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HEATLINES IN LAMINAR BOUNDARY LAYER FLOW

INTRODUCTION-

Heatlines were proposed in 1983 by Kimura and Bejan (1983) as adequate tools for visualization and analysis of convection heat transfer. The first application of heatlines was in the visualization of laminar natural convection in a two dimensional rectangular enclosure heated from the side. The method has since been adopted and extended in several ways in the post-1984 literature. The purpose of heatlines is to provide a new way to visualize the heat transfer in fluid flow. The heatlines in heat transfer are analogous to streamlines in fluid flow. In 2-D fluid flow we defined stream function $$\psi$$ as -


 * $$U = {\partial \psi \over \partial y},\quad V = {\partial \psi \over \partial x} ~$$

& it satisfy, Mass continuity equation
 * $$\frac{\partial U}{\partial x} + \frac{\partial V}{\partial y}=0,~$$

The actual flow is locally parallel to the ? = constant line. Hence instead of using (u,v) directly we use ? = constant streamlines to analyse entire flow field and its characteristics.

In convection, the energy transfer through the flow field is both by thermal diffusion and enthalpy flow. For this we have defined a function H(x,y) known as heatfunction such that the net flow of energy (thermal diffusion  + enthalpy flow) across each H = constant line is zero.

For 2-D flow in Cartesian coordinate, the heat function, H, is given by-


 * $$ \frac{\partial H}{\partial y} = {\rho C_pU(T-T_{ref} )}- K \frac{\partial T}{\partial x} ~$$	        (1)    Net flow of energy in x direction.


 * $$ -\frac{\partial H}{\partial x} = {\rho C_pV(T-T_{ref} )}- K \frac{\partial T}{\partial y} ~$$               (2)    Net flow of energy in y direction.

Such that H(x, y) satisfy energy equation,
 * $$U\frac{\partial T}{\partial x} + V\frac{\partial T}{\partial y}=\alpha(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2})~$$                  (3)
 * $$\rho C_pU\frac{\partial T}{\partial x} + \rho C_pV\frac{\partial T}{\partial y}=K(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2})~$$


 * $$ \frac{\partial [\rho C_pU(T-T_{ref})- K \frac{\partial T}{\partial x}]}{\partial x}+\frac{\partial [\rho C_pV(T-T_{ref})- K \frac{\partial T}{\partial x}]}{\partial y} ~$$

Using equation (1) & (2)
 * $${\partial \over \partial x } \left({\partial H \over \partial y}\right) + {\partial \over \partial y } \left({-\partial H \over \partial x}\right)= 0 ~$$
 * $${\partial^2 H \over \partial x\, \partial y}-{\partial^2 H \over \partial y\, \partial x}= 0 ~$$

Hence Heatlines function H(x,y) satisfy Energy Equation. The reference temperature reference temperature is an arbitrary constant & is the lowest temperature in the boundary layer region.

According to the heat function definition any H = constant "heatline" is locally parallel to the direction of net energy flow through the convection field. The family of H = constant heatline depicts the actual path of the flow of energy through the entire field.

Note that, if the fluid flow subsides (u = v =0), the "heatlines" become identical to the "heat flux lines" used in the study of conduction phenomena. Therefore, as a heat transfer visualization technique the use of heatlines is the convection counterpart of a standard technique (heat flux lines) used in conduction.

METHODOLOGY:

To find heatfunction H(x, y) using equation 1 & 2 over a flat rectangular plate or wall

Case 1- Isothermal Cold wall

Here free stream is warmer and wall is cold i.e. : $$T_{\infty} >T_{o}$$

Taking $$T_{ref} =T_{o}$$     (always smaller one)

