# 13.9.2: Flooding and Reversal Flow

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The limits of one kind the counter–current flow regimes, that is stratified flow are discussed here. This problem appears in nuclear engineering (or boiler engineering) where there is a need to make sure that liquid (water) inserted into the pipe reaching the heating zone. When there is no water (in liquid phase), the fire could melt or damage the boiler. In some situations, the fire can be too large or/and the water supply failed below a critical value the water turn into steam. The steam will flow in the opposite direction. To analyze this situation consider a two dimensional conduit with a liquid inserted in the left side as depicted in Figure 13.13. The liquid velocity at very low gas velocity is constant but not uniform. Further increase of the gas velocity will reduce the average liquid velocity. Additional increase of the gas velocity will bring it to a point where the liquid will flow in a reverse direction and/or disappear (dried out).

Fig. 13.15 A diagram to explain the flood in a two dimension geometry.

A simplified model for this situation is for a two dimensional configuration where the liquid is flowing down and the gas is flowing up as shown in Figure 13.15. It is assumed that both fluids are flowing in a laminar regime and steady state. Additionally, it is assumed that the entrance effects can be neglected. The liquid flow rate, $$Q_L$$, is unknown. However, the pressure difference in the ($$x$$ direction) is known and equal to zero. The boundary conditions for the liquid is that velocity at the wall is zero and the velocity at the interface is the same for both phases $$U_G = U_L$$ or $$\left.\tau_i\right|_G = \left.\tau_i\right|_L$$. As it will be shown later, both conditions cannot coexist. The model can be improved by considering turbulence, mass transfer, wavy interface, etc. This model is presented to exhibits the trends and the special features of counter-current flow. Assuming the pressure difference in the flow direction for the gas is constant and uniform. It is assumed that the last assumption does not contribute or change significantly the results. The underline rational for this assumption is that gas density does not change significantly for short pipes (for more information look for the book "Fundamentals of Compressible Flow'' in Potto book series in the Fanno flow chapter.). The liquid film thickness is unknown and can be expressed as a function of the above boundary conditions. Thus, the liquid flow rate is a function of the boundary conditions. On the liquid side, the gravitational force has to be balanced by the shear forces as

$\dfrac{d\tau_{xy}} {dx} = \rho_L\,g \label{phase:eq:dTauDx}$

The integration of equation (55) results in

$\tau_{xy} = \rho_L\,g\,x + C_1 \label{phase:eq:Taufx}$ The integration constant, $$C_1$$, can be found from the boundary condition where $$\tau_{xy} (x=h) = \tau_i$$. Hence,

$\tau_i = \rho_L\,g\,h + C_1 \label{phase:eq:tauIini}$ The integration constant is then $$C_i = \tau_i - \rho_L\,g\,h$$ which leads to

$\tau_{xy} = \rho_L\,g\,(x-h) + \tau_i \label{phase:eq:tauL}$ Substituting the newtonian fluid relationship into equation (58) to obtained

$\mu_L \dfrac{dU_y}{dx} = \rho_L\,g\,(x-h) + \tau_i \label{phase:eq:Gtau}$ or in a simplified form as

$\dfrac{dU_{y} }{dx} = \dfrac{\rho_L\,g\,(x-h) }{\mu_L} + \dfrac{\tau_i}{\mu_L} \label{phase:eq:GtauS}$ Equation (60) can be integrate to yield

$U_{y} = \dfrac{\rho_L\,g} {\mu_L} \left(\dfrac{x^2}{2} -h\,x \right) + \dfrac{\tau_i\,x}{\mu_L} + C_2 \label{phase:eq:Uyproife}$ The liquid velocity at the wall, [$$U(x=0) =0$$], is zero and the integration coefficient can be found to be

$C_2 = 0 \label{phase:eq:zeroWall}$ The liquid velocity profile is then

Liquid Velocity

$\label{phase:eq:U1proife} U_{y} = d\dfrac{\rho_L\,g} {\mu_L} \left(\dfrac{x^2}{2} -h\,x \right) + \dfrac{\tau_i\,x}{\mu_L}$

The velocity at the liquid–gas interface is

$U_{y}(x=h) = \dfrac{\tau_i\,h}{\mu_L} - \dfrac{\rho_L\,g\,h^2} {2\,\mu_L} \label{phase:eq:Uproife2}$

