# Scale analysis of natural convection

(Difference between revisions)
 Revision as of 18:49, 5 July 2010 (view source)← Older edit Current revision as of 11:36, 10 July 2010 (view source) (2 intermediate revisions not shown) Line 1: Line 1: - While scale analysis cannot provide the exact functions in eq. (6.36), the form of these functions can be provided [[#References|(Bejan, 2004)]]. Let us refer to Fig. 6.1 and consider the governing equations in the thermal boundary layer ($y\sim {{\delta }_{t}}$) for the entire flat plate ($x\sim L$). The thickness of the thermal boundary layer in which the effects of the heated wall are felt is much smaller than the length of the vertical plate, i.e. ${{\delta }_{t}}\ll L$. For the continuity equation (6.17) to be satisfied, the scales of the two terms must be the same:$\frac{u}{L}\sim \frac{v}{{{\delta }_{t}}}$ + ==Scaling== + [[Image:Natural convection over a vertical flat plate.jpg|thumb|300 px|alt=Natural convection over a vertical flat plate (Pr >1) |'''Figure 1 Natural convection over a vertical flat plate (Pr >1).''']] + + The average Nusselt number for natural convection can be expressed as: +
+ ${{\overline{\text{Nu}}}_{L}}=f(\text{G}{{\text{r}}_{L}},\Pr )$ +
+ + While scale analysis cannot provide the exact function in the above expression, the form of these functions can be provided [[#References|(Bejan, 2004)]]. Let us refer to Fig. 1 and consider the governing equations in the thermal boundary layer (''y'' ~ ${{\delta }_{t}}$) for the entire flat plate (''x'' ~ ''L''). The thickness of the thermal boundary layer in which the effects of the heated wall are felt is much smaller than the length of the vertical plate, i.e. ${{\delta }_{t}}\ll L$. For the continuity equation +
+ $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ +
+ to be satisfied, the scales of the two terms must be the same: + +
$\frac{u}{L}\sim \frac{v}{{{\delta }_{t}}} The scale of the velocity component in the y-direction is therefore: The scale of the velocity component in the y-direction is therefore: Line 13: Line 27: which indicates that [itex]v\ll u$ for flow in the thermal boundary layer. The scale of the velocity component in the x-direction, ''u'', is still unknown at this point. which indicates that $v\ll u$ for flow in the thermal boundary layer. The scale of the velocity component in the x-direction, ''u'', is still unknown at this point. - While the left hand side of the energy equation (6.21) shows the effect of advection, the right-hand side shows the effect of diffusion. The scales of the two advective terms on the left hand side of eq. (6.21) are:$u\frac{\partial T}{\partial x}\sim u\frac{\Delta T}{L},\text{ }v\frac{\partial T}{\partial y}\sim v\frac{\Delta T}{{{\delta }_{t}}}\sim u\frac{\Delta T}{L}$ + The energy equation is: + +
+ $u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}$ +
+ + While the left hand side of the energy equation shows the effect of advection, the right-hand side shows the effect of diffusion. The scales of the two advective terms on the left hand side of energy equation are: + +
$u\frac{\partial T}{\partial x}\sim u\frac{\Delta T}{L},\text{ }v\frac{\partial T}{\partial y}\sim v\frac{\Delta T}{{{\delta }_{t}}}\sim u\frac{\Delta T}{L} - which indicates that the scale of the second term on the left hand side of eq. (6.21) is identical to the scale of the first term when the scale of the velocity component in the y-direction is given by eq. (6.37). The temperature difference [itex]\Delta T={{T}_{w}}-{{T}_{\infty }}$ in the above scale analysis represents the scale of the excess temperature, $T-{{T}_{\infty }}$. The scale of the right-hand side of eq. (6.21) is: $\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}}$ + which indicates that the scale of the second term on the left hand side of energy equation is identical to the scale of the first term when the scale of the velocity component in the y-direction is given by eq. (1). The temperature difference $\Delta T={{T}_{w}}-{{T}_{\infty }}$ in the above scale analysis represents the scale of the excess temperature, $T-{{T}_{\infty }}$. The scale of the right-hand side of energy equation is: +
