An Analytical Method for Determining the Convection Heat Transfer Coefficient Between Flowing Fluid and Rock Fracture Walls

Abstract

The convective heat transfer coefficient (HTC) is a useful indicator that characterizes the convective heat transfer properties between flowing fluid and hot dry rock. An analytical method is developed to explore a more realistic formula for the HTC. First, a heat transfer model is described that can be used to determine the general expression of the HTC. As one of the novel elements, the new model can consider an arbitrary function of temperature distribution on the fracture wall along the direction of the rock radius. The resulting Dirichlet problem of the Laplace equation on a semi-disk is successfully solved with the Green’s function method. Four specific formulas for the HTC are derived and compared by assuming the temperature distributions along the radius of the fracture wall to be zeroth-, first-, second-, and third-order polynomials. Comparative verification of the four specific formulas based on the test data shows that the formula A corresponding to the zeroth-order polynomial always predicts stable HTC values. At low flow rates, the four formulas predict similar values of HTC, but at higher flow rates, formulas B and D, respectively, corresponding to the first- and third-order polynomials, predict either too large or too small values of the HTC, while formula C, corresponding to the second-order polynomial, predicts relatively acceptable HTC values. However, we cannot tell which one is the more rational formula between formulas A and C due to the limited information measured. One of the clear advantages of formula C is that it can avoid the drawbacks of the discontinuity of temperature and the singular integral of HTC at the points (±R, 0). Further experimental work to measure the actual temperature distribution of water in the fracture will be of great value. It is also found that the absorbed heat of the fluid, Q, has a significant impact on the prediction results of the HTC. The temperatures at the inlet and the outlet used for Q should be consistent with the assumptions adopted in the derivation of its corresponding HTC formula. A mismatched value of Q might be the reason that some existing HTC formulas predict negative or extremely large HTCs at high flow rates.

Appendix: Solution to Dirichlet Problem of Laplace Equation on a Semi-disk

1.1 Green Function for Semi-disk in a Plane

The Dirichlet problem of Laplace equation on a semi-disk reads,

$$ \left\{ {\begin{array}{*{20}l} {\Delta u = 0\quad {\text{in}}\;\varOmega \, } \hfill \\ {u = \left\{ {\begin{array}{*{20}l} {u_{c} } \hfill & {{\text{on}}\;L_{1} } \hfill \\ {f_{1} (x)} \hfill & {{\text{on}}\;L_{2} } \hfill \\ {f_{2} (x) \, } \hfill & {{\text{on}}\;L_{3} } \hfill \\ \end{array} } \right.} \hfill \\ \end{array} } \right. $$

The materials below originated from Chapter 12 (Asmar 2005). We would like to apply the so-called method of images, which uses basic facts from plane geometry about the circle and reflection, to derive the Green functions. Generally, the Green’s function for $\varOmega \subset R^{2}$ is of the form,

$$ G\left( {x,y,x_{0} ,y_{0} } \right) = \frac{1}{2}\ln \left( {\left( {x - x_{0} } \right)^{2} + \left( {y - y_{0} } \right)^{2} + h\left( {x,y,x_{0} ,y_{0} } \right)} \right) $$

(25)

where h satisfies

$$ \left\{ {\begin{array}{*{20}l} {\Delta \nu = 0} \hfill & {{\text{in}}\;\varOmega } \hfill \\ {\nu = - \frac{1}{2}\ln \left( {\left( {x - x_{0} } \right)^{2} + \left( {y - y_{0} } \right)^{2} } \right)} \hfill & {{\text{on}}\;\partial \varOmega } \hfill \\ \end{array} } \right. $$

(26)

and then

$$ G\left( {x,y,x_{0} ,y_{0} } \right) = 0,\quad {\text{on}}\;\partial \varOmega $$

(27)

Now we introduce a well-known transformation called the Steiner inversion, see Section 12.4 (Asmar 2005). When A is not the origin, the point A* is the inverse of the point A through the circle {(x, y) $\in$ R ²: |x|² + |y|² = R ²}. By definition of the Steiner inversion, A* is the point on the ray from the origin O through A at a distance such that $OA \cdot OA^{*} = R^{2}$.

