5.16.1. Appendix 5.1: IHX Matrix Coefficients

The coefficients in Eq. (5.4-44) for the $j$-th vertical section of the shell in terms of the quantities defined in Section 5.4.2.2 are as follows:

(5.16-1) $$a_{1} \left( j \right) = \left( \rho c \right)_{\text{SH}} d_{\text{SH}} + \theta_{2\text{S}} \Delta t H_{\text{S}}\left( j \right) + \theta_{2\text{S}} \Delta t \frac{\left( hA \right)_{\text{snk}}}{P_{\text{S}}}$$

(5.16-2) $$a_{2} \left( j \right) = - \frac{1}{2} \theta_{2\text{S}} \Delta t H_{\text{S}}\left( j \right)$$

(5.16-3) $$a_{3} \left( j \right) = - \frac{1}{2} \theta_{2\text{S}} \Delta t H_{\text{S}}\left( j \right)$$

(5.16-4) $$a_{4} \left( j \right) = - \Delta t H_{\text{S}}\left( j \right) T_{\text{SH}3}\left( j \right) + \Delta t H_{\text{S}}\left( j \right) {\overline{T}}_{\text{CS}3}\left( j \right) + \frac{\Delta t\left( hA \right)_{\text{snk}} \left\lbrack T_{\text{snk}} - T_{\text{SH}3}\left( j \right) \right\rbrack}{P_{\text{S}}}$$

(5.16-5) $${\overline{T}}_{\text{CP}3} \left( j \right) = \frac{1}{2} \left\lbrack T_{\text{CS}3}\left( j \right) + T_{\text{CS}3}\left( j + 1 \right) \right\rbrack$$

where

$\theta_{2\text{S}}$ = the degree of implicitness for the shell-side coolant channel

$\Delta t$ = the time interval

The coefficients in Eq. (5.4-45) for the $j$-th vertical section of the shell-side coolant for normal flow (downward) are:

(5.16-6) $$\begin{split}e_{1} \left( j \right) &= \frac{1}{2} A_{\text{c}} {\overline{\rho}}_{\text{CS}}\left( j \right) {\overline{c}}_{\text{CS}}\left( j \right) + \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j \right)}{\Delta z\left( j \right)} \theta_{2\text{S}} \left| w_{\text{S}4} \right| \\ &+ \Delta t P_{\text{S}} H_{\text{S}}\left( j \right) \frac{1}{2} \theta_{2\text{S}} + \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j \right) \frac{1}{2} \theta_{2\text{S}}\end{split}$$

(5.16-7) $$e_{2}\left( j \right) = 0$$

(5.16-8) $$e_{3}\left( j \right) = - \Delta t P_{\text{S}} H_{\text{S}}\left( j \right) \theta_{2\text{S}}$$

(5.16-9) $$e_{4}\left( j \right) = 0$$

(5.16-10) $$e_{5}\left( j \right) = - \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j \right) \theta_{2\text{S}}$$

(5.16-11) $$e_{6}\left( j \right) = 0$$

(5.16-12) $$\begin{split}e_{7}\left( j \right) = \frac{1}{2} A_{\text{CS}} {\overline{\rho}}_{\text{CS}}\left( j \right) {\overline{c}}_{\text{CS}}\left( j \right) - \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j \right)}{ \Delta z\left( j \right)} \theta_{2\text{S}} \left| \ w_{\text{S}4} \right| \\ + \Delta t P_{\text{S}} H_{\text{S}}\left( j \right) \frac{1}{2} \theta_{2\text{S}} + \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j \right) \frac{1}{2} \theta_{2\text{S}}\end{split}$$

