heat and mass transfer rk rajput

350 Heat and Mass Transfer

Graetz number, G = Pe - r? Grashoff number, Gr = P ~ P g At L3

p2 3. The following similarities for testing of 'heat transfer equipment' must be ensured between

the model and the prototype: (i) Geometric similarity (ii) Kinematic similarity (iii) Dynamic similarity (iv) Similarity of fluid entry conditions (v) Siniilarity of boundary temperature field.

THEORETICAL QUESTIONS 1. What is dimensional analysis? 2. What are the uses of dimensional analysis? 3. Explain the term dimensional homogeneity. 4. Describe the Rayleigh's method for dimensional analysis. 5. Describe Buckingham's method or x-theorem to formulate a dimensionally homogeneous equation

between the various physical quantities effecting a certain phenomenon. 6. What are dimensionless numbers? 7. Discuss the physical significance of the following dimensionless number Re, Nu, Pr, St, Gr. 8. Show by dimensional analysis for forced convection, Nu = (I (Re, Pr) 9. Show by dimensional analysis for free convection, Nu = (I (Pr, Gr)

10. What are the advantages and limitations of 'Dimensional analysis' ? 11. What do you mean by 'Characteristic length or Equivalent diameter' ? 12. For testing of 'heat transfer equipment' which of the similarities must be ensured between the model and

the prototype ?

'4

Forced Convection 7.1. Laminarflow over aflat plate: Introduction to boundary layer - Boundary layer definitions and

characteristics - Momentum equation for hydrodynamic boundary layer over a flat plate - Blasius solution for laminar boundary layer flows - Von Karman integral momentum equation (Approximate hydrodynamic boundary layer analysis) - Thermal boundary layer - Energy equation of thermal boundary layer over a flat plate - Integral energy equation (Approximate solution of energy equation). 7.2. Laminar tube flow: Development of boundary layer - Velocity distribution - Temperature distribution, 7.3. Turbulent flow over aflat plate; Turbulent boundary layer - Total drag due to laminar and turbulent layers - Reynolds analogy. 7.4 Turbulent tubeflow. 7.5 Empirical correlations - Typical Examples - Highlights -Theoretical Questions - Unsolved Examples.

A. LAMINAR FLOW 7.1. LAMINAR FLOW OVER A FLAT PLATE 7.1.1. Introduction to Boundary Layer

The concept of boundary layer was first introduced by L. Prandtl in 1904 and since then it has been applied to several fluid flow problems.

When a real fluid (viscous fluid) flows past a stationary solid boundary, a layer of fluid ' which comes in contact with the boundary surface, adheres to it (on account of viscosity) and condition of no slip occurs (The no-slip condition implies that the velocity of fluid at a solid boundary must be same as that of boundary itself). Thus the layer of fluid which cannot slip away from the boundary surface undergoes retardation; this retarded layer further causes retardation for the adjacetit layers of the fluid, thereby developing a small region in the immediate vicinity of the boundary surface in which the velocity of the flowing fluid increases rapidly from zero at the boundary surface and approaches the velocity of main stream. The layer adjacent to the boundary is known as boundary layer. Boundary layer is formed whenever there is relative motion between the boundary and the fluid. Since zo = , the fluid exerts a shear stress on the boundary and boundary exerts an equal and opposite force on fluid known as the shear resistance.

According to boundary layer theory the extensive fluid medium around bodies moving in fluids can be divided into following two regions:

( i ) A thin layer adjoining the boundary called the boundary layer where the viscous shear takes place.

(ii) A region outside the boundary layer where the flow behaviour is quite like that of an ideal fluid and the potentialflow theory is applicable.

7.1.1.1 Boundary Layer Definitions and Characteristics Consider the boundary layer formed on a flat plate kept parallel to flow of fluid of velocity

U (Fig. 7.1) (Though the growth of a boundary layer depends upon the body shape, flow over a flat plate aligned in the direction of flow is considered, since most of the flow surface can be approximated to a Jlat plate and for simplicity).


- The edge facing the direction of flow is called leading edge. - The rear edge is called the trailing edge. - Near the leading edge of a flat plate, the boundary layer is wholly laminar. For a laminar

- boundary layer the velocity distribution is parabolic. - The thickness of the boundary layer (6) increases with distance from the leading edge x,

as more and mare fluid is slowed down by the viscous boundary, becomes unstable and breaks into turbulent boundary layer over a transition region.

Laminar --4 Tr?nsi- Turbulentboundary layer -1 boundary layer tlon Y I

1 Fig. 7.1. Boundary layer on a flat plate. For a turbulent boundary layer, if the boundary is smooth, the roughness projections are

covered by a very thin layer which remains laminar, called laminar sublayer. The velocity distribution in the turbulent boundary layer is given by Log law of Prandtl's one-seventh power law.

The characteristics of a boundary layer may be summarised as follows: ( i ) 6 (thickness of boundary layer) increases as distance from leading edge x increases.

(ii) 6 decreases as U increases. (iii) 6 increases as kinematic viscosity (v ) increases.

1 (iv)\,q, = p [f) hence ro decreases as x increases. However, when boundary layer becomes I

turbulent, it shows a sudden increase and then decreases with increasing x. ( v ) When U decreases in the downward direction, boundary layer growth is reduced.

(vi) When U decreases in the downward direction. flow near the boundary is further retarded. boundary layer growth is faster and is susceptible to separation.

(vii) The various characteristics of the boundary layer on flat plate (e.g.. variation of 6, $ or force F) are governed by inertial and viscous forces: hence they are functions of either ux UL - or -. v v

I (viii) 1f UI < 5 x lo5 ... boundary layer is laminar (velocity distribution is parabolic). v ux If - > 5 x 10' ... boundary layer is turbulent on that portion (velocity distribution follows v

Log law or a power law). (ir) Critical value of at which boundary layer changes from laminar to turbulent depends

v on :

- turbulence in ambient flow,

Forced Convection 353

- surface roughness, - pressure gradient, - plate curvature, and - temperature difference between fluid and bounary.

(x) Though the velocity distribution would be a parabolic curve in the laminar sub-layer zone, but in view of the very small thickness we can reasonably assume that velocity distribution is linear and so the velocity gradient can be considered constant.

Boundary layer thickness (6) : The velocity within the boundary layer increases from zero at the boundary surface to the

velocity of the main stream asymptotically. Therefore, the thickness of the boundary layer is arbitrarily defined as that distance from the boundary in which the velocity reaches 99 per cent of the velocity of the free stream (u = 0.99U). It is denoted by the symbol 6 This definition. however. nives an approximate value of the boundary layer thickness and hence 6 is generally

, " m *

termed as nominal thickness of the boundary layer. The boundary layer thicknpss for greater accuracy is defined in terms of certain mathematical

expressions which are the metsure of the boundary layer on the flow. The commonly adopted definitions of the boundary layer thickness are:

1. Displacement thickness (6*) 2. Momentum thickness (0) 3. Energy thickness (6,).

Displacement thickness (a*) : The displacement thickness can be defined as follws: "It is the distance, measured perpendicular to the boundby, by whi

is displaced on account of formation of boundary layer." Or

"It is an additional "wall thickness" that would have to be added to compensate for the reduction in jlow rate on accounibf boundary layer formation. "

The displacement thickness is denoted by 6*. Let fluid of density p flow past a stationary plate with velocity U as shown in Fig. 7.2.

Consider an elementary strip of thickness dy at a distance y from the plate. Assuming unit width, the mass flow per second through the elementary strip

= oudv . . . ( i ) . . Mass flow per second through the elementary strip (unit width) if the plate were not there

= p U.dy . ...( ii)

Boundary layer ___,

___, Stationary plate 2

Fig. 7.2. Displacement thickness.

