New Perspective for Heat Transfer Evaluation During Film Condensation Inside Tubes

If in a tube a three-dimensional control volume is taken inside and it is also considered that the heat flux by conduction cannot be neglected, then, the equation for the conservation of energy applied to the analyzed tube is described by [9]:

$\rho C_{p} \frac{\partial T}{\partial t}=\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right)+\frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y}\right)+\frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right)$    (2)

The energy equation for the fluid is described by:

$\begin{aligned}

\rho c\left[\frac{\partial T}{\partial \tau}+V_{x} \frac{\partial T}{\partial x}+V_{y}\right.&\left.\frac{\partial T}{\partial y}+V_{z} \frac{\partial T}{\partial z}\right] \\

&=k\left(\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}+\frac{\partial^{2} T}{\partial z^{2}}\right)+\mu \Phi

\end{aligned}$     (3)

In Eq. (3), the viscous cutting effects are considered by means of the Schlichting function of viscous dissipation $\Phi$ [4]:

$\Phi=2\left[\left(\frac{\partial V_{x}}{\partial x}\right)^{2}+\left(\frac{\partial V_{y}}{\partial y}\right)^{2}+\left(\frac{\partial V_{z}}{\partial z}\right)^{2}\right]+\left(\frac{\partial V_{x}}{\partial x}+\frac{\partial V_{y}}{\partial y}+\frac{\partial V_{z}}{\partial z}\right)^{2}$     (4)

The viscous dissipation function is considered because it can generate a very important influence on high velocity flows. The compressibility of the flow is present due to the existence of a phase change, for this reason it is necessary to consider the density variations associated with the phase change. The process can be considered as continuous, therefore the continuity equation is:

$\frac{\partial \rho}{\partial \tau}+\frac{\partial\left(\rho V_{x}\right)}{\partial x}+\frac{\partial\left(\rho V_{y}\right)}{\partial y}+\frac{\partial\left(\rho V_{z}\right)}{\partial z}=0$    (5)

In the tube, an elemental section is taken as a control volume (see Figure 1). In this analysis, the flow is assumed to be one-dimensional, so the directional transport effects along the tube (axial) are more important than the radial ones. Therefore, Eqns. (2), (3), (4) and (5) can be simplified to expressions (6), (7), (8) and (9):

One-dimensional energy equation. for the tube

$\rho c_{p} \frac{\partial T}{\partial \tau}=\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right)$    (6)

One-dimensional energy equation. for the fluid

$\rho c\left[\frac{\partial T}{\partial \tau}+V_{x} \frac{\partial T}{\partial x}\right]=k\left(\frac{\partial^{2} T}{\partial x^{2}}\right)+\mu \Phi$    (7)

One-dimensional function of the viscous dissipation term

$\Phi=3\left(\frac{\partial V_{x}}{\partial x}\right)^{2}$    (8)

One-dimensional continuity equation

$\frac{\partial \rho}{\partial \tau}+\frac{\partial\left(\rho V_{x}\right)}{\partial x}=0$    (9)

In Figure 1, assuming that the heat exchange process is stationary, segments PB and AQ are the respective inputs and outputs of the heat flow to the analyzed control volume in the PB and AQ segments the heat transfer process can be described by means of an unknown functional, described by the following relationships:

$\left\{\begin{array}{ll}

x=X_{1}(t) & \text { for } P B \\

x=X_{2}(t) & \text { for } A Q

\end{array}\right.$    (10)

The mathematical axiom of the maximum value generates that the analyzed problem has a univocal and continuous solution throughout its interval. For this purpose, the Eq. (6) is applied, obtaining that the Green differential is:

$\Xi_{(T)}=-a^{2} \frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial T}{\partial t}$    (11)

1.png

Figure 1. Elementary volume used in the analyzed problem

The elemental control volume PAQB is divided into four sub-elements PB, AQ, BQ and AP. Therefore, for each segment, the integration of the differential Eq. (6) and its homogeneous combination to Green's Integral produces as a result a complex integral that includes the sum of four integral relations, one for each sub-element respectively [10, 11]:

$\begin{array}{c}

\int_{P B} \varphi \psi d x-\int_{A Q} \varphi \psi d x+\int_{B Q}[\varphi \psi d x \\

\left.+a^{2}\left(\psi \frac{\partial \varphi}{\partial x}-\varphi \frac{\partial \psi}{\partial x}\right) d t\right] \\

-\int_{A P}\left[\varphi \psi d x+a^{2}\left(\psi \frac{\partial \varphi}{\partial x}-\varphi \frac{\partial \psi}{\partial x}\right) d t\right]=0

\end{array}$       (12)

In Eq. (12) the term $\varphi(x, t)$ is the solution for heat transfer process.

