# Integral solution for laminar and turbulent natural convection

(Difference between revisions)
 Revision as of 03:15, 17 June 2010 (view source) (Created page with ' ===6.4.2 Integral Solution for Laminar and Turbulent Natural Convection=== ====Laminar Flow==== Multiplying the continuity equation (6.17) by ''u'' and adding the resulting equa…')← Older edit Revision as of 15:20, 19 June 2010 (view source)Newer edit → Line 1: Line 1: - - ===6.4.2 Integral Solution for Laminar and Turbulent Natural Convection=== - ====Laminar Flow==== Multiplying the continuity equation (6.17) by ''u'' and adding the resulting equation to the momentum equation (6.20) yields: Multiplying the continuity equation (6.17) by ''u'' and adding the resulting equation to the momentum equation (6.20) yields: Line 11: Line 8: -
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad()$
+
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)$
(6.87) (6.87) Line 18: Line 15: -
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad()$
+
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad(2)$
(6.88) (6.88) - In order to obtain the solution of a natural convection problem, the appropriate velocity and temperature profiles must be utilized together with the integral equations (6.87) and (6.88). The velocity and temperature profiles depend on the thicknesses of the momentum and thermal boundary layers, which in turn depend on the Prandtl number. Assuming the velocity profile is the third degree polynomial function of ''y'' in the boundary layer and using the boundary conditions to determine the unspecified constant, the velocity profile becomes (see Problem 6.8): + In order to obtain the solution of a natural convection problem, the appropriate velocity and temperature profiles must be utilized together with the integral equations (1) and (2). The velocity and temperature profiles depend on the thicknesses of the momentum and thermal boundary layers, which in turn depend on the Prandtl number. Assuming the velocity profile is the third degree polynomial function of ''y'' in the boundary layer and using the boundary conditions to determine the unspecified constant, the velocity profile becomes (see Problem 6.8): -
$\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad()$
+
$\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)$
(6.89) (6.89) Line 32: Line 29: -
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad()$
+
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)$
(6.90) (6.90) - The following analysis will be based on the assumption that the momentum and thermal boundary layers have the same thickness, i.e. ${{\delta }_{t}}=\delta$. Substituting the velocity and temperature profiles into the integral form of the momentum equation (6.87) and energy equation (6.88) yields: + The following analysis will be based on the assumption that the momentum and thermal boundary layers have the same thickness, i.e. ${{\delta }_{t}}=\delta$. Substituting the velocity and temperature profiles into the integral form of the momentum equation (1) and energy equation (2) yields: -
$\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad()$
+
$\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)$
(6.91) (6.91) -
$\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad()$
+
$\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)$
(6.92) (6.92) Line 50: Line 47: -
$U=\delta =0,\text{ }x=0 \qquad \qquad()$
+
$U=\delta =0,\text{ }x=0 \qquad \qquad(7)$
(6.93) (6.93) - which are the initial conditions of eqs. (6.91) and (6.92). It is expected that as ''x'' increases, both ''U'' and $\delta$ should increase. Let us assume that they are functions of ''x'' such that + which are the initial conditions of eqs. (5) and (6). It is expected that as ''x'' increases, both ''U'' and $\delta$ should increase. Let us assume that they are functions of ''x'' such that -
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad()$
+
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)$
(6.94) (6.94) - Substituting eq. (6.94) into eqs. (6.91) and (6.92), one obtains the following: + Substituting eq. (8) into eqs. (5) and (6), one obtains the following: -
$\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad()$
+
$\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)$
(6.95) (6.95) -
$\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad()$
+
$\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)$
(6.96) (6.96) Line 81: Line 78: - which are satisfied only if $m=1/2\text{ and }n=1/4$. This suggests that $U\propto {{x}^{1/2}}$ and $\delta \propto {{x}^{1/4}}$; this is in agreement with the result of the scaling analysis. Substituting back the values of ''m'' and ''n'' into eqs. (6.95) and (6.96), one obtains + which are satisfied only if $m=1/2\text{ and }n=1/4$. This suggests that $U\propto {{x}^{1/2}}$ and $\delta \propto {{x}^{1/4}}$; this is in agreement with the result of the scaling analysis. Substituting back the values of ''m'' and ''n'' into eqs. (9) and (10), one obtains Line 103: Line 100: -
$\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad()$
+
$\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad(11)$
(6.97) (6.97) Line 119: Line 116: - Substituting eq. (6.97) into the above expression yields: + Substituting eq. (11) into the above expression yields: -
$\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad()$
+
$\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad(12)$
(6.98) (6.98) - Figure 6.5 shows the comparison between the integral solution, eq. (6.98), and the similarity solution, eq. (6.82). It can be seen that the integral solution under predicts the local Nusselt number for low Prandtl number but over predicts + Figure 1 shows the comparison between the integral solution, eq. (12), and the similarity solution, eq. (6.82). It can be seen that the integral solution under predicts the local Nusselt number for low Prandtl number but over predicts - [[Image: chapter6g_(5).gif |thumb|400 px|alt= Comparison between integral solution and similarity solutions for natural convection over a heated vertical wall. | Figure 6.5: Comparison between integral solution and similarity solutions for natural convection over a heated vertical wall.]] + [[Image: chapter6g_(5).gif |thumb|400 px|alt= Comparison between integral solution and similarity solutions for natural convection over a heated vertical wall. | Figure 1: Comparison between integral solution and similarity solutions for natural convection over a heated vertical wall.]] Line 138: Line 135: -
