Chapter-6---Natural-convection_2021_Heat-Transfer-Engineering

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6.3 Boundary layer equations and nondimensional numbers 183

Consider the boundary layer forms of these equations (Eqs. 6.22 and 6.24)

	∂u	∂u	∂2 u		(6.44)
u ∂x + v ∂y = ν			∂y2 + gβ(T − T∞ )		(6.44)
u ∂x + v ∂y = ν			∂y2 + gβ(T − T∞ )
		∂T	∂T	∂2 T	(6.45)
		u ∂x + v	∂y = α	∂y2	(6.45)
		u ∂x + v	∂y = α	∂y2

Integrating Eq. (6.44) from 0 to δ in y and Eq. (6.45) from 0 to δT in y, with the understanding that δ = δT, we have

∂u

∂u dy =

∂

∫u

dy +

∫ v

∫ν

u2 dy + ∫ gβ(T − T∞ ) dy

(6.46)

and

∂x

∂y

∫u

∂T dy + ∫ v

∂T dy = ∫α

∂

(6.47)

∂x

∂y

Eq. 6.46 can be written as

∫

∂u2 dy − ∫u

∂u

dy + ∫

∂uv dy − ∫u ∂v dy

∂x

∂y

(6.48)

ν ∂u

+ ∫ gβ(T − T∞ )dy

∂y

Invoking the continuity equation for the sum of the second and fourth terms on the left-hand side and the Newton’s law of viscosity for the first term on the righthand side of the equation, we can rewrite Eq. (6.48) as

d	δ	τ w	δ
	∫(u2 − uu∞ ) dy = −		+ ∫ gβ(T − T∞ ) dy	(6.49)
dx
	0	ρ 0

In arriving at Eq. (6.49), the Leibnitz rule has been used to interchange the order of differential and integral operators and ∂∂x has been replaced by dxd as y dependence

goes, upon integrating (u2 – uu∞) with respect to y from 0 to δ. The velocity gradient
∂u	is also 0 at y = δ. τw is the wall shear stress, given by µ(∂u/∂y)y=0. The integral
∂y
energy equation is exactly same as the one derived in chapter 5 and is given by
		d	δ	∂ T
		d	∫ u(T − T∞ ) dy = −α	∂ T		(6.50)
		dx	∫ u(T − T∞ ) dy = −α	∂ y		(6.50)
		dx	0		y = 0
			0		y = 0

Once these equations are in place, the integral method progresses with the usual assumptions of polynomial profiles for temperature and velocity. We have a special difficul ty here, as unlike forced convection, we do not have a reference velocity. Equation 6.49 can be simplified with the understanding that u subscript infinity is 0 and this then becomes the final form of the momentum integral equation for laminar natural convection.

184 CHAPTER 6 Natural convection

Assuming a linear profile for temperature (the simplest possible) of the form

(T − T )		y
	∞	= a + b		(6.51)
(T	− T )		δ
w	∞

We can determine a and b from the boundary conditions

y = 0,	T = Tw	(6.52)
y = δ ,	T = T∞	(6.53)

On applying these two boundary conditions, the “trial” temperature profile turns out to be

(T − T )		y		(6.54)
	∞	= 1 −		(6.54)
(T	− T )		δ
w	∞

From an earlier discussion on the nature of the velocity profile, it is clear that at

y = 0, u = 0 and at y = δ, u = 0 with u = umax somewhere in between with umax itself being as yet unknown. However, a quick order of magnitude analysis will help us get

a ballpark of umax. Consider the x-momentum equation (Eq. 6.44), and let the scale for u uref. Consider the two terms on the right-hand side of Eq. (6.44). The respective scales are

ν	uref	gβ(Tw − T∞ )			(6.55)
	uref
	δ 2
uref			δ 2	gβ(Tw − T∞ )	(6.56)
			ν

We may replace the term with an equality by introducing a constant “C” with the understanding that “C” will be of the order of unity.

uref	= C	δ 2	gβ(Tw − T∞ )	(6.57)
		ν

Now we can get back to the “trial” profile for velocity. In view of the tricky boundary conditions of u = 0 at y = 0 as well as at y = δ, we assume a cubic profile for velocity as follows.

u		y			y 2		y	3
		=		+ d		+ e			(6.58)
u			δ
	ref				δ		δ

The following boundary conditions are applicable

At, y = 0, u = 0

(6.59)

6.3 Boundary layer equations and nondimensional numbers 185

This is already satisfied by Eq. (6.58)

At y = δ , u = 0

1 + d + e = 0

At y = δ , ∂∂ uy= 0

1 + 2δd + 3δe = 0

1 + 2d = −3e

From Eqs. 6.61 and 6.64, we have e =1 and d = –2.The velocity profile is given by

y 2

y 3

− 2

ref

y 2

1 −

(6.60)

(6.61)

(6.62)

(6.63)

(6.64)

(6.65)

Now we will have to examine what happens if we apply the profile to the momentum equation at the wall, i.e., y = 0 for any x.