If the Green functional is equated to zero, then the source function $\psi=G_{0}(x, t, \xi, \tau)$  is generated. This equation can be expressed in terms of infinite line as:

$\psi=G_{0}(x, t, \xi, t)=\frac{1}{2 \sqrt{\pi a^{2}(t-\tau)^{2}}} e^{-\frac{(x-\xi)^{2}}{4 a^{2}(t-\tau)}}$     (13)

The solution for heat transfer process $\varphi(x, t)$ is obtained in the elementary volume PAQB, assuming that in the inward contour of the control volume the solution of the heat transfer problem is given by $\varphi(x, t+h), \text { where } h>0$. Substituting $x-h=\xi \text { and } t=\tau+h$, then Eq. (13) is transformed to:

$\psi=G_{0}(x, t)=\frac{1}{2 \sqrt{x a^{2} h^{2}}} e^{-\frac{(2 x-h)}{4 a^{2} h}}$    (14)

$\int_{P B} \varphi(x, t) \frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{\frac{(2 x-h)^{2}}{4 a^{2} h}} d x-\int_{A Q} \varphi(x, t) \frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} d x-$

$+\int_{B Q}\left[\varphi(x, t) \frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} d x+a^{2}\left(\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} \frac{\partial \varphi(x, t)}{\partial x}-\varphi(x, t) \frac{\partial\left(\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}}\right)}{\partial x}\right) d t\right]-$         (15)

$+\int_{A P}\left[\varphi(x, t) \frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} d x+a^{2}\left(\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} \frac{\partial \varphi(x, t)}{\partial x}-\varphi(x, t) \frac{\partial\left(\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}}\right)}{\partial x}\right) d t\right]=0$

Applying the limit when $h \rightarrow 0$ in Eq. (14), considering that is a continuous function. The obtained result is substituted into Eq. (12), then, Eq. (15) is generated.

Eq. (15) is very complex, and its solution by traditional integration methods would require considerable time and greater rigor. However, this difficulty can be overcome if numerical solution techniques are used, among them the benefits provided by the finite element method (FEM). For this purpose, we start from the criterion that by means of FEM techniques, it is possible to obtain a weak solution of the problem studied. For this purpose, the minimum energy principle is applied on the control volume PABQ, which is equivalent to finding a minimum of the integral complex shown in Eq. (15). For this purpose the following substitutions are required:

$\bar{\omega} \varphi(x, t) \frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}}=\bar{\omega}$    (16)

or:

$\begin{array}{l}

\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{-\frac{(2 x-h)^{2}}{4 a^{2} h}} \frac{\partial \varphi(x, t)}{\partial x} \\

-\varphi(x, t) \frac{\partial\left(\frac{1}{2 \sqrt{\pi a^{2} h^{2}}} e^{\left.-\frac{(2 x-h)^{2}\quad}{4 a^{2} h}\right)}\right.}{\partial x}=\omega

\end{array}$    (17)

That proves to be equivalent:

$\bar{\omega} \frac{\partial \varphi}{\partial x}-\varphi(x, t) \frac{\partial \bar{\omega}}{\partial x}=\omega$     (18)

Substituting Eq. (18) and Eq. (16), into Eq. (15):

$\begin{array}{c}

\int_{P B} \bar{\omega} d x-\int_{A Q} \bar{\omega} d x+\int_{B Q}\left[\bar{\omega} d x+a^{2}\left(\bar{\omega} \frac{\partial \varphi(x, t)}{\partial x}\right.\right. \\

\left.\left.-\varphi(x, t) \frac{\partial \bar{\omega}}{\partial x}\right) d t\right] \\