(5.16-13) $$\begin{split}e_{8}\left( j \right) = - \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j \right)}{ \Delta z\left( j \right)} \left\{ \theta_{1\text{S}} \left| \ w_{\text{S}3} \right| + \theta_{2\text{S}} \left| w_{\text{S}4} \right| \left\lbrack T_{\text{CS}3}\left( j \right) - T_{\text{CS}3}\left( j + 1 \right) \right\rbrack \right\} \\ + \Delta t P_{\text{S}} H_{\text{S}}\left( j \right) \left\{ T_{\text{SH}3}\left( j \right) - \frac{1}{2} \left\lbrack T_{\text{CS}3}\left( j \right) + T_{\text{CS}3}\left( j + 1 \right) \right\rbrack \right\} \\ + \Delta t S P_{\text{S}} H_{\text{ST}}\left( j \right) \left\{ T_{\text{TU}3}\left( j \right) - \frac{1}{2} \left\lbrack T_{\text{CS}3}\left( j \right) + T_{\text{CS}3}\left( j + 1 \right) \right\rbrack \right\}\end{split}$$

The same coefficients for reversed flow (upward) in the shell-side coolant channel are:

(5.16-14) $$\begin{split}e\left( j \right) = \frac{1}{2} A_{\text{CS}} {\overline{\rho}}_{\text{CS}}\left( j - 1 \right) {\overline{c}}_{\text{CS}}\left( j - 1 \right) + \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j - 1 \right)}{\Delta z \left( j - 1 \right)} \theta_{2\text{S}} \left| w_{\text{S}4} \right| \\ + \Delta t P_{\text{S}} H_{\text{S}}\left( j - 1 \right) \frac{1}{2} \theta_{2\text{S}} + \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j - 1 \right) \frac{1}{2} \theta_{2\text{S}}\end{split}$$

(5.16-15) $$e_{2}\left( j \right) = - \Delta t P_{\text{S}}H_{\text{S}}\left( j - 1 \right) \theta_{2\text{S}}$$

(5.16-16) $$e_{3}\left( j \right) = 0$$

(5.16-17) $$e_{4}\left( j \right) = - \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j - 1 \right) \theta_{2\text{S}}$$

(5.16-18) $$e_{5}\left( j \right) = 0$$

(5.16-19) $$\begin{split}e_{6}\left( j \right) = \frac{1}{2} A_{\text{CS}} {\overline{\rho}}_{\text{CS}}\left( j - 1 \right) {\overline{c}}_{\text{CS}}\left( j - 1 \right) - \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j - 1 \right)}{\Delta z\left( j - 1 \right)} \theta_{2\text{S}} \left| w_{\text{S}4} \right| \\ + \Delta t P_{\text{S}} H_{\text{S}}\left( j - 1 \right) \frac{1}{2} \theta_{2\text{S}} + \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j - 1 \right) \frac{1}{2} \theta_{2\text{S}}\end{split}$$

(5.16-20) $$e_{7}\left( j \right) = 0$$

(5.16-21) $$\begin{split}e_{8}\left( j \right) = - \Delta t \frac{{\overline{c}}_{\text{CS}}\left( j - 1 \right)}{\Delta z\left( j - 1 \right)} \left\{ \left( \theta_{1\text{S}} \left| w_{\text{S}3} \right| + \theta_{2\text{S}} \left| w_{\text{S}4} \right| \right) \left\lbrack T_{\text{CS}3}\left( j \right) - T_{\text{CS}3}\left( j - 1 \right) \right\rbrack \right\} \\ + \Delta t P_{\text{S}} H_{\text{S}}\left( j - 1 \right) \left\{ T_{\text{SS}3}\left( j - 1 \right) - \frac{1}{2} \left\lbrack T_{\text{CS}3}\left( j - 1 \right) + T_{\text{CS}3}\left( j \right) \right\rbrack \right\} \\ + \Delta t S P_{\text{ST}}\left( j - 1 \right) \left\{ T_{\text{TU}3}\left( j - 1 \right) - \frac{1}{2} \left\lbrack T_{\text{CS}3}\left( j - 1 \right) + T_{\text{CS}3}\left( j \right) \right\rbrack \right\}\end{split}$$

The terms $e_{9} \left( j \right)$ and $e_{10} \left( j \right)$ have been added to Eq. (5.4-45) because they appear during the solution of the simultaneous equations. These arrays are set to zero before the solution is begun.

