<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd">
<article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article">
 <front>
  <journal-meta>
   <journal-id journal-id-type="publisher-id">
    wjet
   </journal-id>
   <journal-title-group>
    <journal-title>
     World Journal of Engineering and Technology
    </journal-title>
   </journal-title-group>
   <issn pub-type="epub">
    2331-4222
   </issn>
   <issn publication-format="print">
    2331-4249
   </issn>
   <publisher>
    <publisher-name>
     Scientific Research Publishing
    </publisher-name>
   </publisher>
  </journal-meta>
  <article-meta>
   <article-id pub-id-type="doi">
    10.4236/wjet.2024.124060
   </article-id>
   <article-id pub-id-type="publisher-id">
    wjet-137217
   </article-id>
   <article-categories>
    <subj-group subj-group-type="heading">
     <subject>
      Articles
     </subject>
    </subj-group>
    <subj-group subj-group-type="Discipline-v2">
     <subject>
      Chemistry 
     </subject>
     <subject>
       Materials Science, Engineering
     </subject>
    </subj-group>
   </article-categories>
   <title-group>
    Analysis of the Flow of a Molten Slag in an Open Channel Using Transient and Steady-State Solutions of the Saint-Venant Equations
   </title-group>
   <contrib-group>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Miguel A.
      </surname>
      <given-names>
       Barron
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Joan
      </surname>
      <given-names>
       Reyes
      </given-names>
     </name>
    </contrib>
    <contrib contrib-type="author" xlink:type="simple">
     <name name-style="western">
      <surname>
       Aristeo
      </surname>
      <given-names>
       Garrido
      </given-names>
     </name>
    </contrib>
   </contrib-group> 
   <aff id="affnull">
    <addr-line>
     aDepartamento de Materiales, Universidad Autonoma Metropolitana Azcapotzalco, Ciudad de Mexico, Mexico
    </addr-line> 
   </aff> 
   <pub-date pub-type="epub">
    <day>
     04
    </day> 
    <month>
     09
    </month>
    <year>
     2024
    </year>
   </pub-date> 
   <volume>
    12
   </volume> 
   <issue>
    04
   </issue>
   <fpage>
    976
   </fpage>
   <lpage>
    986
   </lpage>
   <history>
    <date date-type="received">
     <day>
      10,
     </day>
     <month>
      October
     </month>
     <year>
      2024
     </year>
    </date>
    <date date-type="published">
     <day>
      3,
     </day>
     <month>
      October
     </month>
     <year>
      2024
     </year> 
    </date> 
    <date date-type="accepted">
     <day>
      3,
     </day>
     <month>
      November
     </month>
     <year>
      2024
     </year> 
    </date>
   </history>
   <permissions>
    <copyright-statement>
     © Copyright 2014 by authors and Scientific Research Publishing Inc. 
    </copyright-statement>
    <copyright-year>
     2014
    </copyright-year>
    <license>
     <license-p>
      This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/
     </license-p>
    </license>
   </permissions>
   <abstract>
    In recent years, metallurgical slags have been increasingly used as materials for the manufacture of cement, pavement and filling material. The transport of the molten slag to the receiving pots is carried out through open channels. The transient and steady-state flow of a molten slag in a rectangular open channel is numerically analyzed here. For the transient flow, the Saint-Venant equations were numerically solved. For the steady-state flow, the derivatives in time and space in the Saint-Venant equations were set equal to zero and a polynomial of degree 3 is obtained whose roots are the slag height values. It was assumed that the viscosity of the slag has an Arrhenius-type behavior with temperature. Four values of temperature values, namely 1723.15, 1773.15, 1823.15, 18873.15 ˚K, and five values of the angle of inclination of the channel, namely 1, 2, 3, 4, 5 degrees, are considered. Numerical results show that the steady-state values of the height and velocity of the molten slag depend strongly on the temperature of the slag and the angle of inclination of the channel. As the slag temperature and channel angle increase, the value of the steady-state slag height decreases. The value of the steady-state slag velocity increases as the slag temperature and channel inclination angle increase.
