Laminar external natural convection on vertical and horizontal flat plates, over horizontal and vertical cylinders and sphere, as well as plumes, wakes and other types of free flow will be discussed in this section. 1
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), (6.0) and (6.1). Introducing the stream function,, and dimensionless temperature, : T T u, v, y x T T w (6.55) the continuity equation (6.17) is satisfied and the momentum and energy equations (6.0) and (6.1) become: 3 g ( T T 3 w ) (6.56) y yx x y y
y x x y y with the following boundary conditions: (6.57) (6.58) y 0, 1 at y 0 x 0, 0 at y y x (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. Thus, the similarity variable should have the following form: y H( x) where H(x) is an unspecified stretching function. (6.60) 3
The objective now is to reduce eqs. (6.56) and (6.57) to ordinary differential equations. The stream function,, which is function of x and y, can be expressed as function of x and. If the similarity solution exists, one can express the stream function as ( x, ) F ( ) G ( x ) (6.61) 61) where F( ) and G( x) are the similarity function and the stretching function, respectively. The dimensionless temperature can be assumed as a function of only, i.e., ( x, ) ( ) (6.6) The derivatives in eqs. (6.60) (6.6) can be obtained as shown on the next slide: 4
F( ) Gx ( ) F( ) HxGx ( ) ( ) y y F ( )[ H ( x) G( x) H ( x) G ( x)] F ( ) H ( x) G( x) yx F ( ) yh ( x ) G ( x ) F ( ) G ( x ) x yy F ( ) H ( x) G( x) 3 3 F ( ) H ( x) G( x) 3 y ( ) ( ) yh ( x) x x ( ) ( ) H ( x ) y y 5
where the primes for F and θ denote the derivatives with respect to, 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: g ( T T ) G HG G (6.63) 63) w F FF F 3 HG H H H Pr G F 0 H ( ) 0 (6.64) 64) In order to convert eqs. (6.63) and (6.64) to ordinary differential equations with 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: 6
3 HG A const (6.65) G B const H HG const C H Differentiation of eq. (6.65) and division of the resultant equation by H 4 yields the following equation: HG G 3 0 H H which is satisfied if eqs. (6.66) and (6.67) are satisfied. Substituting eq. (6.68) 68) into eq. (6.63), 63) we have g ( Tw T ) G G F FF F 3 HG H 3 H ( ) 0 (6.66) (6.67) (6.68) (6.69) 7
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 g ( T T ) w 1 3 HG G 3 H 8
The corresponding G and H functions then become: 1 Gx ( ) 4 Grx 4 11 H( x) Grx x 4 where Gr x is the local Grashof number defined as: 1/4 1/4 g ( Tw T ) x Gr x 3 (6.70) (6.71) (6.7) which is equivalent to square of the Reynolds number based on the scale of the local l velocity u g ( T T ) x. 0 ( w ) 9
Substituting eqs. (6.70) and (6.71) into eqs. (6.69) and (6.64), the momentum and the energy equations become: F 3FF ( F) 0 (6.73) 3PrF 0 (6.74) The velocity components in the x- and y-directions can be expressed in terms of similarity variables by the following equations: 1/ u Grx F( ) y y x (6.75) Grx 1/4 v [ F( ) 3 F( ) x ] x x x 4 (6.76) which are obtained from eqs. (6.60), (6.61), (6.70) and (6.71). 1/4 10
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( ) F( ) 0, and ( ) 1 at 0 (6.77) F( ) ( ) 0 at (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 dimensionless velocity and temperature for various Prandtl number are shown in Figs. 6.3 and 6.4, respectively. 11
F'()=Ra -1/ x ux/ 1 0.8 0.6 0.4 0. Pr=(0.01,0.5,1,,5, 10,100,1000) 1 0.8 0.6 0.4 0. = =(T-T )/(T w -T ) 1 0.8 0.6 0.4 0. 0 (a) () Velocity yprofiles Pr=(0.01,0.5,1,,5,10,100,1000) 0 1 3 4 5 6 =(Gr x /4) 1/4 y/x (b) Temperature profiles Figure 6.3 Velocity and temperature profile in the boundary layer for external natural convection over a vertical isothermal surface 1 0 0 0 0 0 =(T-T )/(T w -T ) 0 0 0 1 3 4 =Ra 1/4 x y/x (a) Velocity profiles 1 1 Pr=(0.01,0.5,1,,5,10,100,1000) 0.8 0.8 0.6 0.6 0.4 0.4 0. 0. 0 0 0 1 3 4 =Ra 1/4 x y/x (b) Temperature profiles Figure 6.4 Velocity and temperature profiles in the boundary layer based on modified scale. 1
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 The local Nusselt number is x k T 1 Grx k (0) T T 4 y w x 4 y0 1/4 (6.79) h x (6.80) x (0) Nu Gr (Pr)Gr k 1/4 1/4 x x x 13
