Analogies and differences in different transport phenomena

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<math>\frac{\partial ^{2}u^{+}}{\partial y^{+2}}</math>
<math>\frac{\partial ^{2}u^{+}}{\partial y^{+2}}</math>
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Energy
Energy
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<math>u^{+}\frac{\partial \theta }{\partial x^{+}}+v^{+}\frac{\partial \theta }{\partial y^{+}}=\frac{1}{\Pr \operatorname{Re}}\frac{\partial ^{2}\theta }{\partial y^{+2}}</math>
<math>u^{+}\frac{\partial \theta }{\partial x^{+}}+v^{+}\frac{\partial \theta }{\partial y^{+}}=\frac{1}{\Pr \operatorname{Re}}\frac{\partial ^{2}\theta }{\partial y^{+2}}</math>
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Species
Species
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<math>u^{+}\frac{\partial \varphi }{\partial x^{+}}+v^{+}\frac{\partial \varphi }{\partial y^{+}}=\frac{1}{\text{Sc}\operatorname{Re}}\frac{\partial ^{2}\varphi }{\partial y^{+2}}</math>
<math>u^{+}\frac{\partial \varphi }{\partial x^{+}}+v^{+}\frac{\partial \varphi }{\partial y^{+}}=\frac{1}{\text{Sc}\operatorname{Re}}\frac{\partial ^{2}\varphi }{\partial y^{+2}}</math>
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<math>\text{at }y^{+}=0,\text{  u}^{\text{+}}=0,\text{  }\theta \text{=}\varphi \text{=0}</math>
<math>\text{at }y^{+}=0,\text{  u}^{\text{+}}=0,\text{  }\theta \text{=}\varphi \text{=0}</math>
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<math>y^{+}=0,\text{  }v^{+}=v_{w}^{+}</math>
<math>y^{+}=0,\text{  }v^{+}=v_{w}^{+}</math>
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<math>y^{+}=\infty ,\text{  }u^{+}=1,\text{  }\theta =\varphi =1</math>
<math>y^{+}=\infty ,\text{  }u^{+}=1,\text{  }\theta =\varphi =1</math>
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where the dimensionless variables are defined as:
where the dimensionless variables are defined as:
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\end{align}</math>
\end{align}</math>
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Let’s first consider the analogy between momentum and heat transfer. Equations (4.349) and (4.350) and appropriate boundary conditions are the same if Pr = 1.  The solutions for u+ and θ are, therefore, exactly the same if Pr = 1 and one expects to have a relationship between the skin friction coefficient, cf, and heat transfer coefficient, h.
Let’s first consider the analogy between momentum and heat transfer. Equations (4.349) and (4.350) and appropriate boundary conditions are the same if Pr = 1.  The solutions for u+ and θ are, therefore, exactly the same if Pr = 1 and one expects to have a relationship between the skin friction coefficient, cf, and heat transfer coefficient, h.
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<math>\frac{c_{f}}{2}=\frac{\tau _{w}}{\rho U_{\infty }^{2}}=\frac{\mu \left. \frac{\partial u}{\partial y} \right|_{y=0}}{\rho U_{\infty }^{2}}=\frac{\nu \left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{U_{\infty }L}=\frac{\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{\operatorname{Re}}</math>
<math>\frac{c_{f}}{2}=\frac{\tau _{w}}{\rho U_{\infty }^{2}}=\frac{\mu \left. \frac{\partial u}{\partial y} \right|_{y=0}}{\rho U_{\infty }^{2}}=\frac{\nu \left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{U_{\infty }L}=\frac{\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{\operatorname{Re}}</math>
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<math>Nu=\frac{hL}{k}=\frac{-k\left. \frac{\partial T}{\partial y} \right|_{y=0}L}{k\left( T_{\omega }-T_{\infty } \right)}=\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}</math>
<math>Nu=\frac{hL}{k}=\frac{-k\left. \frac{\partial T}{\partial y} \right|_{y=0}L}{k\left( T_{\omega }-T_{\infty } \right)}=\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}</math>
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Since θ = u+ for Pr = 1, one also concludes that
Since θ = u+ for Pr = 1, one also concludes that
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<math>\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}=\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}</math>
<math>\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}=\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}</math>
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Therefore, combining eqs. (4.356) and (4.357) and using eq. (4.358) gives
Therefore, combining eqs. (4.356) and (4.357) and using eq. (4.358) gives
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<math>\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}</math>