-\int_{A P}\left[\bar{\omega} d x+a^{2}\left(\bar{\omega} \frac{\partial \varphi(x, t)}{\partial x}-\right.\right. \\

\left.\left.-\varphi(x, t) \frac{\partial \bar{\omega}}{\partial x}\right) d t\right]=0

\end{array}$     (19)

With the application of FEM techniques, Eq. (19) can be transformed into a local domain, which in turn would be integrated of two linear elements composed of three nodes each (one-dimensional quadratic element). If second order parabolic arcs are used instead of linear segments in the free meshing adjustment, it is possible to obtain a more cautious approximation for the temperature distribution profile. The one-dimensional element used (see Figure 2), has a node at each end and a third node located in the center of the element [9, 10].

2.png

Figure 2. One-dimensional element (quadratic)

For these elements, the form functions are given by:

$N_{1}=-\frac{1}{2} \xi(1-\xi) ; N_{2}=(1+\xi)(1+\xi)$; $N_{3}=\frac{1}{2} \xi(1+\xi)$    (20)

The nodal solutions $N_{1}, N_{2} \text { and } N_{3}$ combined in the form of a product generate a weak function, which allows describing the temperature field and its distribution $d x d \tau$ by conduction along the axial dimension $d x$. Applying the substitution $x-h=\xi$ and assuming the extreme limit when $h=0$, then, this transformation allows us to reach the term a $x=\xi$. Therefore, three quadratic elements are possible, which is equivalent to having three work zones [11]:

$N_{1}=-\frac{1}{2} x+x^{2} ; N_{2}=1+2 x+x^{2}$; $N_{3}=\frac{1}{2} x+x^{2}$     (21)

If the variable dependence with time is unknown, then it is required to use the simplified energy equation, Eq. (3), and substitute it in the parabolic function present in the nodal solution. This technique is necessary for each integral element of the simplified Eq. (19), then:

Segment PB

Entry

$\left(-\frac{1}{2} x+x^{2}\right) V_{X}=a\left(-\frac{1}{2} x+x^{2}\right)^{2}+3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$   (22)

Intermediate

$\left(1-2 x+x^{2}\right) V_{x}=a\left(1-2 x+x^{2}\right)^{2}+3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$     (23)

Exit

$\left(\frac{1}{2} x+x^{2}\right) V_{X}=a\left(\frac{1}{2} x+x^{2}\right)^{2}+3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$     (24)

Segment AQ

Entry

$-\left(-\frac{1}{2} x+x^{2}\right) V_{X}=-a\left(-\frac{1}{2} x+x^{2}\right)^{2}+-3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$    (25)

Intermediate

$-\left(1+2 x+x^{2}\right) V_{x}=-a\left(1+2 x+x^{2}\right)^{2}-3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$      (26)

Exit

$-\left(\frac{1}{2} x+x^{2}\right) V_{X}=-a\left(\frac{1}{2} x+x^{2}\right)^{2}-3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$     (27)

When reviewing the control volume, it is verified that the sides BQ and AP do not include vertical components, however, if they consider the infinite source function, Eq. (14), for which it is required that the variation of physical properties a along the nodal segment are subordinate to the continuity equation, Eq. (9). The density changes as the length of the tube increases, due to the phase change and the compressibility of the vapor. If the process is considered as stationary, then $\partial \rho / \partial \tau=0$, therefore, Eq. (9) is transformed to [12]:

$\frac{\partial\left(\rho V_{x}\right)}{\partial x}=0$   (28)

If vapor density at the inlet is called $\rho_{V}$, and liquid density at the outlet is called $\rho_{L}$. Then, when checking the viscous stress diagram (see Figure 1), two fundamental forces are distinguished, the first associated with the drag of the vapor and the second is linked to the gravitational force, while in the opposite direction the effect of viscous forces (friction) appears. As the entire heat transfer process happens in a continuous, confined and oriented medium, then the velocity variation depends on a characteristic dimension of the duct, so it can be established that the differential Eq. (27) can be transformed to generate the relationship that allows evaluating the velocity profile (differential equation of the velocity profile), and that is also valid for any section of the conduit examined, this is:

$\mu_{L} \frac{\partial^{2} V}{\partial x^{2}}+\left(\rho_{L}+\rho_{V}\right) g=0$    (29)