   </abstract>
   <kwd-group> 
    <kwd>
     Molten Slag
    </kwd> 
    <kwd>
      Open Channel
    </kwd> 
    <kwd>
      Phase-Portrait
    </kwd> 
    <kwd>
      Saint-Venant Equations
    </kwd> 
    <kwd>
      Steady State
    </kwd> 
    <kwd>
      Transient Solution
    </kwd>
   </kwd-group>
  </article-meta>
 </front>
 <body>
  <sec id="s1">
   <title>1. Introduction</title>
   <p>The flow of molten slag in open channels is interesting from both an academic and an industrial point of view. The open channels for the metallurgical industry are very different from those used for water transportation. The former are short in length so that the molten metal or slag does not solidify or reoxidize; They have a steeper slope, and the material from which they are made must withstand the high temperatures of the molten material without degrading. Works on the flow of water in open channels have been published for many decades, however, literature on the transport of molten slag in the metallurgical field is still scarce. In <xref ref-type="bibr" rid="scirp.137217-1">
     [1]
    </xref>, it was studied the fluid flow in the blast furnace trough using both physical model and commercial fluid flow software. In <xref ref-type="bibr" rid="scirp.137217-2">
     [2]
    </xref>, the flow characteristics in the taphole region of a blast furnace trough was analyzed using a 1/5th scale perspex model. Besides, velocities and turbulence intensities were measured by means of laser doppler velocimetry (LDV). In <xref ref-type="bibr" rid="scirp.137217-3">
     [3]
    </xref>, the transient three-dimensional turbulent flow in a blast furnace trough was studied using Computational Fluid Dynamics software (CFD). The effect of blast furnace trough geometry on the slag-metal separation was analyzed using CFD numerical simulation in <xref ref-type="bibr" rid="scirp.137217-4">
     [4]
    </xref>. In <xref ref-type="bibr" rid="scirp.137217-5">
     [5]
    </xref>, a transient 3D CFD model was solved to investigate the features of the multiphase flow in a blast furnace trough. Most of these works perform complex three-dimensional transient numerical simulations using Computational Fluid Dynamics tools. The results of these simulations, given that they resolve particular cases in great detail, are difficult to generalize to understand the basic aspects of fluid flow of molten slag in open channels.</p>
   <p>The flow of molten slag in an open channel is three-dimensional and turbulent in nature. In addition, heat flow and phase changes make the analysis more complicated. However, the Saint-Venant equations are still used to properly describe the one-dimensional flow of liquids in channels under the following conditions: the slope of the channel is relatively small and without abrupt changes, the cross section is uniform, the channel is straight, the pressure distribution is hydrostatic, and the speed is uniform <xref ref-type="bibr" rid="scirp.137217-6">
     [6]
    </xref>. In a recent work <xref ref-type="bibr" rid="scirp.137217-7">
     [7]
    </xref>, the present authors studied the unsteady, non-uniform flow of a molten blast furnace slag in a rectangular open channel through the numerical solution of the Saint-Venant equations. They conclude that, for the temperatures considered, the slag flow in the channel for an angle of 5 degrees is supercritical in nature. However, for an angle of 1 degree the flow is transcritical, that is, it presents a transition from subcritical to supercritical. In this work, the flow of a molten blast furnace slag in a rectangular open channel using transient and stable solutions of the Saint-Venant equations is studied. The effect of the angle of inclination of the open channel and the temperature of the slag on the transient behavior of the height and velocity is analyzed through the numerical solution of the Saint-Venant equations. It is assumed that the viscosity of the slag has an Arrhenius-type behavior with temperature. Saint-Venant equations are solved using an explicit backwards finite difference scheme. Furthermore, the equilibrium points, or steady-state solutions, of the Saint-Venant equations are analytically obtained, finding the roots of a polynomial of third degree obtained. A phase portrait of the flow was constructed from different initial conditions that show the convergence and stability of the steady-state values.</p>
  </sec><sec id="s2">
   <title>2. The Saint-Venant Equations</title>
   <p>These equations consist of a continuity equation and momentum equations that account for gravitational forces and friction <xref ref-type="bibr" rid="scirp.137217-6">
     [6]
    </xref>. They are derived assuming shallow water conditions, which means that the depth of the liquid is small compared to the horizontal dimension of the flows. The Saint-Venant equations are expressed by the following equations:</p>
   <p>Continuity equation:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        b 
      </mi> 
      <mfrac> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          h 
        </mi> 
       </mrow> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          t 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        + 
      </mo> 
      <mfrac> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          Q 
        </mi> 
       </mrow> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          x 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        = 
      </mo> 
      <mn>
        0 
      </mn> 
     </mrow> 
    </math> (1)</p>
   <p>Momentum equation:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          Q 
        </mi> 
       </mrow> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          t 
        </mi> 
       </mrow> 
      </mfrac> 
      <mo>
        + 
      </mo> 
      <mfrac> 
       <mn>
         1 
       </mn> 
       <mi>
         b 
       </mi> 
      </mfrac> 
      <mfrac> 
       <mo>
         ∂ 
       </mo> 
       <mrow> 
        <mo>
          ∂ 
        </mo> 
        <mi>
          x 
        </mi> 
       </mrow> 
      </mfrac> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <msup> 
           <mi>
             Q 
           </mi> 
           <mn>
             2 
           </mn> 
          </msup> 
         </mrow> 
         <mi>
           h 
         </mi> 
        </mfrac> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        + 
      </mo> 
      <mi>
        g 
      </mi> 
      <mi>
        b 
      </mi> 
      <mi>
        h 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <mo>