Using Blasius boundary layer solution and dimensionless variables-


 * $$x^*=\frac{x}{L}\ $$,    $$y^*=\frac{y}{L.Re_{L}^\frac{1}{2}}\ $$ ,   $$\theta=\frac{T-T_o}{T_{\infty}-T_o}\ $$ ,    $$H^*=\frac{H}{\rho C_p.U_{\infty}.(T_{\infty}-T_o).L.Re_{L}^{-\frac{1}{2}}}\ $$ ,    $$\eta=\frac{y.Re_{x}^\frac{1}{2}}{x}\ $$  ,   $$Re_x=\frac{x.U_{\infty}}{\nu}\ $$ ,

$$U=U_{\infty}.f^'(\eta)$$,   $$V=\frac{1}{2}.\sqrt{\frac{\nu.U_{\infty}}{x}}.(\eta.f^'-f)\ $$,    $$\eta=y^*.x^{-\frac{1}{2}}$$ , $$\theta=\phi(\eta,Pr)$$

Inside the boundary layer, temperature gradient in x direction is neglected,
 * $$ \frac{\partial H}{\partial y} = {\rho C_pU(T-T_{ref} )}= {\rho C_p.U_{\infty}.f^'(T-T_{ref} )}  ~$$
 * $$ \frac{\partial H^*}{\partial y^*} = \frac{\rho C_p.U_{\infty}.(T_{\infty}-T_o ).L.Re_{L}^{-\frac{1}{2}}}{L.Re_{L}^{-\frac{1}{2}}}= {\rho C_p.U_{\infty}.f^'(T-T_{ref} )}~$$
 * $$ \frac{\partial H^*}{\partial y^*}=f^'.\theta $$                                                                                   (4)

Also  $$ -\frac{\partial H}{\partial x} = {\rho C_pV(T-T_{ref} )}- K \frac{\partial T}{\partial y} ~$$


 * $$ -\frac{\partial H^*[\frac{\rho C_p.U_{\infty}.(T_{\infty}-T_o ).L.Re_{L}^{-\frac{1}{2}}}{L}]}{\partial x^*} = \frac{1}{2}.\rho C_p.\sqrt{\frac{\nu.U_{\infty}}{x^*.L}}.(\eta.f^'-f)(T-T_o)\}- \frac {k}{L.Re_{L}^{-\frac{1}{2}}.\frac{\partial T}{\partial y^*}} ~$$
 * $$ \frac{\partial H^*}{\partial x^*} = \frac{1}{2}.x^\frac{1}{2}.(\eta.f^'-f)\theta - \frac{1}{Pr}.\frac{\partial\theta }{\partial y^*} ~$$             (5)

Hence, Heatfunction field depends on Prandtl Number Pr

NON DIMENSIONAL FORM OF HEATFUNCTION

Let dimensionless heatfunction,  $$ H^*.(x^*,y^* )= {x^{*\frac{1}{2}}.g[\eta(x^*,y^*)] }$$                                                    (6)

Now Differentiating  eqaution  (6) w.r.t   $$ Y^*	$$
 * $$ \frac{\partial H^*}{\partial y^*}= x^{*\frac{1}{2}}.\frac{\partial g}{\partial \eta}.\frac{\partial \eta}{\partial y^*}=x^{*\frac{1}{2}}.g^'.\frac{\partial (\frac{y^*}{x^{*\frac{1}{2}}})}{\partial y^*} $$
 * $$ \frac{\partial H^*}{\partial y^*}=g^'= g^'[\eta(x^*,y^* )] $$                                     (7)

Similarly differentiating equation (6)  w.r.t	  $$ x^* $$
 * $$ \frac{\partial H^*}{\partial x^*}=\frac{1}{2}.x^{*(-\frac{1}{2})}.g + x^{*\frac{1}{2}}.g^'.\frac{\partial \eta}{\partial x^*}$$
 * $$ \frac{\partial H^*}{\partial x^*}=\frac{1}{2}.x^{*(-\frac{1}{2})}.g + x^{*\frac{1}{2}}.g^'.\frac{-y^*.x^{*(-\frac{3}{2})}}{2}$$
 * $$ -\frac{\partial H^*}{\partial x^*}=\frac{1}{2\sqrt{x}}.(\eta.g^'-g) $$

