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		CHAPTER

Forced convection		5
Forced convection

5.1 Introduction

In the previous chapter, the basic ideas of convection heat transfer and heat transfer coefficient were introduced, along with the full set of governing equations of fluid flow and heat transfer. The physical significance of each of the terms in these equations was elucidated. In this chapter, we first introduce the boundary layer theory through the order of magnitude analysis, followed by nondimensionalization of the governing equations. Next, the integral method is developed for flow over a flat plate. Further, correlations for cylinder and sphere are presented. These are followed by a presentation of forced convection in tubes and ducts. The analytical solution is presented for a simple laminar case, and the chapter ends with a quick introduction to turbulent flow and correlations for these.

5.2 Approximation using order of magnitude analysis

In this section, we look at Prandtl’s famous order of magnitude analysis leading to the boundary layer theory, which is one of the cornerstones of fluid mechanics and convective heat transfer. Even before this analysis, Prandtl argued that near the wall, the product of (∂u / ∂y) cannot be neglected even though µ is very small, as ∂u/∂y will be very high near the wall so that the product of the two is not a quantity that can be neglected. This announced the birth of fluid mechanics.

The governing equations for two-dimensional, laminar, incompressible flow were derived in the previous chapter. For the case of steady flow, the governing equations are

Continuity equation

∂u	+	∂v	= 0	(5.1)
∂x		∂y

x-momentum equation

∂u

+ v

∂u

∂2 u

−

∂P

+ X

∂x

= µ

∂x2

∂y2

∂x

∂y

y-momentum equation

∂v

+ v

∂v

∂2 v

−

∂P

+ Y

ρ u

∂x

= µ

∂x2

∂y2

∂y

Heat Transfer Engineering. http://dx.doi.org/10.1016/B978-0-12-818503-2.00005-8

(5.2)

(5.3)

121

122 CHAPTER 5 Forced convection

FIGURE 5.1

Boundary layer over a flat plate.

We now perform an order of magnitude analysis of these equations by introducing the concept of the boundary layer. We neglect body forces X and Y in the analysis. Before we begin, we need to understand what exactly a boundary layer is. Let us take an example of a flow parallel over a flat plate, as shown in Fig. 5.1.

Over the surface of the plate, at any section, say AA, starting from the leading edge (i.e., at x = 0), we can intuit that the velocity increases from u = 0 at y = 0 to u → u∞ , as y → ∞. If we take a cutoff of 0.99 u∞ , and see the value of y at that point, for a particular x, that value of y is termed as δ, the boundary layer thickness, which should be a function of x, so that without any loss of generality, we denote it as δ (x). The locus of all these δ ' s at various values of x is known as the boundary layer. The region outside the boundary layer is free from the effects of the viscous shear that is present at the wall. The region within the boundary layer is thus the region where a free stream moving at u∞ is “braked” by the presence of the wall. The key quantity that determines the amount of braking and the pressure loss and attendant required pumping power thereof is the dynamic viscosity of the fluid, From the preceding discussion, it is also clear that some sort of simplification to the governing equations should be possible if we can conjecture that δ /x << 1. Stated explicitly, if the plate is, say, 1 m long, δ at x = 1 m is of the order of a few mm, under this assumption. This is a game-changing assumption, and we will see how this results in dramatic simplification of the governing equations. Let us perform an order of magnitude analysis. The scales for length x, y, and velocity u are L, δ, and u∞ , respectively. From the scales of x and y, we have

δ << L

(5.4)

All these scales ensure that x/L, y/δ , u/u∞ vary from 0 to 1. Substituting for the above scales in the continuity equation (Eq. (5.1)) leads to the condition that u∞ /L and v / δ have to be comparable.