$\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}} - The scales of the two sides of the energy equation must be the same:[itex]u\frac{\Delta T}{L}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}}$ + The scales of the two sides of the energy equation must be the same: + +
$u\frac{\Delta T}{L}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}} The above equation can be used to estimate the scale of u as follows: The above equation can be used to estimate the scale of u as follows: Line 41: Line 66: where the scale of the thermal boundary layer thickness, δt, is still unknown at this point. where the scale of the thermal boundary layer thickness, δt, is still unknown at this point. - The respective scales of the two inertial terms on the left-hand side of the momentum equation (6.20) are: + The momentum equation is + + + [itex]u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$ +
+ + The respective scales of the two inertial terms on the left-hand side of the momentum equation (10) are: - $u\frac{\partial u}{\partial x}\sim \frac{{{u}^{2}}}{L}$ +
$u\frac{\partial u}{\partial x}\sim \frac{{{u}^{2}}}{L} - [itex]v\frac{\partial u}{\partial y}\sim v\frac{u}{{{\delta }_{t}}}\sim \frac{{{u}^{2}}}{L}$ +
$v\frac{\partial u}{\partial y}\sim v\frac{u}{{{\delta }_{t}}}\sim \frac{{{u}^{2}}}{L} - The respective scales of the viscosity and buoyancy terms on the right-hand side of eq. (6.20) are: + The respective scales of the viscosity and buoyancy terms on the right-hand side of momentum are: - [itex]\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}\sim \nu \frac{u}{\delta _{t}^{2}}$ +
$\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}\sim \nu \frac{u}{\delta _{t}^{2}} - [itex]g\beta (T-{{T}_{\infty }})\sim g\beta \Delta T$ +
$g\beta (T-{{T}_{\infty }})\sim g\beta \Delta T We can very well see that three forces are at play in the boundary layer region, which are inertia, viscosity and buoyancy forces. Considering the scale of u obtained from eq. (2), the scales of these three forces are We can very well see that three forces are at play in the boundary layer region, which are inertia, viscosity and buoyancy forces. Considering the scale of u obtained from eq. (2), the scales of these three forces are Line 67: Line 98: |} |} - Among these three forces, the buoyancy force is never negligible because without it natural convection would not occur. Therefore, the scale of buoyancy force can be used to measure the importance of the inertial and viscous forces. Dividing eq. (4) by the scale of buoyancy force, [itex]g\beta \Delta T$ + Among these three forces, the buoyancy force is never negligible because without it natural convection would not occur. Therefore, the scale of buoyancy force can be used to measure the importance of the inertial and viscous forces. Dividing eq. (4) by the scale of buoyancy force, + +
$g\beta \Delta T$ , one obtains the following:$\begin{matrix} , one obtains the following:[itex]\begin{matrix} \frac{{{\alpha }^{2}}}{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & \frac{\nu \alpha }{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & 1 \\ \frac{{{\alpha }^{2}}}{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & \frac{\nu \alpha }{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & 1 \\ \text{Inertia} & \text{Viscous} & \text{Buoyancy} \\ \text{Inertia} & \text{Viscous} & \text{Buoyancy} \\ - \end{matrix}$ + \end{matrix}