The velocity can vanish (zero) inside the film in another point which can be obtained from

$0 = \dfrac{\rho_L\,g} {\mu_L} \left(\dfrac{x^2}{2} -h\,x \right) + \dfrac{\tau_i\,x}{\mu_L} \label{phase:eq:UproifeZero}$ The solution for equation (65) is

$\label{phase:eq:UproifeZeroPoint1} \left. x\,\right|_{@ U_L=0} = 2\, h - \dfrac{2\,\tau_i}{\mu_L\,g\,\rho_L}$ The maximum $$x$$ value is limited by the liquid film thickness, $$h$$. The minimum shear stress that start to create reversible velocity is obtained when $$x=h$$ which is

$\label{phase:eq:UproifeZeroPoint} 0 = \dfrac{\rho_L\,g} {\mu_L} \left(\dfrac{h^2}{2} -h\,h \right) + \dfrac{\tau_i\,h}{\mu_L} \ \nonumber \hookrightarrow {\tau_i}_{0} = \dfrac {h\,g\,\rho_L} {2}$ If the shear stress is below this critical shear stress $${\tau_i}_{0}$$ then no part of the liquid will have a reversed velocity. The notation of $${\tau_i}_{0}$$ denotes the special value at which a starting shear stress value is obtained to have reversed flow. The point where the liquid flow rate is zero is important and it is referred to as initial flashing point. The flow rate can be calculated by integrating the velocity across the entire liquid thickness of the film.

$\dfrac{Q}{w} = \int_0^h U_{y}dx = \int_0^h \left[ \dfrac{\rho_L\,g} {\mu_L} \left(\dfrac{x^2}{2} -h\,x \right) + \dfrac{\tau_i\,x}{\mu_L} \right] dx \label{phase:eq:flowRate}$ Where $$w$$ is the thickness of the conduit (see Figure 13.15). Integration equation (68) results in

$\dfrac{Q}{w} = \dfrac{h^2 \left( 3\,\tau_i-2\,g\,h\,{\rho_L} \right) } {6\,\mu_L} \label{phase:eq:QB}$ It is interesting to find the point where the liquid mass flow rate is zero. This point can be obtained when equation (69) is equated to zero. There are three solutions for equation (69). The first two solutions are identical in which the film height is $$h=0$$ and the liquid flow rate is zero. But, also, the flow rate is zero when $$3\,\tau_i=2\,g\,h\,{\rho_L}$$. This request is identical to the demand in which

Shear Stress

$\label{phase:eq:tau_iCriticla} {\tau_i}_{\text{ critical} } = \dfrac{2\,g\,h\,{\rho_L}}{3}$

This critical shear stress, for a given film thickness, reduces the flow rate to zero or effectively "drying'' the liquid (which is different then equation (67)). For this shear stress, the critical upward interface velocity is

Critical Velocity

$\label{phase:eq:UatInterfaceCritical} \left.{U}_{critical} \right|_{interface} = \overbrace{\dfrac{1}{6}}^{\left( \dfrac{2}{3} - \dfrac{1} {2} \right)} \left( \dfrac{\rho_L\,g\,h^2}{\mu_L} \right)$

The wall shear stress is the last thing that will be done on the liquid side. The wall shear stress is

$\left.\tau_{L}\right|_{@ wall} = \mu_L \left. \dfrac{dU}{dx} \right|_{x=0} = \mu_L \left( \dfrac{\rho_L\,g}{\mu_L} \left( \cancelto{0}{2\,x} - h \right) + \overbrace{\dfrac{2\,g\,h\,{\rho_L}}{3}}^{\tau_i} \dfrac{1}{\mu_L} \right)_{x=0} \label{phase:eq:lwallStress}$
Simplifying equation be equal $$g\,h\,\rho_L$$ to support the weight of the liquid. becomes (notice the change of the sign accounting for the direction)

$\left.\tau_{L}\right|_{@ wall} = \dfrac{g\,h\,{\rho_L}}{3} \label{phase:eq:lwallStressF}$
Again, the gas is assumed to be in a laminar flow as well. The shear stress on gas side is balanced by the pressure gradient in the $$y$$ direction. The momentum balance on element in the gas side is