In addition, the boundary conditions for normal shell-side coolant channel flow are

(5.16-22) $$e_{1}\left( jm \right) = 1; \quad e_{2,3,4,5,6,7} \left( jm \right) = 0; \quad e_{8}\left( jm \right) = \Delta T_{\text{CS}}\left( jm \right)$$

For reversed primary channel flow, they are

(5.16-23) $$e_{1}\left( 1 \right) = 1; \quad e_{2,3,4,5,6,7}\left( 1 \right) = 0; quad e_{8}\left( 1 \right) = \Delta T_{\text{CS}}\left( 1 \right)$$

and for both cases, they are

(5.16-24) $$\begin{split}e_{2}\left( 1 \right) = 0; \quad e_{4}\left( 1 \right) = 0; \quad e_{6}\left( 1 \right) = 0 \\ e_{3}\left( jm \right) = 0; \quad e_{5}\left( jm \right) = 0; \quad e_{7}\left( jm \right) = 0\end{split}$$

The coefficients in Eq. (5.4-46) for the $j$-th vertical section of the tube are:

(5.16-25) $$\begin{split}c_{1}\left( j \right) = \left( \rho c \right)_{\text{TU}} \frac{1}{2} \left( P_{\text{ST}} + P_{\text{TT}} \right) d_{\text{TU}} + \Delta t \theta_{2\text{S}} P_{\text{ST}} H_{\text{ST}}\left( j \right) \\ + \Delta t \theta_{\text{ST}} P_{\text{TT}} H_{\text{TT}}\left( j \right)\end{split}$$

(5.16-26) $$c_{2}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{S}} P_{\text{ST}} H_{\text{ST}}\left( j \right)$$

(5.16-27) $$c_{3}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{S}} P_{\text{ST}} H_{\text{ST}}\left( j \right)$$

(5.16-28) $$c_{4}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{T}} P_{\text{TT}} H_{\text{TT}}\left( j \right)$$

(5.16-29) $$c_{5}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{T}} P_{\text{TT}} H_{\text{TT}}\left( j \right)$$

(5.16-30) $$\begin{split}c_{6}\left( j \right) = - \Delta t \left\lbrack P_{\text{ST}} H_{\text{ST}}\left( j \right) + P_{\text{TT}} H_{\text{TT}}\left( j \right) \right\rbrack T_{\text{TU}3}\left( j \right) \\ + \Delta t P_{\text{ST}} H_{\text{ST}}\left( j \right) \frac{1}{2}\left\lbrack T_{\text{CS}3}\left( j \right) + T_{\text{CS}3}\left( j + 1 \right) \right\rbrack \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \frac{1}{2}\left\lbrack T_{\text{CT}3}\left( j \right) + T_{\text{CT}3}\left( j + 1 \right) \right\rbrack\end{split}$$

The coefficients in Eq. (5.4-47) for the $j$-th vertical section of the tube-side coolant for normal flow (upward) are:

(5.16-31) $$\begin{split}f_{1}\left( j \right) = \frac{1}{2} A_{\text{CT}} {\overline{\rho}}_{\text{CT}}\left( j - 1 \right) {\overline{c}}_{\text{CT}}\left( j - 1 \right) + \Delta t \frac{{\overline{c}}_{\text{CT}}\left( j - 1 \right)}{\Delta z\left( j - 1 \right) S} \left| w_{\text{T}4} \right| \theta_{2\text{T}} \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j - 1 \right) \frac{1}{2} \theta_{2\text{T}}\end{split}$$

(5.16-32) $$f_{2}\left( j \right) = - \Delta t P_{\text{TT}} H_{\text{TT}}\left( j - 1 \right) \theta_{2\text{T}}$$

(5.16-33) $$f_{3}\left( j \right) = 0$$

(5.16-34) $$\begin{split}f_{4} \left( j \right) = \frac{1}{2} A_{\text{CT}} {\overline{\rho}}_{\text{CT}}\left( j - 1 \right) - \Delta t \frac{{\overline{c}}_{\text{CT}}\left( j - 1 \right)}{\Delta z \left( j - 1 \right) S} \left| w_{\text{T}4} \right| \theta_{2\text{T}} \\ + \Delta t P_{\text{TT}} H_{\text{TT}} \left( j - 1 \right) \frac{1}{2} \theta_{2\text{T}}\end{split}$$

