Appendix 1. Derivation from Cauchy equation to transient fluid loss
The Cauchy equation of motion is given by:
$$ \rho \frac{\partial \mathbf{v}}{\partial t}+\rho \left(\mathbf{v}\cdot \nabla \right)\mathbf{v}=\nabla \cdot \left(-p\mathbf{I}+\boldsymbol{\uptau} \right)+\rho \mathbf{g} $$
(18)
At the steady state without inertial and gravity effects, we get,
$$ 0=\nabla \left(-p\mathbf{I}+\boldsymbol{\uptau} \right) $$
(19)
In a radial system 1D, the above equation simplifies to, see Panton (
1984):
$$ {\tau}_{rz}=z\frac{\partial p}{\partial r} $$
(20)
Equation (
20) relates the pressure to shear stress. Another equation to relate shear stress to velocity is given by Herschel-Bulkley model, such that,
$$ {\tau}_{rz}={\tau}_0+m{\left(\frac{dv_r}{dz}\right)}^n $$
(21)
Connecting these two Eqs. (
21) and (
20) as,
$$ z\frac{\partial p}{\partial r}={\tau}_0+m{\left(\frac{dv_r}{dz}\right)}^n $$
(22)
Solving the differential equation for the general solution of the velocity profile of the half fracture (i.e.,
z ∈ [0,
w/2]) domain, we get,
$$ {v}_r(z)=\frac{n\left(-\frac{\partial p}{\partial r}\frac{w}{2}+{\tau}_0\right){\left(\frac{\frac{\partial p}{\partial r}\frac{w}{2}+{\tau}_0}{m}\right)}^{1/n}+n\left(-\frac{\partial p}{\partial r}z+{\tau}_0\right){\left(\frac{\frac{\partial p}{\partial r}z-{\tau}_0}{m}\right)}^{\frac{1}{n}}}{\frac{\partial p}{\partial r}\left(n+1\right)} $$
(23)
Equation (
23) is valid along the fracture aperture in the
Z-direction, where two regions, according to the velocity profile, can be distinguished: plug (non-deformed) region and free (deformed) region. We can see from Herschel-Bulkley equation model that the plug region corresponds to zero derivative of the velocity in the
Z-direction, that is,
\( \frac{dv_r}{dz}=0 \). Substituting this condition in the fluid model with pressure Eq. (
22), the following equation of fluid yield stress is obtained,
$$ {z}_{plug}\frac{\partial p}{\partial r}={\tau}_0 $$
(24)
The velocity is defined at the three B.C. as,
$$ {v}_r(z)=\left\{\begin{array}{cc}{v}_{r, plug}(z)& for\kern0.33em z\le {z}_{plug}\\ {}{v}_{r, free}(z)& for\kern0.33em {z}_{plug}<z<\frac{w}{2}\\ {}0& for\kern0.33em z=\frac{w}{2}\end{array}\right. $$
(25)
In Eq. (
5),
zplug is the vertical extension of the plug region;
vr, plug and
vr, free denote the velocities within the plug region and free region, respectively. The last condition is a result of no-slip B.C.