From the above, we can infer that the scale for the velocity in the y direction is

v ≈	u∞δ	(5.5)
	L

Substituting for slenderness assumption δ << L (Eq. (5.4)) in Eq. (5.5), we have

5.2 Approximation using order of magnitude analysis 123

v << u∞

(5.6)

Hence, the v velocity is much less compared to the u velocity. This is the first key result of the scaling or order of magnitude analysis. Furthermore, as the length scales of x and y are L and δ respectively, one can infer that

∂u

(5.7)

∂y

∂x

∂v

(5.8)

∂y

∂x

Moreover, as u

v, this leads to

∂u

∂v

(5.9)

∂y

Looking at the scales for the x-momentum equation term by term, we have

∞

−

∂P

ρ u∞

, u∞

= µ

∂x

(5.10)

Let us look at the right-hand side. Since scales for pressure are as yet not defined by us, we retain the terms as they are.

We know that δ << L, and this leads to

∂2 u >> ∂2 u ∂y2 ∂x2

Hence, the x-momentum equation can be simplified as

	∂u	+ v	∂u	= µ	∂2 u	−	∂P
ρ u	∂x				∂y2		∂x
			∂y

(5.11)

(5.12)

Now, consider the y-momentum equation (Eq. 5.3). The order of the equation is u∞. u∞δ /L2. The order of the x-momentum equation is u∞ .u∞ /L. Therefore, the y-momentum equation has an order of δ /L times the x-momentum equation. However, δ /L << 1 and, in view of this, the y–momentum equation can be neglected in relation to the x-momentum equation. We now have a key piece of information about the pressure, which is as follows.

∂ p	= 0	(5.13)
∂y

Let us now apply the x-momentum equation to the free stream, where

u = u∞		(5.14)
∂u	= 0	(5.15)
∂x


124	CHAPTER 5 Forced convection
				v = 0				(5.16)
			∂2 u			= 0		(5.17)
			∂y2
	In view of all this,
			∂ p	=	dp		= 0.	(5.18)
			∂x		dx
	Now, at any vertical section, say, AA along x (see Fig. 5.1), if we traverse an in-
	strument to measure the variation of pressure, since ∂ p / ∂y = 0 at all points on section
	AA, as we traverse from the free stream to the inside of the boundary layer, the pres-
	sure must be the same. Because dp/dx = 0 at the outside boundary layer with p = p∞ ,
	it stands to reason that at the outer edge of the boundary layer too, p = p∞, and by
	extending the above arguments, ∂ p/ ∂x, dp/dx = 0 within the boundary layer too. So,
	by a clever use of the information that ∂ p/∂y is zero within the boundary layer and
	that p∞ is a constant outside the boundary layer, we have achieved one more major
	simplification. In fact, this is equally profound as now pressure is no longer a vari-
	able in the boundary layer equations. Hence dp/dx = 0 within the boundary layer too.
	Please note this in no way means that p = 0. This simplification simply allows us to
	ignore the pressure gradient term in an analysis of boundary flow over a flat plate. If,
	on the other hand, the free stream velocity u∞ is not a constant, but a function of x,
	as in the case of flow over variable area geometries like a wedge, then dp/dx within
	the boundary can still be found from the dp/dx at the free stream, which itself can
	be determined using the Bernouli’s equation in the free stream. In short, the pressure
	distribution in the free stream is imposed within the boundary layer. This fact is used
	for rapid calculations of net pressure forces in aerodynamic flows.
	The energy equation was given in Chapter 4 as eqn. 4.39. For the case of steady
	flow with no heat generation, it reduces to
		∂T	+	∂T
	ρcp	∂x	+	∂y		= k 2T + µφ		(5.19)
		∂x		∂y

With the understanding that δT is the thermal boundary layer thickness (see

Fig. 5.1) is much less than L, we have

∂T	>>	∂T	(5.20)
∂y		∂x

The thermal boundary layer thickness, δT , is defined as the value of y, where

T = T∞ + 0.01(Tw − T∞ )

With regard to the viscous dissipation term,

∂u	>>	∂v	(5.21)
∂y		∂x

∂u	>>	∂u	(5.22)
∂y		∂x

∂u	>>	∂v	(5.23)
∂y		∂y

5.3 Nondimensionalization of the governing equations

125

In view of this, the viscous dissipation term becomes (∂u/ ∂y)2 .