$\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad()$
+
$\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad(13)$
(6.106) (6.106) Line 145: Line 142: -
$\text{R}{{\text{a}}_{*x}}=\frac{g\beta {q}''{{x}^{4}}}{\alpha \nu k} \qquad \qquad()$
+
$\text{R}{{\text{a}}_{*x}}=\frac{g\beta {q}''{{x}^{4}}}{\alpha \nu k} \qquad \qquad(14)$
(6.107) (6.107) Line 160: Line 157: - if all terms are time-averaged as should be used for turbulent flows. Using the definition of shear stress and heat flux, eqs. (6.87) and (6.88) can be respectively modified as + if all terms are time-averaged as should be used for turbulent flows. Using the definition of shear stress and heat flux, eqs. (1) and (2) can be respectively modified as -
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad()$
+
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(15)$
(6.108) (6.108) -
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad()$
+
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(16)$
(6.109) (6.109) Line 174: Line 171: -
$\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad()$
+
$\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad(17)$
(6.110) (6.110) - where ''U'' is a characteristic velocity for the near wall. Equation (6.110) satisfies all of the velocity boundary conditions. It is further assumed that the velocity and temperature boundary layer thicknesses are the same (${{\delta }_{t}}=\delta$) and the temperature profile is: + where ''U'' is a characteristic velocity for the near wall. Equation (17) satisfies all of the velocity boundary conditions. It is further assumed that the velocity and temperature boundary layer thicknesses are the same (${{\delta }_{t}}=\delta$) and the temperature profile is: -
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad()$
+
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad(18)$
(6.111) (6.111) - which yields $T={{T}_{w}}$ at ''y'' = 0 and $T={{T}_{\infty }}\text{ at }y=\delta$. It should be pointed out that eqs. (6.110) and (6.111) are valid only for $y<\delta$. For large ''y'', one has ''u'' = 0 and $T={{T}_{\infty }}$. + which yields $T={{T}_{w}}$ at ''y'' = 0 and $T={{T}_{\infty }}\text{ at }y=\delta$. It should be pointed out that eqs. (17) and (18) are valid only for $y<\delta$. For large ''y'', one has ''u'' = 0 and $T={{T}_{\infty }}$. - Substituting eqs. (6.110) and (6.111) into eqs. (6.108) and (6.109), the integral equations become: + Substituting eqs. (17) and (18) into eqs. (15) and (16), the integral equations become: Line 200: Line 197: -
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad()$
+
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(19)$
(6.112) (6.112) -
$0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad()$
+
$0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(20)$
(6.113) (6.113) Line 211: Line 208: -
$\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad()$
+
$\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad(21)$
(6.114) (6.114) -
$\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad()$
+
$\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad(22)$
(6.115) (6.115) Line 222: Line 219: -
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad()$
+
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad(23)$
(6.116) (6.116) -
$0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad()$
+
$0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad(24)$
(6.117) (6.117) Line 233: Line 230: -
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad()$
+
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(25)$
(6.118) (6.118) - Substituting the above equations into eqs. (6.116) and (6.117), one finds that the values which satisfy these two equations are $m=0.5$ and $n=0.7$(Problem 6.13). By following the procedure similar to the case of laminar natural convection (Problem 6.14), the local Nusselt number can then be obtained: + Substituting the above equations into eqs. (23) and (24), one finds that the values which satisfy these two equations are $m=0.5$ and $n=0.7$(Problem 6.13). By following the procedure similar to the case of laminar natural convection (Problem 6.14), the local Nusselt number can then be obtained: -
$\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad()$
+
$\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad(26)$
(6.119) (6.119) Line 248: Line 245:
${{\overline{\text{Nu}}}_{L}}$, is related to the local Nusselt number at $x=L$ as follows (see Problem 6.15).
${{\overline{\text{Nu}}}_{L}}$, is related to the local Nusselt number at $x=L$ as follows (see Problem 6.15). - ${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad()$
+ +
${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad(27)$
(6.120) (6.120) Line 257: Line 255: -
${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad()$
+
${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad(28)$
(6.121) (6.121) Line 267: Line 265: -
${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad()$
+
${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad(29)$
(6.122) (6.122) Line 274: Line 272: -
${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad()$
+
${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad(30)$
(6.123) (6.123) - Practically, one can use eq. (6.123) for laminar natural convection and eq. (6.122) for turbulent natural convection. These two correlations can provide better accuracy than the earlier correlation in simpler form. For the case of constant heat flux where the surface temperature, ${T_w}$, increases with increasing ''x'', [[#References|Churchill and Chu (1975)]] suggested that eq. (6.122) is still valid provided the constant 0.492 is changed to 0.437, and ${{\overline{\text{Nu}}}_{L}}$ and ${{Ra}_L}$ must be defined using the averaged wall temperature of the vertical plate. + Practically, one can use eq. (30) for laminar natural convection and eq. (29) for turbulent natural convection. These two correlations can provide better accuracy than the earlier correlation in simpler form. For the case of constant heat flux where the surface temperature, ${T_w}$, increases with increasing ''x'', [[#References|Churchill and Chu (1975)]] suggested that eq. (29) is still valid provided the constant 0.492 is changed to 0.437, and ${{\overline{\text{Nu}}}_{L}}$ and ${{Ra}_L}$ must be defined using the averaged wall temperature of the vertical plate. ==References== ==References==