Substitute Eq. (
24) into the solution of the general velocity profile in Eq. (
23), we get,
$$ {v}_{r, free}(z)=\frac{n}{n+1}\left({z}_{plug}-\frac{w}{2}\right){\left(\frac{\frac{\partial p}{\partial r}\left(\frac{w}{2}-{z}_{plug}\right)}{m}\right)}^{1/n}+\frac{n}{n+1}\left(z-{z}_{plug}\right){\left(\frac{\frac{\partial p}{\partial r}\left(z-{z}_{plug}\right)}{m}\right)}^{\frac{1}{n}} $$
(26)
Use Eq. (
26) to find velocity profile in the plug region (plug velocity) by substituting
\( {z}_{plug}=\frac{\tau_0}{\frac{\partial p}{\partial r}} \) and simplifying, we get,
$$ {v}_{r, plug}(z)=\frac{n}{n+1}\left(\frac{\tau_0}{\frac{\partial p}{\partial r}}-\frac{w}{2}\right){\left(\frac{\frac{w}{2}\frac{\partial p}{\partial r}-{\tau}_0}{m}\right)}^{1/n} $$
(27)
Now, we have two velocity profiles as expressed as,
$$ {\displaystyle \begin{array}{c}{v}_{r, free}(z)=\frac{n}{n+1}\left({z}_{plug}-\frac{w}{2}\right){\left(\frac{\frac{\partial p}{\partial r}\left(\frac{w}{2}-{z}_{plug}\right)}{m}\right)}^{1/n}+\frac{n}{n+1}\left(z-{z}_{plug}\right){\left(\frac{\frac{\partial p}{\partial r}\left(z-{z}_{plug}\right)}{m}\right)}^{\frac{1}{n}}\\ {}{v}_{r, plug}(z)=\frac{n}{n+1}\left(\frac{\tau_0}{\frac{\partial p}{\partial r}}-\frac{w}{2}\right){\left(\frac{\frac{w}{2}\frac{\partial p}{\partial r}-{\tau}_0}{m}\right)}^{1/n}\end{array}} $$
(28)
We transform the velocities into the total volumetric flow rate (flux);
$$ {Q}_{total}={Q}_{plug}+{Q}_{free} $$
(29)
Applying the surface integral of the velocity field for the two regions,
$$ {Q}_{total}=4\pi r\underset{0}{\overset{z_{plug}}{\int }}{v}_{(r) plug} dz+4\pi r\underset{z_{plug}}{\overset{w/2}{\int }}{v}_{(r) free} dz $$
(30)
Substituting, we find the total flux,
$$ {Q}_{total}=\frac{4\pi r}{m^{1/n}}{\left(\frac{dp}{dr}\right)}^{1/n}{\left(\frac{w}{2}-\frac{\tau_0}{\frac{dp}{dr}}\right)}^{1/n+1}\left(\frac{n}{n+1}\frac{\tau_0}{\frac{dp}{dr}}+\frac{n}{2n+1}\left(\frac{w}{2}-\frac{\tau_0}{\frac{dp}{dr}}\right)\right) $$
(31)
Simplifying the above expresses to get,
$$ {Q}_{total}=\frac{4\pi r}{m^{1/n}}{\left(\frac{w}{2}\right)}^{1/n+2}\left(\frac{n}{2n+1}\right){\left(\frac{dp}{dr}\right)}^{1/n}{\left(1-\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^{1/n}\left(1-\left(\frac{n}{n+1}\right)\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}-\left(\frac{n}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^2\right) $$
(32)
To avoid obtaining complex number because of the negative
\( \frac{dp}{dr} \) inside the power term, we rewrite the equation as,
$$ {Q}_{total}^n=\frac{{\left(4\pi r\right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n\left(\frac{dp}{dr}\right)\left(1-\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right){\left(1-\left(\frac{n}{n+1}\right)\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}-\left(\frac{n}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^2\right)}^n $$
(33)
The last term is approximated by using the second-order Taylor expansion, that is,
$$ {Q}_{total}^n=\frac{{\left(4\pi r\right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n\left(\frac{dp}{dr}\right)\left(1-\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)\left(1-\left(\frac{n}{n+1}\right)\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}-\left(\frac{n^2}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^2\right) $$
(34)
Simplifying,