Forced Convection Heat and Mass Transfer

Energy thickness is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in K,E. o f the flowing fluid on account of boundary layer formation. It is denoted by 6,

Reduction of mass flow rate through the elementary strip = p ( U - u ) d y

[The difference (U - u) is called velocity of defect] Total reduction of mass flow rate due to introduction of plate

S

= J P ( u - U ) d~ . . . (iii) 0

(if the fluid is incompressible) Let the plate is displaced by a distance S* and ,velocity of flow for the distance 6* is equal

to the maidfree stream velocity (i.e., U). Then, loss of the mass of the fluidlsec. flowing through the distance S*

J ., - 7

Refer Fig. 7.2. Mass of flow per second through the elementaIy strip = pudy K.E. of this fluid inside the boundary layer

1 1 = - 2 rn u2 - - (pudy) u2 - 2

K.E. of the same mass of fluid before 1 entering the boundary Iayer = 7 (pudy) tJ2

Loss of K.E. through elementary strip 1 1 2 d .,.(i)

= - 2 (pudy) u2 - 5(pudy)u2 = ?pu (U -u ) y = pU6*

. . .(iv) Equating Eqns. (iii) and (iv), we get

6

pU6* = j p ( U - u ) d y 0 A

6 1

;. Total loss of K.E. of fluid = I -; pu (u' - u2) dy , - A 0

Let 8, = distance by which the plate is displaced to cornpia te for the reduction in K.E. Then loss of K.E. through 6, of fluid flowing with velocity 0

1 = I (pU6.) u2 ...( ii)

Equating Eqns. (i) and (ii), we have s

IcpusA 2 u2 = J 0 + p u ( u 2 - u 2 ) d y

Momentum thickness (0) : Momentum thickness is defined as the distance through which the total loss of momentum

per second be equal to if it were passing a stationary plate. It is denoted by 8. It may also be defined as the distance, measured perpendicular to the boundary of the solid

body, by which the bounhry should be displaced to compensate for reduction in momentum of the flowing fluid on account of boundary layer formation.

Refer Fig. 7.2. Mass of flow per second through the elementary strip = p u dy Momentudsec of this fluid inside the boundary layer = p u dy x u = pu2 dy Momentum/sec of the same mass of fluid before entering boundary layer = pu Udy Loss of momentudsec = puUdy - pu2dy = pu (U - u ) dy :. Total loss of momentudsec.

S

= J ~ " ( 0 - u ) d ~ . ..(i) 0

u

Example 7.1. The velocity distribution in the boundoly layer is given by: 3 = i, where u is the velocity at a distance y from the plate and u = U at y = 6, 6 being boundary layer thickness. Find:

(i) The displacement thickness, (ii) The momentum r* thickness,

Let 8 = distance by which plate is displaced when the fluid is flowing with a constant velocity U .

Then loss of momentudsec of fluid flowing through distance 8 with a velocity U = p 8 U 2 . . .(ii)

Equating Eqns. ( i ) and (ii), we have (iii) The energy thickness, and (iv) The value of 8 ' u Y Solution. Velocity distribution : - = 11 6

( i ) The displacement thickness, 6*: II

...( Given)

...I Eqn. (7.111

The momentum thickness is useful.in kinetics.

Heat and Mass Transfer

(ii) The momentum thickness, I : S

I = j: (l - :Idy ...[ Eqn. (7.2)] 0

(iii) The energy thickness, 6, :

F (iv) The value of - . I '

- --

or - - - ---- ' ' I 2 - 3.0 6 - 120 I - 6/6

Example 7.3. I f velocity distribution in laminar boundary layer over a flat plate is assumed U Example 7.2. The velocity distribution in the boundary layer is given by = to be given by second order polynomial u = a + by + c y , determine its form using the necessary

boundary conditions. -- - -- I y2 6 being boundary layer thickness. Solution. Velocity distribution: u = a + by + cyL 2 6 2 lj2' The following boundary conditions must be satisfied:

Calculate the following: (i) At Y = 0, u = 0 (i) The ratio of displacement thickness to boundary layer thickness . . u = a + by + cy2

O = a + O + O ; . a = O (ii) The ratio of momentum thickness to boundary layer thickness (ii) ~t y = 6, u = U

U = b6 + c6' ...( i) . . u 3 y I y 2 Solution. Velocity distribution : - = -- - --

...( Given) du U 2 6 2 6 2 (iii) At y = 6, - = o dy ( i) 6*/6: . . d

= - (a + by + cy2) = b + 2cy = b + 2c6 = 0 ... (dl) 6 6 6 . ~ 1 l - - d y = [ l - ? ~ + l d ( :) [ 2 6 2 @ ) d y i o Substituting the value of b (= - 2 ~ 6 ) from(ii) in (i), we get

U = (- 2c6)6 + cs2 = - 2cs2 + c6' - c6' I

* I

358 Heat and Mass Transfer .

. , Hence form of the velocity distribution is

7.1.2. Momentum Equation for Hydrodynamic Boundary Layer Over a Flat Plate Consider a fluid flowing over a stationary flat plate and the development of hydrodynamic

boundary layer as shown in Fig. 7.3. (a) . In order to derive a differential equation for the boundary layer, let us consider an elemental, two-dimensional control volume (dx x dy x unit depth) within the boundary layer region [enlarged view shown in Fig. 7.3 (b)]. Following assumptions are made:

1. The flow is steady and the fluid is incompressible. 2. The viscosity of the fluid is constant. 3. The pressure variations in the direction perpendicular to the plate are negligible. 4. Viscous-shear forces in the Y-direction are negligible.

- -

5. Fluid is continuous both in space and time. Refer Figure 7.3. (b). Let u = velocity of fluid flow at left hand face AB.

Velocity distribution

_+ Stationary plate

--b

Elemental control volume

U 4 r ' d x

B .t b c (b) '5

Fig. 7.3. Equation of motion for boundary layer.

Forced Convection

au Then, u + - . dx = velocity of fluid flow at the right hand face CD (since the flow velocity ax au

changes in the X-direction at the rate of change given by - and the ax

change in velocity during distance dr will be - . dr . G )I Similarly, let, v = fluid velocity at the bottom face BC,

av then, v + y dy = fluid velocity at the top-face AD.

OY The mass flow rate along X-direction, m, = pu (dy ,X 1) = pu dy ...( 7.4) The change of momentum of the mass mx along X-direction is given by

dM, = m, x change in vebcity in X-direction

The mass flow rate along Y-direction, m, = pv (dr x 1 ) = pvdx ...( 7.6) The change of momentum of the mass my along Y-direction is given by

-

au = pv -. h.dy . ... (7.7)

JY Total viscous force along the X-direction is given by

F, = [(z + 6t) - z] x area au a

= [Ips& + 3-(.-$).dy) - p$] ( h x 1) a 2 ~

= p ---i- a~ . dr.dy . . .(7.8)

Assuming the gravitational forces are balanced by buoyancy forces for equilibrium of the element, we have

Inertia forces = viscous forces au au azu :. p u - . h . d y + p v - . h . d y = p 5 . d x . d y ax a~ a~

or au au p azu u - + v - = - S T ax a~ P ay

or au au a 2 ~ u- + v - = ax ay 'v ...( 7.9) (substituting v = ' P 1

Equation (7.9) is known as equation of motion or momentum equation for hydrodynamic boundary layer. 7.1.3. Blasius Exact Solution for Laminar Boundary Layer Flows

The velocity distribution in the boundary layer can be obtained by solving the equation of

Heat and Mass Transfer Forced Convection

motion for hydrodynamic boundary layer [Eqn. (7.9)]. The following boundary conditions should be satisfied.

(i) At y = 0, u = 0 (ii) At y = 0, v = 0 (iii) At y = =, u = U The Blasius technique for an exact solution of the hydrodynamic boundary layer lies in

the conversion of the following differential equations into a single differential equation.

I au au a 2 ~ The hydrodynamic equation for boundary layer: u - aX"5"p ...( Eqn. 7.9) Continuity equation: du dv - + - = o ax ay Here f is abbreviated as f (q) - Prandtl suggested that the solution of Eqn. (7.9) can be obtained by reducing the number

of variables with the help of magnitude analysis of the boundary layer thickness and transforming the partial differential equation into ordinary differentials.

The inertia forces represented by the left terms, in the Eqn. (7.9), must be balanced by the viscous forces represented by the right terms.

As u 2 v, therefore, we may write as Now, - a ' - . - ( - a & a & ? ? I . u 8f ) ( ) a ay a y m h h a ~ Similarly, @u fl d"f ...( 7.18)

Again,

au u Also as u = U and - = - along A plate length L, therefore, we have ax L'

1 I I From experiments it has been o b s e r v e W velocity profiles at different locations along the

plate are geometrically similar, i.e., they differ only by a stretching factor in the Y-direction.

i u This implies that the dimensionless velocity - can be expressed at any location x as a function U 1 of the dimensionless distance from the wall 1 6'

Substituting the value of 6 from Eqn. (7.1 1) in Eqn. (7.12), we obtain, r 1

where, q = y g d e n o f e s the stretching factor.

