12.16. Appendices

12.16.1. Derivation of the Expression for the Spatial Derivative of the Liquid Temperature at a Liquid-Vapor Interface

The derivation of Eq. (12.5-44), the expression for $\frac{\partial T_{l}}{\partial \xi}\big|_{\xi = 0}$, from the general heat conduction equation involves a somewhat lengthy process which will be described in this appendix. The starting point is Eq. (12.5-39), the general heat conduction equation, written as

(12.16-1) $$\frac{\partial ^{2} T\left( \xi, t' \right)}{\partial \xi^{2}} + \frac{Q\left( \xi, t' \right)}{k_{l}} = \frac{1}{\alpha}\frac{\partial T \left(\xi,t' \right)}{\partial t'}$$

where the nomenclature is defined in Section 12.5. As discussed in Section 12.5, the boundary conditions for Eq. (12.16-1) are $T\left(0, t' \right)$ equal to the vapor temperature and $T \left(\xi, t' \right)$ finite, while the initial condition is that $T\left( \infty, 0 \right)$ is known. If the Laplace transform is applied to Eq. (12.16-1), the resulting equation is

(12.16-2) $$\frac{\partial ^{2} \overline{T} \left( \xi, s \right)}{\partial \xi ^{2}} + \frac{\overline{Q} \left( \xi, s \right)}{k_{l}} = \frac{1}{\alpha}\left(s \overline{T} \left(\xi, s \right) - T\left( \xi, 0 \right) \right)$$

where $\overline{T}$ and $\overline{Q}$ and are the Laplace transforms of $T$ and $Q$, respectively, and $s$ is the transform variable. Eq. (12.16-2) will be somewhat easier to work with if the function $g\left( \xi, s \right)$ is defined as

(12.16-3) $$g\left( \xi, s \right) = -\frac{\overline{Q} \left(\xi, s \right)}{k_{l}}-\frac{T\left( \xi , 0 \right)}{\alpha}$$

so that the transformed equation becomes

(12.16-4) $$\frac{\partial ^{2} \overline{T}}{\partial \xi ^{2}} - \frac{s}{\alpha} \overline{T} = g$$

The arguments $\xi$ and $s$ have been suppressed in Eq. (12.16-4) for simplicity of notation. Eq. (12.16-4) is a second-order partial differential equation in $\xi$, which can be solved in a straightforward manner. First, one additional notational simplification will be introduced by defining the variable $q$ as

(12.16-5) $$q = \sqrt{\frac{s}{\alpha}}$$

which gives

(12.16-6) $$\frac{\partial^{2} \overline{T}}{\partial \xi^{2}} - q^{2} \overline{T} = g$$

The solution to this equation can be divided into a homogeneous solution $\overline{T}_{h}$ and a particular solution $\overline{T}_{p}$

(12.16-7) $$\overline{T} = \overline{T}_{h} + \overline{T}_{p}$$

The homogeneous solution is just the solution to the equation

(12.16-8) $$\frac{\partial^{2} \overline{T}_{h}}{\partial \xi} - q^{2} \overline{T}_{h} = 0$$

so that $\overline{T}_h$ has the form

(12.16-9) $$\overline{T}_{h} = C_{1} \text{exp} \left( q \xi \right) - C_{2} \text{exp} \left(-q \xi \right)$$

where $C_{1}$ and $C_{2}$ are constants to be determined from the boundary conditions.

Finding the functional form of the particular solution is a more complex process than was finding the form of the homogeneous solution. The particular solution satisfies the equation

(12.16-10) $$\frac{\partial^{2} \overline{T}_{p}}{\partial \xi^{2}} - q^{2} \overline{T}_{p} = g$$

which can be rewritten as

(12.16-11) $$\left( \frac{\partial}{\partial \xi} - q \right) \left(\frac{\partial}{\partial \xi} + q \right) \overline{T}_{p} = g$$