<math>\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}</math>
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This relationship between the friction coefficient and Nu is referred to as Reynolds analogy and is appropriate when Pr = 1. If Pr ≠ 1, we already concluded that <math>\frac{\delta _{T}}{\delta }=\Pr ^{-1/3}\text{  for  }0.5\le \Pr \le 10</math> from the similarity solution presented in Section 4.6. Using this information, one can generalize the result of the Reynolds analogy to Pr ≠ 1 by
This relationship between the friction coefficient and Nu is referred to as Reynolds analogy and is appropriate when Pr = 1. If Pr ≠ 1, we already concluded that <math>\frac{\delta _{T}}{\delta }=\Pr ^{-1/3}\text{  for  }0.5\le \Pr \le 10</math> from the similarity solution presented in Section 4.6. Using this information, one can generalize the result of the Reynolds analogy to Pr ≠ 1 by
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<math>\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}\Pr ^{-1/3}</math>
<math>\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}\Pr ^{-1/3}</math>
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Now, we focus on similarities between the heat and mass transfer for comparison of eqs. (4.350) and (4.351) with their appropriate boundary conditions. It is clear that the solution for differential equations (4.350) and (4.351) for θ and <math>\varphi </math> are the same if Sc and Pr are interchanged appropriately.
Now, we focus on similarities between the heat and mass transfer for comparison of eqs. (4.350) and (4.351) with their appropriate boundary conditions. It is clear that the solution for differential equations (4.350) and (4.351) for θ and <math>\varphi </math> are the same if Sc and Pr are interchanged appropriately.
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<math>\text{Nu}_{x}=0.332\operatorname{Re}_{x}^{1/2}\Pr ^{1/3}</math>
<math>\text{Nu}_{x}=0.332\operatorname{Re}_{x}^{1/2}\Pr ^{1/3}</math>
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Therefore, it can also be assumed that the solution of eq. (4.351) for <math>v_{w}=0</math> is
Therefore, it can also be assumed that the solution of eq. (4.351) for <math>v_{w}=0</math> is
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<math>\text{Sh}_{x}=0.332\operatorname{Re}_{x}^{1/2}\text{Sc}^{1/3}</math>
<math>\text{Sh}_{x}=0.332\operatorname{Re}_{x}^{1/2}\text{Sc}^{1/3}</math>
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Combining eqs. (4.361) and (4.362) gives
Combining eqs. (4.361) and (4.362) gives
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<math>\frac{\text{Nu}}{\text{Sh}}\text{=}\frac{\text{Pr}^{\text{1/3}}}{\text{Sc}^{\text{1/3}}}</math>
<math>\frac{\text{Nu}}{\text{Sh}}\text{=}\frac{\text{Pr}^{\text{1/3}}}{\text{Sc}^{\text{1/3}}}</math>
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It should be noted that the convective effect due to vertical velocity at the surface was neglected in predicting h and hm in the above analysis and therefore the analogy presented in (4.363) is for very low mass transfer at the wall. The effect is that the contribution of <math>v_{w}</math> in heat and mass flux was primarily due to diffusion.
It should be noted that the convective effect due to vertical velocity at the surface was neglected in predicting h and hm in the above analysis and therefore the analogy presented in (4.363) is for very low mass transfer at the wall. The effect is that the contribution of <math>v_{w}</math> in heat and mass flux was primarily due to diffusion.
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<math>{q}''_{w}=\rho c_{p}v_{w}\left( T_{w}-T_{\infty } \right)-k\left. \frac{\partial T}{\partial y} \right|_{y=0}</math>
<math>{q}''_{w}=\rho c_{p}v_{w}\left( T_{w}-T_{\infty } \right)-k\left. \frac{\partial T}{\partial y} \right|_{y=0}</math>
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<math>{\dot{m}}''_{w}=\rho \left( \omega _{1,\omega }v_{w}-D_{12}\left. \frac{\partial \omega _{1}}{\partial y} \right|_{y=0} \right)</math>
<math>{\dot{m}}''_{w}=\rho \left( \omega _{1,\omega }v_{w}-D_{12}\left. \frac{\partial \omega _{1}}{\partial y} \right|_{y=0} \right)</math>
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The analogy in momentum, heat, and mass transfer can also be applied to complex and/or coupled transport phenomena problems including phase change and chemical reactions. Obviously, in these circumstances, the simple relationship developed in eq. (4.363) is not applicable. To show the usefulness of this analogy to more complex transport phenomena problems we will apply it to sublimation with a chemical reaction for forced convective flow over a flat plate.