where (Pr) (0) / is the function of Prandtl number. The dependence of on the Prandtl number is evidenced by eq. (6.74) and by Fig. 6.3(b). The values of for various Pr have been obtained numerically by Ostrach (1953). Ede (1964) proposed the following function that correlated the numerical results: 3 Pr (Pr) 1/ 45(1Pr Pr) The local Nusselt number thus becomes: Nu x 3 Pr 4 5(1 Pr 1/ Pr) 1/4 1/4 (Gr Pr) x 1/4 (6.81) 14
which h can also be rewritten in terms of Rayleigh number Nu x 3 Pr 1/ 4 5(1 Pr Pr) 1/4 Ra Equations (6.81) and (6.8) are valid for 0 < Pr <. 1/4 x (6.8) 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 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). 15
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, 004) expressed in eqs. (6.43) and (6.44), which is equivalent to a selection of the stretching functions in the following form: 1 1/ 4 G( ( x ) Ra x Pr 1 1/4 H( x) Ra x x 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. 16
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 h the velocity and temperature profiles constantly shift as Pr changes, and the peak dimensionless velocity is not of order 1. 17
Example 6.1 For the cases that Pr 0 or Pr, the similarity solution can be obtained by transforming eqs. (6.73) and (6.74) using the following new variables: z (Pr) 1 1, F1 F/ z(pr), and 1, where z 1 and z are unknown scaling coefficients. Obtain the similarity equation using the new similarity variables. 18
Solution: To transform eqs. (6.73) and (6.74) to equations in terms of the new similarity variables, the relationship between the derivatives of original and new dimensionless stream function and temperature must be identified. The derivative of the original dimensionless temperature is: d d 1 ( ) 1 ( 1 ) z1 (Pr) d1 d The second order derivative of original dimensionless temperature is: d d1 d ( ) [ ( )] [ 1( 1) z1(pr)] z1(pr) 1( 1) z1(pr) d d d 1 1 19
Substituting the above derivatives and definition of F 1 into eq. (6.74), the energy equation becomes 1z1 3PrF11 z1 z 0 The choice of z 1 and z should be such that the Prandtl number should not appear explicitly in the new energy equation, i.e., z(pr) z1(pr) /(3Pr). The derivatives of the stream function are df d (Pr) 1 df 1 z 1 F ( ) z (Pr) z1(pr) F1 ( 1 ) d d d 3Pr 1 1 3 df d1 z1(pr) F ( ) F 1 ( 1 ) d1 d 3Pr 4 df d1 z1(pr) F( ) F 1( 1) d 1 d 3Pr 0
Substituting the above expressions into eq. (6.73), one obtains the following equation F 1( 1) 3 FF 1 1( 1) [ F1 ( 1 )] 1 0 4 3Pr 9Pr z (Pr) where the three terms on the left-hand side represent the effects of viscosity, inertia, and buoyancy, respectively. The choice of the scaling factor must ensure that the equation is valid under extreme conditions. The results of the scaling analysis require that the viscous effect vanish when Pr 0. On the other hand, the effect of inertia term should vanish when Pr. 1 1
Le Fevre (1956) suggested the following 4 3Pr z 1 (Pr) 1 Pr Thus, the momentum and energy equations become: Pr( F 1 1) 1 FF 1 1 F 1 (6.83) 3 1 F1 1 0 (6.84) Equations (7.83) and (7.84) allow simplification for the cases that Pr 0 or Pr. Numerical solution of the simplified equation for these two cases yields the following results (Le Fevre, 1956): Nu x 1/4 1/4 0.600(GrxPr ) 0.600(Ra xpr) Pr 0 1/4 1/4 0.503(Gr Pr) 0.503Ra Pr (6.85) x x
1/4 Equations (6.8) and (6.85) indicate that hx x. The average Nusselt number over the entire vertical wall, Nu L, is related to the local Nusselt number at x Lby the following. Nu L 4 Nu 3 L (6.86) 3
6.4. Integral Solution for Laminar and Turbulent Natural Convection Laminar Flow Multiplying py the continuity equation (6.17), by u and adding the resulting equation to the momentum equation (6.0) yields: u ( uv ) u g ( T T ) x y y Integrating the above equation with respect to y in the interval of (0, Y), where Y is greater than both and t, one obtains d Y u Y udy g ( TT ) dy dx 0 0 (6.87) y y 0 4
By following a similar il procedure, the integral energy equation can be obtained as follows d dx Y 0 T u ( T T ) dy y y0 (6.88) 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: u y y 1 U (6.89) where U is a characteristic velocity that is a function of x. 5