In order to account for the fact that the vertical camponent of velocity occurs in the boundary layer equation of motion (7.9)' it is essential to define a stream function \y such that,

- -.

au au azu Inserting the values of u, - - - and v from Eqns. (7.15). (7.16), (7.17), (7.18) and a x 9 ay' a2

(7.19) in Eqn. (7.9), we get - V 8f vx ' 4 3 or J7.14 (a)]

The continuous stream function is the mathematical postulation such that its partial differential with respect to x gives the velocity in the Y-direction (generally taken as negative) and, its partial differential with respect to y gives the velocity in the X-direction:


or 2f"'+ ff" = 0 ...( 7.20) which is an ordinary (but non-linear) differential equation for f. The number of primes on f denotes the number of successive derivatives off (q) with respect toy. The physical and transformed boundary conditions are:

Physical boundary conditions Transformed boundary conditions

(ii) At y = 0, v = O At ?l = 0. f = O (iii) At y = w u = U A t q = w , g= f J = I

4

Fig. 7.4. Velocity distribution in boundary layer on flat plate by Howarth

The numerical solution of Eqn. (7.20) with the corresponding values of u and v are plotted in Fig. 7.4, and the results are listed in table 7.1.

The following results are of particular interest: v 1. The single curve I1 shows the variation of normal velocity - It is to be noted that at the u '

outer edge of the boundary layer where q + -, this does not go to zero but approaches the value


Table 7.1. Laminar boundary layer solution for a flat plate

2. The graphlcurve I (i.e., the velocity distribution parallel to the surface) enables us to calculate the parameters : (i) Boundary layer thickness, 6 and (ii) skin friction coefficient, c,. (i) Boundary layer thickness, 6:

The boundary layer thickness 6 is taken to. be the distance from the plate surface to a point U

at which the velocity is within 1% of the asymptotic limit, i.e., - U = 0.99; it occurs at q = 5.0 (Fig. 7.4). Therefore, the value of q at the edge of boundary layer O, = 6) is given by

- 7

where Ux Re, = - v is the local Reynolds number based on distance x from the leading edge of the plate.

(ii) Skin friction coefficient; Cf: The skin friction coefficient (Cf) is defined as the ratio of shear stress zo at the plate to

I the dyMmic head 5 p ~ ' caused by free stream velocity. Thus the local skin friction coefficient Cfi at any value of x is

Heat and Mass Transfer Forced Convection 365

To Cfx = - = 1 1 ...( 7.23) u 2 p2

From Fig. 7.4, the gradient at 11 = 0 is

Let ABCD be a small element of a boundary layer of the boundary layer).

(the edge DC represents the outer edge

layer + flat plate

Fig. 7.5. Momentum equation for boundary layer by Von Karman.

Mass rate of fluid entering through AD

0

Mass rate of fluid leading through BC

= ipudy o 2 l P u d y J h :. Mass rate of fluid entering the control \ -1ume through the surface DC

The average value of the skin friction coefficient Zf can be determined by integrating the local skin friction coefficient Cfx from x = 0 to x = L (where L is the plate length) and then dividing the integrated result by the plate length

= mass rate of fluid through BC - mass rate of fluid through AD

The fluid is entering through DC with a uniform velocity U. where ReL is the Reynolds number based upon total length L of the plate. 7.1.4. Von Karman Integral Momentum equation (Approximate Hydrodynamic Boundary

Layer Analysis)

Momentum rate of fluid entering the control volume in x-direction through AD 6

= I pu2 dy Since it is difficult to obtain the exact solution of hydrodynamic boundary layer [Eqn. (7.9)]

even for as simple geometry as flat plate (moreover the proper similarity variable is not known or does not exist. for many practical shapes), therefore, a substitute procedure entailing adequate accuracy has been developed which is known as "Approximate lntegral Method" and this is based upon a boundary-layer momentum equation derived by Von Karman.

- - --

Momentum rate of fluid leaving the control volume in x-direction through BC 8 r 6 1

Momentum rate of fluid entering the control volume through DC in x-direction r 6 1

von Karman suggested a method based on the momentum equation by the use of which the1 growth of a boundry layer along a flat plate, the wall shear stress and the drag force could be determined (when the velocity distribution in the boundary layer is known). Starting from the beginning of the plate, the method can be used for both laminar and turbulent boundnry layers

Figure 7.5, shows a fluid flowing over a thin plate (placed at zero incidence) with a free stream velocity equal to U. Consider a small length dr of the plate at a distance x from the leading edge as shown in Fig. 7.5 (a); the enlarged view of the small length of the plate is shown in Fig. 7.5 (b) . Consider unit width of plate perpendicular to the direction of flow.

(': velocity = I / )

? L"

:. Rate of change of momentum of control volume

Heat and Mass Transfer Forced Convection 367

= momentum rate of fluid through BC - momentum rate of fluid through AD - momentum rate of fluid through DC.

d ( p is constant for incompressible fluid) 0

As per momentum principle the rate of change of momentum on the control volume ABCD must be equal to the total force on the control volume in the same direction. The only external force acting on the control volume is the shear force acting on the side AB in the direction B to A (Fig. 7.56). The value of this force (drqg force) is given by

A F D = z 0 x d r Thus the total external force in the direction of rate of change of momentum

= - , r 0 x d x ...( 7.27) Equating the Eqns. (7.26) and (7.27), we have

But 6 ! ( 1 - $) dy = momentum thickness (8)

...(7. 28)

...[ 7.28 (a)]

Euation (7.28) is known as Von Karman momentum equation for boundary layerflow, and is used to find out the frictional drag on smooth flat plate for both laminor and turbulent boundary layers. Evidently, this integral equation (7.28) u expresses the wall shear stress,% as a function of the non-dimensional velocity distribution -a u is a point velocity at the boundary layer and U U' is the velocity at the outer edge of boundary layer.

The following boundary conditions must be satisfied for any assumed velocity distribution du (i) At the surface of the plate: y = 0, u = 0, - dY = finite value

(ii) At the outer edge of boundary layer: y = 6, u = U

The shear stress zo for a given velocity profile in laminar, transition or turbulent zone is obtained from Eqn. [7.28 (a)] or [7.28 (b)]. Then drag force on a small distance dr of a plate is given by

A FD = shear stress x area = zo x (B x dx) = zo x B x dr (where B = width of the plate)

1 :. Total drag on the plate of L length L one side.

F D = j ~ ~ D = ] b ~ ~ ~ & o ..(7.29)

1 - The ratio of the shear stress zo to the quantity - pu2 is known as the "Local coefficient 2

of drag" (or coeflcient of skin friction) and is denoted by Cfr

1 - The ratio of the total drag force to the quantity - ~ A U ~ is called 'Average coefficient of 2

drag' and is denote by CD. -

- FD 1. e., Cf = 7 ...( 7.31)

- PA u2 2 where p = mass density of fluid,

A = area of surfacelplate, and U = free stream velocity.

It has been observed through experiments that for laminar boundary layer, the velocity distribution is parabolic and the velocity profiles at different locations u along the plate are

geometrically similar. This means that the dimensionless velocity - can be expressed at any u

locatjon x as a function of the dimensionless distance from the wall, 6'

...[ 7.28 (b)] \ , . ,

, The constants can be evaluated by using the following boundary conditions:

Forced Convection Heat and Mass Transfer

a 2 ~ (i) At y = 0 (wall surface), u = U and - = 0 a9

au (ii) At y = 6 (outer edge of the boundary layer), u = U and - = 0 ay

By applyiug boundary conditions the constants are evaluated which gives the velocity profile as:

, , . ,

In order to determine the boundary layer thickness and average skin-friction coefficient for laminar flow over a flat plate, let us now use this above velocity profile in the Von Karman

U integral equation. Now putting the value of - in Eqn. (7.28), we get u

or 39 d6

To = - pu2 - ...( 7.34) 280 du Newton's law of viscosity, at solid surface, gives

= ($Izo 3 d 3 Y

= 4% p ( 6 ) - 3$) llv=o or 70 = 3CLLI 26 ...( 7.35) From equations (7.34) and (7.35), we have

39 6 3pU % pu2 dx - 26

or = @.%fx 13 pu Since, 6 is a function of x only, integration yields

B B+c - - -

2 1 3 p U By using the boundary condition 6 = 0 at x = 0, we obtain the integration constant C = 0.