If the function $u$ is defined as

(12.16-12) $$u = \frac{\partial \overline{T}_p}{\partial \xi} + q \overline{T}_{p}$$

then Eq. (12.16-11) takes the simpler form

(12.16-13) $$\frac{\partial u}{\partial \xi} - qu = g$$

Eq. (12.16-13) is a simple first-order equation which is easily solved. First, multiply the equation by $e^{-q \xi}$:

(12.16-14) $$\frac{\partial u}{\partial \xi} e^{-q \xi} - que^{-q \xi} = ge^{-q \xi}$$

which can be written more simply as

(12.16-15) $$\frac{\partial}{\partial \xi} \left( u e^{-q \xi} \right) = ge^{-q \xi}$$

Eq. (12.16-15) can then be solved for u by integrating from $0 \text{ to } \xi$,

(12.16-16) $$ue^{-q \xi} = \int_{0}^{\xi} ge^{-q \xi'} d\xi'$$

or

(12.16-17) $$u = e^{q \xi} \int_{0}^{\xi} ge^{-q \xi'} d\xi'$$

If this expression is substituted into Eq. (12.16-12), the following equation for the particular solution results:

(12.16-18) $$\frac{\partial \overline{T}_{p}}{\partial \xi} + q \overline{T}_{p} = e^{q \xi} \int_{0}^{\xi} ge^{-q \xi'} \partial \xi'$$

Eq. (12.16-18) is solved using the same procedure as was employed to solve Eq. (12.16-18) for $u$, only this time, the multiplier is $e^{q \xi}$, giving

(12.16-19) $$\frac{\partial}{\partial\xi}\left( \overline{T}_{p}e^{q\xi} \right) = e^{2q\xi}\int_{0}^{\xi}{ge^{- q\xi''}d\xi''}$$

which gives the following expression for $\overline{T}_{p}$:

(12.16-20) $$\overline{T}_{p} = e^{- q\xi}\int_{0}^{\xi}e^{2q\xi'}\left( \int_{0}^{\xi'}{ge^{- q\xi'}d\xi''} \right)d\xi''$$

Using integration by parts, Eq. (12.16-20) can be modified from a double integral to the difference of two integrals, giving

(12.16-21) $$\overline{T}_{p} = e^{q\xi}\int_{0}^{\xi}{\frac{g}{2q}e}^{- q\xi'}d\xi' - e^{- q\xi}\int_{0}^{\xi}{\frac{g}{2q}e}^{q\xi'}d\xi'$$

The Laplace transform of the liquid temperature is then just the sum of the expression for $\overline{T}_{n}$ from Eq. (12.16-9) and that for $\overline{T}_{p}$ from Eq. (12.16-21):

(12.16-22) $$\begin{split}\overline{T} = \overline{T}_{h} + \overline{T}_{p} \\ = C_{1} \text{exp} \left(q \xi \right) + C_{2} \text{exp}\left(-q \xi \right) + \text{exp}\left( q \xi \right) \int_{0}^{\xi} \frac{g}{2q} \text{exp} \left(-q \xi' \right) d \xi' \\ {} - \text{exp}\left(-q \xi \right) \int_{0}^{\xi} \frac{g}{2q} \text{exp} \left(q \xi' \right) d\xi'\end{split}$$

The constants $C_{1}$ and $C_{2}$ in Eq. (12.16-22) can be evaluated by imposing the boundary conditions stated at the beginning of the appendix. The condition at $\xi = 0$ gives

(12.16-23) $$\overline{T} \left(0 , s \right) = C_{1} + C_{2}$$

while that for $\xi \rightarrow \infty$ requires that

(12.16-24) $$C_{1} + \int_{0}^{\infty} \frac{g}{2q} \text{exp} \left(-q \xi' \right) d \xi' = 0$$

If these are applied to Eq. (12.16-22), the resulting expression for $\overline{T}$ is