Similarly, il l the temperature t profile can be obtained by assuming a second degree polynomial function and the result is: T T y 1 Tw T T (6.90) The following analysis will be based on the assumption that the momentum and thermal boundary layers have the same thickness, i.e. T. Substituting the velocity and temperature profiles into the integral form of the momentum and energy equations yields: 1 d U 1 (6.91) ( U ) g( Tw T ) 105 dx 3 1 d ( U ) 30 dx (6.9) 6
At the leading edge of the vertical plate, the boundary layer thickness is zero and the characteristic velocity U is also zero: U 0, x0 (6.93) which are the initial conditions of eqs. (6.91) and (6.9). It is expected that as the x increases, both U and δ should increase. Let us assume that they are functions of x such that m n U C1x, Cx (6.94) Substituting eq. (7.94) into eqs. (6.91) and (6.9), one obtains the following: 7
m n 1 C1 1 1 CCx x g 1 ( T T w ) C x 105 C 3 m n m n m n m n 1 CC n x x 1 30 C 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: m n1mnn (6.95) (6.96) mn1 n which h are satisfied only if m 1/ and n1/4. This suggests 1/ 1/4 that U x and x ; this is in agreement with the result of scaling analysis. Substituting back the values of m and n into eqs. (6.95) and (6.96), one obtains CC 1 C1 1 g ( Tw T ) C 84 C 3 CC 1 40 C 8
Solving for C 1 and C from the above two equations yields: C C 1/ 1/ 5 0 g ( Tw T ) 5 3 1 1 1/4 1/4 1/ 15 0 g ( Tw T ) 4 16 1 1/ The boundary layer thickness therefore becomes: 0.95 Pr 1/4 393 3.93 Grx Pr (6.97) The local heat transfer coefficient at the surface of the vertical plate can be obtained from eq. (6.3) : h x 1/4 k T k T T w y y0 9
The local l Nusselt number is: hx x Nu x k x 0.6 0.5 0.4 1/4 Nu x /Ra x 0.3 0. 0.1 Integral Solution Similarity Solution 0.0 10-4 10-3 10-10 -1 10 0 10 1 10 10 3 10 4 Figure 6.5 Comparison between integral solution and similarity solutions Pr 30
Substituting eq. (6.97) into the above expression yields: 1/4 1/4 Pr 1/4 Pr 1/4 x x x Nu 0.508 Gr 0.508 Ra 095 0.95 Pr 095 0.95 Pr (6.98) 31
Example 6. The thickness of the momentum and thermal boundary layers are assumed to be the same in the above analysis. For the fluid with Prandtl number greater than 1, the following temperature and velocity profiles can be used: T T y/ t (6.99) u U T w e T T y/ e y/ t 1 e (6.100) where, t and U are unknown functions of x. Find the Nusselt number using integral solution based on the above profiles. 3
Solution: Substituting eqs. (6.99) and (6.100) into eqs. (6.87) and (6.88) and setting Y, the following equations are obtained: where d U q Uq g( T w T ) dx ( q )(1 q ) q d U dx (1 q)(1 q) q(pr) T is the ratio between the thicknesses of momentum and thermal boundary layers. 33
Since q is function of Prandtl number only, the above differential equations can be rewritten as: q d Uq ( U ) g ( T T ) (6.101) ( q)(1 q) dx q 1 d ( U ) (1 q )(1 q ) dx w (6.10) An additional equation is needed since there are three unknowns in eqs. (6.101) and (6.10). Due to the non- slip and non-permeable boundary condition at the vertical wall (u = v = 0 at y = 0), eq. (6.0) in the region near the wall becomes: u 0 g( T ) w T y y0 34
Substituting eqs. (6.99) and (6.100) into the above momentum equation, one obtains: Uq( q) g ( T T ) (6.103) w Equations (6.101) and (6.10) are similar to eqs. (6.91) and 1/ 1/4 (6.9); thus, one can expect that U x and x. By following the similar procedure that we used to solve eqs. (6.91) and (6.9), the local Nusselt number can be obtained: 3 1/4 3 q 1/4 Nu Ra (6.104) x 8( q 1)( q 1/)( q ) where the ratio of the boundary layer thickness can be obtained from the following equation x 35
Pr 5 1/ q 6 q q (6.105) For the case the surface heat flux of the flat plate is uniform, an integral solution can be performed to obtain the following results where Nu x Pr 360 0.8 Pr 1/5 g qx Ra* x k 4 1/5 Ra is the Rayleigh number based on the heat flux. 1/5 * x (6.106) (6.107) 36
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 result will be invalid. 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 t flows. 37