Following the order of magnitude analysis, the governing equations are simpli-

fied as follows for the case of negligible viscous dissipation:

Continuity equation

∂u

∂v

= 0

(5.24)

∂y

∂x

Momentum equation

∂u

∂2 u

∂x

+ v ∂y

ν ∂y2

(5.25)

Energy equation

ρC

∂T

+ v

∂T

∂2 T

∂u 2

(5.26)

∂x

∂y

= k

∂y2

+ µ

∂y

The above equations are referred to as the boundary layer equations. Eqs. (5.24– 5.26) are three equations in three unknowns, u, v, and T , and hence they satisfy closure. Mathematically, Eqs. (5.25) and (5.26) are parabolic in nature and are easier to solve compared to the original N-S equations, and the equation of energy (without simplification).

5.3 Nondimensionalization of the governing equations

We now move on to nondimensionalizing or normalizing the boundary layer equations. To make matters simple, we drop the viscous dissipation term in Eq. 5.26, which becomes significant only for high speed flows. The motivation for this nondimensionalization is to extract the pertinent dimensionless parameters that govern the fluid flow and heat transfer. Pertinent dimensionless parameters can also be obtained using the Buckingham’s Pi theorem.

We now normalize (or nondimensionalize) the governing equations as follows. Let


u+ =			u	(5.27)
u+ =			u	(5.27)
			u
			∞
v+ =			v	(5.28)
v+ =	u			(5.28)
	u
			∞
x+ =			x	(5.29)
x+ =			L	(5.29)
			L
y+ =		y		(5.30)
			L

φ =	(T − T∞ )			(5.31)
φ =	(T	− T	)	(5.31)
	w	∞

126CHAPTER 5 Forced convection

and

p+ =

ρu

∞

Using the above, the continuity equation becomes

∂u+

∂v+

= 0

∂x+

∂y+

The momentum equation becomes

∂u+

ρu2

dp+

∂2 u+

∞

+ +

∞

+ = −

∞

∂x

∂y

L dx

Upon simplification, we get

∂u+

+ v+

∂u+

= −

dp+

∂2 u+

∂x+

∂y+

dx+

u L

∂y+2

∞

where

∞

The x-momentum equation turns out to be

∂u+

+ v+

∂u+

= −

dp+

∂2 u+

∂x+

∂y+

dx+

ReL

∂y+2

Similarly, the boundary layer energy equation becomes

∂φ

+ v+

∂φ

∂2 φ

∂x+

∂y+

ρc

∂y+2

u L

∞

αν

ρc u L

νu L

u∞ L

Re Pr

∞

The energy equation becomes

∂φ

+ v+

∂φ

∂2 φ

∂x+

∂y+

∂y+2

ReL Pr

In the light of the above results, the following becomes evident:

u+ = f	x+ , y+ , Re ,	dp+

			+
1	L	dx

The shear stress at the wall (τ w ) is given by

	∂u
τ w = µ
	∂y y=0

(5.32)

(5.33)

(5.34)

(5.35)

(5.36)

(5.37)

(5.38)

(5.39)

(5.40)

(5.41)

(5.42)

5.3 Nondimensionalization of the governing equations 127

∂u	=	u		∂u+		(5.43)
	=		∞			(5.43)
∂y	y=0	L		∂y+	+	=0
	y=0				y	=0

The local skin friction coefficient is given by

τ w

f ,x

1 ρu∞2

cf ,x =

µu

∂u+

∞

∂y+

ρu∞

cf ,x =

∂u+

∂y

ReL

y+ =0

(5.44)

(5.45)

(5.46)

From the functional form of u+ given in Eq. (5.41)

c		= f		x+ , Re ,	dp+		(5.47)

	f ,x		2			+
				L	dx

For a flat plate	dp+	= 0
	dx+
		cf ,x = f2 (x+ , ReL )
			(5.48)

dp+

For any other given geometry dx+ can be obtained from the Bernouli’s equation.

cf ,x =	2	f2 (x+ , ReL )	(5.49)

	ReL

When we take an average from x = 0 to x = L to obtain the average shear stress and the average skin friction coefficient, the x+ dependence also vanishes!