The analogy in momentum, heat, and mass transfer can also be applied to complex and/or coupled transport phenomena problems including phase change and chemical reactions. Obviously, in these circumstances, the simple relationship developed in eq. (4.363) is not applicable. To show the usefulness of this analogy to more complex transport phenomena problems we will apply it to sublimation with a chemical reaction for forced convective flow over a flat plate.
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<math>\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0</math>
<math>\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0</math>
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<math>u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left( \nu \frac{\partial u}{\partial y} \right)</math>
<math>u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left( \nu \frac{\partial u}{\partial y} \right)</math>
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<math>\frac{\partial }{\partial x}(\rho c_{p}uT)+\frac{\partial }{\partial y}(\rho c_{p}vT)=\frac{\partial }{\partial y}\left( k\frac{\partial T}{\partial y} \right)+{\dot{m}}'''_{o}\text{ }h_{c,o}</math>
<math>\frac{\partial }{\partial x}(\rho c_{p}uT)+\frac{\partial }{\partial y}(\rho c_{p}vT)=\frac{\partial }{\partial y}\left( k\frac{\partial T}{\partial y} \right)+{\dot{m}}'''_{o}\text{ }h_{c,o}</math>
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<math>\frac{\partial }{\partial x}(\rho u\omega _{o})+\frac{\partial }{\partial y}(\rho v\omega _{o})=\frac{\partial }{\partial y}\left( \rho D\frac{\partial \omega _{o}}{\partial y} \right)-{\dot{m}}'''_{o}</math>
<math>\frac{\partial }{\partial x}(\rho u\omega _{o})+\frac{\partial }{\partial y}(\rho v\omega _{o})=\frac{\partial }{\partial y}\left( \rho D\frac{\partial \omega _{o}}{\partial y} \right)-{\dot{m}}'''_{o}</math>
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where <math>{\dot{m}}'''_{0}</math> is the rate of oxidant consumption (kg/m3*s),  hc,0 is the heat released by combustion per unit mass consumption of the oxidant (J/kg), which is different from the combustion heat, and<math>\omega _{0}</math> is the mass fraction of the oxidant in the gaseous mixture.
where <math>{\dot{m}}'''_{0}</math> is the rate of oxidant consumption (kg/m3*s),  hc,0 is the heat released by combustion per unit mass consumption of the oxidant (J/kg), which is different from the combustion heat, and<math>\omega _{0}</math> is the mass fraction of the oxidant in the gaseous mixture.
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\end{matrix}\_at\_y\to \infty </math>
\end{matrix}\_at\_y\to \infty </math>
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{| class="wikitable" border="0"
{| class="wikitable" border="0"
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\end{matrix}\_at\_y=0</math>
\end{matrix}\_at\_y=0</math>
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\end{matrix}</math>
\end{matrix}</math>
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\end{matrix}</math>
\end{matrix}</math>
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<math>\frac{\partial }{\partial x}\left[ \rho u(c_{p}T+\omega _{o}h_{c,o}) \right]+\frac{\partial }{\partial y}\left[ \rho v(c_{p}T+\omega _{o}h_{c,o}) \right]=\frac{\partial }{\partial y}\left[ k\frac{\partial T}{\partial y}+\rho Dh_{c,o}\frac{\partial \omega _{o}}{\partial y} \right]</math>
<math>\frac{\partial }{\partial x}\left[ \rho u(c_{p}T+\omega _{o}h_{c,o}) \right]+\frac{\partial }{\partial y}\left[ \rho v(c_{p}T+\omega _{o}h_{c,o}) \right]=\frac{\partial }{\partial y}\left[ k\frac{\partial T}{\partial y}+\rho Dh_{c,o}\frac{\partial \omega _{o}}{\partial y} \right]</math>
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Considering the assumption that the Lewis number is unity, i.e. <math>Le=\nu /D=1</math>, eq. (4.374) can be rewritten as
Considering the assumption that the Lewis number is unity, i.e. <math>Le=\nu /D=1</math>, eq. (4.374) can be rewritten as
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\end{align}</math>
\end{align}</math>
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\end{matrix}</math>
\end{matrix}</math>
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|-
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<math>\begin{align}