x T w quiescent fluid, u =0 Turbulent gravity Laminar y Figure 6.6 Transition from laminar to turbulent 38
Using the definition iti of shear stress and heat flux, eqs. (6.87) and (6.88) can be respectively modified as d Y Y w udy g ( T T ) dy (6.108) dx 0 0 d dx Y 0 ut ( T ) dy q w c p (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 1/7 4 u y y 1 U (6.110) 39
where U is a characteristic ti 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 thickness are the same and the temperature profile is: T T y 1 T T w (6.111) which yields T T = T T at w at y 0 and y. It should be pointed out that eqs. (6.110) and (6.111) are valid only for y. For large y, one has u = 0 and. T T Substituting eqs. (6.110) and (6.111) into eqs. (6.108) and (6.109), the integral equations become: d 1 /7 8 1 1/7 w U 1 d g ( T ) (1 ) 0 w T d dx 0 1/7 40
d 1 1/7 4 1/7 q UT w ( w T ) 1 1 d dx 0 c where y /. Evaluating the integrals in the above two equations yields: (6.11) d 0.053 U w 0.15 g ( T w T ) dx d q w 0.0366( Tw T ) ( U ) dx c p (6.113) It is generally assumed that t 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 derived for forced convection can be applied: w 0.05 05 0.5 (6.114) U ( U / ) p 41
1/4 0.33 / 3 q w 0.05Pr cut ( T ) U p w Substituting the above expressions into the integral momentum and energy equations, one obtains: d 0.053 0.15 ( ) 0.05 dx 0.75 d 0.67 0.5 U 0.0366 ( U) 0.05Pr 0.5 dx 0.5 U g ( T w T ) 05 0.5 Pr U 1.75 (6.115) (6.116) (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: m n U C x, C x (6.118) 1 4
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. By following the procedure similar to the case of laminar natural convection, the local Nusselt number can then be 1/15 obtained: Pr /5 Nu x 0.095 Ra /3 /5 x (1 0.494Pr ) (6.119) which suggests that the local heat transfer coefficient is proportional to x 0.. The average Nusselt number for the entire vertical plate, Nu L, is related to the local Nusselt number at x L as follows: Nu L 0.834Nu L (6.10) 43
Empirical i 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 Nu L C Ra n L (6.11) which is confirmed by experimental studies. For laminar flow, one can utilize the values of C 059 0.59 and n 1/4. For turbulent flow, C 0.1 and n 1/3 can be used. Churchill and Chu (1975) studied numerous sets of experimental data and recommended the following correlation: 1/6 Ra L Nu L 0.85 9 /16 8/ 7 [1 (0.49/ Pr) ] (6.1) 44
which h covers all Prandtl number and Grashof number between 0.1 and 10 1. For the case of laminar convection, the following correlation yields better result: 1/4 0.67Ra L Nu L 0.68 [1 (0.49 / Pr) 9/16 ] 4/9 (6.13) 45
6.4.3 Natural Convection over Inclined and Horizontal Surface Inclined Flat Surface Depending on whether the wall is heated or cooled and tilted upward or downward, there are four different configurations for convection over an inclined flat plate (see Fig. 6.7). While the effect of inclination thickens the boundary layers for the upper surface of case (a) and the bottom surface of case (b), the layer becomes thinner for bottom surface of case (a) and upper surface of case (b). 46
T w < T g g T w > T (a) heated surface (b) cooled surface Figure 6.7 Natural convection over inclined surface 47
When the inclination angle is moderate ( 60 60 ), the momentum equation for laminar natural convection over an inclined wall is identical to eq. (6.0), except that the gravitational acceleration g must be replaced by the component of the gravitational acceleration along the direction that is parallel to the inclined plate, g cos. For laminar natural convection, the empirical correlations that were developed in the previous subsection, e.g. eq. (6.13), are still valid if the Rayleigh number is defined as 3 g cos ( Tw T ) x Ra x (6.14) heat transfer in turbulent natural convection is not sensitive to the inclination angle, and so eq. (6.1) can be used. 48
Horizontal Flat Surface Due to its complicated nature, analytical solutions for natural convection on a horizontal surface are only available for limited cases. Rotem and Claasen (1969) studied natural convection over a semi-infinite isothermal flat plate with a single leading edge (see Fig. 6.8). quiescent fluid, u =0 y gravity T, ρ Velocity boundary layer, δ u Thermal boundary layer, δ T T w x Figure 6.8 Natural convection over a horizontal flat plate 49