(12.16-25) $$\begin{split} \overline{T}\left(\xi, s \right) = -\text{exp}\left(q \xi \right) \int_{0}^{\infty} \frac{g}{2q}\text{exp} \left( -q \xi' \right) d\xi' + \overline{T}\left(0,s\right) \text{exp} \left(-q \xi \right) \\ + \text{exp} \left(-q \xi \right) \int_{0}^{\infty} \frac{g}{2q} \text{exp}\left(-q \xi' \right) d \xi' + \text{exp}\left(q \xi \right) \int_{0}^{\xi} \frac{g}{2q} \text{exp} \left(-q \xi' \right) d\xi' \\ - \text{exp}\left( -q \xi \right) \int_{0}^{\xi} \frac{g}{2q} \text{exp} \left(q \xi' \right) d\xi'\end{split}$$

Taking the derivative of Eq. (12.16-25) with respect to $\xi$ and evaluating the result at $\xi = 0$ gives

(12.16-26) $$\begin{split}\frac{\partial \overline{T}}{\partial \xi} \bigg|_{\xi = 0} = -q \int_{0}^{\infty} \frac{g}{2q} \text{exp} \left(-q \xi' \right) d \xi' - q \overline{T} \left(0, s \right) \\ -q \int_{0}^{\infty} \frac{g}{2q} \text{exp} \left(-q \xi' \right)d\xi' + \frac{g}{2q} - \frac{g}{2q}\end{split}$$

which, substituting in the functions represented by $q$ and $g$, produces

(12.16-27) $$\begin{split}\frac{\partial \overline{T}}{\partial \xi} \bigg|_{\xi = 0} = - \overline{T} \left(0, s \right) \sqrt{\frac{s}{\alpha}} + \int_{0}^{\infty} \frac{\overline{Q} \left( \xi', s \right)}{k_{l}} \text{exp} \left(-\xi' \sqrt{\frac{s}{\alpha}} \right) d\xi' \\ + \int_{0}^{\infty} \frac{T\left(\xi', 0 \right)}{\alpha} \text{exp} \left(-\xi' \sqrt{\frac{s}{\alpha}} \right) d\xi'\end{split}$$

All that remains now is to take the inverse Laplace transform of Eq. (12.16-27). Representing the inverse transform by $L^{-1}$, the necessary inverse transforms are

(12.16-28) $$L^{-1} \left[ \frac{\partial \overline{T}}{\partial \xi} \bigg|_{\xi = 0} \right] = \frac{\partial T}{\partial \xi} \bigg|_{\xi = 0}$$

(12.16-29) $$\begin{split}L^{-1} \left[ \overline{T} \left(0, s\right) \sqrt{\frac{s}{\alpha}} \right] = L^{-1} \left[\overline{T} \left(0, s \right) \frac{s}{\sqrt{s \alpha}} \right] \\ = L^{-1} \left[s \overline{T} \left(0 , s \right) \frac{1}{\sqrt{\alpha}{\sqrt{s}}} \right] \\ = \int_{0}^{t'} \frac{1}{\sqrt{\pi\alpha}} \frac{1}{\sqrt{t'-\tau}} \left( \frac{\partial T(0,\tau)}{\partial\tau} \right) d\tau\end{split}$$

by convolution

(12.16-30) $$\begin{split}L^{-1} \left[ T \left( \xi, 0 \right) \text{exp} \left(-\xi \sqrt{\frac{s}{\alpha}} \right) \right] = T \left( \xi, 0 \right) L^{-1} \left[ \text{exp} \left(- \frac{\xi}{\sqrt{\alpha}} \sqrt{s} \right) \right] \\ = T \left(\xi, 0 \right) \frac{\xi}{2 \sqrt{\pi \alpha \left(t' \right)^{3}}} \text{exp} \left( \frac{-\xi^{2}}{4 \alpha t'} \right)\end{split}$$

(12.16-31) $$L^{-1} \left[ \overline{Q} \left(\xi, s \right) \text{exp} \left(-\xi \sqrt{\frac{s}{\alpha}} \right) \right] = \int_{0}^{t'} Q\left( \xi, \tau \right) \frac{\xi}{2 \sqrt{\pi \alpha \left(t' - \tau \right)^{3}}} \text{exp} \left(-\frac{\xi^{2}}{4\alpha\left(t' - \tau \right)} \right) d\tau$$