## Revision as of 15:20, 19 June 2010

Multiplying the continuity equation (6.17) by u and adding the resulting equation to the momentum equation (6.20) yields: $\frac{\partial {{u}^{2}}}{\partial x}+\frac{\partial (uv)}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$

Integrating the above equation with respect to y in the interval of (0, Y), where Y is greater than both δ and δt, one obtains: $\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)$

(6.87)

By following a similar procedure, the integral energy equation can be obtained as follows $\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad(2)$

(6.88)

In order to obtain the solution of a natural convection problem, the appropriate velocity and temperature profiles must be utilized together with the integral equations (1) and (2). The velocity and temperature profiles depend on the thicknesses of the momentum and thermal boundary layers, which in turn depend on the Prandtl number. Assuming the velocity profile is the third degree polynomial function of y in the boundary layer and using the boundary conditions to determine the unspecified constant, the velocity profile becomes (see Problem 6.8): $\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)$

(6.89)

where U is a characteristic velocity that is a function of x. Similarly, the temperature profile can be obtained by assuming a second degree polynomial function and the result is (see Problem 6.9): $\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)$

(6.90)

The following analysis will be based on the assumption that the momentum and thermal boundary layers have the same thickness, i.e. δt = δ. Substituting the velocity and temperature profiles into the integral form of the momentum equation (1) and energy equation (2) yields: $\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)$

(6.91) $\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)$

(6.92)

At the leading edge of the vertical plate, the boundary layer thickness is zero and the characteristic velocity U is also zero: $U=\delta =0,\text{ }x=0 \qquad \qquad(7)$

(6.93)

which are the initial conditions of eqs. (5) and (6). It is expected that as x increases, both U and δ should increase. Let us assume that they are functions of x such that $U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)$

(6.94)

Substituting eq. (8) into eqs. (5) and (6), one obtains the following: $\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)$

(6.95) $\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)$

(6.96)