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<math>{q}''_{w}=\frac{\tau _{w}}{U_{\infty }}\left[ (c_{p}T+\omega _{o}h_{c,o})_{w}-(c_{p}T+\omega _{o}h_{c,o})_{\infty } \right]=\frac{\tau _{w}}{U_{\infty }}[c_{p}(T_{w}-T_{\infty })+h_{c,o}(\omega _{o,w}-\omega _{o,\infty})]</math>
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  & {q}''_{w}=\frac{\tau _{w}}{U_{\infty }}\left[ (c_{p}T+\omega _{o}h_{c,o})_{w}-(c_{p}T+\omega _{o}h_{c,o})_{\infty } \right] \\
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& \begin{matrix}
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  {} & =\frac{\tau _{w}}{U_{\infty }}\left[ c_{p}(T_{w}-T_{\infty })+h_{c,o}(\omega _{o,}_{w}-\omega _{o}_{,\infty }) \right] \\
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\end{matrix} \\
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<math>-{q}''_{w}={\dot{m}}''_{f}h_{sv}+{q}''_{\ell }</math>
<math>-{q}''_{w}={\dot{m}}''_{f}h_{sv}+{q}''_{\ell }</math>
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where the two terms on the right-hand side of eq. (4.378) represent the latent heat of sublimation, and the sensible heat required to raise the surface temperature of the solid fuel to sublimation temperature and heat loss to the solid fuel.  
where the two terms on the right-hand side of eq. (4.378) represent the latent heat of sublimation, and the sensible heat required to raise the surface temperature of the solid fuel to sublimation temperature and heat loss to the solid fuel.  
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<math>{\dot{m}}''_{f}=Z\frac{\tau _{w}}{U_{\infty }}</math>
<math>{\dot{m}}''_{f}=Z\frac{\tau _{w}}{U_{\infty }}</math>
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where Z is the transfer driving force or transfer number defined as
where Z is the transfer driving force or transfer number defined as
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<math>Z=\frac{c_{p}(T_{\infty }-T_{w})+h_{c,o}(\omega _{o,\infty }-\omega _{o,w})}{h_{sv}+{q}''_{\ell }/\dot{{m}''}_{f}}</math>
<math>Z=\frac{c_{p}(T_{\infty }-T_{w})+h_{c,o}(\omega _{o,\infty }-\omega _{o,w})}{h_{sv}+{q}''_{\ell }/\dot{{m}''}_{f}}</math>
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Using the friction coefficient gives
Using the friction coefficient gives
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<math>c_{f}=\frac{\tau _{w}}{\rho U_{\infty }^{2}/2}</math>
<math>c_{f}=\frac{\tau _{w}}{\rho U_{\infty }^{2}/2}</math>
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eq. (4.379) therefore becomes
eq. (4.379) therefore becomes
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<math>{\dot{m}}''_{f}=\frac{c_{f}}{2}\rho U_{\infty }Z</math>
<math>{\dot{m}}''_{f}=\frac{c_{f}}{2}\rho U_{\infty }Z</math>
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The surface blowing velocity of the gaseous fuel is then
The surface blowing velocity of the gaseous fuel is then
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<math>v_{w}=\frac{{\dot{m}}''_{f}}{\rho }=\frac{c_{f}}{2}U_{\infty }Z</math>
<math>v_{w}=\frac{{\dot{m}}''_{f}}{\rho }=\frac{c_{f}}{2}U_{\infty }Z</math>
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where the friction coefficient, cf, can be obtained from the solution of the boundary layer flow over a flat plate with blowing on the surface.  The similarity solution of the boundary layer flow problem exists only if the blowing velocity satisfies  
where the friction coefficient, cf, can be obtained from the solution of the boundary layer flow over a flat plate with blowing on the surface.  The similarity solution of the boundary layer flow problem exists only if the blowing velocity satisfies  
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<math>B=\frac{(\rho v)_{w}}{(\rho u)_{\infty }}\operatorname{Re}_{x}^{1/2}</math>
<math>B=\frac{(\rho v)_{w}}{(\rho u)_{\infty }}\operatorname{Re}_{x}^{1/2}</math>
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Combining eqs. (4.383) and (4.384) yields
Combining eqs. (4.383) and (4.384) yields
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<math>B=\frac{Z}{2}\operatorname{Re}_{x}^{1/2}c_{f}</math>
<math>B=\frac{Z}{2}\operatorname{Re}_{x}^{1/2}c_{f}</math>
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Glassman (1987) recommended an empirical form of eq. (4.385) based on numerical and experimental results:
Glassman (1987) recommended an empirical form of eq. (4.385) based on numerical and experimental results:
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<math>B=\frac{\ln (1+Z)}{2.6Z^{0.15}}</math>
<math>B=\frac{\ln (1+Z)}{2.6Z^{0.15}}</math>
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Revision as of 06:26, 7 April 2010

