Sublimation with Chemical Reaction

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During combustion involving a solid fuel, the solid fuel may burn directly or it may be sublimated before combustion. In the latter case – which will be discussed in this subsection – gaseous fuel diffuses away from the solid-vapor surface. 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 1 shows the physical model of the problem under consideration (Kaviany, 2001). 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

 Sublimation with chemical reaction.
Figure 1: Sublimation with chemical reaction.

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.

The conservations of mass, momentum, energy and species of mass in the boundary layer are

\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0 \qquad \qquad(1)


u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left( \nu \frac{\partial u}{\partial y} \right) \qquad \qquad(2)


\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}}{{h}_{c,o}} \qquad \qquad(3)

\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}} \qquad \qquad(4)

where {{{\dot{m}}'''}_{o}} is rate of oxidant consumption (kg/m3-s). hc,o is the heat released by combustion per unit mass consumption of the oxidant (J/kg), which is different from the combustion heat defined in Chapter 3. ωo is mass fraction of the oxidant in the gaseous mixture.

The corresponding boundary conditions of eqs. (1) – (4) are

u\to {{u}_{\infty }}\begin{matrix}   , & T\to {{T}_{\infty }}\begin{matrix}   , & {{\omega }_{o}}\to {{\omega }_{o,\infty }}  \\
\end{matrix}  \\


y\to \infty \qquad \qquad(5)


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


y=0 \qquad \qquad (6)


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} \qquad \qquad(7)


The heat flux at the solid fuel surface is

{{{q}''}_{w}}=-k\frac{\partial T}{\partial y}\begin{matrix}
   , & y=0  \\
\end{matrix} \qquad \qquad(8)


The exact solution of the heat and mass problem described by eqs. (1) – (4) can be obtained using conventional numerical simulation, which is very complex. However, it is useful here to introduce the results obtained by Kaviany (2001) using analogy between momentum and heat transfer. Multiplying eq. (4) by hc,o and adding the result to eq. (3), 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 D{{h}_{c,o}}\frac{\partial {{\omega }_{o}}}{\partial y} \right] \\ 
\end{align} \qquad \qquad(9)

Considering the assumption that Lewis number is unity, i.e., Le = α / D = 1, eq. (9) can be rewritten as

  & \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} \qquad \qquad(10)

which can be viewed as an energy equation with quantity cpT + ωohc,o 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. (8) 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} \qquad \qquad(11)


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] \\ 
 & \begin{matrix}
   {} & =\frac{{{\tau }_{w}}}{{{u}_{\infty }}}\left[ {{c}_{p}}({{T}_{w}}-{{T}_{\infty }})+{{h}_{c,o}}({{\omega }_{o,}}_{w}-{{\omega }_{o}}_{,\infty }) \right]  \\
\end{matrix} \\ 
\end{align} \qquad \qquad(12)


The energy balance at the surface of the solid fuel is

-{{{q}''}_{w}}={{{\dot{m}}''}_{f}}{{h}_{sv}}+{{{q}''}_{\ell }} \qquad \qquad(13)


where the two terms on the right-hand side of eq. (13) 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. (12) and (13) yields the rate of sublimation on the solid fuel surface

{{{\dot{m}}''}_{f}}=Z\frac{{{\tau }_{w}}}{{{u}_{\infty }}} \qquad \qquad(14)


where Z is 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}_{sg}}+{{{{q}''}}_{\ell }}/{{{{m}''}}_{f}}} \qquad \qquad(15)


By using the friction coefficient –

{{C}_{f}}=\frac{{{\tau }_{w}}}{\rho u_{\infty }^{2}/2} \qquad \qquad(16)


eq. (14) becomes

{{{\dot{m}}''}_{f}}=\frac{{{C}_{f}}}{2}\rho {{u}_{\infty }}Z \qquad \qquad(17)


The surface blowing velocity of the gaseous fuel is then

{{v}_{w}}=\frac{{{{{\dot{m}}''}}_{f}}}{\rho }=\frac{{{C}_{f}}}{2}{{u}_{\infty }}Z \qquad \qquad(18)


where the friction coefficient Cf can be obtained from the solution of boundary layer flow over a flat plate with blowing on the surface (Kaviany, 2001; Kays et al., 2004). The similarity solution of the boundary layer flow problem exists only if 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} \qquad \qquad(19)


Combination of eqs. (18) and (19) yields

B=\frac{Z}{2}\operatorname{Re}_{x}^{1/2}{{C}_{f}} \qquad \qquad(20)


Glassman (1987) recommended an empirical form of eq. (20) based on numerical and experimental results:

B=\frac{\ln (1+Z)}{2.6{{Z}^{0.15}}} \qquad \qquad(21)


  & Z=\frac{{{c}_{p}}({{T}_{\infty }}-{{T}_{w}})+{{h}_{c,o}}({{\omega }_{o,\infty }}-{{\omega }_{o,w}})}{{{h}_{sv}}} \\ 
 & \text{     }=\frac{1.063\times (27-727)+12000\times (0.21-0.1)}{500}=0.5257 \\ 

The blowing parameter obtained from eq. (21) is

B=\frac{\ln (1+Z)}{2.6{{Z}^{0.15}}}=\frac{\ln (1+0.5257)}{2.6\times {{0.5257}^{0.15}}}=0.1789

The blowing velocity at the surface is obtained from eq.(19):

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

which can be integrated to yield the average blowing velocity:

  & {{{\bar{v}}}_{w}}=\frac{2{{\rho }_{\infty }}}{{{\rho }_{w}}}B{{\left( {{u}_{\infty }}\nu L \right)}^{1/2}} \\ 
 & \text{    }=\frac{2\times 1.1614}{0.3482}\times 0.1789\times {{\left( 1\times 60.21\times {{10}^{-6}}\times 1 \right)}^{1/2}}=0.009259\text{m/s} \\ 


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