In Eq. (29), $V(x)$ is the velocity through the film, for any value of x. To solve this problem, two boundary conditions are required. On the wall the condition of non-slip fluid is applied, therefore:

$x=0 \text { and } V=0$    (30)

On the film surface, the vapor drag is assumed can be despised. If the function $\delta(x)$ is considered to be the thickness of the film, the required boundary condition will then be as follows:

$x=\delta ; \frac{\partial V}{\partial x}=0$   (31)

The condensate film thickness $\delta(x)$  has not been determined. Considering that the vapor drag is negligible can be a valid assumption on many occasions; however, it is applicable only when the vapor velocity is not very high. By integrating the differential Eq. (29), the following is obtained [13, 14]:

$\frac{\partial V}{\partial x}=-\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}+C_{1}$     (32)

Applying in Eq. (32) the condition of contour given in Eq. (31), then:

$0=-\delta \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}+C_{1}$     (33)

Solving the integration constant $C_{1}$ in Eq. (33):

$C_{1}=\delta \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}$    (34)

Replacing Eq. (34) and the contour condition given in Eq. (31) into Eq. (32)

$\frac{\partial V}{\partial x}=\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(\delta-x)$    (35)

Integrating again the differential Eq. (33) gives:

$V=\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left(\delta x-\frac{x^{2}}{2}\right)+C_{2}$    (36)

Using the boundary condition stated in Eq. (31) and substituting it into Eq. (36) leads to $C_{2}=0$. Reordering Eq. (36) later:

$V=\delta^{2} \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left[\frac{x}{\delta}-\frac{1}{2}\left(\frac{x}{\delta}\right)^{2}\right]$    (37)

Eq. (37) shows that the velocity profile $V(x)$ is parabolic. The velocity will have a maximum value on the surface of the film when $x=\delta$. Therefore, the maximum velocity can be obtained, if the boundary condition given in Eq. (31) is substituted in Eq. (37), then [15]:

$V_{M a x}=\delta^{2} \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{2 \mu_{L}}=\frac{g \delta^{2} \sin \theta}{2 v_{L}} \frac{\left(\rho_{L}-\rho_{V}\right)}{\rho_{L}}$     (38)

Replacing Eq. (35) in Eq. (28):

$\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(\delta-x)=0$    (39)

Therefore, the arbitrary function $\varphi(x, t)$ that was assumed to establish the solution of the thermal exchange problem, takes into account the variation of the velocity profile along the characteristic length of the conduit examined.

If additionally, it is considered that the arbitrary function $\varphi(x, t)$ is similar in its numerical and functional value to Eq. (39), then, in the third and fourth terms of Eq. (19) the arbitrary function $\varphi(x, t)$  can be replaced by Eq. (37). Integrating:

$\begin{array}{l}

\bar{\omega} x+\left\{\frac{\partial}{\partial x}\left[\frac{a^{2} t \bar{\omega}\left(\rho_{L}-\rho_{V}\right) g \sin \theta\left(\delta^{2}-x\right)}{\mu_{L}}\right]+\right. \\

\left.-\frac{a^{2} t\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left(\delta^{2}-x\right) \frac{\partial \bar{\omega}}{\partial x}\right\}-\frac{1}{2} \bar{\omega} x+ \\

+a^{2} t\left[\frac{\bar{\omega} \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left(\delta^{2}-x\right)}{\partial x}-\right. \\

\left.-\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left(\delta^{2}-x\right) \frac{\partial \bar{\omega}}{\partial x}\right]

\end{array}$    (40)

When the velocity is a maximum, then the equality $x=\delta$ is fulfilled. Applying this equality in Eq. (38):

$\begin{aligned}

\bar{\omega} & x^{2}+\frac{a^{2} t \bar{\omega}\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}-\\

&-\bar{\omega} \frac{a^{2} t\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}+\\

+& \frac{a^{2} t \bar{\omega}\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}-\bar{\omega} x+\\

+& \frac{a^{2} t \bar{\omega}\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}=0

\end{aligned}$     (41)

Simplifying Eq. (41)

$\bar{\omega} x^{2}-\bar{\omega} x=0$    (42)