τ w = f3 (ReL ) alone

This is indeed a revelation after so much toil.

Similarly, if we consider the dimensionless form of the energy equation

		dp+
φ =	f4 x+ , y+ , ReL , Pr,
		dx	+

For a flat plate,

dp+

= 0

dx+

h (T

− T

) = −k

∂T

(At the wall, q

= q

)

∂y

∞

y=0

conv

cond

h (Tw

) = −

(T − T

)

∂φ

− T

∞

∂y+ y+ =0

(5.50)

(5.51)

(5.52)

(5.53)

128 CHAPTER 5 Forced convection

h = − kLf

Nux kf	= −	kf
L		L

∂φ

∂ +y y+

∂φ

∂ +y y+

(5.54)

(5.55)

where Nux is the local Nusselt number (the dimensionless heat transfer coefficient) given by hL/kf . The local Nusselt number can also be defined based on x as the length scale. Here, it is based on L for convenience.

Nux = f5 (x+ , ReL , Pr)

(5.56)

The Nusselt number averaged over the length L, known as the mean or average Nusselt number is given by

	= f6 (ReL , Pr)
NuL	= f6 (ReL , Pr)	(5.57)

The above development is profound as it has helped us obtain key dimensionless parameters in convective fluid flow and heat transfer. Even if one is an experimentalist and wants to stay away from computing to solve the Navier-Stokes equations and the energy equation, the nondimensionalization procedure reveals to us the contours of how the skin friction coefficient and Nusselt number correlations will turn out to be.

5.4 Approximate solution to the boundary layer equations

Consider the momentum equation in the x-direction inside the boundary layer (Eq. 5.25)

u	∂u	+ v	∂u	=	∂2 u
	∂x		∂y		ρ ∂y2

Integrating the above equation from 0 to δ in y, we have

∫ δ u	∂udy + ∫ δ v		∂udy = ∫ δ ∂2 u2 dy			(5.58)
0	∂x	0	∂y	0 ρ ∂y
δ	∂u	δ	∂u	µ ∂u		(5.59)

∫0 u	∂xdy + ∫0 v		∂ydy = −	ρ ∂y
					y=0

Eq. (5.59) can be written as

2 ∫ δ u	∂u	dy + ∫ δ	∂(uv)	dy − ∫ δ u	∂v	dy − ∫ δ u	∂u	dy = −	τ w	(5.60)
	∂x				∂y				ρ
0		0 ∂y		0		0	∂x

where τ w is the shear stress at the wall. Eq. (5.60) can be rearranged as given below.

δ		∂u		δ		∂u		∂v	τ w

2∫0	u	∂x	dy + uv	y=δ − ∫0	u		+	dy = −	ρ	(5.61)

					∂x			∂y

5.4 Approximate solution to the boundary layer equations 129

The term within the brackets in Eq. (5.61) is nothing but a statement of the continuity equation and as a consequence is 0. In view of this, Eq. (5.61) becomes

2 ∫0δ u ∂∂uxdy + uv y=δ = − τρw

From the definition of the boundary layer, we know that u = u∞ at y = δ. Now we need to determine the value of v at y = δ.

Let us take recourse to the continuity equation Eq. (5.24)

∂∂ux + ∂∂vy = 0

Integrating the above continuity equation from y = 0 to y = δ, we have

∫ δ	∂u	dy = − ∫ δ	∂v	dy = −v		− v


0 ∂x		0 ∂y			y=δ		y=0

vδ = − ∫0δ ∂∂uxdy

Substituting Eq. (5.64) in Eq. (5.62) we get

2∫0δ u ∂∂uxdy − u∞ ∫0δ ∂∂uxdy = − τρw

∫0δ ∂∂x (u2 − u∞u)dy = − τρw

(5.62)

(5.63)

(5.64)

(5.65)

(5.66)