The continuity it and energy equations are same as the case of vertical plate, eqs. (6.17) and (6.1). The momentum equation is, however, different from that of vertical plate because the direction of gravity is normal to the horizontal plate. The momentum equation is: u u u 1 p p d u v (6.15) x y y x 1 pd g ( T T ) y where p d is the dynamic pressure whose gradient is the driving force for the boundary layer flow. (6.16) 50
Defining i the following similarity il it variables: 1/5 1/5 4/5 Gr Gr Gr y x x 5 x, 5 F( ), pd F( ) G( ) x 5 5 x 5 (6.17) Eqs. (6.17), (6.15), (6.16) and (6.1) become: F 3 F F f G G 0 5 5 (6.18) G (6.19) 3Pr F 0 (6.130) which can be numerically solved to obtain the following local Nusselt number: 1/5 1/4 Nu 0.394Gr Pr (6.131) x x 51
Equation (6.131) is valid for an isothermal surface. For a horizontal surface with constant heat flux, the local Nusselt number is: 1/5 1/4 Nu x 0.501Gr x Pr (6.13) While the above solution is for the case of a heated surface facing upward, it can also be applied to the case of a cold surface facing downward. However, the similarity solution cannot be applied for the case of a hot surface facing downward or a cold surface facing upward. 5
For most applications, the horizontal surface cannot be treated as semi-infinite surface with only one leading edge. Two extreme types of flow situation arise for the horizontal surface with finite size (see Fig. 6.9). If the hot surface of the plate faces upwards or cold surface of the plate faces downwards, a plume is formed whereby the hotter fluid layers rise up or colder fluid mass goes down, respectively. As shown in Fig. 6.9 (a) and (b), the boundary layer starts from the edge of the horizontal surface and the flow leaves the boundary layer as a vertical plume at the central region. 53
T w < T T w > T (a) Hot surface (b) Cold surface Figure 6.9 Natural convection over horizontal surface 54
For the case of upward facing hot isothermal surfaces or downward facing cold isothermal surfaces, the following correlation can be applied: Nu L 0.54Ra 10 Ra 10 0.15Ra 10 Ra 10 1/4 4 7 L L 1/3 7 9 L L (6.133) The characteristic length in the average Nusselt number and Rayleigh number is L = A/P, where A refers to the area of the plate and P is the perimeter. The corresponding correlation for downward facing hot surfaces or upward facing cold surfaces is: Nu 0.7Ra 10 Ra 10 1/ 4 5 10 L L L (6.134) 55
6.4.4 4 Natural Convection over Vertical Cylinder Cylinders and Spheres For the vertical surface of the cylinder, the development of the boundary layer is shown in Fig. 6.10. If the ratio of diameter to height, D/L, is large enough, the correlations for natural convection over avertical flat plate can be used to calculate the natural convection over the vertical cylinder. For fluid with Prandtl number greater than 1, the condition under which the correlation for flat plate is applicable is D 1/4 Ra L (6.135) L Ra L 56
This condition is referred to as the thick cylinder limit. it If 1/4 D/L is less than Ra L, the boundary layer thickness will be comparable to the radius of the cylinder and the effect of surface curvature cannot be neglected. For this case the governing equations must be given in a cylindrical coordinate system: u 1 ( vr) 0 (6.136) x r y u v ( ur) 1 u u r g( T T ) x r r r r r (6.137) T v ( Tr ) 1 T u r (6.138) x r y r r r 57
The boundary conditions for eqs. (6.136) (6.138) are u v0, T Tw, at r R (6.139) u v 0, T T, at r (6.140) The integral momentum and energy equations can be obtained by integrating eqs. (6.136) (6.138) with respect to r in the interval of (R, R + δ): d dx u R R urdrr g ( T T ) rdr R r R r R (6.141) d dx R T r ut ( T ) rdr R R (6.14) r r R 58
Substituting eqs. (6.89) and (6.90) into eqs. (6.141) and (6.14), one obtains the following integral equations: d U (6.143) (8R 3 ) U 840R 70 g( Tw T ) (4 R ) dx d 40 [(7R ) U] dx (6.144) Unlike the case of natural convection over a vertical flat plate, eqs. (6.143) and (6.144) do not suggest that U and δ can be expressed as a simple function of x similar to eq. (6.94). Le Fevre and Ede (1956) assumed that U U 1 U 1 U0 and 0 R R R R (6.145) 59