$$ {Q}_{total}^n=\frac{{\left(4\pi r\right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n\left(\frac{dp}{dr}\right)\left(1-\left(\frac{2n+1}{n+1}\right)\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}+\left(\frac{n-{n}^2}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^2+\frac{n^2}{n+1}{\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^3\right) $$
(35)
By neglecting the higher-order terms, we get,
$$ {Q}_{total}^n=\frac{{\left(4\pi r\right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n\left(\frac{dp}{dr}\right)\left(1-\left(\frac{2n+1}{n+1}\right)\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}+\left(\frac{n-{n}^2}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}\frac{dp}{dr}}\right)}^2\right) $$
(36)
We arrange the pressure gradient term as a polynomial equation of second degree, that is,
$$ 0={\left(\frac{dp}{dr}\right)}^2-\left(\frac{Q_{total}^n}{r^n\frac{{\left(4\pi \right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n}+\left(\frac{2n+1}{n+1}\right)\frac{\tau_0}{\frac{w}{2}}\right)\frac{dp}{dr}+\left(\frac{n-{n}^2}{n+1}\right){\left(\frac{\tau_0}{\frac{w}{2}}\right)}^2 $$
(37)
This equation can be seen as a quadratic equation with unknown
\( \frac{dp}{dr} \). To simplify the expression, we define the following quantities,
$$ {\displaystyle \begin{array}{c}A=\frac{{\left(4\pi \right)}^n}{m}{\left(\frac{w}{2}\right)}^{2n+1}{\left(\frac{n}{2n+1}\right)}^n\\ {}B=\left(\frac{2n+1}{n+1}\right)\frac{\tau_0}{w/2}\\ {}D=\left(\frac{n-{n}^2}{n+1}\right){\left(\frac{\tau_0}{w/2}\right)}^2\end{array}} $$
(38)
Equation (
37) becomes,
$$ 0={\left(\frac{dp}{dr}\right)}^2-\left(\frac{Q_{total}^n}{r^nA}+B\right)\frac{dp}{dr}+D $$
(39)
This equation is a nonhomogeneous, nonlinear first-order ordinary differential equation. The general solution of the polynomial quadratic equation is in the form of
\( \frac{-b\pm \sqrt{b^2-4 ac}}{2a} \). Hence,
$$ \frac{dp}{dr}=\frac{1}{2}\left(B+\frac{Q_{total}^n}{r^nA}\pm \sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) $$
(40)
There are two roots for this equation, from which the positive root is selected, thus,
$$ \frac{dp}{dr}=\frac{1}{2}\left(B+\frac{Q_{total}^n}{r^nA}+\sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) $$
(41)
The pressure at the inlet is
pw, and the pressure at the mud-font is
pf. It should be noted that the interface is moving with time
rf(
t). As a result, we implement a moving boundary condition at the interface only, that is,
$$ \underset{p_w}{\overset{p_f}{\int }} dp=\underset{r_w}{\overset{r_f(t)}{\int }}\frac{1}{2}\left(B+\frac{Q_{total}^n}{r^nA}+\sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) dr $$
(42)
Integrate the left-hand side, and the first two sides on the right-hand side, we get,
$$ {p}_f-{p}_w=\frac{1}{2}\left(\underset{r_w}{\overset{r_f(t)}{\int }} Bdr+\underset{r_w}{\overset{r_f(t)}{\int }}\frac{Q_{total}^n}{r^nA} dr+\underset{r_w}{\overset{r_f(t)}{\int }}\left(\sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) dr\right) $$
(43)
$$ {p}_f-{p}_w=\frac{B\left({r}_f-{r}_w\right)}{2}+\frac{Q_{total}^n\left({r}_f^{1-n}-{r}_w^{1-n}\right)}{2\left(1-n\right)A}+\frac{1}{2}\underset{r_w}{\overset{r_f(t)}{\int }}\left(\sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) dr $$
(44)
To compute the last integral term on the right, we use,
$$ \frac{1}{2}\underset{r_w}{\overset{r_f(t)}{\int }}\left(\sqrt{{\left(B+\frac{Q_{total}^n}{r^nA}\right)}^2-4D}\right) dr=f\left({r}_f(t)\right)-f\left({r}_a\right) $$
(45)
which can be solved as (Choi et al.
