.. _section-A5.1: Appendix 5.1: IHX Matrix Coefficients ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The coefficients in :eq:`eq-5.4-32` for the :math:`j`-th vertical section of the shell in terms of the quantities defined in :numref:`section-5.4.2.2` are as follows: .. math:: :label: A5.1‑1 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}}} .. math:: :label: A5.1‑2 a_{2} \left( j \right) = - \frac{1}{2} \theta_{2\text{S}} \Delta t H_{\text{S}}\left( j \right) .. math:: :label: A5.1‑3 a_{3} \left( j \right) = - \frac{1}{2} \theta_{2\text{S}} \Delta t H_{\text{S}}\left( j \right) .. math:: :label: A5.1‑4 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}}} .. math:: :label: A5.1‑5 {\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 :math:`\theta_{2\text{S}}` = the degree of implicitness for the shell-side coolant channel :math:`\Delta t` = the time interval The coefficients in :eq:`eq-5.4-33` for the :math:`j`-th vertical section of the shell-side coolant for normal flow (downward) are: .. math:: :label: A5.1‑6 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}} .. math:: :label: A5.1‑7 e_{2}\left( j \right) = 0 .. math:: :label: A5.1‑8 e_{3}\left( j \right) = - \Delta t P_{\text{S}} H_{\text{S}}\left( j \right) \theta_{2\text{S}} .. math:: :label: A5.1‑9 e_{4}\left( j \right) = 0 .. math:: :label: A5.1‑10 e_{5}\left( j \right) = - \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j \right) \theta_{2\text{S}} .. math:: :label: A5.1‑11 e_{6}\left( j \right) = 0 .. math:: :label: A5.1‑12 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}} .. math:: :label: A5.1‑13 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\} The same coefficients for reversed flow (upward) in the shell-side coolant channel are: .. math:: :label: A5.1‑14 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}} .. math:: :label: A5.1‑15 e_{2}\left( j \right) = - \Delta t P_{\text{S}}H_{\text{S}}\left( j - 1 \right) \theta_{2\text{S}} .. math:: :label: A5.1‑16 e_{3}\left( j \right) = 0 .. math:: :label: A5.1‑17 e_{4}\left( j \right) = - \Delta t S P_{\text{ST}} H_{\text{ST}}\left( j - 1 \right) \theta_{2\text{S}} .. math:: :label: A5.1‑18 e_{5}\left( j \right) = 0 .. math:: :label: A5.1‑19 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}} .. math:: :label: A5.1‑20 e_{7}\left( j \right) = 0 .. math:: :label: A5.1‑21 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\} The terms :math:`e_{9} \left( j \right)` and :math:`e_{10} \left( j \right)` have been added to :eq:`eq-5.4-33` 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 .. math:: :label: A5.1‑22 \begin{align} e_{1}\left( jm \right) = 1; && e_{2,3,4,5,6,7} \left( jm \right) = 0; && e_{8}\left( jm \right) = \Delta T_{\text{CS}}\left( jm \right) \end{align} For reversed primary channel flow, they are .. math:: :label: A5.1‑23 \begin{align} e_{1}\left( 1 \right) = 1; && e_{2,3,4,5,6,7}\left( 1 \right) = 0; && e_{8}\left( 1 \right) = \Delta T_{\text{CS}}\left( 1 \right) \end{align} and for both cases, they are .. math:: :label: A5.1‑24 \begin{align} e_{2}\left( 1 \right) = 0; && e_{4}\left( 1 \right) = 0; && e_{6}\left( 1 \right) = 0; \\ e_{3}\left( jm \right) = 0; && e_{5}\left( jm \right) = 0; && e_{7}\left( jm \right) = 0 \end{align} The coefficients in :eq:`eq-5.4-34` for the :math:`j`-th vertical section of the tube are: .. math:: :label: A5.1‑25 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) .. math:: :label: A5.1‑26 c_{2}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{S}} P_{\text{ST}} H_{\text{ST}}\left( j \right) .. math:: :label: A5.1‑27 c_{3}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{S}} P_{\text{ST}} H_{\text{ST}}\left( j \right) .. math:: :label: A5.1‑28 c_{4}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{T}} P_{\text{TT}} H_{\text{TT}}\left( j \right) .. math:: :label: A5.1‑29 c_{5}\left( j \right) = - \frac{1}{2} \Delta t \theta_{2\text{T}} P_{\text{TT}} H_{\text{TT}}\left( j \right) .. math:: :label: A5.1‑30 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 The coefficients in :eq:`eq-5.4-35` for the :math:`j`-th vertical section of the tube-side coolant for normal flow (upward) are: .. math:: :label: A5.1‑31 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}} .. math:: :label: A5.1‑32 f_{2}\left( j \right) = - \Delta t P_{\text{TT}} H_{\text{TT}}\left( j - 1 \right) \theta_{2\text{T}} .. math:: :label: A5.1‑33 f_{3}\left( j \right) = 0 .. math:: :label: A5.1‑34 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}} .. math:: :label: A5.1‑35 f_{5}\left( j \right) = 0 .. math:: :label: A5.1‑36 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\} The same coefficients for reversed flow (downward) in the intermediate coolant channel are: .. math:: :label: A5.1‑37 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}} .. math:: :label: A5.1‑38 f_{2}\left( j \right) = 0 .. math:: :label: A5.1‑39 f_{3}\left( j \right) = - \Delta t P_{\text{TT}} H_{\text{TT}}\left( j \right) \theta_{2\text{T}} .. math:: :label: A5.1‑40 f_{4}\left( j \right) = 0 .. math:: :label: A5.1‑41 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}} .. math:: :label: A5.1‑42 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\} The terms for :math:`f_{7} \left( j \right)` and :math:`f_{8} \left( j \right)` have been added to :eq:`eq-5.4-35` 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 .. math:: :label: A5.1‑43 \begin{align} f_{1}\left( 1 \right) = 1; && f_{2,3,4,5}\left( 1 \right) = 0; && f_{6}\left( 1 \right) = \Delta T_{\text{CT}}\left( 1 \right) \end{align} For reversed tube-side channel flow, they are .. math:: :label: A5.1‑44 \begin{align} f_{1}\left( j\max \right) = 1; && f_{2,3,4,5}\left( j\max \right) = 0; && f_{6}\left( j\max \right) = \Delta T_{\text{CT}}\left( j\max \right) \end{align} and for both cases, they are .. math:: :label: A5.1‑45 \begin{align} f_{2}\left( 1 \right) = 0; && f_{4}\left( 1 \right) = 0; \\ f_{1}\left( j\max \right) = 0; && f_{5}\left( j\max \right) = 0 \end{align}