# Integral solution for laminar and turbulent natural convection

(Difference between revisions)
 Revision as of 21:42, 21 June 2010 (view source)← Older edit Revision as of 20:18, 23 June 2010 (view source)Newer edit → Line 9: Line 9:
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)$
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)$
- (6.87) + Line 16: Line 16:
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad(2)$
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=-\alpha {{\left. \frac{\partial T}{\partial y} \right|}_{y=0}} \qquad \qquad(2)$
- (6.88) + Line 23: Line 23:
$\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)$
$\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)$
- (6.89) + Line 30: Line 30:
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)$
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)$
- (6.90) + Line 37: Line 37:
$\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)$
$\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)$
- (6.91) +
$\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)$
$\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)$
- (6.92) + Line 48: Line 48:
$U=\delta =0,\text{ }x=0 \qquad \qquad(7)$
$U=\delta =0,\text{ }x=0 \qquad \qquad(7)$
- (6.93) + Line 55: Line 55:
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)$
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)$
- (6.94) + Line 62: Line 62:
$\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)$
$\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)$
- (6.95) +
$\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)$
$\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)$
- (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: 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: Line 101: Line 100:
$\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad(11)$
$\frac{\delta }{x}=3.93{{\left( \frac{0.952+\Pr }{{{\Pr }^{2}}} \right)}^{1/4}}\text{Gr}_{x}^{-1/4} \qquad \qquad(11)$
- (6.97) + Line 120: Line 119:
$\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad(12)$
$\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad(12)$
- (6.98) + Line 136: Line 135:
$\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad(13)$
$\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad(13)$
- (6.106) + Line 143: Line 142:
$\text{R}{{\text{a}}_{*x}}=\frac{g\beta {q}''{{x}^{4}}}{\alpha \nu k} \qquad \qquad(14)$
$\text{R}{{\text{a}}_{*x}}=\frac{g\beta {q}''{{x}^{4}}}{\alpha \nu k} \qquad \qquad(14)$
- (6.107) + - + is the modified Rayleigh number based on the heat flux. is the modified Rayleigh number based on the heat flux. Line 161: Line 159:
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(15)$
$\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(15)$
- (6.108) +
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(16)$
$\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(16)$
- (6.109) + Line 172: Line 170:
$\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad(17)$
$\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad(17)$
- (6.110) + Line 179: Line 177:
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad(18)$
$\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad(18)$
- (6.111) + Line 198: Line 196:
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(19)$
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(19)$
- (6.112) +
$0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(20)$
$0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(20)$
- (6.113) + Line 209: Line 207:
$\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad(21)$
$\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad(21)$
- (6.114) +
$\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad(22)$
$\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad(22)$
- (6.115) + Line 220: Line 218:
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad(23)$
$0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad(23)$
- (6.116) +
$0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad(24)$
$0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad(24)$
- (6.117) + Line 231: Line 229:
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(25)$
$U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(25)$
- (6.118) + Line 238: Line 236:
$\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad(26)$
$\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad(26)$
- (6.119) + Line 247: Line 245:
${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad(27)$
${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad(27)$
- (6.120) + Line 255: Line 253: -
${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad(28)$
+
${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad(28)$
- (6.121) + Line 265: Line 263:
${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad(29)$
${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad(29)$
- (6.122) + Line 272: Line 270:
${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad(30)$
${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad(30)$
- (6.123) +

## Revision as of 20:18, 23 June 2010

Multiplying the continuity equation $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ from External Natural Convection from Heated Vertical Plate by u and adding the resulting equation to the momentum equation $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$ from External Natural Convection from Heated Vertical Plate 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: $\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=-\nu {{\left. \frac{\partial u}{\partial y} \right|}_{y=0}}+g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy} \qquad \qquad(1)$

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

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): $\frac{u}{U}=\frac{y}{\delta }{{\left( 1-\frac{y}{\delta } \right)}^{2}} \qquad \qquad(3)$

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): $\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}={{\left( 1-\frac{y}{{{\delta }_{t}}} \right)}^{2}} \qquad \qquad(4)$

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: $\frac{1}{105}\frac{d}{dx}({{U}^{2}}\delta )=-\nu \frac{U}{\delta }+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }})\delta \qquad \qquad(5)$ $\frac{1}{30}\frac{d}{dx}(U\delta )=\frac{2\alpha }{\delta } \qquad \qquad(6)$

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

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 $U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(8)$

Substituting eq. (8) into eqs. (5) and (6), one obtains the following: $\frac{2m+n}{105}C_{1}^{2}{{C}_{2}}{{x}^{2m+n-1}}=-\frac{{{C}_{1}}}{{{C}_{2}}}\nu {{x}^{m-n}}+\frac{1}{3}g\beta ({{T}_{w}}-{{T}_{\infty }}){{C}_{2}}{{x}^{n}} \qquad \qquad(9)$ $\frac{m+n}{30}{{C}_{1}}{{C}_{2}}{{x}^{m+n-1}}=\frac{2\alpha }{{{C}_{2}}}{{x}^{-n}} \qquad \qquad(10)$