The above two relations can be true for all x only if the indices of x for all terms in the same equation are the same, i.e. when the following equations hold: \begin{align} & 2m+n-1=m-n=n \\ & m+n-1=-n \\ \end{align}

which are satisfied only if m = 1 / 2 and n = 1 / 4. This suggests that $U\propto {{x}^{1/2}}$ and $\delta \propto {{x}^{1/4}}$; this is in agreement with the result of the scaling analysis. Substituting back the values of m and n into eqs. (9) and (10), one obtains $\frac{C_{1}^{2}{{C}_{2}}}{84}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu +\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}$ $\frac{{{C}_{1}}{{C}_{2}}}{40}=\frac{2\alpha }{{{C}_{2}}}$

Solving for C1 and C2 from the above two equations yields: ${{C}_{1}}=4{{\left( \frac{5}{3} \right)}^{1/2}}\nu {{\left( \frac{20}{21}+\frac{\nu }{\alpha } \right)}^{-1/2}}{{\left[ \frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}} \right]}^{1/2}}$ ${{C}_{2}}=4{{\left( \frac{15}{16} \right)}^{1/4}}{{\left( \frac{20}{21}+\frac{\nu }{\alpha } \right)}^{1/4}}{{\left[ \frac{g\beta ({{T}_{w}}-{{T}_{\infty }})}{{{\nu }^{2}}} \right]}^{-1/4}}{{\left( \frac{\nu }{\alpha } \right)}^{-1/2}}$

The boundary layer thickness therefore becomes: $\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad(11)$

(6.97)

The local heat transfer coefficient at the surface of the vertical plate can be obtained from eq. (6.3): ${{h}_{x}}=-\frac{k}{{{T}_{w}}-{{T}_{\infty }}}{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}=\frac{2k}{\delta }$

The local Nusselt number is: $\text{N}{{\text{u}}_{x}}=\frac{{{h}_{x}}x}{k}=\frac{2x}{\delta }$

Substituting eq. (11) into the above expression yields: $\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad(12)$

(6.98)

Figure 1 shows the comparison between the integral solution, eq. (12), and the similarity solution, eq. (6.82). It can be seen that the integral solution under predicts the local Nusselt number for low Prandtl number but over predicts Figure 1: Comparison between integral solution and similarity solutions for natural convection over a heated vertical wall.

the local Nusselt number for high Prandtl number. At Pr = 10 − 4, the integral solution yields $\text{N}{{\text{u}}_{x}}/\text{Ra}_{x}^{1/4}=0.051$, which is 13% lower than the value of 0.059 obtained from the similarity solution. At Pr = 0.72, which is the Prandtl number for air, the result obtained by the integral solution ( $\text{N}{{\text{u}}_{x}}/\text{Ra}_{x}^{1/4}=0.412$) is 6.8% higher than the similarity solution. It can also be seen that the agreement between the integral and similarity solutions is the best at high Prandtl number. When Pr = 104, the difference between the integral and similarity solution is only 1.5%.

The above discussion is limited to the case where the surface temperature of the vertical plate is uniform. For the case where the surface heat flux of the flat plate is uniform, an integral solution can be performed to obtain the following results (see Problem 6.11): $\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad(13)$

(6.106)

where $\text{R}{{\text{a}}_{*x}}=\frac{g\beta {q}''{{x}^{4}}}{\alpha \nu k} \qquad \qquad(14)$

(6.107)

is the modified Rayleigh number based on the heat flux.

## Contents

#### Turbulent Flow

For small temperature difference between the heated wall and bulk fluid and short vertical plate, the natural convection is laminar and the above similarity and integral solutions are valid. Once the Rayleigh number exceeds a critical value, the natural convection will become turbulent and the above results will be invalid. It was believed that the transition from laminar to turbulent occurs at Rax˜109 until Bejan and Lage (1990) showed that for a wide range of Prandtl number ( $0.001<\Pr <1000$) the criterion for transition from laminar to turbulent is actually Grx˜109. Alternatively, one can say that the transition takes place at $\text{R}{{\text{a}}_{x}}\sim {{10}^{9}}\Pr$. Thus, the critical Rayleigh number for low-Prandtl number fluid is less than 109. For high-Prandtl number fluid, on the other hand, the critical Rayleigh number is higher than 109.

