Similarity Solution for Natural Convection on a Vertical Surface

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6.4.1 Similarity Solution for Natural Convection on a Vertical Surface

The boundary layer-type governing equations for the external convection problem shown in Fig. 6.1 are eqs. (6.17), (6.20) and (6.21). Introducing the stream function ψ, and dimensionless temperature, θ:


u=\frac{\partial \psi }{\partial y},\text{ }v=-\frac{\partial \psi }{\partial x},\text{ }\theta =\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}} \qquad \qquad()

(6.55)


the continuity equation (6.17) is satisfied and the momentum and energy equations (6.20) and (6.21) become:


\frac{\partial \psi }{\partial y}\frac{{{\partial }^{2}}\psi }{\partial y\partial x}-\frac{\partial \psi }{\partial x}\frac{{{\partial }^{2}}\psi }{\partial {{y}^{2}}}=\nu \frac{{{\partial }^{3}}\psi }{\partial {{y}^{3}}}+g\beta ({{T}_{w}}-{{T}_{\infty }})\theta \qquad \qquad()

(6.56)


\frac{\partial \psi }{\partial y}\frac{\partial \theta }{\partial x}-\frac{\partial \psi }{\partial x}\frac{\partial \theta }{\partial y}=\alpha \frac{{{\partial }^{2}}\theta }{\partial {{y}^{2}}} \qquad \qquad()

(6.57)


with the following boundary conditions:


\frac{\partial \psi }{\partial y}=\frac{\partial \psi }{\partial x}=0,\text{ }\theta =1\text{    at }y=0 \qquad \qquad()

(6.58)


\frac{\partial \psi }{\partial y}=\frac{\partial \psi }{\partial x}=0,\text{ }\theta =0\text{    at }y\to \infty \qquad \qquad()

(6.59)


The idea behind the similarity solution is that the velocity and temperature profiles in different x in the boundary layers are geometrically similar, differing only by a stretching factor in the x-direction (Kays et al., 2005). Thus, the similarity variable should have the following form:


\eta =y\cdot H(x) \qquad \qquad()

(6.60)


where H(x) is an unspecified stretching function. The objective now is to reduce eqs. (6.56) and (6.57) to ordinary differential equations. The stream function, ψ, which is a function of x and y, can be expressed as a function of x and η. If the similarity solution exists, one can express the stream function as


\psi (x,\eta )=\nu F(\eta )\cdot G(x) \qquad \qquad()

(6.61)


where F(η) and G(x) are the similarity function and the stretching function, respectively. The dimensionless temperature can be assumed as a function of η only, i.e.,


\theta (x,\eta )=\theta (\eta ) \qquad \qquad()

(6.62)


The derivatives in eqs. (6.60)–(6.62) can be obtained as shown below


\frac{\partial \psi }{\partial y}=\nu {F}'(\eta )\frac{\partial \eta }{\partial y}G(x)=\nu {F}'(\eta )H(x)G(x)


\frac{\partial \psi }{\partial x}=\nu {F}'(\eta )y{H}'(x)G(x)+\nu F(\eta )\cdot {G}'(x)


\frac{{{\partial }^{2}}\psi }{\partial y\partial x}=\nu {F}'(\eta )[{H}'(x)G(x)+H(x){G}'(x)]+\nu {F}''(\eta )\eta {H}'(x)G(x)


\frac{{{\partial }^{2}}\psi }{\partial {{y}^{2}}}=\nu {{F}'}'(\eta )\cdot {{H}^{2}}(x)\cdot G(x)


\frac{{{\partial }^{3}}\psi }{\partial {{y}^{3}}}=\nu {F}'''(\eta )\cdot {{H}^{3}}(x)\cdot G(x)


\frac{\partial \theta }{\partial x}={\theta }'(\eta )\frac{\partial \eta }{\partial x}={\theta }'(\eta )y{H}'(x)


\frac{\partial \theta }{\partial y}={\theta }'(\eta )\frac{\partial \eta }{\partial y}={\theta }'(\eta )H(x)


where the primes for F and θ denote the derivatives with respect to η, while the primes for G and H denote the derivatives with respect to x. Substituting the above derivatives into eqs. (6.56) and (6.57) and considering eq. (6.60), one obtains:


{F}'''+\frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}{{H}^{3}}G}\theta +\frac{{{G}'}}{H}F{F}''-\left[ \frac{{H}'G}{{{H}^{2}}}+\frac{{{G}'}}{H} \right]{{({F}')}^{2}}=0 \qquad \qquad()

(6.63)


{\theta }''+\Pr \frac{{{G}'}}{H}F{\theta }'=0 \qquad \qquad()

(6.64)


In order to convert eqs. (6.63) and (6.64) to ordinary differential equations with η as the sole independent variable, all functions of x must be cancelled, which is possible only if the following combination of H and G and their derivatives are satisfied:


{{H}^{3}}G=A=\text{const} \qquad \qquad()

(6.65)


\frac{{{G}'}}{H}=B=\text{const} \qquad \qquad()

(6.66)


\frac{{H}'G}{{{H}^{2}}}=C=\text{const} \qquad \qquad()

(6.67)