At y = 0, we know that u = 0, v = 0, and so the momentum equation reduces to

∂2 u	gβ(T	− T	)	(6.66)
∂ y2 = −	w	∞
	ν
	ν

Differentiating Eq. (6.65) twice and substituting in Eq. (6.66) (at y = 0),

−	4uref	= −		gβ(T	− T	)	(6.67)
				w	∞		(6.67)
	δ 2			ν
uref =		δ 2	gβ(Tw		− T∞ )		(6.68)
		4ν

A quick sanity check is in order. An estimate of uref from Eq. (6.57) is very similar to the one obtained in Eq. (6.68), with C = 14. This confirms that we committed

no “grave error” in assuming the cubic profile and working out the constants of the profile, as discussed above.

We now have a choice to proceed with Eqs. 6.57 or 6.68 for uref. However, if we proceed with Eq. (6.68), we are left with only one unknown which is δ and we have two Eqs. 6.49 and 6.50. This leads to a mathematical “uneasiness” (technically an overdetermined system). and so we stick with Eq. (6.57) so that we have two unknowns C and δ and two integral equations!

186 CHAPTER 6 Natural convection

Substituting for the velocity profile in the momentum integral equation together with form of uref, we get the following

y 2

y 4

uref

∫uref .

−

dy = −ν

+ gβ(Tw − T∞ )∫

−

On simplifying Eq. (6.69), we get

y 2

y 4

gβ(Tw − T∞ )

∫δ

−

dy =

− C

(6.69)

(6.70)

We now integrate the term within the integrand on the left hand side and simplify as follows.

gβ(Tw

− T∞ )

∫δ

−

(6.71)

− C

gβ(Tw

− T∞ )

−

5δ

6δ

7δ

3δ

(6.72)

− C

On simplifying, after a bit of algebra to get rid of the delta on the right hand side of the above equation, we get

1 C2	gβ(T	− T )δ 3	dδ	=	1	− C

21 ν 2			dx		2
	w	∞

Upon integrating Eq. (6.73) and substituting for δ = 0 at x = 0, we get

	4		1	C2				−1
δ		= 84		− C			gβ(Tw − T∞ )	x
δ		= 84		− C	ν	2	gβ(Tw − T∞ )	x
			2		ν

(6.73)

(6.74)

Please note that Eq. (6.74) confirms that the velocity boundary layer thickness is

proportional to the buoyancy term raised to	−	1	and x1/4. This revalidates the cou-
		4

pling between the momentum and thermal boundary layers. We can now substitute for δ in the integral energy equation and use this equation to determine the constant C. Upon doing this we can evaluate

qw = −k	∂T	=	−k(Tw	− T∞ )	(6.75)

	∂y		δ
		y=0

with δ having been determined qW, and hence hx and Nux can be determined. One can intuit now that the presence of ν in the expression for δ and that of α in the integral equation confirms that the Prandtl number Pr given by ν/α must appear in some

6.3 Boundary layer equations and nondimensional numbers 187

avatar in C and should eventually get into the expression for Nux. Let us now complete the remaining part of this fascinating journey of the integral method to solve the boundary layer equations.

Consider the integral energy equation given by Eq. (6.50)

∂T

∫u(T − T∞ ) dy = −α

∂y

y = 0

Substituting for the velocity and temperature profiles, we have

1 Cgβ(T

− T )

∫

∞

−

(Tw − T∞ )d

= α (Tw − T∞ )

Cgβ(T − T )

y 3

∞

∫

1 −

(Tw − T∞ )d

∫

1 −

= ∫ x(1 − x)

where

x =

1 y

∫

1 −

= ∫ x(1 −

3x + 3x

− x

)dx

0 δ

On integrating and simplifying

∫

−

Consequent upon the evaluation of the integral we have

Cgβ(T − T ) d δ 3

∞

dx 20

Cgβ(T − T )

d δ 3

= α

∞

dx 20

3Cgβ(T − T ) d δ 4

= 20α

∞

δ 4

20αν

3Cgβ(Tw

− T∞ )

(6.76)

(6.77)

(6.78)

(6.79)

(6.80)

(6.81)

(6.82)

(6.83)

(6.84)

(6.85)

188 CHAPTER 6 Natural convection

Integrating Eq. (6.85) we get

δ 4 =		80αν			x + A
δ 4 =	3Cgβ(T		− T	)	x + A
	3Cgβ(T		− T	)
		w	∞
	At x = 0, δ = 0
		A = 0
		80α xν				1/ 4
δ =
δ =
		3Cgβ(Tw − T∞ )