Using equation 5
 * $$\frac{1}{2\sqrt{x}}.(\eta.g^'-g)=\frac{1}{2}.x^\frac{1}{2}.(\eta.f^'-f)\theta - \frac{1}{Pr}.\frac{\partial\theta }{\partial y^*}=\frac{1}{2}.x^\frac{1}{2}.(\eta.f^'-f)\theta - \frac{1}{Pr}.\frac{\partial\theta }{\partial \eta}.\frac{\partial\eta }{\partial y^*} $$
 * $$\eta.g^'-g=(\eta.f^'-f).\theta-2.\frac{\theta^'\sqrt{x^*}}{Pr}.\frac{1}{x^{*(\frac{1}{2})}}$$
 * $$\eta.g^'-g=(\eta.f^'-f).\theta-2.\frac{\theta^'}{Pr}$$                                                              (8)

Using equation (4),(7) & (8)
 * $$\eta.f^'.\theta-g=\eta.f^'.\theta-f.\theta-2.\frac{\theta^'}{Pr}$$
 * $$g=f.\theta+2.\frac{\theta^'}{Pr}$$                                                                                    (9)

Hence, using eequation (6) & (9)
 * $$H^*(x^*,\eta)=x^{*\frac{1}{2}}.[f(\eta).\theta(\eta)+2.\frac{\theta^'(\eta)}{Pr}]$$                                   (10)

$$ H^*$$ value increases with increase in $$ x^{*\frac{1}{2}} $$       (along the wall as cold wall absorbs energy).



Put 	?=0 	as	at 	y=0 & so  y*=0     (at  tip of  boudary layer )
 * $$H^*(x^*,0)=2.\frac{\theta^'(0)}{Pr}.x^{*\frac{1}{2}}$$                          (only function of  x*)                 (11)

so, ?'(?=0) is only function of Pr

Using	: $$\theta^'(0) = 0.332 Pr^{\frac{1}{3}}$$              ? Pr>0.5 &	: $$\theta^'(0) = 0.564 Pr^{\frac{1}{2}}$$              ? Pr<<0.5

Further, at the End of Wall  i.e.  x=L  or x*=1


 * $$H^*(1,0)=2.\frac{\theta^'(0)}{Pr}$$          ( 12)

This is the maximum value of H*(1,0), Dimensionless Heat Function



RELATION BETWEEN NUSSELT NUMBER AND HEATFUNCTION

Putting  ?'(0)    value in equation  (12)


 * $$H^*(1,0) = 0.664 Pr^{-\frac{2}{3}}$$              ? Pr>0.5             (13)(a)


 * $$H^*(1,0) = 0.128 Pr^{-\frac{1}{2}}$$              ? Pr<<0.5            (13)(b)

Nusselt number for laminar flow over a flat plate is given by-


 * $$\overline{\mathrm{Nu}}\ = 0.664\, \mathrm{Re}_L^{1/2}\, \mathrm{Pr}^{-1/3}, (\mathrm{Pr} > 0.5) $$


 * $$\overline{\mathrm{Nu}}\ = 0.128\, \mathrm{Re}_L^{1/2}\, \mathrm{Pr}^{-1/2}, (\mathrm{Pr} << 0.5) $$


 * $$\mathrm{Nu}\ =H^*(1,0).Pr.Re_L^{\frac{1}{2}}$$          (14)

using : $$H^*(x^*,\eta)=H^*(x^*,y^*)$$


 * $$H^*(1,0) = \frac{x=L,y=0}{\rho C_p.U_{\infty}.(T_{\infty}-T_o).L.Re_{L}^{-\frac{1}{2}}}$$

NUSSELT NUMBER    $$\overline{Nu}=\frac{H(L,0).Pr.Re_{L}^{\frac{1}{2}}}{\rho C_p.U_{\infty}.(T_{\infty}-T_o).L.Re_{L}^{-\frac{1}{2}}}=\frac{H(L,0).Pr.Re_{L}}{\rho C_p.U_{\infty}.(T_{\infty}-T_o).L}$$


