5.16.6. Appendix 5.6: Optional Eulerian Solution for Pipe Temperatures

As mentioned in Section 5.4.1, a Eulerian calculation can be used to speed up the pipe temperature calculation if the coolant moves more than two nodes in a time step. This Eulerian speed-up has only been implemented for flow in the nominal direction; if flow reversal has occurred in a pipe, then the Eulerian calculation is not used.

For the Eulerian calculation, Eq. (5.4-1) for the coolant is replaced by

(5.16-56) $$\rho_{\text{c}} c_{\text{c}}A_{\text{c}} \frac{\partial \text{T}_{\text{c}}}{\partial \text{t}} + wc_{\text{c}} \frac{\partial \text{T}_{\text{c}}}{\partial \text{z}} = P_{\text{er}} h_{\text{wc} }\left( T_{\text{w}} - T_{\text{c}} \right) + q_c'$$

Eq. (5.4-2) is still used for the wall. Finite differencing of Eq. (5.16-56) gives

(5.16-57) $$\begin{split}\rho_{\text{c}} c_{\text{c}}A_{\text{c}}\left\lbrack \frac{T_{\text{c}6\text{j}} + T_{\text{c}6\text{j} + 1} - T_{\text{c}5\text{j}} - T_{\text{c}5\text{j}+1}}{2\delta t} \right\rbrack \\ + \frac{wc_{\text{c}}}{\Delta z} \left\lbrack \theta_{1} \left( T_{\text{c}5\text{j} + 1} - T_{\text{c}5\text{j}} \right) + \theta_{2}\left( T_{\text{c}6\text{j} + 1} - T_{\text{c}6\text{j}} \right) \right\rbrack = \\ P_{\text{er}} h_{\text{wc}} \left\{ \theta_{1} \left\lbrack T_{\text{w}5\text{j}} - \frac{T_{\text{c}5\text{j}} + T_{\text{c}5\text{j} + 1}}{2} \right\rbrack \right. \\ \left. + \theta_{2} \left\lbrack T_{\text{w}6\text{j}} - \frac{T_{\text{c}6\text{j}} + T_{\text{c}6\text{j} + 1}}{2} \right\rbrack \right\} +q_{cj}'\end{split}$$

Similarly, finite differencing of Eq. (5.4-2) gives

(5.16-58) $$\begin{split}\frac{M_{\text{w}} C_{\text{w}}}{\delta t} \left( T_{\text{w}6\text{j}} - T_{\text{w}5\text{j}} \right) = P_{\text{er}} h_{\text{wc}} \left\lbrack \theta_{1} \left( \frac{T_{\text{c}5\text{j}} + T_{\text{c}5\text{j} + 1}}{2} - T_{\text{w}5\text{j}} \right) \right. \\ \left. + \theta_{2} \left( \frac{T_{\text{c}6\text{j}} + T_{\text{c}6\text{j} + 1}}{2} - T_{\text{w}6\text{j}} \right) \right\rbrack \\ + \left( hA \right)_{\text{snk}} \left( T_{\text{snk}} - \theta_{1} T_{\text{w}5\text{j}} - \theta_{2} T_{\text{w}6\text{j}} \right) + q_{wj}'\end{split}$$

Eq. (5.16-58) can be rewritten as

(5.16-59) $$T_{\text{w}6\text{j}} = B_{\text{w}0\text{j}} + B_{\text{w}1\text{j}} \left( T_{\text{c}6\text{j}} + T_{\text{c}6\text{j} + 1} \right)$$

where

(5.16-60) $$B_{\text{w}1\text{j}} = \frac{P_{\text{er}} h_{\text{wc}} \theta_{2}\delta t}{2 d_{\text{w}}}$$