$\dfrac{d{\tau_{xy}}_G}{dx} = \dfrac{dP}{dy} \label{phase:eq:GasTau}$

The pressure gradient is a function of the gas compressibility. For simplicity, it is assumed that pressure gradient is linear. This assumption means or implies that the gas is incompressible flow. If the gas was compressible with an ideal gas equation of state then the pressure gradient is logarithmic. Here, for simplicity reasons, the linear equation is used. In reality the logarithmic equation should be used ( a discussion can be found in "Fundamentals of Compressible Flow'' a Potto project book). Thus, equation (74) can be rewritten as

$\dfrac{d{\tau_{xy}}_G}{dx} = \dfrac{\Delta P}{\Delta y} = \dfrac{\Delta P}{L} \label{phase:eq:GasTauA}$ Where $$\Delta y = L$$ is the entire length of the flow and $$\Delta P$$ is the pressure difference of the entire length. Utilizing the Newtonian relationship, the differential equation is

$\dfrac{d^2U_{G}}{dx^2} = \dfrac{\Delta P}{\mu_G\,L} \label{phase:eq:d2Udx}$ Equation (76) can be integrated twice to yield

$U_{G} = \dfrac{\Delta P}{\mu_G\,L} \, x^2 + C_1\, x + C_2 \label{phase:eq:Gvelocity}$ This velocity profile must satisfy zero velocity at the right wall. The velocity at the interface is the same as the liquid phase velocity or the shear stress are equal. Mathematically these boundary conditions are

$U_{G}(x=D) = 0 \label{phase:eq:UGDzero}$ and

$\label{phase:eq:UGhUL} \begin{array}{rcrr} U_{G}(x=h) &= U_L(x=h) & \;(a) & \qquad or \ \tau_G (x=h) &= \tau_L (x=h) & \;(b) & \end{array}$ Applying B.C. (78) into equation (77) results in
$\label{phase:eq:C2ini} U_G = 0 = \dfrac{\Delta P}{\mu_G\,L} \, D^2 + C_1\, D + C_2 \Longrightarrow C_2 = - \dfrac{\Delta P}{\mu_G\,L} \, D^2 + C_1\, D$ Which leads to

$U_{G} = \dfrac{\Delta P}{\mu_G\,L} \, \left(x^2 -D^2\right) + C_1\, \left( x - D \right) \label{phase:eq:GvelocityIMProved}$ At the other boundary condition, equation (79)(a), becomes

$\dfrac{\rho_L\,g\,h^2}{6\,\mu_L} = \dfrac{\Delta P}{\mu_G\,L} \, \left(h^2 -D^2\right) + C_1\, \left( h - D \right) \label{phase:eq:}$ The last integration constant, $$C_1$$ can be evaluated as

$C_1 = \dfrac{\rho_L\,g\,h^2}{6\,\mu_L \, \left( h - D \right)} - \dfrac{\Delta P \, \left(h +D\right) }{\mu_G\,L} \label{phase:eq:C1L}$ With the integration constants evaluated, the gas velocity profile is

$U_{G} = \dfrac{\Delta P}{\mu_G\,L} \, \left(x^2 -D^2\right) + \dfrac{\rho_L\,g\,h^2 \left( x - D\right) }{6\,\mu_L \left( h - D \right)} - \dfrac{\Delta P \, \left(h +D\right) \left( x - D\right) }{\mu_G\,L} \label{phase:eq:GvelocityIMProvedF}$ The velocity in equation (84) is equal to the velocity equation (64) when ($$x=h$$). However, in that case, it is easy to show that the gas shear stress is not equal to the liquid shear stress at the interface (when the velocities are assumed to be the equal). The difference in shear stresses at the interface due to this assumption, of the equal velocities, cause this assumption to be not physical. The second choice is to use the equal shear stresses at the interface, condition (79)(b). This condition requires that

$\mu_G \dfrac{dU_G}{dx} = \mu_L \dfrac{dU_L}{dx} \label{phase:eq:tauLtauG}$ The expressions for the derivatives are

$\overbrace{\dfrac{2\,h\,\Delta P}{L} + \mu_G \, C_1}^{gas\;side} = \overbrace{\dfrac{ 2\, g\,h\,\rho_L} {3}}^{liquid\;side} \label{phase:eq:dTauLG}$ As result, the integration constant is

$C_1 = \dfrac{ 2\, g\,h\,\rho_L} {3\,\mu_G} - \dfrac{2\,h\,\Delta P}{\mu_G\,L} \label{phase:eq:C1tau}$ The gas velocity profile is then