Eq. (42) turns out to be identical to the term that describes the input discretization in the nodal distribution, therefore, the remaining nodal combinations (two) are applicable to the two finite elements that cover the horizontal segments, and the node effect initial in the third and fourth terms of Eq. (19) can be neglected. For this reason, it can be established that these initial nodes are stationary or origin points, therefore:

Segment BQ

Entry

$V_{X}=a x+3 \mu x\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$     (43)

Intermediate

$\begin{array}{l}

\left(1+2 x+x^{2}\right) V_{x}=a x\left(1+2 x+x^{2}\right)^{2}+ \\

+3 \mu x\left(\frac{\partial V_{X}}{\partial x}\right)^{2}

\end{array}$    (44)

Exit

$\left(\frac{1}{2} x+x^{2}\right) V_{X}=a x\left(\frac{1}{2} x+x^{2}\right)^{2}+3 \mu x\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$   (45)

Segment AP

Entry

$-V_{X}=-a x-3 \mu\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$    (46)

Intermediate

$\begin{array}{c}

-\left(1+2 x+x^{2}\right) V_{X}=-a x\left(1+2 x+x^{2}\right)^{2}+ \\

-3 \mu x\left(\frac{\partial V_{X}}{\partial x}\right)^{2}

\end{array}$    (47)

Exit

$-\left(\frac{1}{2} x+x^{2}\right) V_{x}=-a x\left(\frac{1}{2} x+x^{2}\right)^{2}-3 \mu x\left(\frac{\partial V_{X}}{\partial x}\right)^{2}$    (48)

Substituting Eq. (43) into Eqns. (35) and (37) gives:

$\begin{aligned}

\delta^{2} \frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\left[\frac{x}{\delta}-\frac{1}{2}\left(\frac{x}{\delta}\right)^{2}\right]=a x+\\

+3 \mu x\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(\delta-x)\right]^{2}

\end{aligned}$    (49)

$\begin{array}{l}

\frac{x \delta\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}-\frac{x^{2}\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{2 \mu_{L}} \\

=a x+3 \mu x\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(\delta-x)\right]^{2}

\end{array}$     (50)

Grouping and reducing similar terms in Eq. (50):

$(\delta-x)=a+\left(3 \mu x^{2}+\frac{\delta^{2}}{x}-2 \delta\right)\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\right]$    (51)

As x is the characteristic dimension, then, $x=r=d / 2$. Substituting this term into Eq. (51):

$\begin{aligned}

\left(\delta-\frac{d}{2}\right)=& a+\left[3 \mu\left(\frac{d}{2}\right)^{2}+\frac{2 \delta^{2}}{d}-2 \delta\right] . \\

&\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\right]

\end{aligned}$   (52)

Solving for the diameter in Eq. (52)

$d=2 \delta-2 a-\left(1.5 \mu d^{2}+\frac{4 \delta^{2}}{d}-4 \delta\right)\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\right]$    (53)

Eq. (53) is a transcendent type Equation, and allows obtaining the required duct diameter for a pre-established flow of condensate through a horizontal tube, in order to avoid fluid stagnation. In a horizontal tube $\theta=0^{\circ} ; \text { therefore, } \sin \theta=0$ and the Eq. (53) can be simplified, obtaining:

$d=2 \delta-2 a$    (54)

or:

$x=2(\delta-a)$   (55)

If V_X≈V Max is considered, Eq. (38) now becomes [16]:

$V_{X}=\frac{g d^{2} \sin \theta}{2 v_{L}} \frac{\left(\rho_{L}-\rho_{V}\right)}{\rho_{L}}$    (56)

The velocity gradient is described by Eq. (35), in which the velocity profile was assumed for the analysis as parabolic. If additionally, it is considered that the maximum velocity is obtained in the intermediate node, then Eqns. (55), (56) and (57) are substituted in the intermediate segment BQ (Eq. (44), obtaining Eq. (57):

Segment BQ (Intermediate)

$\begin{array}{l}

\left(1+4 \delta-4 a+4 \delta^{2}-8 a \delta+4 a^{2}\right) \frac{g \delta^{2} \sin \theta}{2 v_{L}} \frac{\left(\rho_{L}-\rho_{V}\right)}{\rho_{L}} \\