            ∂ 
          </mo> 
          <mi>
            h 
          </mi> 
         </mrow> 
         <mrow> 
          <mo>
            ∂ 
          </mo> 
          <mi>
            x 
          </mi> 
         </mrow> 
        </mfrac> 
        <mo>
          + 
        </mo> 
        <msub> 
         <mi>
           S 
         </mi> 
         <mi>
           f 
         </mi> 
        </msub> 
        <mo>
          − 
        </mo> 
        <msub> 
         <mi>
           S 
         </mi> 
         <mn>
           0 
         </mn> 
        </msub> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <mn>
        0 
      </mn> 
     </mrow> 
    </math> (2)</p>
   <p>In the above equations, h and Q are the height of the liquid and the volumetric flow rate, respectively, t is time and x is the space coordinate. Besides, b is the width of the channel, g is the gravitational acceleration, 
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mi>
        tan 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mi>
         θ 
       </mi> 
       <mo>
         ) 
       </mo> 
      </mrow> 
     </mrow> 
    </math> is the channel slope, θ is the inclination angle of the channel, and S<sub>f</sub> is the friction slope which can be estimated from the Darcy-Weisbach formula <xref ref-type="bibr" rid="scirp.137217-8">
     [8]
    </xref>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mi>
         f 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         f 
       </mi> 
       <mi>
         D 
       </mi> 
      </msub> 
      <mfrac> 
       <mrow> 
        <msup> 
         <mi>
           u 
         </mi> 
         <mn>
           2 
         </mn> 
        </msup> 
       </mrow> 
       <mrow> 
        <mn>
          8 
        </mn> 
        <mi>
          g 
        </mi> 
        <mi>
          h 
        </mi> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (3)</p>
   <p>where f<sub>D</sub> is the frictional resistance coefficient and u is the liquid velocity. The value of f<sub>D</sub> depends on the type of flow <xref ref-type="bibr" rid="scirp.137217-8">
     [8]
    </xref>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         f 
       </mi> 
       <mi>
         D 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mi>
         C 
       </mi> 
       <mrow> 
        <mi>
          R 
        </mi> 
        <mi>
          e 
        </mi> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> for laminar flow (4)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         f 
       </mi> 
       <mi>
         D 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <mn>
          0.223 
        </mn> 
       </mrow> 
       <mrow> 
        <mi>
          R 
        </mi> 
        <msup> 
         <mi>
           e 
         </mi> 
         <mrow> 
          <mn>
            0.25 
          </mn> 
         </mrow> 
        </msup> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> for transition flow (5)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         f 
       </mi> 
       <mi>
         D 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <msup> 
       <mrow> 
        <mrow> 
         <mo>
           [ 
         </mo> 
         <mrow> 
          <mn>
            2 
          </mn> 
          <mi>
            log 
          </mi> 
          <mrow> 
           <mo>
             ( 
           </mo> 
           <mrow> 
            <mfrac> 
             <mrow> 
              <mn>
                2 
              </mn> 
              <mi>
                h 
              </mi> 
             </mrow> 
             <mrow> 
              <msub> 
               <mi>
                 e 
               </mi> 
               <mi>
                 r 
               </mi> 
              </msub> 
             </mrow> 
            </mfrac> 
           </mrow> 
           <mo>
             ) 
           </mo> 
          </mrow> 
          <mo>
            + 
          </mo> 
          <mn>
            1.74 
          </mn> 
         </mrow> 
         <mo>
           ] 
         </mo> 
        </mrow> 
       </mrow> 
       <mrow> 
        <mo>
          − 
        </mo> 
        <mn>
          2 
        </mn> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> for turbulent flow (6)</p>
   <p>C is a constant, Re is the Reynolds number, and e<sub>r</sub> is the rugosity of the wall channel. Re is defined as follows:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        R 
      </mi> 
      <mi>
        e 
      </mi> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           D 
         </mi> 
         <mi>
           h 
         </mi> 
        </msub> 
        <mi>
          u 
        </mi> 
        <mi>
          ρ 
        </mi> 
       </mrow> 
       <mi>
         μ 
       </mi> 
      </mfrac> 
     </mrow> 
    </math> (7)</p>
   <p>In which D<sub>h</sub> is the hydraulic diameter, and ρ and μ are the density and the viscosity of the liquid, respectively. D<sub>h</sub> is defined as</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         D 
       </mi> 
       <mi>
         h 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <mn>
          4 
        </mn> 
        <mi>
          b 
        </mi> 
        <mi>
          h 
        </mi> 
       </mrow> 
       <mrow> 
        <mi>
          b 
        </mi> 
        <mo>
          + 
        </mo> 
        <mn>
          2 
        </mn> 
        <mi>
          h 
        </mi> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (8)</p>
   <p>In order to introduce the effect of viscosity and temperature of the liquid in the Saint-Venant equations, the velocity u is solved from Equation (7) and substituted into Equation (3):</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mi>
         f 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           f 
         </mi> 
         <mi>
           D 
         </mi> 
        </msub> 
       </mrow> 
       <mrow> 
        <mn>
          8 
        </mn> 
        <mi>
          g 
        </mi> 
        <mi>
          h 
        </mi> 
       </mrow> 
      </mfrac> 
      <msup> 
       <mrow> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mfrac> 
           <mrow> 
            <mi>
              R 
            </mi> 
            <mi>
              e 
            </mi> 
            <mi>
              μ 
            </mi> 
           </mrow> 
           <mrow> 
            <msub> 
             <mi>
               D 
             </mi> 
             <mi>
               h 