which can be expressed in terms of dimensionless parameters as [[#References|(Bejan, 2004)]]: which can be expressed in terms of dimensionless parameters as [[#References|(Bejan, 2004)]]: Line 96: Line 129: |} |} - is the Rayleigh number that is related to the Grashof number by + is the Rayleigh number that is related to the Grashof number by $\text{R}{{\text{a}}_{L}}=\text{G}{{\text{r}}_{L}}\Pr$. Therefore, the relative importance of inertia and viscous forces depends on the Prandtl number, which is a property of the fluid. Thus, if the Prandtl number is high ($\Pr \gg 1$), the inertia term will be negligible and the viscosity term will balance the buoyancy term, whereas if the Prandtl number is low enough ($\Pr \ll 1$) – as for liquid metals – then the inertia term is considerable and balances the buoyancy term in steady state. The scale analysis for high- and low-Prandtl number fluids is presented in detail below. - $\text{R}{{\text{a}}_{L}}=\text{G}{{\text{r}}_{L}}\Pr$ + - . Therefore, the relative importance of inertia and viscous forces depends on the Prandtl number, which is a property of the fluid. Thus, if the Prandtl number is high ($\Pr \gg 1$), the inertia term will be negligible and the viscosity term will balance the buoyancy term, whereas if the Prandtl number is low enough ($\Pr \ll 1$) – as for liquid metals – then the inertia term is considerable and balances the buoyancy term in steady state. The scale analysis for high- and low-Prandtl number fluids is presented in detail below. + ==High Prandtl Number Fluids== + + For the case of high Prandtl number fluids, the kinematic viscosity of the fluid is much greater than its thermal diffusivity, so viscous force has a significant effect on the momentum equation and the effect of inertial force is negligible. The viscous force in fact balances the buoyancy force, and so from eq. (5), one obtains the following: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ ${{\delta }_{t}}\sim L\text{Ra}_{L}^{-1/4}$ +
+ |{{EquationRef|(7)}} + |} + + Substituting eq. (7) into eqs. (2) and (3), the scales of the velocity components become: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $u\sim \frac{\alpha }{L}\text{Ra}_{L}^{1/2}$ +
+ |{{EquationRef|(8)}} + |} + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $v\sim \frac{\alpha }{L}\text{Ra}_{L}^{1/4}$ +
+ |{{EquationRef|(9)}} + |} + + The scale of the heat transfer coefficient can be obtained by analyzing the following equation +
${{h}_{x}}=-\frac{k}{{{T}_{w}}-{{T}_{\infty }}}{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}$
+ i.,e., +
${{h}_{x}}=\frac{-k{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}}{{{T}_{w}}-{{T}_{\infty }}}\sim \frac{k(\Delta T/{{\delta }_{t}})}{\Delta T}=\frac{k}{{{\delta }_{t}}}$
+ + The scale of Nusselt number is therefore: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $\text{Nu}=\frac{{{h}_{x}}L}{k}\sim \frac{L}{{{\delta }_{t}}}\sim \text{Ra}_{L}^{1/4}$ +
+ |{{EquationRef|(10)}} + |} + + which will be confirmed by exact solution. + + For the case of high Prandtl number fluids the momentum boundary layer thickness, δ, is much greater than the thermal boundary layer thickness, δt (see Fig. 1). While the above scale analysis indicated that flow within the thermal boundary layer is dominated by balance between the buoyancy force and the viscous force, buoyancy force does not exist beyond the thermal boundary layer. Therefore, the velocity that is developed outside the thermal boundary layer but within the momentum boundary layer results from the viscous drag of the thermal boundary layer. Therefore, the flow between the thermal and momentum boundary layers is dominated by the balance between the viscous force and the inertial force. Since the thermal boundary layer is very thin for the case of $\Pr \gg 1$, there exists a balance between inertia and viscosity in the momentum equation over the entire momentum boundary layer. + + Refer to Fig. 1 once again and consider the momentum equation of the momentum boundary layer (''y'' ~ $\delta$) for the entire flat plate (''x'' ~ ''L''). The thickness of the momentum boundary layer is much smaller than the length of the vertical plate, i.e. $\delta \ll L$. The force balance in the momentum boundary layer between inertial and viscous forces gives us: + +
$\frac{{{u}^{2}}}{L}\sim \nu \frac{u}{{{\delta }^{2}}}$