=\left(2 a \delta-2 a^{2}\right) \cdot\left(1+4 \delta-4 a+4 \delta^{2}-8 a \delta+4 a^{2}\right)^{2} \\

+(6 \mu \delta-6 \mu a)\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(2 a-\delta)\right]^{2}

\end{array}$    (57)

Reducing similar terms in Eq. (57):

$\begin{array}{c}

\frac{\left(\rho_{L}-\rho_{V}\right) g \delta^{2} \sin \theta}{2 \mu_{L}} \\

\frac{(6 \mu \delta-6 \mu a)\left[\frac{\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}(2 a-\delta)\right]^{2}}{\left(1+4 \delta-4 a+4 \delta^{2}-8 a \delta+4 a^{2}\right)}+ \\

-\frac{\left(1+4 \delta-4 a+4 \delta^{2}-8 a \delta+4 a^{2}\right)}{=\left(2 a \delta-2 a^{2}\right)}

\end{array}$     (58)

Eq. (58) is a weak solution for the nodal discretization developed to establish the nodal distribution of the condensate at the midpoint of the horizontal control volume. When the velocity is lower, it is satisfied that $\delta \approx 0$, therefore, Eq. (58) becomes:

$\frac{(6 \mu a)\left[\frac{2 a\left(\rho_{L}-\rho_{V}\right) g \sin \theta}{\mu_{L}}\quad\right]^{2}}{\left(1-4 a+4 a^{2}\right)}=2 a^{2}$     (59)

From Eq. (55):

$a=\frac{x}{2}-\delta=\frac{d}{4} ; \quad \Rightarrow \quad x=\frac{d}{2} ; \delta=0$    (60)

Substituting Eq. (60) into Eq. (59) and simplifying:

$\frac{\left[d\left(\rho_{L}-\rho_{V}\right) g \sin \theta\right]^{2}}{\sqrt{6 \mu}\left(\mu_{L} d^{2}-\mu_{L} d\right)^{2}}=0$     (61)

The basic problem is studied for vertical installations, therefore, for horizontal tubes $\theta=90^{\circ}$, then $\sin \theta=1$. Transforming conveniently in Eq. (61) and applying radicals’ properties, then:

$\frac{\sqrt{g d\left(\rho_{L}-\rho_{V}\right)}}{6 \mu}=0$   (62)

where, 6μ is the viscous dissipation function [17]. The viscous dissipation term is the quotient between the density of the vapor and the mass flow:

$6 \mu=G / \rho_{V}^{0.5}$   (63)

The effect generated by variation of the thermodynamic vapor quality $x^{\prime}$, is included in the viscous dissipation criteria [9-11], then Eq. (61) is transformed to:

$6 \mu=G x / \rho_{V}^{0.5}$    (64)

Replacing Eq. (64) into Eq. (62):

$\frac{\sqrt{g d \rho_{V}\left(\rho_{L}-\rho_{V}\right)}}{x G}=0$    (65)

The dimensionless parameter shown in Eq. (65) was generated with the use of weak formulations, applying FEM techniques. The dimensional parameter given by Eq. (65) is known as dimensionless velocity (1/J_g) and is an essential and required element for the description and prediction of film condensation inside tubes [18-20].

$J_{g}=\frac{x G}{\sqrt{g d \rho_{V}\left(\rho_{L}-\rho_{V}\right)}}$    (66)

For vertical and inclined tubes, the application of weak solutions leads to a result identical to that obtained in Eq. (66) [21-23]. A criterion currently accepted is the use of the Shah parameter to define the prevailing condensation regime [8], establishing that three types of thermal regimes are possible inside vertical and inclined tubes. The parameter given by Shah is described by means of the following expression:

$Z=\left(\frac{1-x}{x}\right)^{0.8} \operatorname{Pr}_{L}^{0.4}$    (67)

Condensation inside tubes is governed and directly conditioned by two dimensionless parameters, Eqns. (66) and (67) [24-26]. The finite element evaluated, for the case of vertical and inclined geometric configuration, is made up of three nodes; therefore, it is required to identify the three regions generated for this purpose. However, for horizontal configurations, the intermediate node is suppressed, causing only two regions to be required. To define the validity range of each zone for the case of inclined and vertical tubes, (see Figure 1) it is assumed that the flow enters the elementary section through segment PB and leaves this section through segment QA.