(5.16-61) $$d_{\text{w}} = M_{\text{w}} C_{\text{w}} + \theta_{2} \delta t \left\lbrack P_{\text{er}} h_{\text{wc}} + \left( hA \right)_{\text{snk}} \right\rbrack$$

and

(5.16-62) $$\begin{split}B_{\text{w}0\text{j}} = \left\{ M_{\text{w}} C_{\text{w}} T_{\text{w}5\text{j}} + P_{\text{er}} h_{\text{wc}} \theta_{1} \delta t \left\lbrack \frac{T_{\text{c}5\text{j}} + T_{\text{c}5\text{j} + 1}}{2} - T_{\text{w}5\text{j}} \right\rbrack \right. \\ \left. + \left( hA \right)_{\text{snk}} \delta t \left( T_{\text{snk}} - \theta_{1} T_{\text{w}5\text{j}} \right) + q_{wj}' \delta t \right\} \big/ d_{\text{w}}\end{split}$$

Similarly, Eq. (5.16-57) can be rewritten as

(5.16-63) $$T_{\text{c}6\text{j} + 1} = B_{\text{c}0\text{j}} + B_{\text{c}1\text{j}} T_{\text{c}6\text{j}} + B_{\text{c}2\text{j}} T_{\text{w}6\text{j}}$$

where

(5.16-64) $$\begin{split}B_{\text{c}0\text{j}} = \left\{ \Delta z \rho_{\text{c}} c_{\text{c}} A_{\text{c}} \left( T_{\text{c}5\text{j}} + T_{\text{c5j} + 1} \right) + 2\theta_{1} \delta t w c_{\text{c}} \left( T_{\text{c}5\text{j}} - T_{\text{c}5\text{j} + 1} \right) \right. \\ \left. + 2 \theta_{1} \delta t \Delta z P_{\text{er}} h_{\text{wc}} \left\lbrack T_{\text{w}5\text{j}} - \frac{T_{\text{c}5\text{j}} + T_{\text{c}5\text{j} + 1}}{2} \right\rbrack + 2\delta t \Delta z q_{cj}' \right\} \big/ d_{\text{c}}\end{split}$$

(5.16-65) $$d_{\text{c}} = \rho_{\text{c}} c_{\text{c}} A_{\text{c}} \Delta z + 2\theta_{2} \delta t \left( wc_{\text{c}} + \frac{P_{\text{er}} h_{\text{wc}}}{2} \Delta z \right)$$

(5.16-66) $$B_{\text{c}1\text{j}} = - \left\{ \rho_{\text{c}} c_{\text{c}} A_{\text{c}} \Delta z + 2\theta_{2} \delta t \left( \frac{P_{\text{er}} h_{\text{wc}}}{2} \Delta z - wc_{\text{c}} \right) \right\} \big/ d_{\text{c}}$$

and

(5.16-67) $$B_{\text{c}2\text{j}} = \frac{2\theta_{2} \delta t\Delta z P_{\text{er}} h_{\text{wc}}}{d_{\text{c}}}$$

Eq. (5.16-59) and Eq. (5.16-63) can be combined to give

(5.16-68) $$T_{\text{c}6\text{j} + 1} = B_{\text{cc}0\text{j}} + B_{\text{cc}1\text{j}} T_{\text{c}6\text{j}}$$

where

(5.16-69) $$B_{\text{cc}0\text{j}} = \frac{B_{\text{c}0\text{j}} + B_{\text{c}2\text{j}} B_{\text{w}0\text{j}}}{1 - B_{\text{c}2\text{j}} B_{\text{w}1\text{j}}}$$

and

(5.16-70) $$B_{\text{cc}1\text{j}} = \frac{B_{\text{c}1\text{j}} + B_{\text{c}2\text{j}} B_{\text{w}1\text{j}}}{1 - B_{\text{c}2\text{j}} B_{\text{w}1\text{j}}}$$

Note that $B_{\text{cc}0\text{j}}$ and $B_{\text{cc}1\text{j}}$ can be calculated before the temperatures at the end of the sub-interval are known.

The pipe inlet temperature at the end of the step is used to set the first coolant temperature, $T_{\text{c}61}$. The code marches along the pipe, using Eq. (5.16-68) to calculate $T_{\text{c}6\text{j}+1}$ after $T_{\text{c}6\text{j}}$ has been calculated. After the coolant temperatures have been calculated, Eq. (5.16-59) is used to calculate the wall temperatures.