In particular, if A = (x ₀,y ₀) ∈ {(x, y) $\in$ R ²: |x|² + |y|² ≤ R ²}, then

$$ \left( {x_{0}^{*} ,y_{0}^{*} } \right) = \frac{{R^{2} }}{{\left| {x_{0} ,y_{0} } \right|^{2} }}\left( {x_{0} ,y_{0} } \right) = \frac{{R^{2} }}{{\left| {x_{0} } \right|^{2} + \left| {y_{0} } \right|^{2} }}\left( {x_{0} ,y_{0} } \right) $$

(28)

Let

$$ G_{1} \left( {x,y,x_{0} ,y_{0} } \right) = \frac{1}{2}\ln \left[ {\frac{{\left( {x - x_{0} } \right)^{2} + \left( {y - y_{0} } \right)^{2} }}{{\left( {x - x_{0}^{*} } \right)^{2} + \left( {y - y_{0}^{*} } \right)^{2} }}} \right] - \frac{{\ln \sqrt {x_{0}^{2} + y_{0}^{2} } }}{R} $$

(29)

Then

$$ G_{1} \left( {x,y,x_{0} ,y_{0} } \right),\quad {\text{on}}\,L_{1} = \left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} = R^{2} ,y > 0} \right\} $$

(30)

The reflection of $\left( {x_{0} ,y_{0} } \right)$ about x-axis (y = 0) is $\left( {x_{0} , - y_{0} } \right),$ let

$$ \begin{aligned} & G\left( {x,y,x_{0} ,y_{0} } \right) = G_{1} \left( {x,y,x_{0} ,y_{0} } \right) - G_{1} \left( {x,y,x_{0} , - y_{0} } \right) \\ & \quad = \frac{1}{2}\ln \left[ {\frac{{\left( {x - x_{0} } \right)^{2} + \left( {y - y_{0} } \right)^{2} }}{{\left( {x - x_{0}^{*} } \right)^{2} + \left( {y - y_{0}^{*} } \right)^{2} }}} \right] - \frac{1}{2}\ln \left[ {\frac{{\left( {x - x_{0} } \right)^{2} + \left( {y + y_{0} } \right)^{2} }}{{\left( {x - x_{0}^{*} } \right)^{2} + \left( {y + y_{0}^{*} } \right)^{2} }}} \right] \\ \end{aligned} $$

(31)

then

$$ \left. {G\left( {x,y,x_{0} ,y_{0} } \right)} \right|_{y = 0}\,=\,0 $$

which, by virtue of (29), implies

$$ G\left( {x,y,x_{0} ,y_{0} } \right) = 0,\quad {\text{on}}\; \partial \varOmega $$

(32)

with $\varOmega = \left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} < R^{2} ,y > 0} \right\},$ and therefore $G\left( {x,y,x_{0} ,y_{0} } \right)$ is the unique Green’s function determined by the region $\varOmega$. The uniqueness of Green’s function is from Theorem 3, Section 12.3, in the work by Asmar (2005).

We separate the boundary of $\varOmega$ into two parts: One is a semicircle $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} = R^{2} ,y > 0} \right\},$ and the other is $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} \le R^{2} ,y = 0} \right\}.$ When we deal with the problem in a semicircle, it is convenient to express Green’s function in polar coordinates. Let

$$ \left\{ {\begin{array}{*{20}l} {A = \left( {x_{0} ,y_{0} } \right) = \left( {r_{0} \cos \phi ,\sin \phi } \right)} \hfill \\ {A^{*} = \left( {\frac{{R^{2} }}{{r_{0} }}\cos \phi ,\frac{{R^{2} }}{{r_{0} }}\sin \phi } \right)} \hfill \\ {P = \left( {r\cos \theta ,r\sin \theta } \right)} \hfill \\ \end{array} } \right. $$

(33)

By (29)

$$ G_{1} \left( {x,y,x_{0} ,y_{0} } \right) = G_{1} \left( {r,\theta ,r_{0} ,\phi } \right) = \frac{1}{2}{ \ln }\left[ {R^{2} \frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta - \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta - \phi } \right)}}} \right] $$

Similarly

$$ G_{1} \left( {x,y,x_{0} , - y_{0} } \right) = G_{1} \left( {r,\theta ,r_{0} , - \phi } \right) = \frac{1}{2}\ln \left[ {R^{2} \frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta + \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta + \phi } \right)}}} \right] $$

Therefore

$$ \begin{aligned} G\left( {r,\theta ,r_{0} ,\phi } \right) & = G_{1} \left( {r,\theta ,r_{0} ,\phi } \right) - G_{1} \left( {r,\theta ,r_{0} , - \phi } \right) \\ & = \frac{1}{2}\ln \left[ {\frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta - \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta - \phi } \right)}}} \right] - \frac{1}{2}\ln \left[ {\frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta + \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta + \phi } \right)}}} \right] \\ \end{aligned} $$

(34)

Now we turn to the case $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right| \le R,y = 0} \right\},$ and it follows from (28) that

$$ \left\{ {\begin{array}{*{20}l} {\left( {x - x_{0}^{*} } \right)^{2} + \left( {y - y_{0}^{*} } \right)^{2} = \frac{{\left[ {x\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} - R^{2} x_{0} } \right]^{2} }}{{\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} }} + \frac{{\left[ {y\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} - R^{2} y_{0} } \right]^{2} }}{{\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} }}} \hfill \\ {\left( {x - x_{0}^{*} } \right)^{2} + \left( {y + y_{0}^{*} } \right)^{2} = \frac{{\left[ {x\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} - R^{2} x_{0} } \right]^{2} }}{{\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} }} + \frac{{\left[ {y\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} + R^{2} y_{0} } \right]^{2} }}{{\left( {x_{0}^{2} + y_{0}^{2} } \right)^{2} }}} \hfill \\ \end{array} } \right. $$