@-B or 62= 7 -

140x 2 x~ . . 2 1 3 p U 13 PU

This can be expressed in the non-dimensional form as

...( 7.36) X

x U where Re, = is the Reynolds number based on distance x from the leading edge of the plate.

U r- Further, in order to estimate the value of zo, substituting the value of 6 Eqn. (7.36) in Eqn.

(7.39, we get

20 = 3CLI/ + 4A4 on simplification, we get 6 K' p@ 0.646

=o- 2 7iq' Therefore, the local skin friction coefficient.

Average value of skin friction coefficient, 1 0.646 dx

- L j L cfx*=- J - - E f - 0 L o d $ z i f i

where ReL = k@! is Reynolds number based on total length L of the flate. u

J

Mass pow through boundary layer The mass flow rate per unit width through the boundary layer is calculated assuming parabolic

velocity distribution in the boundary layer as 3 1

-- 370 Heat and Mass Transfer , Forced Convection

Example 7.4. The velocity profile for laminar boundary is in the form given below:

Find the thickness of boundary layer at the endof the plate 1.5 m long and 1 m wide when placed in water flowing with a velocity of 0.12 d s . Calculate the value of coefJicienr of drag also.

Take p for water = 0.001 Ws/rn2 U Solution. Velocity distribution - - u - 2 6 ) - br

The length of the plate, L = 1.5 m The width of the plate, B = 1 m Free stream velocity, U = 0.12 m/s p for water = 0.001 Ns/m2

Thickness of the boundary layer, 6: Reynolds number at the end of the plate (i.e., at a distance o

is given by

...( Given)

f 1.5 m from the leading edge)

Since ReL < 5 x lo5, therefore, this is case of laminar boundary layer. Thickness of boundary layer at a distance of 1.5 m is given by

Coefficient of drag Cf :

Example 7.5. Air is flowing over a smooth flat plate with a velocity of 12 m/s. The velocity profile is in the form:

The length of the plate is 1.1 m and width 0.9mJf laminar boundary layer exists upto a value of Re = 2 x 16 and kinematic viscosity of air is 0.15 stokes, find:

(i) The maximum distance from the leading edge upto which laminar boundary layer exists, and

(ii) The maximum thickness of boundary layer. U Solution. Velocity distribution:

Velocity of air, U = 12 m/s Length of plate, L = l.lm Width of plate, B = 0.9m Reynolds number upto which laminar boundary exists, Re = 2 x lo5 Kinematic viscosity of air, v = 0.15 stokes = 0.15 x lo4 m2/s

( i ) The maximum distance from the leading edge upto which laminar boundary layer exists, X:

Ux or 2 x lo5 = 12 x x Re, = - v 0.15 x lo4

(ii) The maximum thickness of boundary layer, 6 : For the given velocity profile, the maximum thickness of boundary layer is given by

5.48 x =

Example 7.6. A plate of length 750 mm and width 250 mm has been placed longitudinally in a stream of crude oil which flows with a velocity of 5 m/s. I f the oil has a specific gravity of 0.8 and kinematic viscosity of 1 stoke, calculate:

( i ) Boundary layer thickness at the middle of plate, (ii) Shear stress at the middle of plate, and

(iii) Friction drag on one side of the plate. Solution. Length of the plate, L = 750 mm = 0.75 m

Width of the plate, B = 250 mm = 0.25 m Velocity of oil, U = 5 mls Specific gravity of oil = 0.8 Kinematic viscosity of oil, v = 1 stoke = 1 x lod4 m2/s

( i ) Boundary layer thickness at middle of the plate 6: Reynold number,

( '.' At the middle of plate, x = 0.792 = 0.375 m) Since Re, < 5 x lo5, therefore, boundary layer is of laminar character and Blasius solution

gives

(ii) Shear stress at the middle of plate, T,: According to Blasius, the local coefficient of drag (C,J is given by

By definition,

3x2 Heat and Mass Transfer

(iii) Friction drag on one side of the plate, FD:

As the boundary layer is laminar even at the trailing edge, therefore, the average drag (friction) coefficient,

:. Firction drag, 1 FD = x - U2 x area of plate on one side -' 2 P

I 1

= 6.858 x x - x (0.8 x 1000) x 52 x 0.75 x 0.25 2 = 12.86 N

(Note : If velocity profile is not given in the problem, but boundary layer is laminar then Blasius's solution is used.).

Example 7.7. Atnzospheric air at 20C is flowing parallel to a flat plate at a velocity of 2.8 4 s . Assuming cubic velocity profile and using exact Blasius solution, estimate the boundary layer thickness and the local coeficient of drag (or skin friction) at x = 1.2 m from the leading edge of the plate. Also find the deviation of the approximate solution from the exact solution.

Take the kinenmtic viscosity of air at 20C = 15.4 x lo4 m2/s Solution. Velocity of air, U = 2.8 m/s

Distance from the leading edge of the plate, x = 1.2 m

Reynolds number

Blasius solution:

Boundary layer thickness, 6 = -& = = 0.01285 m = 12.85 mm 2 18 x 16 Local coefficient of drag, 0.664

= 0.001422 cfx = K = xlOI Approximate solution (with assumption of cubic velocity profile):

The approximate solution deviates from the exact solution by

Derivation for 6: 12.85 - 11.92 x 100 = 7.34% 12.85 Drviation for

Example 7.8. Air is flowing over a flat plate 5 m long and 2.5 m wide with a v&ocity of 4 m/s at 15C. I f p = 1.208 kg/m3 alzd v = 1.47 x 1 ~ ' m2/s, wlculate:

( i ) Length of plate over which the boundary layer is laminar, and thickness of thk'boundary layer (laminar). i

(ii) Shear stress at the location where boundary layer ceases to be laminar, andt,,.. : (iii) Total drag force on the both sides on that portion of plate where boundary layer is

laminar.

Solution. Length of the plate, L = 5 m Width of the plate, B = 2.5 m

. .

Velocity of air, U = 4 mls Density of air, p = 1.208 kg/m3 Kinematic viscosity of air, v = 1.47 x 1w5 m2/s

( i ) Length of plate over which the boundary layer is laminar: Reynolds number uL 4 x 5 = 1 . 3 6 1 ~ 1 @ - Re, = - - V 1 .47x10-~ Hence on the front portion, boundary layer Ux is laminar and on the rear, it is turbulent.

Re, = - = 5 x 1 d v

Hence the boundary layer is laminar on 1.837 m length of the plate. Thickness of the boundary layer (laminar), 6

(ii) Shear stress at the location where boundary layer ceases to be laminar, g : Local coefficient of drag, = O.CM0939 Cf, = K i F

1 1 b = CfxXgp u ~ = o . w ~ ~ ~ ~ x ~ x I . ~ o ~ ~4~ = 0.00907 ~ / m ~

(iii) Total drag force on both sides of plate, FD 1 : F ~ = Z X C X - ~ A U ~

f 2 - 1.328

where. Cf = average coefficient of drag (or skin friction) = ,- = 1.878 x l(r3 5 x 105 and A = area of the plate = 1.837 x 2.5 = 4.59 m2

1 . .

Fo = 2 x 1.878 x x 5 x 1.208 x 4.59 x 42 = 0.167 N Example 7.9. Airflows over a plate 0.5 m long and 0.6 wide with a velocity of 4 d s . The

u - = sin (: i) u

- If p = 1 . Z l k g / d and v = 0.1 5 x 1 m2/s, calculate: ( i ) Boundary layer thickness at the end of the plate, (ii) Shear stress at 250 mm from the leading edge, and (iii) Drag force on one side of theptate. Solution. Length of plate, L = 0.5 m

Width of plate, B = 0.6 m Velocity of air, U = .4 mls


Density of air, p = 1.24 kg/m3 Kinematic viscosity of air, v = 0.15 x m2/s

Velocity profile, , u = sin (f 3 ( i ) Boundary layer thickness at the end of the plate, 6:

Reynolds number, ux ' 4 x 0e5 = 1 . 3 3 ~ 10s Rex = - = V 0.15 x lo4

Since Rex c 5 x loS, therefore, the boundary layer is laminar over the entire length of the plate.