+ + where the vertical velocity component, u, is induced by the thin thermal boundary layer. Substituting eq. (8) into the above equation yields the scale of the momentum boundary layer: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $\delta \sim L\text{Ra}_{L}^{-1/4}{{\Pr }^{1/2}}$ +
+ |{{EquationRef|(11)}} + |} + + Comparing eqs. (11) and (7) gives us the following relationship between the thicknesses of the momentum and thermal boundary layers: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $\frac{\delta }{{{\delta }_{t}}}\sim {{\Pr }^{1/2}}>1$ +
+ |{{EquationRef|(12)}} + |} + + It is evident that the greater the Prandtl number, the greater the ratio of δ/δt. This means that the region of unheated fluid which is being driven vertically by the heated layer due to viscous action is thicker for fluids with a higher Prandtl number. + + ==Low Prandtl Number Fluids == + + In the case of low Prandtl number, eq. (5) indicates that inertial force has a significant effect on the momentum equation, while the effect of viscous force is negligible. The momentum equation requires that the inertia term to be balanced by the buoyancy term, i.e., + + {| class="wikitable" border="0" + |- + | width="100%" |
+ ${{\delta }_{t}}\sim L{{\text{(R}{{\text{a}}_{L}}\Pr )}^{-1/4}}$ +
+ |{{EquationRef|(13)}} + |} + + Substituting eq. (13) into eqs. (2) and (3), the scales of the velocity components become: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $u\sim \frac{\alpha }{L}{{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/2}}$ +
+ |{{EquationRef|(14)}} + |} + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $v\sim \frac{\alpha }{L}{{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/4}}$ +
+ |{{EquationRef|(15)}} + |} + + [[Image:Natural convection over a vertical flat plate 2.jpg|thumb|400 px|alt=Natural convection over a vertical flat plate (Pr < 1) |'''Figure 2 Natural convection over a vertical flat plate (Pr < 1).''']] + + The scale of the Nusselt number is: + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $\text{Nu}\sim \frac{L}{{{\delta }_{t}}}\sim {{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/4}}$ +
+ |{{EquationRef|(16)}} + |} + + Since the Prandtl number is much less than 1, the thermal boundary layer is driven upwards by buoyancy and is restrained by inertia. Beyond the thermal boundary layer region, where there is no heating effect and consequently no buoyancy effect, the fluid is at rest and there is no motion whatsoever there. Therefore, the velocity boundary layer, ''δ'', must end with the thermal boundary layer, δt. Due to the no slip boundary condition at the wall there should be a velocity peak somewhere between the wall and the thermal boundary layer. Consider a thickness of the velocity boundary layer from the wall up to the point of maximum velocity, ${{\delta }_{u}}$. The effect of inertia is negligible within this thin layer ${{\delta }_{u}}$ – referred to as the shear layer (Bejan, 2004) – and the buoyancy force is balanced by the viscous force: + +
$\nu \frac{u}{\delta _{u}^{2}}\sim g\beta \Delta T$
+ + Substituting eq. (14) into the above equation, the scale of the shear layer can be determined as + + {| class="wikitable" border="0" + |- + | width="100%" |
+ ${{\delta }_{u}}\sim L\text{Gr}_{L}^{-1/4}$ +
+ |{{EquationRef|(17)}} + |} + + The ratio between the thicknesses of the shear layer and the thermal boundary layer can be obtained by dividing eq. (17) by eq. (13): + + {| class="wikitable" border="0" + |- + | width="100%" |
+ $\frac{{{\delta }_{u}}}{{{\delta }_{t}}}\sim {{\Pr }^{1/2}}<1$ +
+ |{{EquationRef|(18)}} + |} + + It should be pointed out that the thickness of the shear layer, δu, is different from the momentum boundary layer, ''δ'', which is approximately equal to the thermal boundary layer thickness for the case where the Prandtl number is less than 1. + + + ==References== + + Bejan, A., 2004, ''Convection Heat Transfer'', 3rd ed., John Wiley & Sons, New York. + + Faghri, A., Zhang, Y., and Howell, J. R., 2010, ''Advanced  Heat and Mass Transfer'', Global Digital Press, Columbia, MO.