Most early work in theoretically predicting the heat and/or mass transfer in both laminar and turbulent flow cases was done using the analogy between momentum, heat, and mass and by predicting the approximate results for the heat and/or mass transfer coefficient from the momentum transfer or skin friction coefficient. Clearly there are severe limitations in using this simple approach; however, it is beneficial to understand the advantages and similarities for physical and mathematical modeling as well as the constraints involving this approach. We present this analogy for the classic problem of heat and mass transfer over a flat plate in this section. Its application to more complicated geometries and boundary conditions as well as turbulent flow is not proven and caution should be taken in applying this approach to other cases.

Mass, momentum, and heat transfer in a laminar boundary layer
Figure 1: Mass, momentum, and heat transfer in a laminar boundary layer.

As presented before, a flat plate at constant wall temperature Tw is exposed to a free stream of constant velocity U∞, temperature T∞ and mass fraction ω1,∞. Due to binary diffusion it can be presented with the following dimensionless conservation boundary layer equations and boundary conditions corresponding to Fig. 1. Continuity

\frac{\partial u^{+}}{\partial x^{+}}+\frac{\partial v^{+}}{\partial y^{+}}=0

(1)

Momentum

u^{+}\frac{\partial u^{+}}{\partial x^{+}}+v^{+}\frac{\partial u^{+}}{\partial y^{+}}=\frac{1}{\operatorname{Re}} \frac{\partial ^{2}u^{+}}{\partial y^{+2}}

(2)

Energy

u^{+}\frac{\partial \theta }{\partial x^{+}}+v^{+}\frac{\partial \theta }{\partial y^{+}}=\frac{1}{\Pr \operatorname{Re}}\frac{\partial ^{2}\theta }{\partial y^{+2}}

(3)

Species

u^{+}\frac{\partial \varphi }{\partial x^{+}}+v^{+}\frac{\partial \varphi }{\partial y^{+}}=\frac{1}{\text{Sc}\operatorname{Re}}\frac{\partial ^{2}\varphi }{\partial y^{+2}}

(4)

\text{at }y^{+}=0,\text{  u}^{\text{+}}=0,\text{  }\theta \text{=}\varphi \text{=0}