We know,

= 0.00657 m = 6.57 mm (ii) Shear stress at 250 mm from the leading edge, 5:

...[ Eqn. (7.30)J But, Cf.= -=-- 0.654 0.654 0.654

= 0.002533

0.15 x

(iii) Drag force on one side of the plate, F, :

where 1.31 CD = = - 0.003592 1.33 x loS -

and A = area of the plate = L x B = 0.5 x 0.6 = 0.3 m2

. . 1 FD = 0.003592 x 7 x 1.24 x 0.3 x 4' = 0.01069 N

7.1.5. Thermal Boundary Layer Whenever a flow of fluid takes place past a heated or cold surface, a temperature field is

set up in the field next to the surface. If the surface of the plate is hotter than fluid, the temperature

Limit of thermal boundary layer

Free stream

Y

t r Plate surface (I,)

t --* I+ Laminar Turbulent

Fig. 7.6. Thermal boundary layer formed during flow of cool fluid over a warm plate.


distribution will be as shown in the Fig. 7.6. The zone or this layer wherein the temperature fiefd exists is called the thermal boundary layer. Due to the exchange of heat between the plate and the fluid, temperature gradient occurs/results.

The thermal boundary layer thickness (6,,,) is arbitrarily defined as the distance y from the plate surface at which

It c-4 - " 7 f I I f I ! Temperature profile

Fig. 7.7. Thermal boundary layer formed during flow of warm fluid over a cool plate. Figure 7.7, shows the shape of the thermal boundary layer when the free stream temperature

t, is above the plate surface temperature t,. The thermal boundary layer concept is analogous to hydrodynamic boundary layer; the

parameters which affect their growth are however different. Whereas the velocity profile of the hydrodynamic boundary layer depends mainly on the fluid viscosity. the temperature profile of the thermal boundary layer depends upon the viscosity, velocity of flow, specific heat and thermal conducitivity of the fluid. The relative magnitude of 6 and 6th are affected by the thermo-physical properties of the fluid; the governing parameter, however, is h e non-dimensional Prandtel number,

pr = 7. ( i ) 6th = 6 ........ when Pr = 1; (ii) 4, < 6 ........ P r > 1 (iii) 6 , > 6 ........ when Pr < 1.

7.1.6. Energy Equation of Thermal Boundary Layer Over a Flat Plate Figure 7.8 (a) shows a hot fluid flowing over a cool flat plate, and development of the

thermal boundary layer. In order to derive an energy equation, consider control volume (dr x dy x unit depth) in the boundary layer so that end effects are neglected. The enlarged view of this control volume is shown in Fig. 7.8 (b) in which the quantities of energy entering and leaving have been indicated.

Involving principle of conservation of energy for the steady state condition, we have: Heat energy convected (E,,,,) through the control volume in X and Y directions + heat

energy conducted (Econd,) through the control volume in X and Y directions + heat generated due to fluid friction (viscous heat generation) in the conuol volume = 0.

But as the rate of temperature change in the X-directidn is small and can be neglected the conservation of energy becomes:

Heat energy convected in X and Y directions + heat energy conducted in Y-direction + viscous S t heat generation = 0.

or d(E,,), + d(EcOnv.), + d(ECod), + viscous heat generation = 0 ...( 7.41) (i) (ii) (iii) (iv)


boundary layer

Plate surface (r,>

Viscous heat (Ec~nv.!v +dy generation

. t

(iii) The heat conduction in tbe y-direction: d(Ecod), = (E~md.)~ - (~cotuf.h+dy

at

. ..(7.44)

the element). = [Shear stress (r) x area upon which it acts)]

Fig. 7.8. Energies entering and leaving the control volume. (i) The Energy convected in X-direction:

(Econ,.)x = mass x specific heat x temperature = [pu(dy x l)] c,t = (pu dy) c,,t

-

... neglecting the product of small quantities . an

(ii) The energy convected in Y-direction: + . The net energy cor+vected in Y-direction, " L

. -

x distance travelled

= [,$ (dr x l)] x ($ dy) Substituting the values in Eqn. (7.411, we get

. . .(7.46)

Form the continuity equation for two-dimensional flow, we have au av -+-=0: thus the Eqn. (7.46) reduces to ---

% ,, at at n a2t p (?= u- + v- = - ... (7.47) ' s + P . , ay ax a~ PC, ay

Equation (7.47) is the requid di#erential energy qUUt0n f0rflDw PaR a fLY p*1te. If heat genemhon is neglected [when the value of U is relatively 10" and difference of temperature bewen the swam and the plate is s d l (of the order of 40c)1, fhe energ Cqudlon reduces

at k a2t 2 t at +,-=--=a u - &=

...( 7.48) ax ay PC, a 3

fi may noted, the energy equation is similar to the momentum equation. Further the dimensions of kinematic viscosity v and thermal diffusivity a are the same.

The auation (7.48) has been derived with the following assumptiom: 1. Steady incompressible flow.

t, - ts - - are constant. 2. m e properties of the fluids evaluated at the film temperature $ - 2


3. The body forces, viscous heating and conduction in the flow direction are negligible. Pohlhausen solution for the 'Energy equation':

at at By using the following variables the energy equation u - + v - = a- can be recast ax JY ay2 into an ordinary differential equation as follows: "1 .

(Stretching factor) = Y 45, yf (StreaIn function) = f (ll), and Also, the values of the &locity components u and v already calculated earlier are:

u = u & 4 ...[ Eqn. (7.15)l .

... [Eqn. (7.19)J 4

Further, from temperature parameter 0 (non-dimensional) defined above. we have

and

or

Also,

t = t,# + (t, - t,) e at - - ae

ax ax - (t- - tJ - = (tm - t#) -

atl ax at ax J7.50) at - - ae ay

" , h - 0,. - ts) - = (t- - t,) - JY atl ay

at

or - - 8 t tJ&e $2 - (1.. - 4) - vxdq2

bsening the above values in the energy equation, we get: at at a2t

u - + v - = a - ax a~ ay2

U d28 = a (t, - t,) - -Z

vx 4 ...( 7.53) After simplification and arrangement of the above equation, we obtaL

Forced Convection

.' Pr (Randtl number) = Thus the partial differential equation (7.48) has been converted into ordinary differential

equation. The boundary conditions to the satisfied are: At t = t,, y - 0 At t = t,, y = m At q = 0, e(q) = ) values in terms of new variable ...( 7.55) ~t q = - em) = 1 The solution obtained by Pohlhausen for energy equation is given by:

de The factor fr) represents the dimensionless dope of the temperature profile at the \Wh=o

surface where q = 0; its value can be obtained by applying the boundary condition at r\ = w, 801) = 1. Thus,

. ,,

Evidently the dimensionless slope is a function of Prandtl number and the calculations made by Prandtl gave the following result :

For 0.6 < Pr 15, [ = 0.332 (PI)'" =o

Figure 7.9 shows the values of 0 (dimensionless temperature distribution) plotted for various values of Pr (Prandtl number).

- The curve for Pr = 0.7 is typical for air and several other gases. - The curve for Pr = 1 is the same as that of curve I in Fig. 7.4. - These curves also enable us to determine the thickness of thermal boundary layer 6,h and,

local average heat transfer coefficients h. Thermal

Fdrced Convection Heat and Mass Trapfpr

Thickness of thermal boundary layer, 6, : Case I: When Pr = 1.

- -

in Eqn. (7.63). we obtain Substituting for - : \'I'. k 1i2 pr)in

= hx i t s - tJ = 0.332 ; (t, - tm)(Rex) A

k hx = 0.332 ; ( ~ e ~ ) l " (pr)'" ...( 7.64 or

Since y = 6, at the outer edge of thermal boundary layer, therefore, 7

h2-X 112 pr)ln NU, = - = 0.332 (Re,) ( ...( 7.65) or k ... (In non-dimensional form) This equation shows that for Pr = 1, the thickness of thermal boundary layer, 6, is equal

[where hx = local convective heat transfer coefficient; Nu. = local value of Nusselt number (... at a distance x from the leading edge

to hydrodynamic boundary layer, 6. Case 11. When Pr < 1.

,- -7 A

of the plate]. The average heat transfer coefficient is given by

This equation shows that for Pr < 1, 6, > 6 Case 111. When Pr > 1.

\ /

If we compare the Eqns. (7.64) and (7.66), we find that -

h = 2hx

This equation shows that for Pr > 1, ath < 6 Pohlhausen has suggested that the following relation is general may be between ~ 1 1 the results in Eqns. (7.64), (7.65) and (7.68) valid for Pr > 0e5.