## Scaling

Figure 1 Natural convection over a vertical flat plate (Pr >1).

The average Nusselt number for natural convection can be expressed as:

${{\overline{\text{Nu}}}_{L}}=f(\text{G}{{\text{r}}_{L}},\Pr )$

While scale analysis cannot provide the exact function in the above expression, the form of these functions can be provided (Bejan, 2004). Let us refer to Fig. 1 and consider the governing equations in the thermal boundary layer (y ~ δt) for the entire flat plate (x ~ L). The thickness of the thermal boundary layer in which the effects of the heated wall are felt is much smaller than the length of the vertical plate, i.e. ${{\delta }_{t}}\ll L$. For the continuity equation

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

to be satisfied, the scales of the two terms must be the same:

$\frac{u}{L}\sim \frac{v}{{{\delta }_{t}}}$

The scale of the velocity component in the y-direction is therefore:

 $v\sim \frac{{{\delta }_{t}}}{L}u$ (1)

which indicates that $v\ll u$ for flow in the thermal boundary layer. The scale of the velocity component in the x-direction, u, is still unknown at this point.

The energy equation is:

$u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}$

While the left hand side of the energy equation shows the effect of advection, the right-hand side shows the effect of diffusion. The scales of the two advective terms on the left hand side of energy equation are:

$u\frac{\partial T}{\partial x}\sim u\frac{\Delta T}{L},\text{ }v\frac{\partial T}{\partial y}\sim v\frac{\Delta T}{{{\delta }_{t}}}\sim u\frac{\Delta T}{L}$

which indicates that the scale of the second term on the left hand side of energy equation is identical to the scale of the first term when the scale of the velocity component in the y-direction is given by eq. (1). The temperature difference $\Delta T={{T}_{w}}-{{T}_{\infty }}$ in the above scale analysis represents the scale of the excess temperature, $T-{{T}_{\infty }}$. The scale of the right-hand side of energy equation is:

$\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}}$

The scales of the two sides of the energy equation must be the same:

$u\frac{\Delta T}{L}\sim \alpha \frac{\Delta T}{\delta _{t}^{2}}$

The above equation can be used to estimate the scale of u as follows:

 $u\sim \alpha \frac{L}{\delta _{t}^{2}}$ (2)

The scale of v can be obtained by substituting eq. (2) into eq. (1):

 $v\sim \frac{\alpha }{{{\delta }_{t}}}$ (3)

where the scale of the thermal boundary layer thickness, δt, is still unknown at this point.

The momentum equation is

$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$

The respective scales of the two inertial terms on the left-hand side of the momentum equation (10) are:

$u\frac{\partial u}{\partial x}\sim \frac{{{u}^{2}}}{L}$
$v\frac{\partial u}{\partial y}\sim v\frac{u}{{{\delta }_{t}}}\sim \frac{{{u}^{2}}}{L}$

The respective scales of the viscosity and buoyancy terms on the right-hand side of momentum are:

$\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}\sim \nu \frac{u}{\delta _{t}^{2}}$

$g\beta (T-{{T}_{\infty }})\sim g\beta \Delta T$

We can very well see that three forces are at play in the boundary layer region, which are inertia, viscosity and buoyancy forces. Considering the scale of u obtained from eq. (2), the scales of these three forces are

 $\begin{matrix} \frac{{{\alpha }^{2}}L}{\delta _{t}^{4}}, & \frac{\nu \alpha L}{\delta _{t}^{4}}, & g\beta \Delta T \\ \text{Inertia} & \text{Viscous} & \text{Buoyancy} \\ \end{matrix}$ (4)

