!>$, $<\!{x}\!>$, $h_{\ell,d}$, and z_d. For completeness we give a brief description of the Matlab functions in the list below.

>

Initializing function which has to be called before the main function is called.

>

Main function which sets up the input quantities and calls sub-functions as appropriate in order to calculate the solution.

>

Calculates the void fraction for a given z coordinate.

>

Returns the flow quality calculated by the Levy profile fit formula.

>

Function which is used to solve for the point of void departure, z_d. Corresponds to the equation (9.5).

>

Returns the enthalpy at void departure, $h_{\ell,d}$, according to the Saha-Zuber correlation (6.93).

>

Calculates the linear heat generation rate, q'(z). The function corresponds to equation (9.2).

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Calculates the mixture quality at position z according to (9.3).

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Calculates the two-phase friction multiplier according to the Jones correlation (6.85).

>

Function which is used to calculate the single-phase Darcy-Weisbach friction factor according to the Colebrook interpolation formula (6.81).

We will calculate the solution for two different power levels. One high power case in which the maximum linear heat generation rate, q₀, is given by

$$q_{0,H} = \frac{2.5}{2.44} {\mbox{ MW}}/{\mbox{m}}$$ (9.13)

and one low power case in which q₀ is given by

$$q_{0,L} = \frac{0.5}{2.44} {\mbox{ MW}}/{\mbox{m}}$$ (9.14)

In both cases the physical properties are takes as (corresponds to water at saturation, $T_{\mbox{\protect\scriptsize sat}} = 280 {\mbox{ }}\mbox{${}^\circ\mbox{C}$}$)

$$\begin{array}{ll} \rho_\ell = 750.70 {\mbox{ kg}}/{\mbox{m... ...7.5 {\mbox{ $\mu$kg}}/({\mbox{m}}\cdot{\mbox{s}}) \end{array}$$ (9.15)

and the geometric quantities are given by (8 x 8 GE fuel element)

$$\begin{array}{ll} A_c = 10.085 \cdot 10^{-3} {\mbox{ m}}^2... ...p}} = [ 0.54, 0.54, 0.54, 0.54, 0.54] & {\mbox{ }} \end{array}$$ (9.16)

The boundary conditions are taken as

$$\begin{array}{l} h_{\ell,0} = 1.22 \cdot 10^{6} {\mbox{ J}}/{\mbox{kg}} \\ p_0 = 7.1 \cdot 10^6 {\mbox{ Pa}} \end{array}$$ (9.17)

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