\the thermal and hydrodynamic boundary layers: 7.1.7. lote& Energy aquation (Approximate ~olution of enera equation) 6 6, = -

... (7.62) Consider a volume shown in Fig. 7.10. Assume that p, cp and (thermo-plastic

(PrV3 Hydrodynamic 7 boundary layer . , The local and average heat transfer coefficients: At the surface of the plate, since there is no fluid motion and the heat transfer can occur

u, t , only through conduction, the heat flux may be written as

Qh Thermal boundary I

B I f I--- #"#-- ' I

I

- I I Control I

.mlnrne I

4 From the relation 7.63, we may develop (i.e., surface temperature gradient) as

- 4

~-xoL .__ ) l Q c k e n - 4

Thin flat plate

I+----x -. _ rr 9 n 1 , + - ~ - ~ 1 ener~V eauation - control volume.

382 Heat and Mass Transfer. ,

propertids) of fluid remain constant within the operating range of the temperature, and the heating of the plate commences at a distance xo from the leading edge of the plate (so that the boundary layer initiates at x = xo and develops and grows beyond that). For unit width of the plate we have:

H Mass of fluid entering through face AB = jo pu dy

...( 7.69) H a "

Mass of fluid leaving through face CD = 4 pu dy + [Io pu d y ] d r ...( 7.70) :. Mass of fluid entering the control volume through face BC

= [jOH Pu dy + $ {joH pu dy) dx] - joH pu dy = $ [joH pu dy] dx -47.71; Heat influx through the face AB,

Q, = mass x specific heat x temperature

Heat efflux through the face CD,

Heat (energy) influx through the face BC (which is outside thermal boundary layer and there the temperature is constant at t,),

Heat conducted into the control volume through face AD,

The energy balance for the element is given by

After simplification and rearrangement, we have

Equation (7.76) is the integral equation for the boundary layer for constant properties and constant free stream temperature t,.

If the viscous work done within the element is considered, then Eqn. (7.76) becomes

[when d r dy = viscous work done within the element ... Eqp. (7.8)] P C , 0 ay2

Usually the viscous dissipation term is very small and is neglected (and may be considered only when velocity of flow field becomes very large). Exp-ion for the convective heat transfer coefficient for laminar flow over a flat blab:

In order to derive an expression for convective heat transfer coefficient for laminar flow

Forced Convection

over a flat plate (that has an unheated starting length xO), let US use cubic velocity and temperature distributions in the integral boundary layer energy equation as follows:

he cubic velocity profile within the boundary layer is of the fom; ...[ Eqn. (7.33)l

U The conditions which are satisfied by the temperature distribution within the boundary layer

are:

(i) At y = 0, t = ts

- , r

W e temperature distribution takes the following fom:

-=--

... (7.79)

B,, putting the proper values of velocity distribution and temperature distribution into the

integrind would be zero). 6th

After putting a = and earlying out the integration, Eqn. (7.80) gets reduced ... (7.81)

\-,, - Neglecting the term involving r4 (because c 6, r < 11, we have

3 d = - 20 U (t, - t,) 5 (62)

Fu~her , from Eqn. (7.77). we kave

386 Heat and Mass Transfer . Forced Convection ,

(v) Thickness of the boundary layer, (vi) Local convective heat transfer coeflcient,

(vii) Average convective heat transfer coeflcient, (viiil Rate of heat transfer by convection, (k) Total drag force on the plate, and (x) Total mass flow rate through the boundary.

Solution. Given: U = 3mls. x = 280 mm = 0.28 m, p = 1.1374 kg/m3, k = 0.02732 W/mC, cP = 1.005 W g K , v = 16.768 x 10d m2/s.

-

(vii) Average convective heat transfer eoeff~cient, h:

/

(viii) Rate of heat transfer by convection, Qco,,: -

Let us first amxtain the type of the flow, whether laminar or turbulent. ux

- Re, = - - = 5.0 x 104 V 16.768 x lo4

Since Re, < 5 ' ~ lo5, hence flow in laminar.

(ix) Total drag force on the plate, FD: FD = q, x area of plate on one side upto 0.28 m

= 0.01519 x 0.28 x 0.28 = 0.00119 N (Ans.) (x) Total mass flow 5 rate through the boundary, m :

m = - pU (6, - 61) 8

(0 Boundary layer thickness at x = 0.28 m, 6:

or 5 x 0.28 6 = 7 = 0.00626 m or 6.26 mm 5 x lo4 (2) Local friction coefficient, C'.: x 1.1374 x 3(0.00626 - 0) = 0.01335 kgls (Ans.)

= 8 Example 7.11. Air at atmospheric pressure and 200C flows over a plate with a velocity

of 5 d s . The plate is 15 mm wide and is maintained at a temperature of 120C. Calculate the thicknesses of hydrodynamic and thermal boundaiy layers and the local heat transfer coefficient at a distance of 0.5 m from the leading edge. Assume that /low is on one side of the plate. or 0.664 4 = - = 0.002969

5 x lo4 (iii) Average friction coefficient, C' p = 0.815 kg/rn3; p = 24.5 x 1 r 6 iVs/m2, Pr = 0.7. k = 0.0364 W/m K . (AMIE Summer, 1997)

Pr = 0.7; k = 0.0364 W / m K. Let us first ascertain the type of flow, whether laminar or tubulent. or

- = 0.005939 c f = x = z F (iv) Shearing stress due to friction, To :

Since Re, < 5 x Id, hence flow is laminar. Boundary layer thickness at x = 0.5 m, 6 :

= 0.002969 x 32 = 0.01519 N / ~ ~ 2 (v) Thickness of thermal boundary layer, &,, : 5x 5 x 0.5 6 = - = - = 8.669 x lo-) m or 8.669 mm (Ans.) f ie" 4iZE

Thickness of thermal boundary layer, at x = 0.5 m, St.: 6 6, = - (Pr) I r n ...[ Eqn. (7.62) s t h = - - = 8.669 9.763 mm (Ans.) ( P ~ ) ' / ~ - (0.7)"~

- 0.00626

- - = 0.00705 m or 7.05 mm (0.7)'" (vi) Local convective heat transfer coefficient, 4 : Local heat transfer coefficient, h, : . . . [Eqn. (7.64)l

k hx = a332 - (Re,)'" (pr)'" X

= 0.332 x x (83l63)ln x (0.7)'" = 6.189 w/m2 K (Ans.) ...[ Eqn. (7.64)] -.-

Example 7.12. Air at atmospheric pressure and 40C flows with a velociry of (I= 5 m/s over a 2 m long f k ~ plde whose surface is kept at a uniform temperature of 120C. Determine

Heat and Mass'Transfer

the average heat transfer coeflcient over the 2 m length of the plate. Also find out the rate of heat transfer between the plate and the air per I m width of the plate. [Air at 1 arm. and 80C, v = 2.107 x 10- m2/s, k = 0.03025 W/m K, Pr = 0.69651 (AMIE Winter, 1998)

Solution. Given : t, = 40C; U = 5 m/s; L = 2 m, t, = 120C; B = 1 m

The properties of air at mean bulk temperature of

v = 2.107 x 10- m2/s; k = 0.03025 W/m K; Pr = 0.6965. Average heat transfer coefficient, h :

Assuming Re,,= 5 x lo5, the flow is laminar. Using exact solution, the average Nusselt number is given by

= 0.664 (~e,)'" ( ~ r ) - ...[ Eqn. (7.68)]

or hL = 0.664 (4.746 x 1@)ln (0.6965)'" = 405.48 k

Rate of heat transfer, Q : Q = ZA, ( t , - t,)

= 6.133 x (2 x 1 ) (120 - 40) = 981.28 W (Ans.) Example 7.13. Air at 27C and I barflows over a plate at a speed of 2 m/s. ( i ) Calculate the boundary layer thickness at 400 mm from the leading edge of the plate.

Find the massflow rate per unit width of the plate. For air p = 19.8 x 10- kg/ms at 27C.

(ii) If the plate is maintained at 60C, calculate the heat transferred per hour. The properties of air at mean temperature of (27 + 6 0 ) d = 43.5' C are given below :

v = 17.36 x 10- m2/s; k = 0.02749 W/mC c,=lOOBJ/kgK; R=287Nm/kgmK; Pr=0.7. (M.U.)