Inserting the equalities above into (31), we have

$$ \begin{aligned} G\left( {x,y,x_{0} ,y_{0} } \right) & = \frac{1}{2}\ln \left[ {\frac{{\left( {x - x_{0} } \right)^{2} + \left( {y - y_{0} } \right)^{2} }}{{\left[ {x\left( {x_{0}^{2} + y_{0}^{2} } \right) - R^{2} x_{0} } \right]^{2} + \left[ {y\left( {x_{0}^{2} + y_{0}^{2} } \right) - R^{2} y_{0} } \right]^{2} }}} \right] \\ & \quad - \frac{1}{2}{ \ln }\left[ {\frac{{\left( {x - x_{0} } \right)^{2} + \left( {y + y_{0} } \right)^{2} }}{{\left[ {x\left( {x_{0}^{2} + y_{0}^{2} } \right) - R^{2} x_{0} } \right]^{2} + \left[ {y\left( {x_{0}^{2} + y_{0}^{2} } \right) + R^{2} y_{0} } \right]^{2} }}} \right]. \\ \end{aligned} $$

(35)

1.2 Analytical Solution for Arbitrary Boundary Functions

We restate Theorem 2, Section 12.3 (Asmar 2005), as follows.

Theorem 1

Suppose that u is harmonic in $\varOmega$ and continuous on its boundary $\varGamma$ . Then for $\left( {x,y} \right) \in \varOmega$, we have

$$ u\left( {x,y} \right) = \frac{1}{2\pi }\int_{\varGamma } {u\left( {x_{0} ,y_{0} ,x,y} \right)} \frac{\partial G}{\partial n}\left( {x_{0} ,y_{0} ,x,y} \right){\text{d}}s\left( {x_{0} ,y_{0} } \right) $$

(36)

where G is the Green’s function for $\varOmega .$

Here we can have a different version from the original one of Asmar (2005) because the fact $G\left( {x,y,x_{0} ,y_{0} } \right) = G\left( {x_{0} ,y_{0} ,x,y} \right)$ is employed. In general, if we write $u\left( {x,y} \right) = u\left( {r_{0} \cos \phi ,r_{0} \sin \phi } \right),$ the outward normal vector $n = \left( {\cos \phi ,\sin \phi } \right)$ on the circle $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} = R^{2} } \right\}$. Therefore

$$ \frac{\partial u}{\partial n} \equiv \left( {u_{x} ,u_{y} } \right) \cdot n = u_{x} \cos \phi + u_{y} \sin \phi = \frac{\partial }{{\partial r_{0} }}\left[ {u\left( {r_{0} \cos \phi ,r_{0} \sin \phi } \right)} \right] $$

Now we calculate $\frac{\partial u}{\partial n}\left( {x_{0} ,y_{0} ,x,y} \right)$ on the circle $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} = R^{2} } \right\}$ as

$$ \frac{\partial u}{\partial n}\left( {x_{0} ,y_{0} ,x,y} \right){\text{d}}s\left( {x_{0} ,y_{0} } \right) = \frac{\partial }{{\partial r_{0} }}\left[ {G\left( {r,\theta ,r_{0} ,\phi } \right)} \right]|_{{r_{0} = R}} R{\text{d}}\phi $$

where ${\text{d}}s = R{\text{d}}\phi$ is used.

Calculations show that

$$ \frac{1}{2}\frac{\partial }{{\partial r_{0} }}\ln \left[ {R^{2} \frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta - \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta - \phi } \right)}}} \right]|_{{r_{0} = R}} = \frac{1}{R}\frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta - \phi } \right)}} $$

and similarly

$$ \frac{1}{2}\frac{\partial }{{\partial r_{0} }}\ln \left[ {R^{2} \frac{{r^{2} + r_{0}^{2} - 2rr_{0} \cos \left( {\theta + \phi } \right)}}{{r^{2} r_{0}^{2} + R^{4} - 2rr_{0} R^{2} \cos \left( {\theta + \phi } \right)}}} \right]|_{{r_{0} = R}} = \frac{1}{R}\frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta + \phi } \right)}} $$

By Theorem 1, we have

$$ \begin{aligned} & \int_{{\left| {x_{0} } \right|^{2} + \left| {y_{0} } \right|^{2} = R^{2} ,y > 0}} {u\left( {x_{0} ,y_{0} } \right)} \frac{\partial G}{\partial n}\left( {x_{0} ,y_{0} ,x,y} \right){\text{d}}s\left( {x_{0} ,y_{0} } \right) \\ & \quad = \int_{0}^{\pi } {u\left( \phi \right)} \left[ {\frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta - \phi } \right)}} - \frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta + \phi } \right)}}} \right]{\text{d}}\phi \\ \end{aligned} $$