(5)

y^{+}=0,\text{   }v^{+}=v_{w}^{+}

(6)

y^{+}=\infty ,\text{   }u^{+}=1,\text{   }\theta =\varphi =1

(7)

where the dimensionless variables are defined as:

\begin{align}
  & u^{+}=\frac{u}{U_{\infty }},\text{   }\theta =\frac{T-T_{w}}{T_{\infty }-T_{w}}\text{,   }\varphi =\frac{\omega _{1}-\omega _{1,w}}{\omega _{1,\infty }-\omega _{1,w}}, \\ 
 & v^{+}=\frac{v}{U_{\infty }},\text{   }y^{+}=\frac{y}{L},\text{   }x^{+}=\frac{x}{L},\text{   }\operatorname{Re}=\frac{U_{\infty }L}{\nu } \\ 
\end{align}

(8)

Let’s first consider the analogy between momentum and heat transfer. Equations (4.349) and (4.350) and appropriate boundary conditions are the same if Pr = 1. The solutions for u+ and θ are, therefore, exactly the same if Pr = 1 and one expects to have a relationship between the skin friction coefficient, cf, and heat transfer coefficient, h.

\frac{c_{f}}{2}=\frac{\tau _{w}}{\rho U_{\infty }^{2}}=\frac{\mu \left. \frac{\partial u}{\partial y} \right|_{y=0}}{\rho U_{\infty }^{2}}=\frac{\nu \left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{U_{\infty }L}=\frac{\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}}{\operatorname{Re}}

(9)

Nu=\frac{hL}{k}=\frac{-k\left. \frac{\partial T}{\partial y} \right|_{y=0}L}{k\left( T_{\omega }-T_{\infty } \right)}=\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}

(10)

Since θ = u+ for Pr = 1, one also concludes that

\left. \frac{\partial \theta }{\partial y^{+}} \right|_{y^{+}=0}=\left. \frac{\partial u^{+}}{\partial y^{+}} \right|_{y^{+}=0}

(11)

Therefore, combining eqs. (4.356) and (4.357) and using eq. (4.358) gives

\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}

(12)

This relationship between the friction coefficient and Nu is referred to as Reynolds analogy and is appropriate when Pr = 1. If Pr ≠ 1, we already concluded that \frac{\delta _{T}}{\delta }=\Pr ^{-1/3}\text{  for  }0.5\le \Pr \le 10 from the similarity solution presented in Section 4.6. Using this information, one can generalize the result of the Reynolds analogy to Pr ≠ 1 by

\frac{c_{f}}{2}=\frac{\text{Nu}}{\operatorname{Re}}\Pr ^{-1/3}

(13)

Now, we focus on similarities between the heat and mass transfer for comparison of eqs. (4.350) and (4.351) with their appropriate boundary conditions. It is clear that the solution for differential equations (4.350) and (4.351) for θ and \varphi are the same if Sc and Pr are interchanged appropriately. We already know the solution for eq. (4.350) for vw = 0 from the similarity solution for 0.5 ≤ Pr ≤ 10

\text{Nu}_{x}=0.332\operatorname{Re}_{x}^{1/2}\Pr ^{1/3}

(14)

Therefore, it can also be assumed that the solution of eq. (4.351) for vw = 0 is

\text{Sh}_{x}=0.332\operatorname{Re}_{x}^{1/2}\text{Sc}^{1/3}

(15)

Combining eqs. (4.361) and (4.362) gives

\frac{\text{Nu}}{\text{Sh}}\text{=}\frac{\text{Pr}^{\text{1/3}}}{\text{Sc}^{\text{1/3}}}

(16)

It should be noted that the convective effect due to vertical velocity at the surface was neglected in predicting h and hm in the above analysis and therefore the analogy presented in (4.363) is for very low mass transfer at the wall. The effect is that the contribution of vw in heat and mass flux was primarily due to diffusion.