Solution. Given : t = 27C; p = 1 bar, U = 2 m/s; x = 400 rnm = 0.4 m (i) Boundary layer thickness, 6 :

4.64 x Boundary layer thickness, 6 = - * 4.64 x 0.4 6 = ,-J===- = 0.00857 m or 8.57 mm (Ans.) 46869

The mass flow rate per metre width is given by,

... assumed Now,

5 = z x 1.16 x 2 x 0.00857 = 0.01242 kgh (Ans.) -. 8

Note, ~f h e mass added in the boundary is to be calculated when h e fluid moves from X , 10 + along the main flow direction then it is given by

S Am =: PU (8, - 8,)

where 6 , and l& are the boundary layer thicknesses at X l and q. -

(ii) Heat transferred per hour, Q : -

-

15'76 3600 = 416.74 kl/h (Ans.)

- 1000

Example 7.14. Air at 1 bar and at a temperature of 30C (Y. = 0.06717 kghrn) flows at a speed of I.2ds over a fiat plate. Determine the boundary layer thickness at distance of So mm and 500 mm from the leading edge of the plate. Also, calculate the mass entrainment bemeen these m o sections. Assume the parabolic velocity distribution as: - = -

Solution. Given : t.. = 30c, )I = 0.06717 kglhm, U = 1.2 mls Boundary layer thicknesses:

E-= 1 x lo5 The density of air,

= 1.15 kg/m3 = RT 287 x (30 + 273)


4 . 6 4 ~ Boundary layer thickness, 6 , = - ... [Eqn. (7.36)]

6 , = 4.64 x 0.25 = 0.00853 m or 8.53 mm (Ans.)

.I18490

Re, = 1.15 x 1.2 x 0.5 x 3600 0.067 17 = 36981

.. Boundary layer thickness, 62 = 4.64 x 0.5 = 0.01206 m = 12.06 mm (Ans.) Gzi-

Mass entrainment: The mass flow rate at any position in the boundary layer is given by

:. The mass entrainment between the two sections be., at x = 250 mm and x = 500 mm.

= 3.045 x kg/s = 10.96 kg/h (Ans.) Example 7.15. Air at 20C is flowing over a flat plate which is 200 mm wide and 500 mm

long. The plate is maintained at 100C. Find the heat loss per hour from the plate if the air is flowing parallel to 500 mm side with 2m4 velocity. What will be the efect on heat transfer if the flow is parallel to-200 mm side.

The properties of dir at (100 + 20)D = 60C are: v = 18.97 x 1@ m2/s, k = 0.025 W/mC and Pr = 0.7. (M.U.) . --, Solution. U = 2 m/s, v = 18.97 x lo4 m2/s, k = 0.025 W/mC and Pr = 0.7.

Heat loss per hour from the plate, Q: Case I. When the flow is parallel to 500 mm side:

-

hL N u = - - - 0.664 (Re,)'" (pr)'" k ...[ Eqn. (7.68)]

- where, Re, = - - uL OS5 = 127 x lo4 V 18.97 x 10"

I I . . Q = ~ s ( t s - t , ) = 6 . 7 6 7 ~ ( 0 . 5 ~ 0 . 2 ) ( l W - 2 0 ) = 5 4 . 1 4 W (Am.) Case 11. When the flow is parallel to 200 mm side: Re, = 2 x 0.2 = 2.11 x lo4 18.97 x lo6

. . h = - - ''O2' 0.2 x 0.664 x (2.11 x 1 0 ~ ) " ~ (0.7)"~ = 10.7 w/m2~c (Ans.)

. . Q = 3; xA, ( t , " -t,) = 10.7 x (0.2 x 0.5) x (100- 20) = 85.6W (Ans.) Example 7.16. In a certain glass making process, a square plate of glass 1 m2 area and 3

mm thick heated uniformly to 90C is cooled by air at 2U'Cflowing over both sides parallel to the plate at 2 m/s. Calculate the initial rate of cooling the plate.

Neglect temperature gradient in the glass plate and consider only forced connection. Take for glass : p = 2500 kg/m3 and cp = 0.67 W / k g K Take the following properties of air :

- .

p = 1.076 kg/mJ; cp = 1008 JAcg K, k = 0,0286 W/mO C and p = 19.8 x l(T6 N-$/mL. (N.U., 1997)

Solution. Given : A = 1 m2; t, = 90C; t, = 2QC, U = 2 d s . 7 Plate at 90C -

Fig. 7.11 Initial rate o f cooling : The average heat transfer co-efficient for the air flow parallel to the plate is given by

-

- hL Nu = - K = 0.664 ( ~ e , ) " ~ (pr)'l3 .[Eqn. 7.68)]

(valid for Pr > 0.5)

Substituting the values in the abovgeqn, we get -

hL= k 0.664 x ( 1 .087x 10")"~ x (0.698)]/~ = 194.19

The heat flow (Q) from both sides of the platet is given by : ~ = 2 h A ( t , - t J = 2 ~ 5 . 5 5 ~ 1 x ( 9 0 - 2 0 ) = 7 7 7 ~

The heat lost by the plate instantaneously is given by : Q=mc,,At=777 E

where m = (area x thicknes$ x p

/

Example 7.17. A flat plate, l m wide and 1.5 m Long is to be maintained at 90C in air with a free stream temperature of 10C. Determine the velocity with which air must flow over

392 Heat and Mass Transfer - f

flat plate along 1.5 m side so that the rate of energy dissipation from the plate is 3.75 kW. Take the following properties of air at 50C:

p = 1.09 kg/m3, k = 0.028W/mC, cp = 1.007 kJ/kgC, p = 2.03 x I @ kg/m-s Pr = 0.7. (M.U.)

Solution, Given : L = 1.5 m, B = lm, t, = 90C, t, = 10C, Q = 3.75 kW Properties of air at (90 + 10)/2 = 50C : p = 1.09 kg/m3, k = 0.028 W/mC, cp = 1.007

kT/kgC, p = 2.03 x lo-' kg/m-s Free stream velocity, U: The heat flow from the plate to air is given by . s ' <

Q = j; A, (t, - t,) k

where h = - x 0.664 (~e,)'" (pr)ln L ...[ Eqn. (7.68)]

or U = 100 m/s (Ans.) Example 7.18. Air at 20C and at atmospheric pressure jlows over aflat plate at a velocity

of 1.8 4 s . If the length of the plate is 2.2 m and is maintained at 100C, calculate the heat transfer rate per unit width using (i) exact and (ii) approximate methods.

The properties of air at mean bulk temperature of (100 + 20)D = 60C are: p = 1.06 kg/m3, cp = 1 .OO5 kJ& OC, k = 0.02894 W/mC, Pr = 0.696 v = 18.97 x 106 m2/s

Solution. Given : t, = 20C, t, = 100C, U = 1.8 mls, L = 2.2 m, B = 1 m Heat transfer rab per unit width:

Reynolds number,

Since Reynolds number is less than 5 x 10' hence flow is laminar. (i) Using exact solution: The average Nusselt number is given by

-

Nu = 0.664 (Re, ) I n ( ~ r ) ' " . . . [Eqn. (7.68)J

;. Heat transfer rate from the plate, Q = A, (t, - t,) = 3.536 x (2.2 x \l) (100 - 20) = 62234 W (Am.)

(ii) Using approximate solution: -

= && = 0.646 (Re, )'" (pr)lD -

k

or = 0.646 (2.087 x 1 0 ' ) ' ~ (0.696)'" = 261.53 k - 261.53 k 261.53 X 0.02894 = 3.44 Wlm2 OC

or h=-= L 2.2

r- -

(i) The average skin friction, CI : UL 1.8 x 0.75 = 84375

Reynolds number, - Re, = --$ - 1.6 x lo4

laminar in nature.

(ii) The average shear stress, T,: 1 -

rw = -pu2 X Cf 2 = & x 1.165 x 1.g2 x 0.004572 = 0.008629 ~ l m ' (Ans.)