The above inverse transforms can be found in any good handbook which lists inverse Laplace transforms. Taking the inverse transform of Eq. (12.16-27) and using the above expressions for the individual inverse transforms finally produces Eq. (12.5-44):

(12.16-32) $$\begin{split}\frac{\partial T \left(t' \right)}{\partial \xi} \bigg|_{\xi = 0} = -\frac{1}{\sqrt{\pi \alpha}} \int_{0}^{t'} \frac{\partial T \left(0, \tau \right)}{\partial \tau} \frac{1}{\sqrt{t' - \tau}} d\tau \\ -\frac{T\left(0,0\right)}{\sqrt{\alpha \pi t'}} + \int_{0}^{\infty} d\xi \int_{0}^{t'} \frac{\xi Q \left(\xi, \tau \right) \text{exp} \left[-\frac{\xi^{2}}{4 \alpha \left(t' - \tau \right)} \right]}{2k_{l}\sqrt{\pi \alpha \left(t' - \tau \right)^{3}}} d \tau \\ + \int_{0}^{\infty} \frac{T\left( \xi, 0 \right) \xi \text{exp}\left[ -\frac{\xi^{2}} {4 \alpha t'} \right]}{2 \alpha \sqrt{\pi \alpha \left(t' \right)^{3}}} d\xi\end{split}$$

12.16.2. Gaussian Elimination Solution of Linearized Vapor-Pressure-Gradient Equations

The series of equations represented by Eq. (12.6-168) is solved as shown in the steps below. Recall that Eq. (12.6-165) $a_{2,J2 - 1} = a_{2,J1} = 1$.

Forward Reduction

1. Divide the second equation by $b_{2,J1}$:

$$b_{i,J1} \rightarrow b_{i,J1}/b_{2,J1}, i = 1 \dots 4 \quad g_{J1} \rightarrow g_{J1}/b_{2,J1}.$$

2. Divide the third equation by $c_{2,J1}$:

$$c_{i,J1} \rightarrow c_{i,J1}/c_{2,J1}, i = 1 \dots 4 \quad h_{J1} \rightarrow h_{J1}/c_{2,J1}$$

3. Subtract the first equation from the second:

$$b_{i,J1} \rightarrow b_{i,J1} - a_{1,J1} \quad b_{2,J1} \rightarrow 0 \quad g_{J1} \rightarrow g_{J1} - d_{J1}$$

4. Subtract the first equation from the third:

$$c_{1,J1} \rightarrow c_{1,J1} - a_{1,J1} \quad c_{2,J1} \rightarrow 0 \quad h_{J1} \rightarrow h_{J1} - d_{J1}$$

5. Divide the second equation by $b_{1,J1}$:

$$b_{i,J1} \rightarrow b_{i,J1}/b_{1,J1}, i=1 \dots 4 \quad g_{J1} \rightarrow g_{J1}/b_{1,J1}$$

6. Divide the third equation by $c_{1,J1}$:

$$c_{i,J1} \rightarrow c_{i,J1} / c_{1,J1}, i = 1 \dots 4 \quad h_{J1} \rightarrow h_{J1} / c_{1,J1}$$

7. Subtract the second equation from the third:

$$c_{1,J1} \rightarrow 0 \quad c_{3,J1} \rightarrow c_{3,J1} - {\widetilde{b}}_{3,J1} \quad c_{4,J1} \rightarrow c_{4,J1} - {\widetilde{b}}_{4,J1}$$

$$h_{J1} \rightarrow h_{J1} - {\widetilde{g}}_{J1}$$

Repeat Steps 8-15 for $j=J1 + 1, J1 + 2, \dots, J2-1$. Then proceed to Step 16.