Substituting the above expression into eqs. (6.143) and (6.144) and grouping the terms for the same order of R, equations for U 0, δ 0, U 1, δ 1, and so on, can be obtained: d 8 (6.146) 0 ( U 00) 840 U 0 80 g ( T w T ) 0 dx d 7 0 ( U 00) 40 (6.147) dx d d dx dx 0 (8 0 U0 8 1 U0 16 0 U0 U1 ) 1 ( 0 U0 ) 3 840U 170 g ( T w T )(8 0 1 0 ) (6.148) d d (6.149) 0 (0U0 71U0 7 0U1) 7 1 ( 0U0) 0 dx dx 60
It can be seen that t eqs. (6.146) and (6.147) have a form similar to eqs. (6.91) and (6.9). Thus, one can assume that: m n U A x, B x 0 0 0 0 and one can use eqs. (6.146) and (6.147) to obtain m1/ and n1/4 and: A 80 80(0 1Pr), B B0 7 g ( Tw T ) 4 0 0 0 (6.150) By following a similar procedure, one can use eqs. (6.148) and (6.149) to obtain 3/4 1/ U Ax, Bx where A 1 1 1 1 4(656 315Pr) (7 315Pr) B, B 7(64 63Pr) 35(64 63Pr) 1 1 B0 0 (6.151) 61
If one only considers the first two terms in eq. (6.145), the boundary layer thickness becomes: 1/ 1/4 B 1x B (6.15) 0x R The local heat transfer coefficient at the surface of the vertical cylinder is: h x k T k T T r w r R and the local Nusselt number is: Nu x hx x k x 6
Substituting eq. (6.15) to the above expression, one obtains Nu x 1/4 7Grx Pr 80(0 1Pr) 3/4 80(0 1Pr) 7 315Pr 1 7Grx Pr 35(64 63Pr) Since the second term in the denominator of the above expression is much less than 1, Le Fevre and Ede (1956) suggested that the above expression for local Nusselt number can be simplified to: Nu x 1/4 7Grx Pr (7 315Pr) 5(0 1Pr) 35(64 63Pr) x R x R (6.153) 63
The average Nusselt number can then be obtained as: 1/4 4 7RaL Pr 4(7 315Pr) L Nu L 3 5(0+1Pr) 1P 35(64 63Pr)D D (6.154) where the characteristic length in Nu L and Ra L is the height of the cylinder. 64
Example 6.3 A cone with the surface temperature of Tw is immersed into aquiescent fluid at a temperaturet of T ( Tw T ). A boundary layer develops from the tip of the cone and thickens as x increases (see Fig. 6.11). Assuming that the thicknesses of the momentum and thermal boundary layers are the same, obtain the local Nusselt number. δ L T x,u w T 0 Figure 7.11 Boundary layer for natural convection over vertical cone y, v 65
Solution: The dimensionless continuity, momentum and energy equations are: ( UR) ( VR) 0 X Y U U 1 U U V R GrL X Y R Y Y Pr u V R x y R Y r where the dimensionless variables are defined as: x y r ul vl T T X, Y, R,, U, V, L L L L T T and R is related to X and Y by R X sin Y cos w (6.155) (6.156) (6.157) 66
The boundary conditions of eqs. (6.155) (6.157) are: U U 0, 1, at Y 0 (6.158) U V 0, 0, at Y (6.159) The integral momentum and energy equations can be obtained by integrating eqs. (6.155) (6.157) with respect to Y in the interval of (0, ): d B A U1 X 1 cot dx A X D X U CGrL X1 cot CX Y Y 0 (6.160) d F X EPr U1X 1 cot dx E X Y Y 00 (6.161) 67
where U 1 is a dimensionless i characteristic ti velocity. The other variables are defined as A 1 ( U / U ) d 0 1, 1 1 1 B U U d C d ( / 1 ), 0, 0 D d, 0 1 1 E ( U / U ) d, F ( U / U ) d where Y / 0 1 0 1. The effect of curvature is reflected by terms that contain ( / X )cot. For the cases that the effect of curvature is negligible, eqs. (6.160) and (6.161) respectively become d X U A ( U X ) CGr X (6.16) dx 1 Gr L Y d X EPr ( U1X ) dx Y Y Y 0 0 (6.163) 68
Assuming the velocity and temperature t in the boundary layer are described by the following equations: 3 11 1 U U which satisfy eqs. (6.158) and (6.159), eqs. (6.16) and (6.163) become: d UX (6.164) A dx ( U X ) 84Gr X 5 1 1 L d dx ( UX) 1 84 X Pr (6.165) 69
Since eqs. (6.164) and (6.165) have a form similar il to eqs. (6.91) and (6.9), one can use a similar approach to 1/4 obtain: X a 0 Gr L 336 Gr X (6.166) 1/ 1/ U1 Gr L X 7a (6.167) 0 Pr where 1/4 1008(3 7 Pr) a 0 49Pr (6.168) To consider the effect of curvature, Alamgir (1979) suggested replacing ( / X )cot in eqs. (6.160) and (6.161) by the following: 1/4 / X cot ( / X)cot 4a Gr cot 0 L 70