2018; Vidunas
2008; Ismail and Pitman
2000),
$$ {\displaystyle \begin{array}{c}f\left({r}_f(t)\right)-f\left({r}_w\right)=\\ {}\frac{F_1\left(\alpha, \beta, \beta^{\prime },\gamma, \chi, \delta \right){r}_f(t)\sqrt{{\left(B+\frac{Q_{total}^n}{r_f{(t)}^nA}\right)}^2-4D}}{{}^2\sqrt{\frac{B{Q}_{total}^n+A{r}_f{(t)}^n\left({B}^2-4D\right)-2A\sqrt{\frac{Q_{total}^nD}{A^2}}}{r_f{(t)}^nA\left({B}^2-4D\right)}}\sqrt{\frac{B{Q}_{total}^n+A{r}_f{(t)}^n\left({B}^2-4D\right)-2A\sqrt{\frac{Q_{total}^nD}{A^2}}}{r_f{(t)}^nA\left({B}^2-4D\right)}}}\\ {}\frac{F_1\left(\alpha, \beta, \beta^{\prime },\gamma, \chi, \delta \right){r}_w\sqrt{{\left(B+\frac{Q_{total}^n}{r_f{(t)}^nA}\right)}^2-4D}}{{}^2\sqrt{\frac{B{Q}_{total}^n+{Ar}_w^n\left({B}^2-4D\right)-2A\sqrt{\frac{Q_{total}^{2n}D}{A^2}}}{r_w^nA\left({B}^2-4D\right)}}\sqrt{\frac{B{Q}_{total}^n+A{r}_w^n\left({B}^2-4D\right)+2A\sqrt{\frac{Q_{total}^{2n}D}{A^2}}}{r_w^nA\left({B}^2-4D\right)}}}\end{array}}- $$
(46)
Appell’s hypergeometric function of the first kind is a solution of the integral term. It is a double hypergeometric series
F1(
α,
β,
β',
γ,
χ,
δ), where,
$$ {\displaystyle \begin{array}{c}\beta =\beta \hbox{'}=-\frac{1}{2}\\ {}\gamma =\frac{n-1}{n}\\ {}\chi =-\frac{Q_{total}^n{r}^{-n}}{AB+2A\sqrt{D}}\\ {}\delta =-\frac{Q_{total}^n{r}^{-n}}{AB-2A\sqrt{D}}\end{array}} $$
(47)
We substitute back these values into Eq. (
46) to get,
$$ {\displaystyle \begin{array}{c}f\left({r}_f(t)\right)-f\left({r}_w\right)=\\ {}\frac{F_1\left(-\frac{1}{n},\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{n-1}{n},\frac{Q_{total}^n{r}_f{(t)}^{-n}}{AB+2\sqrt{D}},\frac{Q_{total}^n{r}_f{(t)}^{-n}}{AB-2\sqrt{D}}\right){r}_f\sqrt{{\left(B+\frac{Q_{total}^n}{r_f{(t)}^nA}\right)}^2-4D}}{\sqrt[2]{\left(1+\frac{B{Q}_{total}^n}{r_f{(t)}^nA\left({B}^2-4D\right)}-\frac{2{Q}_{total}^n\sqrt{D}}{r_f{(t)}^nA\left({B}^2-4D\right)}\right)\left(1+\frac{B{Q}_{total}^n}{r_f{(t)}^nA\left({B}^2-4D\right)}+\frac{2{Q}_{total}^n\sqrt{D}}{r_f{(t)}^nA\left({B}^2-4D\right)}\right)}}-\\ {}\frac{F_1\left(-\frac{1}{n},\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{n-1}{n},-\frac{Q_{total}^n{r}_w^{-n}}{AB+2\sqrt{D}},\frac{Q_{total}^n{r_w}^{-n}}{AB-2\sqrt{D}}\right){r}_w\sqrt{{\left(B+\frac{Q_{total}^n}{{r_w}^nA}\right)}^2-4D}}{\sqrt[2]{\left(1+\frac{B{Q}_{total}^n}{{r_w}^nA\left({B}^2-4D\right)}-\frac{2{Q}_{total}^n\sqrt{D}}{r_w^nA\left({B}^2-4D\right)}\right)\left(1+\frac{B{Q}_{total}^n}{{r_w}^nA\left({B}^2-4D\right)}+\frac{2{Q}_{total}^n\sqrt{D}}{{r_w}^nA\left({B}^2-4D\right)}\right)}}\end{array}} $$
(48)
We now have an analytical solution of pressure as a function of radial distance
r. Equations (
48), (
45), and (
44) are used when the total flow rate entering the fracture is known. Otherwise, if the total flow rate is unknown, we can use the total flow rate as,
$$ {Q}_{total}=\frac{dV_m}{dt} $$
(49)
Mud-loss volume can be found for radial flow as,
$$ {V}_m=\pi w\left({r}_f{(t)}^2-{r_w}^2\right) $$
(50)
Substituting Eq. (
50) into (
49) and differentiating,
$$ {Q}_{total}=2\pi {wr}_f(t)\frac{dr_f(t)}{dt} $$
(51)
Finally, the mathematical problem of lost circulation in a smooth horizontal fracture is solved analytically. Equations (
51), (
48), (
45), and (
44) are the solution of mud-loss flow front
rf(
t) as a function of time
t, based on the given parameters of mud rheology (
n,
m,
τ0), differential pressure (
pf,
pw), and fracture aperture (
w).