(6.86)

(6.87)

(6.88)

(6.89)

Eqs. 6.74 and 6.89 are both expressions for the boundary layer thickness. Equating the two, we can solve for C as

84	1	− C		C2			gβ(T		− T )	−1
						2
	2				ν
84	1	− C		1		=	80 1


	2		C					3 Pr
C =		Pr
					20
	2	Pr +			63

Substituting in Eq. (6.74)

	80αxν
x =

	3Cgβ(Tw − T∞ )

	80xν	2	2		+	20
	80xν		2	Pr	+
δ 4 =						63
						63
	3[Pr gβ					T]

On simplifying, we get

δx = 2.7[Pr + 0.317]1/4 .Pr−1/2Gr−1/4

The heat transfer at the wall is given by

qw = − k	∂ T	= k	(Tw − T∞ )
qw = − k	∂ y	= k	δ
		y=0	Nux .k
qw = hx .(Tw − Tδ ) =			Nux .k	(Tw − T∞ )
qw = hx .(Tw − Tδ ) =			x	(Tw − T∞ )
			x

(6.90)

(6.91)

(6.92)

(6.93)

(6.94)

6.4 Empirical correlations for natural convection 189

On equating 6.93 and 6.94, we get the expression for the local Nusselt number, Nux as

Nux	=		x			(6.95)
Nux	=					(6.95)
		δ
Substituting for δ from Eq. (6.92)
(1/4 )				Pr	(1/4 )	(6.96)
Nux = 0.37(Rax )
Nux = 0.37(Rax )
		0.317 + Pr

Following the developments outlined in chapter 5, we can relate the average Nusselt number NuL from x = 0 to x = L, to the local Nusselt number, Nux, as

	4	(6.97)
NuL =	4	(6.97)
NuL =	3	NuL
	3

(The derivation of the above expression is given as an exercise in problem 6.5 at the end of this chapter).

In Eq. 6.97 NuL is the local Nusselt number at x = L.

	0.25	Pr	0.25
NuL = 0.49(RaL )			(6.98)


		0.317 + Pr

If we employ a quadratic profile for temperature instead of the linear temperature profile and a cubic profile for velocity, as assumed in the above development, we get the following expression for average Nusselt number as

	0.25	Pr	0.25
NuL = 0.678(RaL )			(6.99)


		0.952 + Pr

For gases, Eq. (6.98) is good enough under 20% error, though Eq. (6.99) is more accurate. The derivation of Eq. (6.99) is given as an exercise at the end of the chapter.

6.4 Empirical correlations for natural convection

The development in the previous section has historical significance, as it exemplifies our early efforts to conquer the problem of natural convection.

Churchill and Chu looked at all the experimental data on natural convection over a vertical flat plate up to that point in time and, with guidance about the form of the Nusselt number correlation from the integral solution, proposed the now widely used Churchill-Chu correlation (Churchill and Chu, 1975a) given below

		= 0.68 + 0.67Ra1/4	1 +	0.492	9/16 − 4 /9
Nu	L	= 0.68 + 0.67Ra1/4	1 +	0.492	9/16 − 4 /9	(6.100)
	L	L
				Pr

This works for RaL up to 1012 and is a very useful engineering result. Please see Lienhard and Lienhard (2020) for a fuller discussion of the development of a heat transfer correlation for natural convection on a vertical surface. All properties in

190CHAPTER 6 Natural convection

the above correlation are to be evaluated at a mean temperature Tmean = (Tw + T∞)/2 except for β. For gases, β = 1/T and the temperature T has to be T∞. For liquids, β is available in charts or tables in books and online resources. For problems involving a large ∆T, we might face issues like errors introduced due to the Boussinesq approximation and also those introduced due to the assumptions of constant properties.

However, if (TW – T∞) is, say, 40 or 50 °C, and we use the mean temperature of Tmean, we should get results that are accurate to within a few percentage points with respect to a full-blown variable property solution. In this day and age, we can actually solve the variable property case on an open-source or commercial software and check the error involved in the constant property assumption, all by ourselves. For a flat plate, RaL ≥ 109 is taken to be the criterion for transition to turbulent natural convection.

Natural convection from vertical cylinders

If one encounters a tall vertical cylinder such that Hδ 1, where δ is the largest

boundary layer thickness in the cylinder and H is the height of the cylinder, then the correlation for the vertical flat plate holds good. However, if this assumption does not hold, either because of a thick boundary layer or a short cylinder, correlations for the enhanced heat transfer due to curvature need to be considered. Please refer to Lienhard and Lienhard (2020) for more information on this case.

Natural convection from a heated sphere

For Pr > 0.7 and RaD < 1011, Churchill proposed the following correlation (Churchill, 1983).