From Leibniz rule, the differential and integral signs can be interchanged. Additionally, the equation for boundary layer thickness becomes only a function in just x, consequent upon integration across y. In view of the above,

d	∫ δ u(u − u∞ )dy = −	τ w	(5.67)
dx		ρ
	0

The above equation is known as the integral momentum equation. Consider the energy equation for negligible viscous dissipation

u	∂T	+ v	∂T	= α	∂2 T	(5.68)
u	∂x	+ v	∂y	= α	∂y2

Integrating the above equation from 0 to δT in y, we have

δT

∂T

δT

∂T

δT

∂2T

∫0

∂x

+ ∫0

∂y

dy = ∫0

∂y

2 dy

(5.69)

∫0δT u

∂T dy + ∫0δT v

∂T dy = −α

∂T

(5.70)

∂x

∂y

y=0

δT ∂uT

∂T

∫0 ∂x

dy + vT

y=δT − vT

y=0

= −α

∂y

y=0

(5.71)

130 CHAPTER 5 Forced convection

Ty = δT	= T∞ and v at δT is given by Eq. (5.64) with changed upper limit of integration,
as given below.
	∫0δT ∂uT dy − T∞ ∫0δT ∂udy = −α		∂T
	∫0δT ∂uT dy − T∞ ∫0δT ∂udy = −α		∂T	(5.72)
	∂x	∂x	∂y	y=0
				y=0

δT ∂(uT − uT		)		∂T
∫0	∞		dy = −α
	∂x			∂y

(5.73)

y=0

Using the Leibniz rule and changing ∂/∂x to be d /dx, we get the following equation.

d	δT u (T − T		)dy = −α	∂T	(5.74)

dx ∫0		∞		∂y
					y=0

Eq. (5.74) is known as the integral energy equation. The momentum integral equation (Eq. (5.67)) can also be written as

d	∫ δT u (u − u∞ )dy = −ν	∂u	(5.75)
dx		∂y
	0		y=0

Similarly, the integral form of the species transport equation can be derived as

d δ	∂ρA	(5.76)

dx ∫0 u (ρA − ρA,∞ )dy = −DAB	∂y
dx ∫0 u (ρA − ρA,∞ )dy = −DAB	∂y	y=0
		y=0

Eq. (5.76) is useful in the study of convective mass transfer, where DAB is the binary diffusion coefficient and ρA is the density of the medium that is flowing and is under consideration.

Solution to integral momentum and energy equations with trial velocity and temperature profiles

Consider a linear profile for velocity

u		y
	= a + b		(5.77)
u
		δ
∞

The next step is to apply the boundary conditions on the assumed velocity profiles to obtain the constants in the assumed profile. The boundary conditions applicable for the problem under consideration are:

At y = 0, u = 0	(5.78)
At y = δ, u = u∞	(5.79)

Substituting these boundary conditions into Eq. (5.77), we get a = 0 and b = 1 Hence, the profile assumed in Eq. (5.77) becomes

u	=	y	(5.80)
u∞
	δ

5.4 Approximate solution to the boundary layer equations 131

Rearranging Eq. (5.75) and substituting Eq. (5.80) into it, we get

∫

−

u∞

∞

∫

u∞

1−

y=δ

−

∞

2δ

3δ

y=0

∞ dx 6

u∞

(5.81)

(5.82)

(5.83)

(5.84)

(5.85)

Integrating Eq. (5.85), we get the following equation.

δ2

ν x + A

(5.86)

∞

At x = 0, δ = 0 and so A = 0. Therefore, the solution becomes

δ2

(5.87)

12 = Rex

3.464

(5.88)

Rex

A cubic profile for velocity yields

4.64

(5.89)

x =

Rex

The exact solution to this by Blasius (1921), using the technique of similarity transformation (similar to the one we saw in unsteady conduction)

δ = 4.92

x Rex

(5.90)

As expected, the assumption of a cubic profile for velocity gives a solution close to the exact solution of Blasius.

We now assume a cubic profile for temperature in order to get an estimate of the thermal boundary layer thickness, δT .

T − T		y		y 2		y 3
∞	= a + b		+ c		+ d		(5.91)
Tw − T∞
		δT		δT		δT

132 CHAPTER 5 Forced convection

There are four constants in Eq. (5.91), so we need four boundary conditions to determine the constants.