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

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

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: $\text{N}{{\text{u}}_{x}}=0.508{{\left( \frac{{{\Pr }^{2}}}{0.952+\Pr } \right)}^{\text{1/4}}}\text{Gr}_{x}^{1/4}=0.508{{\left( \frac{\Pr }{0.952+\Pr } \right)}^{\text{1/4}}}\text{Ra}_{x}^{1/4} \qquad \qquad(12)$

Figure 1 shows the comparison between the integral solution, eq. (12), and the similarity solution, eq. $\text{N}{{\text{u}}_{x}}=\frac{3}{4}{{\left[ \frac{2\Pr }{5(1+2{{\Pr }^{1/2}}+2\Pr )} \right]}^{1/4}}\text{Ra}_{x}^{\text{1/4}}$ from Similarity Solution for Natural Convection on a Vertical Surface. 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): $\text{N}{{\text{u}}_{x}}=\frac{2}{{{360}^{1/5}}}{{\left( \frac{\Pr }{0.8+\Pr } \right)}^{\text{1/5}}}\text{Ra}_{*x}^{1/5} \qquad \qquad(13)$

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

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 $\frac{d}{dx}\int_{0}^{Y}{{{u}^{2}}}dy=g\beta \int_{0}^{Y}{(T-{{T}_{\infty }})dy}-\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(15)$ $\frac{d}{dx}\int_{0}^{Y}{u(T-{{T}_{\infty }})}dy=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(16)$

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): $\frac{u}{U}={{\left( \frac{y}{\delta } \right)}^{1/7}}{{\left( 1-\frac{y}{\delta } \right)}^{4}} \qquad \qquad(17)$

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: $\frac{T-{{T}_{\infty }}}{{{T}_{w}}-{{T}_{\infty }}}=1-{{\left( \frac{y}{\delta } \right)}^{1/7}} \qquad \qquad(18)$

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: $0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -\frac{{{\tau }_{w}}}{\rho } \qquad \qquad(19)$ $0.0366({{T}_{w}}-{{T}_{\infty }})\frac{d}{dx}(U\delta )=\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \qquad \qquad(20)$

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: $\frac{{{\tau }_{w}}}{\rho {{U}^{2}}}=\frac{0.0225}{{{(\rho U\delta /\mu )}^{0.25}}} \qquad \qquad(21)$ $\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}U({{T}_{w}}-{{T}_{\infty }})}=0.0225{{\left( \frac{\nu }{U\delta } \right)}^{1/4}}{{\Pr }^{-2/3}} \qquad \qquad(22)$

Substituting the above expressions into the integral momentum and energy equations, one obtains: $0.0523\frac{d}{dx}\left( {{U}^{2}}\delta \right)=0.125g\beta ({{T}_{w}}-{{T}_{\infty }})\delta -0.0225{{\nu }^{0.25}}\frac{{{U}^{1.75}}}{{{\delta }^{0.25}}} \qquad \qquad(23)$ $0.0366\frac{d}{dx}(U\delta )=0.0225{{\Pr }^{-0.67}}{{\nu }^{0.25}}\frac{{{U}^{0.75}}}{{{\delta }^{0.25}}} \qquad \qquad(24)$

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: $U={{C}_{1}}{{x}^{m}},\text{ }\delta ={{C}_{2}}{{x}^{n}} \qquad \qquad(25)$

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: $\text{N}{{\text{u}}_{x}}=0.0295\frac{{{\Pr }^{1/15}}}{{{(1+0.494{{\Pr }^{2/3}})}^{2/5}}}\text{Ra}_{x}^{2/5} \qquad \qquad(26)$

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> ${{\overline{\text{Nu}}}_{L}}=0.834\text{N}{{\text{u}}_{L}} \qquad \qquad(27)$

#### 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: ${{\overline{\text{Nu}}}_{L}}=C\text{Ra}_{L}^{n} \qquad \qquad(28)$

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: ${{\overline{\text{Nu}}}_{L}}={{\left\{ 0.825+\frac{\text{0}\text{.387Ra}_{L}^{1/6}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{8/27}}} \right\}}^{2}} \qquad \qquad(29)$

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: ${{\overline{\text{Nu}}}_{L}}=0.68+\frac{\text{0}\text{.67Ra}_{L}^{1/4}}{{{[1+{{(0.492/\Pr )}^{9/16}}]}^{4/9}}} \qquad \qquad(30)$

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.

## References

Bejan, A, and Lage, J.L., 1990, “Prandtl Number Effect on the Transition in Natural Convection along a Vertical Surface,” ASME Journal of Heat Transfer, Vol. 112, pp. 787 – 790.

Churchill, S. W., and Chu, H.H.S., 1975, “Correlating Equations for Laminar and Turbulent Free Convection from a Vertical Plate,” Int. J. Heat Mass Transfer, Vol. 18, pp. 1323-1329.

Eckert, E.R.G., and Jackson, T.W., 1951, “Analysis of Turbulent Free Convection Boundary Layer on a Flat Plate,” NACA Report 1015.

McAdams, W.H., 1954, Heat Transmission, 3rd Ed. McGraw-Hill, New York, NY.

Warner, C.Y., and Arpaci, V.S., 1968, “An Experimental Investigation of Turbulent Natural Convection in Air at Low Pressure along a Vertical Heated Flat Plate,” Int. J. Heat Mass Transfer, Vol. 11, pp. 397-406.