Numerical Simulation of the Multiphase Flow in the Rheinsahl–Heraeus (RH) System

Abstract

Knowledge of gas–liquid multiphase flow behavior in the Rheinsahl–Heraeus (RH) system is of great significance to clarify the circulation flow rate, decarburization, and inclusion removal with a reliable description. Thus, based on the separate model of injecting gas behavior, a novel mathematical model of multiphase flow has been developed to give the distribution of gas holdup in the RH system. The numerical results show that the predicted circulation flow rates, the predicted flow velocities, and the predicted mixing times agree with the measured results in a water model and that the predicted tracer concentration curve agrees with the results obtained in an actual RH system. With a lower lifting gas flow rate, the rising gas bubbles are concentrated near the wall; with a higher lifting gas flow rate, gas bubbles can reach the center of the up-snorkel. A critical lifting gas flow rate is used to obtain the maximum circulation flow rate.

Appendix

Derivation for relative velocity in horizontal and vertical direction (Eqs. [21] and [22]):

In horizontal direction, Eq. [20] is written as follows:

$$ \left( {{\text{m}}_{\text{g}} + {\frac{ 1}{ 2}}V_{\text{g}} \rho_{\rm l} } \right){\frac{{du_{\text{g,l}}^{r} }}{dt}} = - {\frac{ 1}{ 2}}A_{\text{g}} \rho_{\text{g}} {\text{C}}_{\text{D}} u_{\text{g,l}}^{r2} $$

(A1)

We introduce the variables as follows:

$$ {\text{A}}^{*} = {\text{m}}_{\text{g}} + {\frac{ 1}{ 2}}V_{\text{g}} \rho_{\text{l}} $$

(A2)

$$ {\text{B}}^{*} = {\frac{ 1}{ 2}}A_{\text{g}} \rho_{\text{g}} {\text{C}}_{\text{D}} $$

Furthermore,

$$ u_{\text{g,l}}^{r} = {\frac{dr}{dt}} $$

(A3)

Substituting Eqs. [A2] and [A3] into Eq. [A1], we obtained the following:

$$ {\frac{{du_{\text{g,l}}^{r} }}{dr}} = - {\frac{{{\text{B}}^{*} }}{{{\text{A}}^{*}}}} \times u_{\text{g,l}}^{r} $$

(A4)

The boundary condition at r = 0 is as follows:

$$ u_{\text{g,l}}^{r} \left| {_{r = 0} } \right. = u_{{{\text{g,l}}, 0}}^{r} $$

(A5)

Integrating Eq. [A4], the horizontal component of relative velocity is as follows:

$$ u_{{\rm g,l}}^{r} = u_{{\rm g,l}, 0}^{r} \times { \exp }\left( { - {\frac{{{\text{B}}^{ *} }}{{{\text{A}}^{ *} }}} \times r} \right) = u_{{\rm g,l}, 0}^{r} \times { \exp }\left( { - {\frac{{ 3 {\text{C}}_{\text{D}} \rho_{\rm l} }}{{ 2d_{g} ( 2\rho_{\rm g} + \rho_{\rm l} )}}} \times r} \right) $$

(A6)

Here, because the liquid velocity at the wall is zero, the relative velocity of gas and liquid is equal to the gas velocity at the wall nodes, and then Eq. [21] is obtained as follows:

$$ u_{\text{g,l}}^{r} = u_{{{\text{g}}0}} \times { \exp }\left( { - {\frac{{ 3 {\text{C}}_{\text{D}} \rho_{\text{l}} }}{{ 2d_{g} ( 2\rho_{\text{g}} + \rho_{\text{l}} )}}} \times r} \right) $$

(21)

In vertical direction, Eq. [20] is written as follows:

$$ \left( {{\text{m}}_{\rm g} + {\frac{ 1}{ 2}}V_{\rm g} \rho_{\rm l} } \right){\frac{{du_{{\rm g,l}}^{z} }}{dt}} = - {\frac{ 1}{ 2}}A_{\text{g}} \rho_{\text{g}} {\text{C}}_{\text{D}} {u_{{{\text{g}},{\text{l}}}}^{z}}^2 + \rho_{\rm l} V_{\text{g}} g - \rho_{\rm g} V_{\text{g}} g $$

(A7)

Furthermore,

$$ u_{\text{g,l}}^{z} = {\frac{dz}{dt}} $$

(A8)

Substituting Eq. [A8] into Eq. [A7], we obtained the following:

$$ {\frac{{du_{\text{g,l}}^{z} }}{dz}} = {\text{m}}^{ *} u_{\text{g,l}}^{z} + {\text{n}}^{ *} {\frac{1}{{u_{\text{g,l}}^{z} }}} $$

(A9)

Here, ${\text{m}}^{ *} = - {\frac{{{\text{B}}^{ *} }}{{{\text{A}}^{ *} }}}$ and ${\text{n}}^{*} = {\frac{{\left( {\rho_{\rm l} - \rho_{\rm g} } \right)V_{\rm g} }}{{{\text{A}}^{*} }}}$. Note that Eq. [A9] is a Bernoulli differential equation. So we introduce the following variable:

$$ {\text{U}}^{ *} = {u_{\text{g,l}}^{z}}^2 $$

(A10)

and transform Eq. [A9] into a linear differential equation as follows:

$$ {\frac{1}{2}} {\cdot} {\frac{{d{\text{U}}^{ *} }}{dz}} = {\text{m}}^{ *} {\text{U}}^{ *} + {\text{n}}^{ *} $$

(A11)

Furthermore, the boundary condition for Eq. [A9] at z = z _o is $u_{{{\text{g}},{\text{l}}}}^{z} \left| {_{z = 0} } \right. = 0$. Thus, the boundary condition for Eq. [A11] at z = z _o is

$$ {\text{U}}^{ *} \left| {_{{z = z_{\text{o}} }} } \right. = 0 $$

(A12)

Integrating Eq. [A11], we obtain the following:

$$ {\text{U}}^{ *} = - {\frac{{{\text{n}}^{ *} }}{{{\text{m}}^{ *} }}}\left[ {1 - \exp \left( {{\text{m}}^{ *} \times (z - z_{\text{o}} )} \right)} \right] $$

(A13)

Substituting Eqs. [A2] and [A10] into Eq. [A13], Eq. [22] is as follows:

$$ u_{{\rm g,l}}^{z} = \sqrt {{\frac{{ 4(\rho_{\text{l}} - \rho_{\text{g}} )gd_{\text{g}} }}{{ 3\rho_{\text{l}} {\text{C}}_{\text{D}} }}}\left[ { 1 - { \exp }\left( { - {\frac{{ 3 {\text{C}}_{\text{D}} \rho_{\text{l}} }}{{ 2d_{g} ( 2\rho_{\text{g}} + \rho_{\text{l}} )}}} \times (z - z_{\text{o}} )} \right)} \right]} $$

(22)