Among these three forces, the buoyancy force is never negligible because without it natural convection would not occur. Therefore, the scale of buoyancy force can be used to measure the importance of the inertial and viscous forces. Dividing eq. (4) by the scale of buoyancy force,

gβΔT , one obtains the following:$\begin{matrix} \frac{{{\alpha }^{2}}}{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & \frac{\nu \alpha }{g\beta \Delta T{{L}^{3}}}{{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}, & 1 \\ \text{Inertia} & \text{Viscous} & \text{Buoyancy} \\ \end{matrix}$

which can be expressed in terms of dimensionless parameters as (Bejan, 2004):

 $\begin{matrix} {{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}\text{Ra}_{L}^{-1}{{\Pr }^{-1}}, & {{\left( \frac{L}{{{\delta }_{t}}} \right)}^{4}}\text{Ra}_{L}^{-1}, & 1 \\ \text{Inertia} & \text{Viscous} & \text{Buoyancy} \\ \end{matrix}$ (5)

where

 $\text{R}{{\text{a}}_{L}}=\frac{g\beta ({{T}_{w}}-{{T}_{\infty }}){{L}^{3}}}{\nu \alpha }$ (6)

is the Rayleigh number that is related to the Grashof number by $\text{R}{{\text{a}}_{L}}=\text{G}{{\text{r}}_{L}}\Pr$. Therefore, the relative importance of inertia and viscous forces depends on the Prandtl number, which is a property of the fluid. Thus, if the Prandtl number is high ($\Pr \gg 1$), the inertia term will be negligible and the viscosity term will balance the buoyancy term, whereas if the Prandtl number is low enough ($\Pr \ll 1$) – as for liquid metals – then the inertia term is considerable and balances the buoyancy term in steady state. The scale analysis for high- and low-Prandtl number fluids is presented in detail below.

## High Prandtl Number Fluids

For the case of high Prandtl number fluids, the kinematic viscosity of the fluid is much greater than its thermal diffusivity, so viscous force has a significant effect on the momentum equation and the effect of inertial force is negligible. The viscous force in fact balances the buoyancy force, and so from eq. (5), one obtains the following:

 ${{\delta }_{t}}\sim L\text{Ra}_{L}^{-1/4}$ (7)

Substituting eq. (7) into eqs. (2) and (3), the scales of the velocity components become:

 $u\sim \frac{\alpha }{L}\text{Ra}_{L}^{1/2}$ (8)
 $v\sim \frac{\alpha }{L}\text{Ra}_{L}^{1/4}$ (9)

The scale of the heat transfer coefficient can be obtained by analyzing the following equation

${{h}_{x}}=-\frac{k}{{{T}_{w}}-{{T}_{\infty }}}{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}$

i.,e.,

${{h}_{x}}=\frac{-k{{\left( \frac{\partial T}{\partial y} \right)}_{y=0}}}{{{T}_{w}}-{{T}_{\infty }}}\sim \frac{k(\Delta T/{{\delta }_{t}})}{\Delta T}=\frac{k}{{{\delta }_{t}}}$

The scale of Nusselt number is therefore:

 $\text{Nu}=\frac{{{h}_{x}}L}{k}\sim \frac{L}{{{\delta }_{t}}}\sim \text{Ra}_{L}^{1/4}$ (10)

which will be confirmed by exact solution.

For the case of high Prandtl number fluids the momentum boundary layer thickness, δ, is much greater than the thermal boundary layer thickness, δt (see Fig. 1). While the above scale analysis indicated that flow within the thermal boundary layer is dominated by balance between the buoyancy force and the viscous force, buoyancy force does not exist beyond the thermal boundary layer. Therefore, the velocity that is developed outside the thermal boundary layer but within the momentum boundary layer results from the viscous drag of the thermal boundary layer. Therefore, the flow between the thermal and momentum boundary layers is dominated by the balance between the viscous force and the inertial force. Since the thermal boundary layer is very thin for the case of $\Pr \gg 1$, there exists a balance between inertia and viscosity in the momentum equation over the entire momentum boundary layer.