             </mi> 
            </msub> 
            <mi>
              ρ 
            </mi> 
           </mrow> 
          </mfrac> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
       <mn>
         2 
       </mn> 
      </msup> 
     </mrow> 
    </math> (9)</p>
   <p>Since viscosity is a function of temperature, this variable is considered implicitly in the solution of the Saint-Venant equations.</p>
  </sec><sec id="s3">
   <title>3. Steady-State Solutions of the Saint-Venant Equations</title>
   <p>The steady-state solutions of the Saint-Venant equations are obtained by setting the time and space derivatives in Equations (1-2) to zero. This implies that</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mi>
         f 
       </mi> 
      </msub> 
      <mo>
        − 
      </mo> 
      <msub> 
       <mi>
         S 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mn>
        0 
      </mn> 
     </mrow> 
    </math> (10)</p>
   <p>By algebraically manipulating Equations (8-10) it is obtained the following expression for the steady-state value of h, namely h<sub>s</sub>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mfrac> 
       <mrow> 
        <msup> 
         <mi>
           b 
         </mi> 
         <mn>
           2 
         </mn> 
        </msup> 
        <msubsup> 
         <mi>
           h 
         </mi> 
         <mi>
           s 
         </mi> 
         <mn>
           3 
         </mn> 
        </msubsup> 
       </mrow> 
       <mrow> 
        <msup> 
         <mrow> 
          <mrow> 
           <mo>
             ( 
           </mo> 
           <mrow> 
            <mi>
              b 
            </mi> 
            <mo>
              + 
            </mo> 
            <mn>
              2 
            </mn> 
            <msub> 
             <mi>
               h 
             </mi> 
             <mi>
               s 
             </mi> 
            </msub> 
           </mrow> 
           <mo>
             ) 
           </mo> 
          </mrow> 
         </mrow> 
         <mn>
           2 
         </mn> 
        </msup> 
       </mrow> 
      </mfrac> 
      <mo>
        − 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           f 
         </mi> 
         <mi>
           D 
         </mi> 
        </msub> 
       </mrow> 
       <mrow> 
        <mn>
          128 
        </mn> 
        <mi>
          g 
        </mi> 
        <msub> 
         <mi>
           S 
         </mi> 
         <mn>
           0 
         </mn> 
        </msub> 
       </mrow> 
      </mfrac> 
      <msup> 
       <mrow> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mfrac> 
           <mrow> 
            <mi>
              R 
            </mi> 
            <mi>
              e 
            </mi> 
            <mi>
              μ 
            </mi> 
           </mrow> 
           <mi>
             ρ 
           </mi> 
          </mfrac> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
       <mn>
         2 
       </mn> 
      </msup> 
      <mo>
        = 
      </mo> 
      <mn>
        0 
      </mn> 
     </mrow> 
    </math> (11)</p>
   <p>Expanding and regrouping terms in the previous equation, one obtains a polynomial of degree 3 for h<sub>s</sub>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <msubsup> 
       <mi>
         h 
       </mi> 
       <mrow> 
        <msup> 
         <mrow></mrow> 
         <mi>
           s 
         </mi> 
        </msup> 
       </mrow> 
       <mn>
         3 
       </mn> 
      </msubsup> 
      <mo>
        + 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <msubsup> 
       <mi>
         h 
       </mi> 
       <mrow> 
        <msup> 
         <mrow></mrow> 
         <mi>
           s 
         </mi> 
        </msup> 
       </mrow> 
       <mn>
         2 
       </mn> 
      </msubsup> 
      <mo>
        + 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         3 
       </mn> 
      </msub> 
      <msub> 
       <mi>
         h 
       </mi> 
       <mi>
         s 
       </mi> 
      </msub> 
      <mo>
        + 
      </mo> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         4 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mn>
        0 
      </mn> 
     </mrow> 
    </math> (12)</p>
   <p>where</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         1 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <msup> 
       <mi>
         b 
       </mi> 
       <mn>
         2 
       </mn> 
      </msup> 
     </mrow> 
    </math> (13)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         2 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mo>
        − 
      </mo> 
      <mn>
        4 
      </mn> 
      <mi>
        K 
      </mi> 
     </mrow> 
    </math> (14)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         3 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mo>
        − 
      </mo> 
      <mn>
        4 
      </mn> 
      <mi>
        b 
      </mi> 
      <mi>
        K 
      </mi> 
     </mrow> 
    </math> (15)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         A 
       </mi> 
       <mn>
         4 
       </mn> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mo>
        − 
      </mo> 
      <msup> 
       <mi>
         b 
       </mi> 
       <mn>
         2 
       </mn> 
      </msup> 
      <mi>
        K 
      </mi> 
     </mrow> 
    </math> (16)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        K 
      </mi> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           f 
         </mi> 
         <mi>
           D 
         </mi> 
        </msub> 
       </mrow> 
       <mrow> 
        <mn>
          128 
        </mn> 
        <mi>
          g 
        </mi> 
        <msub> 
         <mi>
           S 
         </mi> 
         <mn>
           0 
         </mn> 
        </msub> 
       </mrow> 
      </mfrac> 
      <msup> 
       <mrow> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mfrac> 
           <mrow> 
            <mi>
              R 
            </mi> 
            <mi>
              e 
            </mi> 
            <mi>
              μ 
            </mi> 
           </mrow> 
           <mi>
             ρ 
           </mi> 
          </mfrac> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
       <mn>
         2 
       </mn> 
      </msup> 
     </mrow> 
    </math> (17)</p>
   <p>The values of the liquid velocity in steady state corresponding to the values of h<sub>s</sub> are obtained in this way:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         u 
       </mi> 