8. Divide by $c_{3,j-1}$:

$$c_{4,j-1} \rightarrow c_{4,j-1} / c_{3, j-1} \quad c_{3, j-1} \rightarrow 1 \quad h_{j-1} \rightarrow h_{j-1}/c_{3,j-1} \quad \left(c_{1,j-1} = 0 \quad c_{2, j - 1} = 0\right)$$

9. Divide by $b_{1,j}$:

$$b_{i, j} \rightarrow b_{i, j} / b_{1, j}, i = 1 \dots 4 \quad g_{j} \rightarrow g_{j} / b_{1,j}$$

10. Divide by $c_{1,j}$:

$$c_{i,j} \rightarrow c_{i,j} / c_{1,j}, i = 1 \dots 4 \quad h_{j} \rightarrow h_{j} / c_{1,j}$$

11. Subtract step 8 from step 9:

$$b_{2,j} \rightarrow b_{2,j} - c_{4,j-1} \quad b_{1,j} \rightarrow 0 \quad g_{j} \rightarrow g_{j} - h_{j-1}$$

12. Subtract step 8 from step 10:

$$c_{2,j} \rightarrow c_{2,j} - c_{4, j - 1} \quad c_{1, j} \rightarrow 0 \quad h_{j} \rightarrow h_{j} - h_{j - 1}$$

13. Divide by $b_{2,j}$:

$$b_{i,j} \rightarrow b_{i,j} / b_{2, j}, i = 3,4 \quad b_{2,j} \rightarrow 1 \quad g_{j} \rightarrow g_{j} / b_{2,j}$$

14. Divide by $c_{2,j}$:

$$c_{i,j} \rightarrow c_{i,j} / c_{2, j}, i = 3,4 \quad c_{2,j} \rightarrow 1 \quad h_{j} \rightarrow h_{j} / c _{2,j}$$

15. Subtract step 13 from step 14:

$$c_{i,j} \rightarrow c_{i,j} - b_{i,j}, i = 3, 4 \quad c_{2,j} \rightarrow 0 \quad h_{j} \rightarrow h_{j} - g_{j}$$

Continue

16. Divide by $c_{4,J2-1}$:

$$c_{3, J2 - 1} \rightarrow c_{3, J2 - 1} / c_{4, J2 - 1} \quad c_{4, J2 - 1} \rightarrow 1 \quad h_{J2 - 1} \rightarrow h_{J2 - 1} / c_{4, J2 - 1}$$

17. Subtract:

$$a_{1, J2 - 1} \rightarrow a_{1, J2 - 1} - c_{3, J2 - 1} \quad a_{2, J2 - 1} \rightarrow 0 \quad d_{J2 - 1} \rightarrow d_{J2 - 1} - h_{J2 - 1}$$

Back Substitution

18. Solve for upper interface pressure:

$$\Delta p^{i}(J2) = d_{J2 - 1} /a_{1,J2 - 1}$$

19. Solve for upper interface flow:

$$\Delta W^{i} (J2) = h_{J2 - 1} - c_{3, J2 - 1} \Delta p^{i} (J2)$$

$$\Delta W (J2 - 1) = \widetilde{g}_{J2 - 1} - \widetilde{b}_{3, J2 - 1} \Delta p^{i} (J2) - \widetilde{b}_{4, J2 - 1} \Delta W^{i} (J2)$$

$$\Delta p (J2 - 1) = h_{J2 - 1} - c_{4, J2 - 1} \Delta W (J2 - 1)$$

Do steps 22 and 23 for $j = J2-2, J2-3, \ldots, J1 + 1$.

$$\Delta W(j) = g_{j} - b_{3,j} \Delta p(j+1) - b_{4, j} \Delta W(j+1)$$

$$\Delta p(j) = h_{j} - c_{4,j} \Delta W(j)$$

24. Solve for lower interface pressure:

$$\Delta p^{i}(J1) = \widetilde{g}_{J1} - \widetilde{b}_{3, J1} \Delta p(J1+1) - \widetilde{b}_{4, J1} \Delta W(J1+1)$$

25. Solve for lower interface flow:

$$\Delta W^{i}(J1) = d_{J1} - a_{1,J1} \Delta p^{i} (J1)$$