The advantage of the integral solution is that it also works for turbulent flow. The integral momentum and energy equations (6.87) and (6.88) are applicable

if all terms are time-averaged as should be used for turbulent flows. Using the definition of shear stress and heat flux, eqs. (1) and (2) can be respectively modified as $\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(15)$

(6.108) $\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(16)$

(6.109)

The velocity and temperature profiles in the boundary layer must be correctly determined so as to reflect the behavior of the turbulent boundary layer. A velocity profile constructed as follows gives a good description of the velocity distribution in natural convection for turbulent flow over a vertical flat plate (Eckert and Jackson, 1951): $\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad(17)$

(6.110)

where U is a characteristic velocity for the near wall. Equation (17) satisfies all of the velocity boundary conditions. It is further assumed that the velocity and temperature boundary layer thicknesses are the same (δt = δ) and the temperature profile is: $\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad(18)$

(6.111)

which yields T = Tw at y = 0 and $T={{T}_{\infty }}\text{ at }y=\delta$. It should be pointed out that eqs. (17) and (18) are valid only for y < δ. For large y, one has u = 0 and $T={{T}_{\infty }}$.

Substituting eqs. (17) and (18) into eqs. (15) and (16), the integral equations become: $\frac{d}{dx}\left[ {{U}^{2}}\delta \int_{0}^{1}{{{\eta }^{2/7}}{{\left( 1-\eta \right)}^{8}}}d\eta \right]=g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \int_{0}^{1}{(1-{{\eta }^{1/7}})d\eta }-\frac{{{\tau }_{w}}}{\rho }$ $\frac{d}{dx}\left[ U({{T}_{w}}-{{T}_{\infty }})\delta \int_{0}^{1}{{{\eta }^{1/7}}{{\left( 1-\eta \right)}^{4}}}\left( 1-{{\eta }^{1/7}} \right)d\eta \right]=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}$

where η = y / δ. Evaluating the integrals in the above two equations yields: $0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(19)$

(6.112) $0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(20)$

(6.113)

At this point, we have two equations and four unknowns (U, δ, q''w, τw). Thus, assumptions have to be made regarding the forms of the expressions for τw and q''w. It is generally assumed that the flow near the wall in a turbulent natural convective boundary layer is similar to that of a turbulent forced convection so that the expressions for τw and q''w derived for forced convection can be applied: $\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad(21)$

(6.114) $\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad(22)$

(6.115)

Substituting the above expressions into the integral momentum and energy equations, one obtains: $0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad(23)$

(6.116) $0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad(24)$

(6.117)

The boundary layer can be assumed to be turbulent from the leading edge of the surface, so the solutions to the above equations should be of the following form: $U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(25)$

(6.118)

Substituting the above equations into eqs. (23) and (24), one finds that the values which satisfy these two equations are m = 0.5 and n = 0.7(Problem 6.13). By following the procedure similar to the case of laminar natural convection (Problem 6.14), the local Nusselt number can then be obtained: $\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad(26)$

(6.119)

which suggests that the local heat transfer coefficient is proportional to x0.2. The average Nusselt number for the entire vertical plate, ${{\overline{\text{Nu}}}_{L}}$, is related to the local Nusselt number at x = L as follows (see Problem 6.15). <center> ${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad(27)$

(6.120)

#### Empirical Correlation

The above analyses for laminar and turbulent natural convection over a vertical flat plate suggest that the average Nusselt number can be expressed in the following format: ${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad(28)$

(6.121)

which is confirmed by experimental studies. For laminar flow ( $\text{G}{{\text{r}}_{L}}\le {{10}^{9}}$), one can utilize the values of C = 0.59 and n = 1 / 4. For turbulent flow ( GrL > 109), C = 0.1 and n = 1 / 3 can be used (McAdams, 1954; Warner and Arpaci, 1968).

Churchill and Chu (1975) studied numerous sets of experimental data and recommended the following correlation: ${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad(29)$

(6.122)

which covers all Prandtl number and Grashof number between 0.1 and 1012. For the case of laminar convection (Grx < 109), the following correlation yields better results: ${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad(30)$

(6.123)

Practically, one can use eq. (30) for laminar natural convection and eq. (29) for turbulent natural convection. These two correlations can provide better accuracy than the earlier correlation in simpler form. For the case of constant heat flux where the surface temperature, Tw, increases with increasing x, Churchill and Chu (1975) suggested that eq. (29) is still valid provided the constant 0.492 is changed to 0.437, and ${{\overline{\text{Nu}}}_{L}}$ and RaL must be defined using the averaged wall temperature of the vertical plate.