(37)

Because $0 \le \theta ,\phi \le \pi ,$ we see that

$$ \frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta - \phi } \right)}} - \frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta + \phi } \right)}} \ge 0 $$

(38)

Since the outward normal vector on $\left\{ {\left( {x,y} \right) \in R^{2} :\;\left| x \right|^{2} + \left| y \right|^{2} \le R^{2} ,y = 0} \right\}$ is $\left( {0, - 1} \right),$ then

$$ \frac{\partial G}{\partial n}|_{{y_{0} = 0}} = - \frac{\partial G}{{\partial y_{0} }}|_{{y_{0} = 0}} $$

and calculation shows by (35) that

$$ \frac{\partial G}{\partial n}|_{{y_{0} = 0}} = - \frac{\partial G}{{\partial y_{0} }}|_{{y_{0} = 0}} = \frac{2y}{{\left( {x - x_{0} } \right)^{2} + y^{2} }} - \frac{{2yR^{2} }}{{\left( {xx_{0} - R^{2} } \right)^{2} + \left( {yx_{0} } \right)^{2} }} $$

Notice that

$$ R^{2} - 2xx_{0} + \left( {x^{2} + y^{2} } \right)\frac{{x_{0}^{2} }}{{R^{2} }} - \left[ {x_{0}^{2} - 2xx_{0} + \left( {x^{2} + y^{2} } \right)} \right] = \left( {R^{2} - x_{0}^{2} } \right)\left( {1 - \frac{{x^{2} + y^{2} }}{{R^{2} }}} \right) \ge 0 $$

We see that

$$ \frac{\partial G}{\partial n}|_{{y_{0} = 0}} \ge 0 $$

which, together with (38), yields

$$ \frac{\partial G}{\partial n}|_{\partial \varOmega } \ge 0 $$

and, from (36)

$$ u\left( {x,y} \right) \ge 0 \quad {\text{if}}\; u|_{\partial \varOmega } \ge 0 $$

By Theorem 1, we have

$$ \begin{aligned} & \int_{{\left\{ {\left| {x_{0} } \right|^{2} + \left| {y_{0} } \right|^{2} = R^{2} ,y = 0} \right\}}} {u\left( {x_{0} ,y_{0} } \right)} \frac{\partial G}{\partial n}\left( {x_{0} ,y_{0} ,x,y} \right){\text{d}}s\left( {x_{0} ,y_{0} } \right) \\ & \quad = \int_{ - R}^{0} {f_{1} \left( x \right)} \left[ {\frac{2y}{{\left( {x - x_{0} } \right)^{2} + y^{2} }} - \frac{{2yR^{2} }}{{\left( {xx_{0} - R^{2} } \right)^{2} + \left( {yx_{0} } \right)^{2} }}} \right]{\text{d}}x_{0} \\ & \quad \quad+ \int_{0}^{R} {f_{2} \left( x \right)} \left[ {\frac{2y}{{\left( {x - x_{0} } \right)^{2} + y^{2} }} - \frac{{2yR^{2} }}{{\left( {xx_{0} - R^{2} } \right)^{2} + \left( {yx_{0} } \right)^{2} }}} \right]{\text{d}}x_{0} \\ \end{aligned} $$

(39)

From (36), (37), and (39), we have the explicit solution to the Dirichlet problem of Laplace equation on a semi-disk in plane

$$ \begin{aligned} u\left( {x,y} \right) & = \frac{1}{2\pi }\int_{0}^{\pi } {u\left( \phi \right)} \left[ {\frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta - \phi } \right)}} - \frac{{R^{2} - r^{2} }}{{r^{2} + R^{2} - 2rR\cos \left( {\theta + \phi } \right)}}} \right]{\text{d}}\phi \\ & \quad + \int_{ - R}^{0} {f_{1} \left( x \right)} \left[ {\frac{2y}{{\left( {x - x_{0} } \right)^{2} + y^{2} }} - \frac{{2yR^{2} }}{{\left( {xx_{0} - R^{2} } \right)^{2} + \left( {yx_{0} } \right)^{2} }}} \right]{\text{d}}x_{0} \\ & \quad + \int_{0}^{R} {f_{2} \left( x \right)} \left[ {\frac{2y}{{\left( {x - x_{0} } \right)^{2} + y^{2} }} - \frac{{2yR^{2} }}{{\left( {xx_{0} - R^{2} } \right)^{2} + \left( {yx_{0} } \right)^{2} }}} \right]{\text{d}}x_{0} . \\ \end{aligned} $$

(40)