In this case, eqs. (6.160) and (6.161) become: where 1 (1 0.6 a0) d ( U1 X) 84(1 0.5 a0) X5 UX dx d X Pr(1 0.5 a0) ( U1X) 84 dx 1/4 cot Gr L is a curvature parameter. Equations (6.170) and (6.171) can be solved to obtain: X a Gr L 1 1/4 (6.169) (6.170) (6.171) (6.17) 71
where a The local wall heat flux is: U 336 Gr X 1/ 1/ 1 L 7a1 Pr(1 0.5 a0 ) 3(1 0.6 a ) 7(1 0.5 a ) Pr 1008 (1 0.5 )[7(1 0.5 )Pr] 0 0 1 a0 a0 y 0 0 1/4 T Tw T Tw T q k k k y L L The average heat flux is: q 1 A A qda where A t L sin is the total surface of the cone, and da x sindx. t (6.173) (6.174) (6.175) (6.176) 7
The average Nusselt number can be obtained as: Nu L hl ql k k( T T ) Substituting eqs. (6.175) and (6.176) into the above expressions, the final result becomes: w 4Gr 16Gr Nu L X dx a 1/4 1/4 L 1 3/4 L 0 1 7a 1 (6.177) 73
Horizontal Cylinder and Sphere When a horizontal cylinder or sphere with temperature T w is immerse in a fluid with temperature T ( Tw T ), a boundary layer develops along the curved surface. The boundary layer thickness is a function of the angle, as shown in Figure 6.1. The local Nusselt numbers along the surfaces of a cylinder and sphere are shown in Fig. 6.13. Although the integral solution can yield results all the way to the top where 180 and Nu 0, the result beyond 165 is no longer applicable because boundary layer separation occurred and plume flow takes place. 74
Based on the integral solution, Merk and Prins recommended the following empirical correlation for natural convection over a horizontal cylinder and sphere: 1/4 Nu D CRa D (6.178) where the characteristic length in the average Nusselt number and Grashof number is the diameter of the cylinder or sphere. The constant C is function of Prandtl number and can be found from Table 6.1 75
Table 6.1 Constant in the empirical correlation for natural convection over horizontal cylinder and sphere Prandtl number 0.7 1.0 10.0 100.0 Cylinder 0.436 0.456 0.50 0.53 0.53 C Sphere 0.474 0.497 0.576 0.59 0.595 δ t T w D T r Figure 6.1 Boundary layer for natural convection over horizontal cylinder or sphere 76
(a) horizontal cylinder (b) Sphere Figure 6.13 Local Nusselt number for natural convection over horizontal cylinder or sphere (Merk and Prins, 1954b) 77
Practically, the empirical correlations based on experimental results are more useful. Churchill and Chu recommended the following correlation for horizontal cylinders: 1/6 0.387Ra D D (6.179) Nu 0.6 [1 (0.559/ Pr) ] 9 /16 8/ 7 which has the same form as the correlation for vertical plate, eq. (6.1), except the characteristic length has been changed from vertical plate height to the diameter of the cylinder. Equation covers (6.179) all Prandtl number and Rayleigh number between 0.1 and 10 1. 78
For natural convection over a sphere, Churchill (1983) suggested that the following correlation can best fit the available experimental data: 1/4 0.589Ra D Nu D (6.180) [1 (0.469/ Pr) 9/16 ] 4/9 where the average Nusselt number, Nu D, and the Rayleigh number, Ra D, are based on the diameter of the sphere. Equation (6.180) is valid for the case where Pr 0.7 and the Rayleigh number is less than 10 11. 79
6.4.5 Free Boundary Flow For some applications, thermally induced flow occurs without presence of a solid wall and is referred to as free boundary flow. When a point or line heat source is immersed into a bulk fluid, the fluid near the heat source is heated, becomes lighter and rises to form a plume [see Fig. 6.14 (a)]. The plume generated by a point heat source is axisymmetric, but a line heat source will create a two-dimensional plume. A thermal is a column of rising air near the ground due to the uneven solar heating of the ground surface. The lighter air near the ground rises and cools due to expansion [see Fig. 6.14(b)]. 80
(a) plume Figure 7.14 Examples of free boundary flows (b) thermal Consider the two-dimensional free boundary flow induced by a line heat source (see Fig. 6.15). In the coordinate system shown in Fig. 6.15, the continuity, momentum and energy equations are the same as those for natural convection near a vertical flat plate, viz. eqs. (6.17), (6.0) and (6.1), respectively. 81