       <mi>
         s 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           Q 
         </mi> 
         <mn>
           0 
         </mn> 
        </msub> 
       </mrow> 
       <mrow> 
        <mi>
          b 
        </mi> 
        <msub> 
         <mi>
           h 
         </mi> 
         <mi>
           s 
         </mi> 
        </msub> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (18)</p>
   <p>The Froude numbers are calculated using the steady-state values of height and velocity through the expression</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        F 
      </mi> 
      <mi>
        r 
      </mi> 
      <mo>
        = 
      </mo> 
      <mfrac> 
       <mrow> 
        <msub> 
         <mi>
           u 
         </mi> 
         <mi>
           s 
         </mi> 
        </msub> 
       </mrow> 
       <mrow> 
        <msub> 
         <mrow> 
          <msqrt> 
           <mrow> 
            <mi>
              g 
            </mi> 
            <mi>
              h 
            </mi> 
           </mrow> 
          </msqrt> 
         </mrow> 
         <mi>
           s 
         </mi> 
        </msub> 
       </mrow> 
      </mfrac> 
     </mrow> 
    </math> (19)</p>
   <p>For Fr &lt; 1 the flow regime is subcritical, gravitational forces are dominant and the flow has large depths and small velocities. For Fr &gt; 1 the flow regime is supercritical, the inertial forces are dominant and the flow has small depths and large velocities <xref ref-type="bibr" rid="scirp.137217-9">
     [9]
    </xref>. For Fr = 1 the flow regime is critical and gravity waves may arise. The critical depth h<sub>c</sub> is the flow depth at a section of the channel where the flow is critical. It is calculated in this way <xref ref-type="bibr" rid="scirp.137217-10">
     [10]
    </xref>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         h 
       </mi> 
       <mi>
         c 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <msup> 
       <mrow> 
        <mrow> 
         <mo>
           ( 
         </mo> 
         <mrow> 
          <mfrac> 
           <mrow> 
            <msubsup> 
             <mi>
               Q 
             </mi> 
             <mrow> 
              <msup> 
               <mrow></mrow> 
               <mn>
                 0 
               </mn> 
              </msup> 
             </mrow> 
             <mn>
               2 
             </mn> 
            </msubsup> 
           </mrow> 
           <mrow> 
            <msup> 
             <mi>
               b 
             </mi> 
             <mn>
               2 
             </mn> 
            </msup> 
            <mi>
              g 
            </mi> 
           </mrow> 
          </mfrac> 
         </mrow> 
         <mo>
           ) 
         </mo> 
        </mrow> 
       </mrow> 
       <mrow> 
        <mrow> 
         <mn>
           1 
         </mn> 
         <mo>
           / 
         </mo> 
         <mn>
           3 
         </mn> 
        </mrow> 
       </mrow> 
      </msup> 
     </mrow> 
    </math> (20)</p>
   <p>and the critical velocity is given by</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <msub> 
       <mi>
         u 
       </mi> 
       <mi>
         c 
       </mi> 
      </msub> 
      <mo>
        = 
      </mo> 
      <msqrt> 
       <mrow> 
        <mi>
          g 
        </mi> 
        <msub> 
         <mi>
           h 
         </mi> 
         <mi>
           c 
         </mi> 
        </msub> 
       </mrow> 
      </msqrt> 
     </mrow> 
    </math> (21)</p>
  </sec><sec id="s4">
   <title>4. Numerical Solution of the Saint-Venant Equations</title>
   <p>Saint-Venant equations (1)-(2) are numerically solved using an explicit backwards finite difference scheme <xref ref-type="bibr" rid="scirp.137217-11">
     [11]
    </xref> <xref ref-type="bibr" rid="scirp.137217-12">
     [12]
    </xref>. Initial conditions and boundary conditions for h and Q are considered as follows:</p>
   <p>Initial conditions for h and Q:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        h 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mn>
          0 
        </mn> 
        <mo>
          , 
        </mo> 
        <mi>
          x 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         h 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
     </mrow> 
    </math> (22)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        Q 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mn>
          0 
        </mn> 
        <mo>
          , 
        </mo> 
        <mi>
          x 
        </mi> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         Q 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
     </mrow> 
    </math> (23)</p>
   <p>Boundary condition for h:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        h 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          t 
        </mi> 
        <mo>
          , 
        </mo> 
        <mn>
          0 
        </mn> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         h 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
     </mrow> 
    </math> (24)</p>
   <p>Boundary condition for Q: a half-sinusoidal perturbation pulse with amplitude A, period T and duration t<sub>p</sub> is imposed at the inlet of the channel to analyze the transient response <xref ref-type="bibr" rid="scirp.137217-11">
     [11]
    </xref>:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        Q 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          t 
        </mi> 
        <mo>
          , 
        </mo> 
        <mn>
          0 
        </mn> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         Q 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
      <mo>
        + 
      </mo> 
      <mi>
        A 
      </mi> 
      <mi>
        sin 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <mn>
            2 
          </mn> 
          <mi>
            π 
          </mi> 
          <mi>
            t 
          </mi> 
         </mrow> 
         <mi>
           T 
         </mi> 
        </mfrac> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        , 
      </mo> 
      <mi>
        t 
      </mi> 
      <mo>
        ≤ 
      </mo> 
      <msub> 
       <mi>
         t 
       </mi> 
       <mi>
         p 
       </mi> 
      </msub> 
     </mrow> 
    </math> (25)</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        Q 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mi>
          t 
        </mi> 
        <mo>
          , 