Refer to Fig. 1 once again and consider the momentum equation of the momentum boundary layer (y ~ δ) for the entire flat plate (x ~ L). The thickness of the momentum boundary layer is much smaller than the length of the vertical plate, i.e. $\delta \ll L$. The force balance in the momentum boundary layer between inertial and viscous forces gives us:

$\frac{{{u}^{2}}}{L}\sim \nu \frac{u}{{{\delta }^{2}}}$

where the vertical velocity component, u, is induced by the thin thermal boundary layer. Substituting eq. (8) into the above equation yields the scale of the momentum boundary layer:

 $\delta \sim L\text{Ra}_{L}^{-1/4}{{\Pr }^{1/2}}$ (11)

Comparing eqs. (11) and (7) gives us the following relationship between the thicknesses of the momentum and thermal boundary layers:

 $\frac{\delta }{{{\delta }_{t}}}\sim {{\Pr }^{1/2}}>1$ (12)

It is evident that the greater the Prandtl number, the greater the ratio of δ/δt. This means that the region of unheated fluid which is being driven vertically by the heated layer due to viscous action is thicker for fluids with a higher Prandtl number.

## Low Prandtl Number Fluids

In the case of low Prandtl number, eq. (5) indicates that inertial force has a significant effect on the momentum equation, while the effect of viscous force is negligible. The momentum equation requires that the inertia term to be balanced by the buoyancy term, i.e.,

 ${{\delta }_{t}}\sim L{{\text{(R}{{\text{a}}_{L}}\Pr )}^{-1/4}}$ (13)

Substituting eq. (13) into eqs. (2) and (3), the scales of the velocity components become:

 $u\sim \frac{\alpha }{L}{{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/2}}$ (14)
 $v\sim \frac{\alpha }{L}{{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/4}}$ (15)
Figure 2 Natural convection over a vertical flat plate (Pr < 1).

The scale of the Nusselt number is:

 $\text{Nu}\sim \frac{L}{{{\delta }_{t}}}\sim {{\text{(R}{{\text{a}}_{L}}\Pr )}^{1/4}}$ (16)

Since the Prandtl number is much less than 1, the thermal boundary layer is driven upwards by buoyancy and is restrained by inertia. Beyond the thermal boundary layer region, where there is no heating effect and consequently no buoyancy effect, the fluid is at rest and there is no motion whatsoever there. Therefore, the velocity boundary layer, δ, must end with the thermal boundary layer, δt. Due to the no slip boundary condition at the wall there should be a velocity peak somewhere between the wall and the thermal boundary layer. Consider a thickness of the velocity boundary layer from the wall up to the point of maximum velocity, δu. The effect of inertia is negligible within this thin layer δu – referred to as the shear layer (Bejan, 2004) – and the buoyancy force is balanced by the viscous force:

$\nu \frac{u}{\delta _{u}^{2}}\sim g\beta \Delta T$

Substituting eq. (14) into the above equation, the scale of the shear layer can be determined as

 ${{\delta }_{u}}\sim L\text{Gr}_{L}^{-1/4}$ (17)

The ratio between the thicknesses of the shear layer and the thermal boundary layer can be obtained by dividing eq. (17) by eq. (13):

 $\frac{{{\delta }_{u}}}{{{\delta }_{t}}}\sim {{\Pr }^{1/2}}<1$ (18)

It should be pointed out that the thickness of the shear layer, δu, is different from the momentum boundary layer, δ, which is approximately equal to the thermal boundary layer thickness for the case where the Prandtl number is less than 1.

## References

Bejan, A., 2004, Convection Heat Transfer, 3rd ed., John Wiley & Sons, New York.

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.