yv y, T 0 T x, u heat source Figure 7.15 Physical model of natural convection from a line heat source The boundary conditions at y can still be described by eqs. (6.4) and (6.5). However, the boundary conditions at the centerline, x 0, should be changed to: g u y T 0, v0 at y0 y (6.181) 8
which indicates that the velocity component in the x- direction and the temperature at the centerline are at their respective maxim. Gebhart et al. assumed that the difference between the centerline temperature and the bulk fluid temperature is the following power law form: n T (6.18) 0 T Nx where the index n will be determined using the overall energy balance. The continuity, momentum, and energy equations can be transformed to the following ordinary differential equations via the same similarity variables for natural convection over a vertical flat plate in Section 6.4.1 and considering eq. (6.18). F (3 n) FF (n)( F) 0 (6.183) 83
Pr[( n3) F4 nf] 0 (6.184) Equations (6.183) and (6.184) are applicable to any natural convection problem that satisfies eq. (6.18). To get the ordinary differential equations that are specific for natural convection induced by a line heat source, proper values of N and n, as well as appropriate boundary conditions, must be specified. The energy convected across any horizontal plane in the plume is q c u( T T ) dy p (6.185) which is identical to the intensity of the line heat source and it can be expressed in terms of similarity il it variables 5n g N 3 4 q 4c (6.186) pn x Fd 4 84
Since the line heat source is the only source of heating, the above should be independent from x which can be true only if n is equal to -3/5. Therefore, the ordinary differential equations for natural convection over a line heat source respectively become: 1 4 F FF ( F ) 0 (6.187) 5 5 1 Pr F 0 5 (6.188) which are subject to the following boundary conditions: F( ) F( ) 0, and ( ) 1 at 0 (6.189) F( ) ( ) 0 at (6.190) 85
Equations (6.187) (6.190) can be solved numerically and the results can be shown in Fig. 6.16. In contrast to Fig. 6.3 that shows the velocity at the wall is zero, the velocity at the centerline is at its maximum. With known intensity of the line heat source, the constant N in eq. (6.18) can be obtained from eq. (6.186): where N 1/5 4 q 4 4 64 g c (6.191) pi I F( ) ( ) d (6.19) is a function of the Prandtl number. The values of I calculated at Pr = 0.7, 1.0, 6.7, and 10.0 by Gebhart et al. are 1.45, 1.053, 0.407 and 0.38, respectively. 86
Figure 6.16 Velocity and temperature distributions of natural convection from a line heat source (a) velocity, (b) temperature (Gebhart et al., 1970; reproduced with permission from Elsevier). 87
Example 6.4 Analyze natural convection over a point heat source to obtain the appropriate ordinary differential equations and the corresponding boundary conditions. 88
Solution: The natural convection in this case will be axisymmetric and the governing equations must be given in a cylindrical coordinate system. Equations (6.136) (6.138) are applicable for this case with the following boundary conditions: u T 0, v0 at r 0 (6.193) r r u v 0, T T (6.194) at r Similar to the case of line heat source, the centerline temperature can also be expressed as a power function of x as indicated by eq. (6.18). Introducing Stokes stream function: 1 1 u, v r r r x (6.195) 89
and defining the following similarity variables: r 1/4 T T Gr x, xf( ), x T T 0 (6.196) the following ordinary differential equations are obtained: F F 1 n ( F) ( F 1) 0 (6.197) ( ) Pr( FnF) 0 The intensity of the point heat source is: n1 q c u ( T T ) rdr c Nx F p d 0 p 0 q and it must be independent from x. (6.198) (6.199) 90
Therefore, the value of n must be -1 and the ordinary differential equations for natural convection over a point heat source become: F F ( F 1) 0 (6.00) ( ) Pr( F) 0 (6.01) The velocity components in the x- and r- directions are related to the dimensionless variables by the following equations: 1/4 Gr F u x, v Gr F F x (6.0) x x 91
which can be substituted into eqs. (6.193) and (6.194) to obtain the following boundary conditions for eqs. (6.00) and (6.01): F 0, F0, 0 and 1 at 0 F 0 at (6.03) (6.04) 9