        </mo> 
        <mn>
          0 
        </mn> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         Q 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
      <mo>
        , 
      </mo> 
      <mi>
        t 
      </mi> 
      <mo>
        &gt; 
      </mo> 
      <msub> 
       <mi>
         t 
       </mi> 
       <mi>
         p 
       </mi> 
      </msub> 
     </mrow> 
    </math> (26)</p>
   <p>where Q<sub>0</sub> = 0.038 m<sup>3</sup>/s, A = 0.75Q<sub>0</sub>, T = 5 s and t<sub>p</sub> = 5 s.</p>
   <p>Parameters of the rectangular open channel considered in the numerical simulation are shown in <xref ref-type="table" rid="table1">
     Table 1
    </xref> <xref ref-type="bibr" rid="scirp.137217-13">
     [13]
    </xref>.</p>
   <table-wrap id="table1">
    <label>
     <xref ref-type="table" rid="table1">
      Table 1
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.137217-"></xref>Table 1. Open channel parameters <xref ref-type="bibr" rid="scirp.137217-13">
       [13]
      </xref>.</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td class="custom-bottom-td custom-top-td acenter" width="55.14%"><p style="text-align:center">Parameter</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="44.86%"><p style="text-align:center">Value</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="55.14%"><p style="text-align:center">Initial height (h<sub>0</sub>)</p></td> 
      <td class="custom-top-td acenter" width="44.86%"><p style="text-align:center">0.15 m</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="55.14%"><p style="text-align:center">Volumetric flow rate (Q<sub>0</sub>)</p></td> 
      <td class="acenter" width="44.86%"><p style="text-align:center">0.038 m<sup>3</sup>/s</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="55.14%"><p style="text-align:center">Channel width (b)</p></td> 
      <td class="acenter" width="44.86%"><p style="text-align:center">0.20 m</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="55.14%"><p style="text-align:center">Initial velocity (u<sub>0</sub>)</p></td> 
      <td class="acenter" width="44.86%"><p style="text-align:center">1.2667 m/s</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="55.14%"><p style="text-align:center">Channel length (L)</p></td> 
      <td class="acenter" width="44.86%"><p style="text-align:center">10.0 m</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="55.14%"><p style="text-align:center">Inclination angle (θ)</p></td> 
      <td class="acenter" width="44.86%"><p style="text-align:center">5 degrees</p></td> 
     </tr> 
     <tr> 
      <td class="custom-bottom-td acenter" width="55.14%"><p style="text-align:center">Initial velocity (u<sub>0</sub>)</p></td> 
      <td class="custom-bottom-td acenter" width="44.86%"><p style="text-align:center">1.27 m/s</p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>The discretized Saint-Venant equations are solved for a simulation time of 20 s, with setting C = 24 and Re = 4 × 10<sup>3</sup>. A channel length of 10 m is assumed and 500 nodes were employed with Δx = 0.02 m. A time step Δt = 1 × 10<sup>−</sup><sup>3</sup> s is established <xref ref-type="bibr" rid="scirp.137217-11">
     [11]
    </xref>.</p>
  </sec><sec id="s5">
   <title>5. Results and Comments</title>
   <p>Five values of the channel inclination angle are considered in the numerical simulations, namely 1, 2, 3, 4 and 5 degrees. As in <xref ref-type="bibr" rid="scirp.137217-11">
     [11]
    </xref>, four temperature values are assumed: 1450˚C, 1500˚C, 1550˚C, and 1600˚C, corresponding to 1723.15, 1773.15, 1823.15, and 1873.15 ˚K, respectively. For the numerical simulations, a molten blast furnace slag is considered, the physical properties of which are described in <xref ref-type="bibr" rid="scirp.137217-14">
     [14]
    </xref>. An Arrhenius-type dependence of the viscosity with temperature is assumed:</p>
   <p>
    <math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> 
      <mi>
        μ 
      </mi> 
      <mo>
        = 
      </mo> 
      <msub> 
       <mi>
         μ 
       </mi> 
       <mn>
         0 
       </mn> 
      </msub> 
      <mi>
        exp 
      </mi> 
      <mrow> 
       <mo>
         ( 
       </mo> 
       <mrow> 
        <mfrac> 
         <mrow> 
          <mi>
            E 
          </mi> 
          <mi>
            a 
          </mi> 
         </mrow> 
         <mrow> 
          <mi>
            R 
          </mi> 
          <mi>
            T 
          </mi> 
         </mrow> 
        </mfrac> 
       </mrow> 
       <mo>
         ) 
       </mo> 
      </mrow> 
     </mrow> 
    </math> (27)</p>
   <p>Above μ<sub>0</sub> is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the absolute temperature (˚K). Viscosity parameters shown in <xref ref-type="table" rid="table2">
     Table 2
    </xref> are taken from <xref ref-type="bibr" rid="scirp.137217-14">
     [14]
    </xref>.</p>
   <table-wrap id="table2">
    <label>
     <xref ref-type="table" rid="table2">
      Table 2
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.137217-"></xref>Table 2. Viscosity parameters of the blast furnace slag, taken from <xref ref-type="bibr" rid="scirp.137217-14">
       [14]
      </xref>.</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td class="custom-bottom-td custom-top-td acenter" width="50.00%"><p style="text-align:center">Parameter</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="50.00%"><p style="text-align:center">Value</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="50.00%"><p style="text-align:center">μ<sub>0</sub>, kg/(m.s)</p></td> 
      <td class="custom-top-td acenter" width="50.00%"><p style="text-align:center">9.469 × 10<sup>−</sup><sup>5</sup></p></td> 
     </tr> 
     <tr> 
      <td class="custom-bottom-td acenter" width="50.00%"><p style="text-align:center">Ea, J/mol</p></td> 
      <td class="custom-bottom-td acenter" width="50.00%"><p style="text-align:center">1.940 × 10<sup>5</sup></p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>
    <xref ref-type="table" rid="table3">
     Table 3
    </xref> shows the Froude numbers for the different combinations of temperature and inclination angle of the channel.</p>
   <table-wrap id="table3">
    <label>
     <xref ref-type="table" rid="table3">
      Table 3