At y = 0, T = Tw	(5.92)
At y = δT ,T = T∞	(5.93)

At y = δT ,	∂T	(5.94)
	∂y = 0

At y = 0, the energy equation is satisfied at the wall. On the wall, u = 0 and v = 0, so the energy equation reduces to

∂2 T	= 0	(5.95)
∂y2

Plugging these boundary conditions into Eq. (5.91), we get the values of a, b, c, d equal to be 1, − 3/2, 0, 1/2, respectively.

Hence, the cubic temperature profile turns out to be

(T − T

)

y 3

θ =

∞

= 1−

(Tw − T∞ )

δT

Substituting Eq. (5.96) into the integral energy equation Eq. (5.74), we get

δT

1 y

(Tw − T∞ )dy = α

(Tw − T∞ )

∫0

u 1

−

2 δT

2δT

δT

(5.96)

(5.97)

At this juncture, in order to get an estimate of δT closer to experiments, we would like to use a cubic profile for velocity, and we also know that δ /x = 4.64/ Rex for this profile. (This is given as an exercise problem at the end of this chapter.) Now, if we assume a cubic profile for velocity we get

u	=	3	y		−	1	y 3		(5.98)

u∞
		2		δ		2		δ

Substituting Eq. (5.98) in Eq. (5.97) Upon integrating Eq. (5.99),

d δT

3 y

9 y2

3 y4

1 y3

3 y4

1 y6

3α

u∞

−

(5.99)

dx ∫0

2 δ

4 (δδT )

2δT

4 δδT

4 δ δT

3α

3δT

−

9δT

3δT

−

δT

3δT

−

δT

(5.100)

4δ

12δδT

20δδT3

8δ 3

20δ 3δT

28(δδ

δT

∞

δT2

−

δT4

3α

(5.101)

280

δ 3

dx 20

∞

5.4 Approximate solution to the boundary layer equations 133

Now, we consider a situation where the following holds.

	δT		(5.102)
	δ < 1		(5.102)
	δ < 1

In view of this, the higher order term

δT4

can be neglected in Eq. (5.101) which

then reduces to

δ 3

d δT2

10α

u∞δT

(5.103)

Let δT /δ = ε, Eq. (5.103) then can be written as

(ε

δ)

10α

= u δε

(5.104)

∞

dε2

ε2

dδ

10α

(5.105)

δε

∞

Substituting Eq. (5.89) in Eq. (5.105),

ν x dε 2

+ ε

ν d x

10α

(5.106)

4.64

u∞ dx

u∞

4.64

u∞ xν ε

ε x

dε 2

0.464

(5.107)

4ε2 x

dε

+ε 3

0.928

(5.108)

dε 3

0.928

(5.109)

3 x dx +ε

The solution to the above differential equation is

= Ax

−3/4

0.928

(5.110)

To get A, we need a boundary condition. At x = 0, ε 3 is finite. Therefore A = 0

ε3 =

0.928

(5.111)

ε = δ

= 0.975Pr

−1/3

δT

(5.112)

Substituting for δ /x = 4.64/

Rex we have

= 4.64 × 0.975 Re

(5.113)

δT

−1/2

–1/3

= 4.52 Re

−1/ 2

δT

–1/3

(5.114)

134 CHAPTER 5 Forced convection

The local heat transfer coefficient can now be determined as

q = h(Tw − T∞ ) = −k		∂T		(5.115)
We can define a local Nusselt number as		∂y		y=0


Nux =	hx		(5.116)
	kf

The Nusselt number is actually the dimensionless heat transfer coefficient, as already mentioned.

Nux kf										∂θ
	(T	− T	) = −k				(T	− T	)		(5.117)
x	(T	− T	) = −k				(T	− T	)		(5.117)
x	w	∞				f	w	∞		∂y	y=0
											y=0
		Nux		=	3x						(5.118)
		Nux		=	2δT						(5.118)
					2δT
	Nu		x	= 0.332 Re1/2 Pr1/3							(5.119)
			x					x

From Nux , we get an expression for hx , integrate this from x = 0 to x = L, and average it over L to get an expression for hL (i.e., hL = 1/L ∫0L hx dx).