{q}''_{w}=\rho c_{p}v_{w}\left( T_{w}-T_{\infty } \right)-k\left. \frac{\partial T}{\partial y} \right|_{y=0}

(17)

{\dot{m}}''_{w}=\rho \left( \omega _{1,\omega }v_{w}-D_{12}\left. \frac{\partial \omega _{1}}{\partial y} \right|_{y=0} \right)

(18)

The analogy in momentum, heat, and mass transfer can also be applied to complex and/or coupled transport phenomena problems including phase change and chemical reactions. Obviously, in these circumstances, the simple relationship developed in eq. (4.363) is not applicable. To show the usefulness of this analogy to more complex transport phenomena problems we will apply it to sublimation with a chemical reaction for forced convective flow over a flat plate. During combustion involving a solid fuel, the solid fuel may either burn directly or be sublimated before combustion. In the latter case – which will be discussed in this subsection – gaseous fuel diffuses away from the solid-vapor interface. Meanwhile, the gaseous oxidant diffuses toward the solid-vapor interface. Under the right conditions, the mass flux of vapor fuel and the gaseous oxidant meet and the chemical reaction occurs at a certain zone known as the flame. The flame is usually a very thin region with a color dictated by the temperature of combustion. Figure 4.32 shows the physical model of the problem under consideration. The concentration of the fuel is highest at the solid fuel surface, and decreases as the location of the flame is approached. The gaseous fuel diffuses away from the solid fuel surface and meets the oxidant as it flows parallel to the solid fuel surface. Combustion occurs in a thin reaction zone where the temperature is the highest, and the latent heat of sublimation is supplied by combustion. The combustion of solid fuel through sublimation can be modeled as a steady-state boundary layer type flow with sublimation and chemical reaction. To model the problem, the following assumptions are made: 1. The fuel is supplied by sublimation at a steady rate. 2. The Lewis number is unity, so the thermal and concentration boundary layers have the same thickness. 3. The buoyancy force is negligible.

Sublimation with chemical reaction (Kaviany, 2001)
Figure 2: Sublimation with chemical reaction (Kaviany, 2001).

The conservation equations for mass, momentum, energy, and species in the boundary layer are

\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0

(19)

u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left( \nu \frac{\partial u}{\partial y} \right)

(20)

\frac{\partial }{\partial x}(\rho c_{p}uT)+\frac{\partial }{\partial y}(\rho c_{p}vT)=\frac{\partial }{\partial y}\left( k\frac{\partial T}{\partial y} \right)+{\dot{m}}'''_{o}\text{ }h_{c,o}

(21)


\frac{\partial }{\partial x}(\rho u\omega _{o})+\frac{\partial }{\partial y}(\rho v\omega _{o})=\frac{\partial }{\partial y}\left( \rho D\frac{\partial \omega _{o}}{\partial y} \right)-{\dot{m}}'''_{o}

(22)

where {\dot{m}}'''_{0} is the rate of oxidant consumption (kg/m3*s), hc,0 is the heat released by combustion per unit mass consumption of the oxidant (J/kg), which is different from the combustion heat, andω0 is the mass fraction of the oxidant in the gaseous mixture. The corresponding boundary conditions of eqs. (4.366) to (4.369) are

u\to U_{\infty }\begin{matrix}
   , & T\to T_{\infty }\begin{matrix}
   , & \omega _{o}\to \omega _{o,\infty }  \\
\end{matrix}  \\
\end{matrix}\_at\_y\to \infty

(23)

u=0\begin{matrix}
   , & v=\frac{{\dot{m}}''_{f}}{\rho }\begin{matrix}
   , & \frac{\partial \omega _{o}}{\partial y}=0  \\
\end{matrix}  \\
\end{matrix}\_at\_y=0

(24)

where {\dot{m}}''_{f} is the rate of solid fuel sublimation per unit area (kg/m2?s) and ρ is the density of the mixture. The shear stress at the solid fuel surface is

\tau _{w}=\mu \frac{\partial u}{\partial y}\begin{matrix}
   , & y=0  \\
\end{matrix}

(25)

The heat flux at the solid fuel surface is

{q}''_{w}=-k\frac{\partial T}{\partial y}\begin{matrix}
   , & y=0  \\
\end{matrix}

(26)

The exact solution of the heat and mass problem described by eqs. (4.366) to (4.369) can be obtained using numerical simulation. It is useful here to introduce the results obtained by Kaviany (2001) using an analogy between momentum, heat, and mass transfer. Multiplying eq. (4.369) by hc,0 and adding the result to eq. (4.368), one obtains