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.137217-"></xref>Table 3. Froude numbers.</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.66%"><p style="text-align:center">T (˚K)</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 1˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 2˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.66%"><p style="text-align:center">θ = 3˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 4˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 5˚</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="16.66%"><p style="text-align:center">1723.15</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">1.62518</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">2.71534</p></td> 
      <td class="custom-top-td acenter" width="16.66%"><p style="text-align:center">3.59150</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">4.35770</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">5.04688</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="16.66%"><p style="text-align:center">1773.15</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">2.85603</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">4.57268</p></td> 
      <td class="acenter" width="16.66%"><p style="text-align:center">5.93154</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">7.11095</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">8.16503</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="16.66%"><p style="text-align:center">1823.15</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">4.62335</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">7.24903</p></td> 
      <td class="acenter" width="16.66%"><p style="text-align:center">9.28030</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">11.02499</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">12.58445</p></td> 
     </tr> 
     <tr> 
      <td class="custom-bottom-td acenter" width="16.66%"><p style="text-align:center">1873.15</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">6.86626</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">10.93795</p></td> 
      <td class="custom-bottom-td acenter" width="16.66%"><p style="text-align:center">13.91215</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">16.43151</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">18.67344</p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>During the numerical simulations, the height, velocity and volumetric flow rate of the molten slag at the channel discharge were monitored. The results are depicted in <xref ref-type="fig" rid="figFigures 1">
     Figures 1
    </xref>, <xref ref-type="fig" rid="figFigures 2">
     Figures 2
    </xref> and <xref ref-type="fig" rid="figFigures 3">
     Figures 3
    </xref>, in which h<sub>c</sub> = 0.1544 m and u<sub>c</sub> = 1.2306 m/s. <xref ref-type="fig" rid="fig1">
     Figure 1
    </xref> shows that for channel inclination angles of 1, 2 and 3 degrees and T = 1723.15 ˚K the slag height shows transient peaks, due to the half-sinusoidal pulse, above the critical height. Subsequently, the height decreases below the critical height of its steady-state value. For all channel inclination angles considered, <xref ref-type="fig" rid="fig1">
     Figure 1
    </xref> shows that the steady-state slag height is less than the critical height, represented by the dashed line. Furthermore, according to <xref ref-type="table" rid="table3">
     Table 3
    </xref>, the value of the Froude number is always greater than unity for all possible combinations of angles and temperatures. <xref ref-type="fig" rid="fig2">
     Figure 2
    </xref> shows the dynamic behavior of the slag velocity for T = 1723.15 ˚K which, after transient peaks, finally reaches its steady-state value above the critical value shown in the dashed line. The above means that, in any case, the flow is considered supercritical in nature.</p>
   <p>The time evolution of the volumetric flow rate for T = 1723.15 K and different channel angles at the channel discharge are shown in <xref ref-type="fig" rid="fig3">
     Figure 3
    </xref>. The transient peaks that appear are due to the half-sinusoidal pulse imposed as a boundary condition at the channel input. This pulse has a duration of 5 s, so after this time, the volumetric flow values return to their original value Q<sub>0</sub>.</p>
   <fig id="fig1" position="float">
    <label>Figure 1</label>
    <caption>
     <title>Figure 1. Transient behavior of the slag height at the channel discharge for different values of the channel inclination angle at T = 1723.15 ˚K.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId72.jpeg?20241106041158" />
   </fig>
   <fig id="fig2" position="float">
    <label>Figure 2</label>
    <caption>
     <title>Figure 2. Transient behavior of the slag velocity at the channel discharge for different values of the channel inclination angle at T = 1723.15 ˚K.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId73.jpeg?20241106041158" />
   </fig>
   <fig id="fig3" position="float">
    <label>Figure 3</label>
    <caption>
     <title>Figure 3. Transient behavior of the slag volumetric flow rate at the channel discharge for different values of the channel inclination angle at T = 1723.15 ˚K.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId74.jpeg?20241106041158" />
   </fig>
   <p>The polynomial given by Equation (12) has one real root and two imaginary roots. For example, for θ = 5 degrees, T = 1723.15 ˚K, μ = 0.72 kg/(m.s), ρ = 2700 kg/m<sup>3</sup>, Re = 4000 and f<sub>D</sub> = 24/Re, the corresponding polynomial has the following roots: i) 0.0525, ii) −0.0231 + 0.0255i, iii) −0.0231 − 0.0255i. <xref ref-type="table" rid="table4">
     Table 4
    </xref> shows the steady-state values of the molten slag height corresponding to the real values of the polynomial roots.</p>
   <table-wrap id="table4">
    <label>
     <xref ref-type="table" rid="table4">
      Table 4
     </xref></label>
    <caption>
     <title>
      <xref ref-type="bibr" rid="scirp.137217-"></xref>Table 4. Steady-state molten slag heights (m).</title>
    </caption>
    <table class="MsoTableGrid custom-table" border="0" cellspacing="0" cellpadding="0"> 
     <tr> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.66%"><p style="text-align:center">T (˚K)</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 1˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 2˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.66%"><p style="text-align:center">θ = 3˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 4˚</p></td> 
      <td class="custom-bottom-td custom-top-td acenter" width="16.67%"><p style="text-align:center">θ = 5˚</p></td> 
     </tr> 
     <tr> 