Appendix 2. Particular solution for Bingham plastic fluids
Here, we show how the proposed solution can be derived mathematically for Bingham plastic fluids, when
n = 1. Let us take Eq. (
36), after approximating the last term, by applying the same approach for Bingham plastic fluid condition (
n = 1),
$$ {Q}_{total}=\frac{\pi {rw}^3}{6m}\left(\frac{dp}{dr}-\frac{3{\tau}_0}{w}\right) $$
(52)
Integrating,
$$ \underset{p_w}{\overset{p_f}{\int }} dp=\underset{r_w}{\overset{r_r(t)}{\int }}\left(\frac{6{mQ}_{total}}{\pi {rw}^3}+\frac{3{\tau}_0}{w}\right) dr $$
(53)
The above equation becomes,
$$ {p}_f-{p}_w=\frac{12{mr}_f(t)}{w^2}\frac{dr_f(t)}{dt}\ln \left(\frac{r_f(t)}{r_w}\right)+\frac{3{\tau}_0}{w}\left({r}_f(t)-{r}_w\right) $$
(54)
Rearranging,
$$ \frac{dr_f(t)}{dt}=\frac{\left({p}_f-{p}_w\right)-\frac{3{\tau}_0}{w}\left({r}_f(t)-{r}_w\right)}{\frac{12{mr}_f(t)}{w^2}\ln \left(\frac{r_f(t)}{r_w}\right)} $$
(55)
Equation (
55) is the final solution of Bingham plastic fluid model of mud invasion front
rf(
t) as a function of time
t. The equation can be derived in different forms, which is consistent with the dimensionless form by Liétard et al. (
2002). The defined dimensionless groups are,
$$ {\displaystyle \begin{array}{c}\alpha =\frac{3{r}_w}{w}\left(\frac{\tau_0}{p_f-{p}_w}\right)\\ {}\beta ={\left(\frac{w}{r_w}\right)}^2\frac{\left({p}_f-{p}_w\right)}{12m}\\ {}{r}_D=\frac{r_f}{r_w},{t}_D=\beta t\end{array}} $$
(56)
Implementing into Eq. (
55) to generate a dimensionless differential form,
$$ \frac{dr_D(t)}{dt_D}=\frac{1-\alpha \left({r}_D(t)-1\right)}{r_D(t)\ln \left({r}_D(t)\right)} $$
(57)
This is an ordinary differential equation that can be solved analytically by using inverse function theorem (IFT) in such a way that any function
f(
x) is both differentiable and invertible. Assuming
y =
f−1(
x) is the inverse of
f(
x) which all
x satisfying
f ' (
f−1(
x)) ≠ 0. Thus, the derivative,
$$ \frac{dy}{dx}=\frac{d}{dx}\left({f}^{-1}(x)\right)={\left({f}^{-1}\right)}^{\hbox{'}}(x)\frac{1}{f\hbox{'}\left({f}^{-1}(x)\right)} $$
(58)
Equation (
57) becomes,
$$ \frac{dt_D}{dr_D(t)}=\frac{r_D(t)\ln \left({r}_D(t)\right)}{1-\alpha \left({r}_D(t)-1\right)} $$
(59)
Providing the closed-form solution after applying initial condition at
tD = 0 and
rD = 1, we get,
$$ {t}_D=-\frac{4}{\alpha}\left(1-{r}_D+\ln \left({r_D}^{r_D}\right)\right)+4\left(\frac{1}{\alpha^2}+\frac{1}{\alpha}\right)\left[L{i}_2\left(\frac{\alpha }{1+\alpha}\right)-L{i}_2\left(\frac{\alpha {r}_D}{1+\alpha}\right)-\ln \left({r}_{Df}\right)\ln \left(1-\frac{\alpha {r}_D}{1+\alpha}\right)\right] $$
(60)
In the above equation, the polylogarithmic function
Li2 is shown as a solution of Eq. (
59), which is the analytical solution of a dimensionless mud invasion front
rD as a function of dimensionless time
tD for Bingham plastic fluid model. Furthermore, an analytical solution was provided as well (Liétard et al.
2002).