(5.16-35) $$f_{5}\left( j \right) = 0$$

(5.16-36) $$\begin{split}f_{6}\left( j \right) = - \Delta t \frac{{\overline{c}}_{\text{CT}}\left( j - 1 \right)}{\Delta z\left( j - 1 \right) S} \left\{ \left( \left| w_{\text{T}3} \right| \theta_{1\text{T}} + \left| w_{\text{T}4} \right| \theta_{2\text{T}} \right) \left\lbrack T_{\text{CT}3}\left( j \right) - T_{\text{CT}3}\left( j - 1 \right) \right\rbrack \right\} \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j - 1 \right) \left\{ T_{\text{TU}3}\left( j - 1 \right) - \frac{1}{2} \left\lbrack T_{\text{CT}3}\left( j - 1 \right) + T_{\text{CT}3}\left( j \right) \right\rbrack \right\}\end{split}$$

The same coefficients for reversed flow (downward) in the intermediate coolant channel are:

(5.16-37) $$\begin{split}f_{1}\left( j \right) = \frac{1}{2} A_{\text{CT}} {\overline{\rho}}_{\text{CT}}\left( j \right) {\overline{c}}_{\text{CT}}\left( j \right) + \Delta t \frac{{\overline{c}}_{\text{I}}\left( j \right)}{\Delta z\left( j \right) S} \left| \ w_{\text{T}4} \right| \theta_{2\text{T}} \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \theta_{2\text{T}}\end{split}$$

(5.16-38) $$f_{2}\left( j \right) = 0$$

(5.16-39) $$f_{3}\left( j \right) = - \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \theta_{2\text{T}}$$

(5.16-40) $$f_{4}\left( j \right) = 0$$

(5.16-41) $$\begin{split}f_{5}\left( j \right) = \frac{1}{2} A_{\text{CT}} {\overline{\rho}}_{\text{CT}}\left( j \right) {\overline{c}}_{\text{CT}}\left( j \right) - \Delta t \frac{{\overline{c}}_{\text{CT}}\left( j \right)}{\Delta z\left( j \right) S} \left| \ w_{\text{T}4} \right| \theta_{2\text{T}} \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \frac{1}{2} \theta_{2\text{T}}\end{split}$$

(5.16-42) $$\begin{split}f_{6}\left( j \right) = - \Delta t \frac{{\overline{c}}_{\text{CT}}\left( j \right)}{\Delta z\left( j \right) S} \left\{ \left( \left| w_{\text{T}3} \right| \theta_{1\text{T}} + \left| w_{\text{T}4} \right| \theta_{2\text{T}} \right) \left\lbrack T_{\text{CT}3}\left( j \right) - T_{\text{CT}3}\left( j + 1 \right) \right\rbrack \right\} \\ + \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \left\{ T_{\text{TU}3}\left( j \right) - \frac{1}{2} \left\lbrack T_{\text{CT}3}\left( j \right) + T_{\text{CT}3}\left( j + 1 \right) \right\rbrack \right\}\end{split}$$

The terms for $f_{7} \left( j \right)$ and $f_{8} \left( j \right)$ have been added to Eq. (5.4-47) because they appear during the solution of the simultaneous equations. These arrays are also set to zero before the solution is begun.

Also, the boundary conditions for normal tube-side coolant channel flow are

(5.16-43) $$f_{1}\left( 1 \right) = 1; \quad f_{2,3,4,5}\left( 1 \right) = 0; \quad f_{6}\left( 1 \right) = \Delta T_{\text{CT}}\left( 1 \right)$$

For reversed tube-side channel flow, they are

(5.16-44) $$f_{1}\left( j\max \right) = 1; \quad f_{2,3,4,5}\left( j\max \right) = 0; \quad f_{6}\left( j\max \right) = \Delta T_{\text{CT}}\left( j\max \right)$$

and for both cases, they are

(5.16-45) $$\begin{split}f_{2}\left( 1 \right) = 0; &\quad f_{4}\left( 1 \right) = 0; \\ f_{1}\left( j\max \right) = 0; &\quad f_{5}\left( j\max \right) = 0\end{split}$$