 * $$\overline{Nu}=\frac{\overline{h}.L}{K}$$      &               $$\overline{h}=\frac{\ddot{Q}}{T_{\infty}-T_o} $$

$$\frac{\ddot{Q}.L}{K.(T_{\infty}-T_o)}=\frac{H(L,0)}{K.(T_{\infty}-T_o)}$$

Hence,   $$\ddot{Q}.L  =  H(L,0) $$                                                                           (15)

Total value of heat function at the end of wall is equal to total Heat-transfer through the wall. per unit length.


 * $$g{\eta}=f.\theta+2.\frac{\theta^'}{Pr}$$

(convection part)+ (conduction part)

Note:- 	At a particular  value of Y  : $$\eta \propto {\frac{1}{\sqrt{x}}}$$

So with increase in x value ? decreases (?x   ?      ?.?)

Fig. 2 shows the pattern of heatlines in the boundary layer region  for  Pr = 0.72 fluid such as air. The boundary layer region is drawn in terms of x* vs  ? with ? as large as 5 so that u approaches 99% of the free stream value U. The heatlines are drawn for constant values of $$H^*Pr^{\frac{2}{3}}$$  in accordance with equation (13a), i.e. for the purpose of showing that the : $$H^*Pr^{\frac{2}{3}} =0.664 $$      line passes through the trailing edge of the wall.

The same heatlines are shown in Figure 3, in the corresponding Cartesian frame of x*  vs  y*. The heatlines show the actual path of the energy absorbed by the wall. They are perpendicular to the wall because at y* = 0 the heat transfer is by pure conduction and the wall is an isotherm. The heatlines that ultimately cross the wall $$00.664 $$  originate from the flow region situated immediately upstream of the tip. Note that when the heatlines enter the boundary layer region they are tilted away from the cold wall. Their direction turns toward the wall, which they eventually cross. This change in direction occurs, approximately, at the half-point of the distance between the tip and the point where the heatline enters the wall. The heatlines that cross the wall are more crowded near the tip than farther downstream. heat flux q” that is proportional to $$x^{0.5}$$





Case 2- Isothermal Warm wall

Here free stream is cold and wall is hot i.e.  : $$T_{o}>T_{\infty}$$

Taking $$T_{ref} =T_{\infty}$$     (always smaller one)


 * $$H^*(x^*,y^*)=\frac{H}{\rho C_p.U_{\infty}.(T_{\infty}-T_o).L.Re_{L}^{-\frac{1}{2}}}\ $$

using similar analysis, healines function found to be-


 * $$H^*(x^*,y^*)=H^*(x^*,\eta)=x^{*-\frac{1}{2}}.[f(\eta).(\theta(\eta)-1)+2.\frac{\theta^'(\eta)}{Pr}]$$

H* value decreases with increase in x*. This is due to fact that now hot wall releases energy into fluid.

When the wall is colder than the fluid, the dimensionless heat function H* increases in the x direction (Fig. 3) because the wall absorbs energy. Figures 4 and 5 show the pattern of heatlines near the hot wall, when the Prandtl number is 0.72. These figures must be compared with Figs. 2 and 3 to see the difference between a wall that releases heat and one that absorbs heat. Figure 4 shows the heatlines in the scaled boundary layer region 0 < ? < 5, while Fig.5 shows the same pattern in the corresponding Cartesian frame. The heatlines point in the y* direction as they emerge from the wall ; later, they are swept downstream by the flow. Their higher density near the tip indicates higher heat fluxes.

CONCLUSIONS

(a) The path of convective heat transfer from a hot fluid to a cold wall differs totally from the path followed by convection from a hot wall to a cold fluid. Compare, for example, Fig. 3 with Fig. 5.

(b) The energy that is absorbed eventually by a cold wall originates from the fluid that approaches the leading edge of wall. The heatlines then turn and become perpendicular as they enter the wall. Examine, for example, Fig. 3.