To define the validity range of zone 1, two input nodes in segments PB and AQ must be taken into account. For the first, Eqns. (35), (38) and (55) are substituted in Eq. (22), while for the second we proceed in the same way with Eqns. (35), (38) and (55), those that are substituted in Eq. (25). The two new expressions obtained are added and subsequently the value of this sum is equated with Eq. (64), solving this equality as a function of dimensionless velocity.

To define the validity range of zone 2, two intermediate nodes in the PB and AQ segments must be taken into account. For the first, Eqns. (35), (38) and (55) are substituted in Eq. (23), while for the second one proceeds in the same way with Eqns. (35), (38) and (55), those that are substituted in Eq. (26). The two new expressions obtained are added and subsequently the value of this sum is equated with Eq. (64), solving this equality as a function of dimensionless velocity.

To define the validity range of zone 3, two output nodes must be taken into account in the PB and AQ segments. For the first, Eqns. (35), (38) and (55) are substituted in Equation (24), while for the second one proceeds in the same way with Eqns. (35), (38) and (55), those that are substituted in Eq. (27). The two new expressions obtained are added and subsequently the value of this sum is equated with Eq. (64), solving this equality as a function of dimensionless velocity.

The mathematical transformations required in the three previous paragraphs turn out to be extremely voluminous and complex, therefore, in the paper only the final results of the mathematical operations performed will be given [27-29]:

For vertical and inclined tubes

Zone 1                                                      $J_{g} \geq \frac{1}{2.37 Z+0.728}$      (68)

Zone 2                          $0.927 e^{\left(-0.0868 Z^{-1.165}\right)}<J_{g}<\frac{1}{2.37 Z+0.728}$      (69)

Zone 3                                    $J_{g} \leq 0.927 e^{\left(-0.0868 Z^{-1.165}\right)}$     (70)

To define the validity range of each zone, in the case of horizontal tubes (see Figure 1), it is assumed that the flow enters the control volume through the BQ segment and leaves it through the AP segment.

To establish the validity range of Zone 1, two input nodes are defined in the BQ and AP segments. For the first, Eqns. (35), (38) and (55) are replaced in Eq. (43), while for the second, Eqns. (36), (39) and (56) are replaced in Eq. (46). The two new relations obtained are added later and the result of the sum is equated with Eq. (64), solving this equality as a function of dimensionless velocity.

To establish the validity range of zone 2, two input nodes are defined in the BQ and AP segments. For the first, Eqns. (35), (38) and (55) are replaced in Eq. (45), while for the second; Eqns. (36), (39) and (56) are replaced in Eq. (48). The two new relations obtained are added later and the result of the sum is equated with Eq. (64), solving this equality as a function of dimensionless velocity.

The mathematical transformations required in the two previous paragraphs turn out to be extremely voluminous and complex, therefore, in the paper only the final results of the mathematical operations performed will be given [30-32]:

For horizontal pipes

Zone 1                                                      $J_{g} \leq 0.979(Z+0.262)^{-0.618}$     (71)

Zone 2                                                     $J_{g}>0.979(Z+0.262)^{-0.618}$                (72)

A new model for heat transfer calculations in film condensation inside tubes was obtained. For this purpose, a modeling process based on FEM techniques was used, with two basic dimensionless groups being determined as guiding principles of the heat transfer process. The analysis process allows to obtain a new improved model to evaluate the heat transfer inside tubes during film condensation. This new model is valid for horizontal, vertical end inclined tubes.

The proposed model was correlated with available experimental data and it was also verified with other relationships existing in the literature, and that have been provided by other authors, detecting a better fit in the proposed model, with a mean deviation of 13.4% for horizontal tubes and 14.9% for vertical and inclined tubes.

It has been shown that the model obtained provides favorable results for flow values from 3 to 850 kg/m²s, pipe diameters from 2 to 50 mm and in the range of reduced pressures from 0.0008 to 0.91. Available data from 22 fluids were used for this purpose, including various organic compounds, water and refrigerants. In the compilation of these data, the works of a total of 33 authors of recognized prestige in the research area were consulted.