\frac{\partial }{\partial x}\left[ \rho u(c_{p}T+\omega _{o}h_{c,o}) \right]+\frac{\partial }{\partial y}\left[ \rho v(c_{p}T+\omega _{o}h_{c,o}) \right]=\frac{\partial }{\partial y}\left[ k\frac{\partial T}{\partial y}+\rho Dh_{c,o}\frac{\partial \omega _{o}}{\partial y} \right]

(27)

Considering the assumption that the Lewis number is unity, i.e. Le = ν / D = 1, eq. (4.374) can be rewritten as

\begin{align}
  & \frac{\partial }{\partial x}\left[ \rho u(c_{p}T+\omega _{o}h_{c,o}) \right]+\frac{\partial }{\partial y}\left[ \rho v(c_{p}T+\omega _{o}h_{c,o}) \right] \\ 
 & =\frac{\partial }{\partial y}\left[ \rho \alpha \frac{\partial }{\partial y}(c_{p}T+\omega _{o}h_{c,o}) \right] \\ 
\end{align}

(28)

which can be viewed as an energy equation with quantity cpT + ω0hc,0) as a dependent variable. Since \partial \omega _{o}/\partial y=0 at y = 0, i.e., the solid fuel surface is not permeable for the oxidant, eq(4.373) can be rewritten as

{q}''_{w}=-\rho \alpha \frac{\partial }{\partial y}(c_{p}T+\omega _{o}h_{c,o})\begin{matrix}
   , & y=0  \\
\end{matrix}

(29)

Analogy between surface shear stress and the surface energy flux yields

{q}''_{w}=\frac{\tau _{w}}{U_{\infty }}\left[ (c_{p}T+\omega _{o}h_{c,o})_{w}-(c_{p}T+\omega _{o}h_{c,o})_{\infty } \right]=\frac{\tau _{w}}{U_{\infty }}[c_{p}(T_{w}-T_{\infty })+h_{c,o}(\omega _{o,w}-\omega _{o,\infty})]

(30)

The energy balance at the surface of the solid fuel is

-{q}''_{w}={\dot{m}}''_{f}h_{sv}+{q}''_{\ell }

(31)

where the two terms on the right-hand side of eq. (4.378) represent the latent heat of sublimation, and the sensible heat required to raise the surface temperature of the solid fuel to sublimation temperature and heat loss to the solid fuel. Combining eqs. (4.377) and (4.378) yields the rate of sublimation on the solid fuel surface

{\dot{m}}''_{f}=Z\frac{\tau _{w}}{U_{\infty }}

(32)

where Z is the transfer driving force or transfer number defined as

Z=\frac{c_{p}(T_{\infty }-T_{w})+h_{c,o}(\omega _{o,\infty }-\omega _{o,w})}{h_{sv}+{q}''_{\ell }/\dot{{m}''}_{f}}

(33)

Using the friction coefficient gives

c_{f}=\frac{\tau _{w}}{\rho U_{\infty }^{2}/2}

(34)

eq. (4.379) therefore becomes

{\dot{m}}''_{f}=\frac{c_{f}}{2}\rho U_{\infty }Z

(35)

The surface blowing velocity of the gaseous fuel is then

v_{w}=\frac{{\dot{m}}''_{f}}{\rho }=\frac{c_{f}}{2}U_{\infty }Z

(36)

where the friction coefficient, cf, can be obtained from the solution of the boundary layer flow over a flat plate with blowing on the surface. The similarity solution of the boundary layer flow problem exists only if the blowing velocity satisfies v_{w}\propto x^{-1/2} . In this case, one can define a blowing parameter as

B=\frac{(\rho v)_{w}}{(\rho u)_{\infty }}\operatorname{Re}_{x}^{1/2}

(37)

Combining eqs. (4.383) and (4.384) yields

B=\frac{Z}{2}\operatorname{Re}_{x}^{1/2}c_{f}

(38)

Glassman (1987) recommended an empirical form of eq. (4.385) based on numerical and experimental results:

B=\frac{\ln (1+Z)}{2.6Z^{0.15}}

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