      <td class="custom-top-td acenter" width="16.66%"><p style="text-align:center">1723.15</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">0.11179</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">0.07944</p></td> 
      <td class="custom-top-td acenter" width="16.66%"><p style="text-align:center">0.06583</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">0.05787</p></td> 
      <td class="custom-top-td acenter" width="16.67%"><p style="text-align:center">0.05247</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="16.66%"><p style="text-align:center">1773.15</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.07686</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.05613</p></td> 
      <td class="acenter" width="16.66%"><p style="text-align:center">0.04712</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.04176</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.03808</p></td> 
     </tr> 
     <tr> 
      <td class="acenter" width="16.66%"><p style="text-align:center">1823.15</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.05544</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.04128</p></td> 
      <td class="acenter" width="16.66%"><p style="text-align:center">0.03498</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.03117</p></td> 
      <td class="acenter" width="16.67%"><p style="text-align:center">0.02854</p></td> 
     </tr> 
     <tr> 
      <td class="custom-bottom-td acenter" width="16.66%"><p style="text-align:center">1873.15</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">0.04146</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">0.03130</p></td> 
      <td class="custom-bottom-td acenter" width="16.66%"><p style="text-align:center">0.02670</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">0.02389</p></td> 
      <td class="custom-bottom-td acenter" width="16.67%"><p style="text-align:center">0.02194</p></td> 
     </tr> 
    </table>
   </table-wrap>
   <p>
    <xref ref-type="fig" rid="fig4">
     Figure 4
    </xref> shows the steady-state slag height as a function of temperature for different inclination angles of the open channel. It can be seen that as the temperature increases, the steady-state value of the slag height decreases. The same inverse dependence is noted with the angle of inclination: as the channel increases its inclination, the stable height of the slag decreases. The opposite occurs with the velocity of the slag, as is seen in <xref ref-type="fig" rid="fig5">
     Figure 5
    </xref>: As the temperature and the angle of inclination increase, the velocity in the steady state increases. This may be due to two factors: i) As the temperature increases, the viscosity of the slag decreases, which in turn decreases the resistance to flow, and ii) as the angle of inclination of the channel increases, the vertical component of the gravitational acceleration is increased, and consequently the slag velocity is also increased.</p>
   <fig id="fig4" position="float">
    <label>Figure 4</label>
    <caption>
     <title>Figure 4. Steady-state height as a function of slag temperature for different channel inclination angles.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId75.jpeg?20241106041158" />
   </fig>
   <fig id="fig5" position="float">
    <label>Figure 5</label>
    <caption>
     <title>Figure 5. Steady-state velocity as a function of slag temperature for different channel inclination angles.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId76.jpeg?20241106041158" />
   </fig>
   <p>Finally, numerical simulations were carried out using different initial conditions in height and velocity, keeping the temperature constant at 1723.15 ˚K and the channel inclination angle at 1 degree. The results are shown in the form of a phase portrait in <xref ref-type="fig" rid="fig6">
     Figure 6
    </xref>, where the convergence of all trajectories towards the corresponding steady state can be seen. This steady state, in which h<sub>s</sub> = 0.111 m and its corresponding u<sub>s</sub> = 0.038/(0.2 × 0.11) = 1.712 m/s, coincides with that reported in <xref ref-type="table" rid="table4">
     Table 4
    </xref>.</p>
   <fig id="fig6" position="float">
    <label>Figure 6</label>
    <caption>
     <title>Figure 6. Phase-portrait of the flow for a temperature of 1725.15 ˚K and a channel angle of 1 degree.</title>
    </caption>
    <graphic mimetype="image" position="float" xlink:type="simple" xlink:href="https://html.scirp.org/file/1561607-rId77.jpeg?20241106041158" />
   </fig>
  </sec><sec id="s6">
   <title>6. Conclusions</title>
   <p>The transient and steady-state flow of a molten slag in an open channel is numerically analyzed. For the transient flow a half-sinusoidal pulse on the volumetric flow rate is imposed as a boundary condition at the channel input and the Saint-Venant equations were solved numerically. For the steady-state flow, the derivatives in time and space in the Saint-Venant equations were set equal to zero, and a polynomial of degree 3, whose roots are the slag height values, was obtained. It was assumed that the viscosity of the slag has an Arrhenius-type behavior with temperature. Four values of temperature and five values of the angle of inclination of the channel are considered. Based on the numerical results obtained, the following conclusions emerge:</p>
   <p>The time trajectories of the height and velocity of the slag present temporary peaks due to the half-sinusoidal pulse imposed at the channel entrance, but subsequently reach a steady state value. The steady-state values of the height and velocity of the molten slag depend strongly on the temperature of the slag and the angle of inclination of the channel. As the slag temperature and channel angle increase, the value of the steady-state slag height decreases. The value of the steady-state slag velocity increases as the slag temperature and channel inclination angle increase. Finally, a phase portrait of the flow shows the convergence of different initial conditions towards the corresponding steady state in the range of values considered.</p>
   <p>The results reported in this work are limited to the one-dimensional analysis of the flow of a molten slag in a rectangular channel under the influence of only two variables, namely the angle of inclination of the channel and the temperature of the slag. Further work should include additional variables such as different channel geometries, higher Reynolds numbers and slag density. The results shown in this work, since they are numerical in nature, should be verified through experimental data.</p>
  </sec>
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