Differentiation of eq. (6.65) and division of the resultant equation by H4 yields the following equation:


3\frac{{H}'G}{{{H}^{2}}}+\frac{{{G}'}}{H}=\text{0} \qquad \qquad()

(6.68)


which is satisfied if eqs. (6.66) and (6.67) are satisfied. Substituting eq. (6.68) into eq. (6.63), we have


{F}'''+\frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}{{H}^{3}}G}\theta +\frac{{{G}'}}{H}F{F}''-\frac{2}{3}\frac{{{G}'}}{H}{{({F}')}^{2}}=0 \qquad \qquad()

(6.69)

Therefore, satisfaction of eqs. (6.65) and (6.66) is sufficient to ensure that eqs. (6.63) and (6.64) become ordinary differential equations. Although any constants A and B in eqs. (6.65) and (6.66) will transform eqs. (6.69) and (6.64) into ordinary differential equations, the proper choice of values of these two constants will yield ordinary differential equations with simple forms. Let us choose


\frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}{{H}^{3}}G}=1


\frac{{{G}'}}{H}=3


The corresponding G and H functions then become:


G(x)=4{{\left( \frac{1}{4}\text{G}{{\text{r}}_{x}} \right)}^{1/4}} \qquad \qquad()

(6.70)


H(x)=\frac{1}{x}{{\left( \frac{1}{4}\text{G}{{\text{r}}_{x}} \right)}^{1/4}} \qquad \qquad()

(6.71)


where Grx is the local Grashof number defined as:


\text{G}{{\text{r}}_{x}}=\frac{g\beta ({{T}_{w}}-{{T}_{\infty }}){{x}^{3}}}{{{\nu }^{2}}} \qquad \qquad()

(6.72)


which is equivalent to the square of the Reynolds number based on the scale of the local velocity {{u}_{0}}=\sqrt{g\beta ({{T}_{w}}-{{T}_{\infty }})x}.

Substituting eqs. (6.70) and (6.71) into eqs. (6.69) and (6.64), the momentum and the energy equations become:

{F}'''+\theta +3F{F}''-2{{({F}')}^{2}}=0 \qquad \qquad()

(6.73)


{\theta }''+3\Pr F{\theta }'=0 \qquad \qquad()

(6.74)


The velocity components in the x- and y-directions can be expressed in terms of similarity variables by the following equations:


u=\frac{\partial \psi }{\partial y}=\frac{\partial \psi }{\partial \eta }\frac{\partial \eta }{\partial y}=\frac{2\nu }{x}Gr_{x}^{1/2}{F}'(\eta ) \qquad \qquad()

(6.75)


v=-\frac{\partial \psi }{\partial x}=-\frac{\partial \psi }{\partial \eta }\frac{\partial \eta }{\partial x}=\frac{\nu }{x}{{\left( \frac{\text{G}{{\text{r}}_{x}}}{4} \right)}^{1/4}}[\eta {F}'(\eta )-3F(\eta ){{x}^{1/4}}] \qquad \qquad()

(6.76)


which are obtained from eqs. (6.60), (6.61), (6.70) and (6.71). Substituting the above expressions into eqs. (6.58) and (6.59), the boundary conditions for eqs. (6.73) and (6.74) can be obtained as follows


F(\eta )={F}'(\eta )=0\text{ and }\theta (\eta )=1\text{    at }\eta =0 \qquad \qquad()

(6.77)


{F}'(\eta )=\theta (\eta )=0\text{      at }\eta \to \infty  \qquad \qquad()

(6.78)


Equations (6.73) and (6.74) are coupled nonlinear ordinary equations with boundary conditions specified at different η and they thus make external natural convection a boundary value problem. The original numerical solution was obtained by Ostrach (1953) for a wide range of Prandtl numbers from 0.01 to 1000. Figure 6.3 shows the dimensionless velocity and temperature for various Prandtl numbers obtained by using a shooting method with the Range-Kutta method. As the Prandtl number increases, the maximum velocity in the boundary layer decreases, and the location at which the peak velocity occurs shifts to smaller η. As shown in Fig. 6.3(a), the thickness of the momentum boundary layer decreases with increasing Prandtl number. Similarly, the thickness of the thermal boundary layer also decreases with increasing Prandtl number, as indicated by Fig. 6.3(b). The absolute value of the dimensionless temperature gradient also increases with increasing Prandtl number.

After the dimensionless temperature in the boundary layer is obtained, the local heat transfer coefficient at the surface of the vertical plate can be obtained from eq. (6.3) as follows:


{{h}_{x}}=-\frac{k}{{{T}_{w}}-{{T}_{\infty }}}{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}=-k{\theta }'(0)\frac{1}{x}{{\left( \frac{\text{G}{{\text{r}}_{x}}}{4} \right)}^{1/4}} \qquad \qquad()

(6.79)


The local Nusselt number is


\text{N}{{\text{u}}_{x}}=\frac{{{h}_{x}}x}{k}=-\frac{{\theta }'(0)}{\sqrt{2}}\text{Gr}_{x}^{1/4}=\phi (\Pr )\text{Gr}_{x}^{1/4} \qquad \qquad()