$U_{G} = \dfrac{\Delta P}{\mu_G\,L} \, \left(x^2 -D^2\right) + \left( \dfrac{ 2\, g\,h \,\rho_L} {3 \, \mu_G} - \dfrac{2\,h\,\Delta P}{\mu_G\,L} \right) \left( x - D \right) \label{phase:eq:UGtau}$ The gas velocity at the interface is then

$\left. U_{G} \right|_{@ x=h} = \dfrac{\Delta P}{\mu_G\,L} \, \left(h^2 -D^2\right) + \left( \dfrac{ 2\,g\,h \,\rho_L} {3\,\mu_G} - \dfrac{2\,h\,\Delta P}{\mu_G\,L} \right) \left( h - D \right) \label{phase:eq:UatInterfacetau}$
This gas interface velocity is different than the velocity of the liquid side. The velocity at interface can have a "slip'' in very low density and for short distances. The shear stress at the interface must be equal, if no special effects occurs. Since there no possibility to have both the shear stress and velocity on both sides of the interface, different thing(s) must happen. It was assumed that the interface is straight but is impossible. Then if the interface becomes wavy, the two conditions can co–exist. The wall shear stress is

$\left.\tau_{G}\right|_{@ wall} = \mu_G \left. \dfrac{dU_G}{dx} \right|_{x=D} = \mu_G \left( \dfrac{\Delta P \, 2\, x}{\mu_G\, L} + \left( \dfrac{ 2\,g\,h \,\rho_L} {3\,\mu_G} - \dfrac{2\,h\,\Delta P}{\mu_G\, L} \right) \right)_{x=D} \label{phase:eq:gwallStress}$ or in a simplified form as

$\left.\tau_{G}\right|_{@ wall} = \dfrac{2\,\Delta P \, \left( D -h \right) }{L} + \dfrac{ 2\,g\,h \,\rho_L} {3} \label{phase:eq:gwallStressF}$

## The Required Pressure Difference

Fig. 13.16 General forces diagram to calculated the in a two dimension geometry.

The pressure difference to create the flooding (drying) has to take into account the fact that the surface is wavy. However, as first estimate the waviness of the surface can be neglected. The estimation of the pressure difference under the assumption of equal shear stress can be applied. In the same fashion the pressure difference under the assumption the equal velocity can be calculated. The actual pressure difference can be between these two assumptions but not must be between them. This model and its assumptions are too simplistic and the actual pressure difference is larger. However, this explanation is to show magnitudes and trends and hence it provided here. To calculate the required pressure that cause the liquid to dry, the total balance is needed. The control volume include the gas and liquid volumes. Figure 13.16 describes the general forces that acts on the control volume. There are two forces that act against the gravity and two forces with the gravity. The gravity force on the gas can be neglected in most cases. The gravity force on the liquid is the liquid volume times the liquid volume as

$F_{gL} = \rho\,g\,\overbrace{h\,L}^{\dfrac{Volme}{w} } \label{phase:eq:FgL}$

The total momentum balance is (see Figure 13.16)

$F_{gL} + \overbrace{L}^{\dfrac{A}{w} }\, {\tau_w}_{G} = \overbrace{L}^{\dfrac{A}{w} }\,{\tau_w}_ {L} + \overbrace{D\,\Delta P}^{\text{force due to pressure}} \label{phase:eq:Lfriction}$ Substituting the different terms into (93) result in

$\rho\,g\,L\,h + L\, \left( \dfrac{2\,\Delta P \, \left( D -h \right) }{L} + \dfrac{ 2\,g\,h \,\rho_L} {3} \right) = L \,\dfrac{g\,h\,\rho_L}{3} + {D\,\Delta P} \label{phase:eq:Gfriction}$ Simplifying equation (94) results in

$\dfrac{4\,\rho\,g\,L\,h}{3} = \left( 2\,h -D \right) \Delta P \label{phase:eq:DeltaP}$ or

$\Delta P = \dfrac{4\,\rho\,g\,L\,h}{3 \, \left( 2\,h -D \right) } \label{phase:eq:DeltaPF}$ This analysis shows far more reaching conclusion that initial anticipation expected. The interface between the two liquid flowing together is wavy. Unless the derivations or assumptions are wrong, this analysis equation (??) indicates that when $$D>2\,h$$ is a special case (extend open channel flow).