From this, we get an expression for the average Nusselt number NuL as follows:

NuL

= 0.664 Re1/x 2 Pr1/3

(5.120)

This is valid for fluids for which Pr >1. We now see why this is so. We know that δT /δ = (0.975/Pr)1/3. We have assumed earlier that δ /δT < 1. Hence for this assumption to be true Pr >1. The above expression is also reasonable for gases with Pr ≈ 1.

Integral method for fluids with Pr < 1

For liquid metals like liquid sodium, potassium, and mercury, the Prandtl number Pr <<1. These are highly conducting liquids with α >>ν and have excellent heat transport properties and are used in nuclear reactors to remove fission heat.

Consider the velocity and thermal boundary layers for flow and heat transfer over a heated flat plate. The boundary layers are shown in Fig. 5.2.

From Fig. 5.2, it is clear that for the most part, within the thermal boundary layer u = u∞ ( δT >> δ). Using the same cubic profile for temperature, as before, we have

	d	∫		3 y
u∞			0δT 1−

	dx			2 δT

y 3

3α

− T

)dy =

− T

)

δT

∞

2δT

∞

δT

3α

−

δ +

∞ dx

2δT

(5.121)

(5.122)

5.4 Approximate solution to the boundary layer equations 135

FIGURE 5.2

The velocity and thermal boundary layers in low-Pr fluids.

u 3

dδT

3α

∞ 8 dx

2δT

δT

dδT

4α

u∞

dδT2

8α

u∞

Integrating once, we get

δT2

8αx

+ A

∞

At x = 0, δT = 0, and so A = 0

δT

8αx

u∞

Multiplying and dividing the RHS by

ν,

δT

8αν x

2.82x

u∞ν

Rex .Pr

The product of Rex .Pr is known as the Peclet number, Pe.

δT = 2.82

x Pex

= − ∂θ

Nux x ∂

y y=0

= − −3

Nux x δ

2 T

(5.123)

(5.124)

(5.125)

(5.126)

(5.127)

(5.128)

(5.129)

(5.130)

(5.131)

136 CHAPTER 5 Forced convection

Substituting for δT in the above expression

	3		Pex	1/2
Nux =		x		= 0.532Pex	(5.132)
	2		2.82x

By using a procedure similar to what was done for Pr >1 fluids, we get the final form of the expression for the average Nusselt number as

NuL

= 1.064 Pe1/2L

(5.133)

From all the above developments, it is clear that the Nusselt number for forced convection, under steady state, takes up the following functional form

NuL

= aReb Prc

(5.134)

For the case of liquid metals, the form of the correlation gets further simplified with b = c, thereby introducing a new dimensionless number called the Peclet number. Analytical and semianalytical treatments are possible, as shown above, for flow over a flat plate. However, for other geometries, empirical correlations based on carefully done experiments are available in literature. Forms of these correlations are guided by theory, as mentioned above.

Example 5.1: Consider a steady, laminar, incompressible, two-dimensional external flow of air (at 30 °C) over a flat plate. The velocity of air is 5 m/s. The length of the plate is 1 m. The plate is maintained at temperature of 90 °C. Under steady state conditions, estimate the average heat transfer coefficient from the plate. Also, calculate the heat transfer from the plate for unit width of the plate.

Solution:

First, we determine the Reynolds number,

Re = uν∞ L

The properties of the fluid are evaluated at the mean film temperature

Tf	=	Tw + T∞
Tf	=	2
		2
Tf	=	90 + 30
Tf		2
Tf	= 60 C

Substituting the values of properties at 60 °C (from Table 5.2),

Re =	5 ×1
	18.82 ×10−6

Re = 2.66 ×105 < 5 ×105 (More about this criterion is in Section 5.5.) The flow is laminar.

Now using the correlation for the external flow over a flat plate

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