(6.80)



 Velocity and temperature profile in the boundary layer for external natural convection over a vertical isothermal surface
Figure 6.3: Velocity and temperature profile in the boundary layer for external natural convection over a vertical isothermal surface


where \phi (\Pr )=-{\theta }'(0)/\sqrt{2} is a function of Prandtl number. The dependence of φ on the Prandtl number is evidenced by eq. (6.74) and by Fig. 6.3(b). The values of φ for various Pr have been obtained numerically by Ostrach (1953). Ede (1964) proposed the following function that correlated the numerical results:

\phi (\Pr )=\frac{3}{4}{{\left[ \frac{2{{\Pr }^{2}}}{5(1+2{{\Pr }^{1/2}}+2\Pr )} \right]}^{1/4}}

The local Nusselt number thus becomes:


\text{N}{{\text{u}}_{x}}=\phi (\Pr )\text{Gr}_{x}^{1/4}=\frac{3}{4}{{\left[ \frac{2\Pr }{5(1+2{{\Pr }^{1/2}}+2\Pr )} \right]}^{1/4}}{{\text{(G}{{\text{r}}_{x}}\text{Pr)}}^{\text{1/4}}} \qquad \qquad()

(6.81)


which can also be rewritten in terms of Rayleigh number


\text{N}{{\text{u}}_{x}}=\frac{3}{4}{{\left[ \frac{2\Pr }{5(1+2{{\Pr }^{1/2}}+2\Pr )} \right]}^{1/4}}\text{Ra}_{x}^{\text{1/4}} \qquad \qquad()

(6.82)


Equations (6.81) and (6.82) are valid for 0<\Pr <\infty . As is demonstrated above, the similarity solution is obtained as a consequence of the geometrical similarity of the velocity and temperature profiles in the boundary layers, i.e., the velocity and temperature profiles vary with x according to the stretching functions, G(x) and H(x) . If the geometrical

 Velocity and temperature profiles in the boundary layer based on modified scale
Figure 6.4: Velocity and temperature profiles in the boundary layer based on modified scale.

similarity exists, the selection of the stretching functions is not unique, and the resulting solutions based on different choices, as long as they all satisfy eqs. (6.65)–(6.67), are equivalent to the solution based on the stretching functions expressed in eqs. (6.70) and (6.71). However, a choice that better represents the physics of the problem will lead to the results presented in a physically more meaningful way. Figure 6.4 was obtained by modifying the numerical results shown in Fig. 6.3 by incorporating the modified stretching function for fluids of Pr > 1 (Bejan, 2004) expressed in eqs. (6.43) and (6.44), which is equivalent to a selection of the stretching functions in the following form


G(x)=\frac{1}{\Pr }\text{Ra}_{x}^{1/4}


H(x)=\frac{1}{x}\text{Ra}_{x}^{1/4}


It is clear in Fig. 6.4 that, in the limit Pr, the temperature profiles collapse onto a single curve, while the dimensionless velocity peak for fluids of Pr > 1 is consistently a number of order 1, showing that the velocity peak falls in the thermal boundary layer. Furthermore, as Pr increases, the velocity profile extends farther and farther into the isothermal fluid. All these features are anticipated by Fig. 6.1 and support the scale analysis, but cannot be seen from Fig. 6.3, in which the velocity and temperature profiles constantly shift as Pr changes, and the peak dimensionless velocity is not of order 1.


Failed to parse (PNG conversion failed;

check for correct installation of latex, dvips, gs, and convert): \Pr ({{{F}'''}_{1}}+{{\theta }_{1}})+{{\theta }_{1}}+{{F}_{1}}{{{F}''}_{1}}=\frac{2}{3}{F}'_{1}^{2} \qquad \qquad()

(6.83)


{{{\theta }''}_{1}}+{{F}_{1}}{{{\theta }'}_{1}}=0 \qquad \qquad()

(6.84)


Equations (6.83) and (6.84) allow simplification for the cases that \Pr \to 0 or \Pr \to \infty (see Problem 6.4). Numerical solution of the simplified equation for these two cases yields the following results (Le Fevre, 1956):


\text{N}{{\text{u}}_{x}}=\left\{ \begin{matrix}
   0.600{{(\text{G}{{\text{r}}_{x}}{{\Pr }^{2}})}^{1/4}}=0.600{{(\text{R}{{\text{a}}_{x}}\Pr )}^{1/4}} & \text{as Pr}\to 0  \\
   0.503{{(\text{G}{{\text{r}}_{x}}\Pr )}^{1/4}}=0.503\text{Ra}_{x}^{1/4}\text{          } & \text{as Pr}\to \infty   \\
\end{matrix} \right. \qquad \qquad()

(6.85)


Equations (6.82) and (6.85) indicate that {{h}_{x}}\propto {{x}^{-1/4}}. The average Nusselt number over the entire vertical wall,

{{\overline{\text{Nu}}}_{L}}, is related to the local Nusselt number at x = L by the following (see Problem 6.6): {{\overline{\text{Nu}}}_{L}}=\frac{4}{3}\text{N}{{\text{u}}_{L}} \qquad \qquad()

(6.86)

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