NuD = 2	+		0.589Ra0.25
					D
		1 +		0.469	9	/16	4 /9	(6.101)
		1 +		0.469	9	/16		(6.101)
				Pr
				Pr

Natural convection from horizontal cylinders

Churchill and Chu have presented correlations by looking at existing data (Churchill and Chu, 1975b). The correlation is given below.

			0.518Ra1/4D
NuD = 0.36	+		0.518Ra1/4D				(6.102)
NuD = 0.36	+	1 +		0.559	9/16	4 /9	(6.102)
		1 +		0.559	9/16
				Pr
				Pr

This is valid for RaD ≤ 109. For RaD > 109, the following correlation, applicable for turbulent flow, is useful.

1/6

RaD

(6.103)

NuD = 0.60

+ 0.387

+ {

0.559

}

9/16

16/9

6.4 Empirical correlations for natural convection 191

Please note that the key difference between Eqs. 6.102 and 6.103 is the Rayleigh number exponents. While the Nusselt number scales as Ra1/4D for laminar flow, the exponent in the case of turbulent flow is 0.33, and the usually accepted cutoff for transition to turbulence in natural convection is that RaL or RaD ≥ 109, as the case may be!

Natural convection from other geometries

Correlation tables from handbooks or online resources may be referred to for correlations involving natural convection from enclosures, heated horizontal plate facing upward/downward, and for other geometries of specific interest. Properties of air are given in Table 6.3.

Example 6.1: An aluminum plate having ρ = 2700 kg/m3, k = 205 W/mK, Cp= 900 J/kgK is heated to a temperature of 400 K. At this point in time (t = 0), the heating is stopped and the plate begins to cool in still air at 310K. The plate is 50 cm tall and 30 cm wide and has a thickness of 10 mm. The surface of the plate has an emissivity of 0.25, which is independent of the temperature.

(a)Develop an expression for the rate of change of the plate temperature assuming spatial isothermality of the plate, neglecting surface radiation.

(b)Rework the expression obtained in (a) if radiation is considered (Temperature of the surroundings for radiation can be considered to be 37 °C).

(c)Determine initial rate of cooling when plate temperature is 400K, when (i) radiation is neglected and (ii) radiation is considered.

(d)Comment on the contribution of radiation in this problem.

(e)Is the assumption of spatial isothermality valid (i) without consideration of radiation and (ii) with consideration of radiation?

Solution Given data

Density ρ = 2700 kg/m3; thermal conductivity k = 205 W/mK; specific heat Cp = 900 J/kgK; initial temperature of the plate (TS) = 400 K; ambient temperature (T∞) = 310 K; surface emissivity (ε) = 0.25; dimensions of the plate = 0.5 × 0.3 × 0.01 (all in m).

(a) Neglecting radiation effects, from energy balance,

−ρC

= h.(2 A ).(T − T

)

∞

= −

h.(2 As ).(T − T∞ ) dT

= −

h.(2 As ).(Ts − T∞ )

ρCpV

t=0

ρCpV

(b) Considering the radiation, from energy balance,

− ρC

= h(2 A )(T − T

)+ σ (2 A )(T 4

− T 4 )

∞

(2 As )

− T∞ )+ σ (T

= −

ρCpV

h(Tw

− T∞ )

(2 As )

= −

h(Tw

− T∞ )+ σ (Ts

− T∞ )

t=0

ρCpV

192 CHAPTER 6 Natural convection

ν = 2.073 × 10–5 m2/s, α = 2.931 × 10 –5 m2/s, k = 0.02984 W/mK, Pr = 0.707.

β =	1	= 3.22 × 10− 3K − 1,

	T
	∞
RaL =	gβ(T		− T )L3
		s	∞
		ν α

RaL =	9.81 × 3.22 × 10− 3 × (400 − 310) × 0.53 × 1010
			2.073 × 2.931
RaL = 5.85 × 108

From Eq. (6.100)

		= 0.68 + 0.67Ra1/4	1 +	0.492	9/16 − 4 /9
Nu	L	= 0.68 + 0.67Ra1/4	1 +	0.492	9/16 − 4 /9
	L	L
				Pr

1/4

0.492

9 /16 − 4 /9

NuL = 0.68 + 0.67 (5.85 10

)

0.707

= 80.65

NuL

= 80.65

0.02984

0.5

hc = 4.81W /m2 K.

(i) Initial rate of cooling of the plate when neglecting radiation

∂ T

=− 0.0356 K /s

∂t t =0

(ii)Initial rate of cooling of the plate when considering radiation

∂ T	= −	2 × 103		× 664.87

∂ t	t =0		2700 × 900 ×10

∂ T	= − 0.055			K /s

∂ t
	t =0

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