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:

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