L

(iii) The ratio of average shear stress to the shear stress at the trailing edge: The skin friction coefficient at the trailing edge (x = L),

0.664 Cfx =

... [Eqn. (7.24)J

;. Shear stress at the trailing edge, 1

r , = 5 p d Cgi = x 1.165 x 1.8~ x 0.002286 = 0.004314 ~ 1 n - i ~ 2

Heat and Mass TrAhsfer

Example 7.20. Air at 3UCJlows with a velocity of 2.8 m/s over a plate 1000 mm (length) x 600 mm (width) x 25 mm (thickness). The top sugace of the plate is maintained at 90C. I f the thermal conductivity of the plate material is 25W/mC, calculate :

( i ) Heat lost by the plate; (ii) Bottom temperature of the plate for the steady state condition. The thermo-physical properties of air at mean film temperature (90 + 30)Q = 60C are:

p = 1.06 kg/m3, c,, = 1.005 kJAg K, k = 0.02894 W/mC, v = 18.97 x I@ m2/s, Pr = 0.696. Solution. Given : t, = 30C, ts = 90C, U = 2.8 d s , kplate = 25 W/mC, L = 1000 mm =

lm, B = 600 mm = 0.6 m, 6 = 25 mm = 0.025m. ( i) Heat lost by the plate: Reynolds number at the trailing edge,

Since Reynolds number is less than 5 x lo5, hence flow is laminar throughout the length,

h~ Nu = 0.664 (ReL )'I2 (pr)'I3 = - k ...[ Eqn. (7.68)]

Nu x k 0.664 ( ~ e , ) " ~ (pr)'l3 x k or Ti (average heat transfer coefficient) = - - L - L

= 6.542 w/m2 "C :. Heat lost by the plate, Q = h A, (t, - t,) or Q = 6.542 x (1.0 x 0.6) (90 - 30) = 235.5 W (Ans.) (ii) Bottom temperature of the plate, t,: Heat lost by the plate Q (calculated above) must be conducted through the plate, hence

exchange from top to bottom surface is

tb = 90 + 0.025 x 235.5 25 (1.0 x 0.6) = 90.39"C (Ans.)

Example 7.21. Air at 30C and at atmospheric pressure flows at a velocity, of 2.2 m/s over a plate maintained at 90C. The length and the width of the plate &re 900 mm and 450 mm respectively. Using exact solution, calculate the heat transfer rate from,

( i ) first half of the plate, (ii) full plate, and (iii) next half of the plate. The properties of air at mean bulk temperature (90 + 30)Q = 60C are: p = 1.06 kg/m3, y = 7.211 kghm, v = 18.97 x I@ m2/s, Pr = 0.696, k = 0.02894 W/mC. Solution. Given : t, = 30Q, U = 2.2 mls, t, = 90C, L = 900mm = 0.9m, B = 450 mm

= 0.45 m.

Forced Convection

( i ) Heat transfer rate from first half of the plate: For first half of the plate,

Since Re < 5 x lo5 hence the flow is laminar. The local Nusselt number is given by,

Nu, = 0.332 (~e,) '" ( ~ r ) " ' ...I Eqn. 7.651

But,

Average heat transfer coefficient, = 2h, = 2 x 4.322 = 8.644 wlm' "C :. Heat transfer rate from first half of the plate,

Q = h A, (t, - 2,) = 8.644 x (0.45 x 0.45) (90 - 30) = 105 W (Ans.)

(ii) Heat transfer rate from full plate: For full plate, x = L = 0.9 m

The heat transfer rate - from entire plate, Q, = h A, (t, - t,) = 6.1 13 x (0.9 x 0.45) x (90 - 30) = 148.54 W (Ans.)

(iii) Heat transfer rate from next half of the plate: Heat transfer rate from the next half of the plate

= Q, - Q, = 148.54 - 105 = 43.54 W (Ans.) Example 7.22. Castor oil at 25C flows at a velocity of 0.1 m/s past a flat plate, in a

certain process. I f the plate is 4.5 m long and is maintained at a un@rm temperature of 9jC, calculate the following using exact solution:

( i ) The hydrodynamic and thermal boundary layer thicknesses on one side of the plate, (ii) The total drag force per unit width on one side of the plate, (iii) The local heat transfer coeflcient at the trailing edge, and (iv) The heat transfer rate. The thermo-physical properties of oil at mean film temperature of (95 + 25)R = 60C are:

p = 956.8 kg/m3; a = 7.2 x I@ m2/s; k = 0.213 W/mC; v = 0.65 x I @ m2/s. Solution. Given : t, = 25"C, t, = 95"C, L = 4.5m, U = 0.1 mls.

&g Heat and Mass Transfer Forced Convection (iii) Local and average convective heat transfer coeflcients; (iv) Heat transfer rate from both sides for unit width of the plate, Iv) Mass entrainment in the boundary layer,

( i ) The hydrodynamic and thermal boundary layer thicknesses, 6, tith: Reynolds number at the end of the plate,

\ ,

(vi) The skin friction coefficient. Assume cubic velocity profile and approximate method. The thermo-physical properties of air at mean film temperature (60 + 2Oyr' = 40C are: p = 1.128 kg/rn3, v = 16.96 x 1@ m2/s, k = 0.02755 W/mC, PI = 0.699. Solution. Given: t, = 20C, t, = 60C, U = 4.5 mls.

Ux-

Since Reynolds number is less than 5 x lo5, hence the flow is laminar in nature. The hydrodynamic boundary layer thickness,

= 0.2704 m or 270.4 mm (Ans.) ...[ Eqn. (7.22)]

The thermal boundary layer thidkness, according to Pohlhausen, is given by: Re, x v 5 x id x 16.96 x lo4 =

or x , = = U 4.5

(where x, = distance from the leading edge at which the flow in the boundary layer changes from where Pr (Prandtl number) = (0'65 lob4 = 902.77

a 7.2 x lo4 laminar to turbulent). ( i ) Thickness of hydrodynamic layer, 6: m e thickness of hydrodynamic layer for cubic velocity profile is given by

i 4.64 x,

6 = 7 ...[ Eqn. (7.36)] 0'2704

= 0.02798 m or 27.98 mm (Ans.) 6th = (902.77)'"

(ii) The - total drag force per unit width on one side of the plate FD: The average skin friction coefficient is given by,

...[ Eqn. (7.25)1 (ii) Thickness of thermal boundary layer, The thermal boundary layer is given by

0.975 6 ...[ Eqn. (7.8911 1 The drag force, FD = x - u2 x area of plate (for one side)

f 2 P ...[ Eqn. (7.31)]

6th = !?.975 * 0.01234

= 0.01355 m or 13.55 mm (Ans.) (0.699)"~

1 or FD = 0.01596 x y x 956.8 x 0.12 x (4.5 x 1 ) = 0.3436 N per meter width (iii) The local heat transfer coefficient at the trailing edge, h, (at x = L):

h x Nu, = = 0.332 (~e,)'" (pr)'I3 k

\-- I

The Nusselt number at x = x, is given by ln pr)113 NIL = 0.332 (Re,) ( ...[ Eqn. i'1.65)I = 0.332 x (6923)In (902.77)'" = 266.98

. . . [Eqn. (7.65)] or h, = 266.98 - - 266'98 0a213 = 12.64 wlm2 OC (Ans,)

X 4.5 h X x , Nu, X k Nu, = or h, = - Put k XC (iv) The heat transfer rate, Q: Q = z A, (t, - t,) -

where h = 2h, = 2 x 12.64 = 2 5 . 2 8 ~ l m ~ 0 ~ Eqn. (7.6711 ,. Q = 25.28 X (4.5 x 1) (95 - 25) = 7963.2 W (Am.) Example 7.23. Air at 20C and at atmospheric pressure jlows at a velociry of 4.5 m/s past

a jlat plate with a sharp leading edge. The entire plate sur$ace is maintained at a temperature of 60C. Assuming that the transition occurs at a critical Reynolds number of 5 x 16, find the distance from the leading edge at which the flow in the boundaly layer changes from laminar to turbulent, At the location, calculate the following :

(i) Thickness of hydrodynumic layer; (ii) Thickness of therml boundary layer;

Average heat transfer coefficient, x=Lf hxak

xc - - -

I h = 2hc = 2 x 3.05 = 6.1 w/m2'C (Am.)

(iv) H a t transfer rate from both sides for unit width of the plaki Q : Q = z (us) At =. 6.1 (2 x 1.88 x 1) (60 - 20) = 917.U W (Am)

Heat and MassqTransfer

(v) Mass entrainment in the boundary layer, in:

Here, 6, = 0 at x = 0 and 6, = 0.01234 m at x = x, = 1.88 m . .

5 m = - x 1.128 X 4.5 (0.01234 - 0 ) = 0.039 kgls or 140.4 kglh (Ans.) 8

(vi) The skin friction coefficient, C': ...[ Eqn. (7.24)1

0.646 CB = d- = 9.136 X lo4 (Ans.) 5 x 10' Example 7.24. A stream of water at 20C ( p = 1.205 kg/m3, p = 0.06533 kg/hm)Jows at

a velocity of 1.8 4 s over a plate 0.6 m long and placed at zero angle of incidence. Using exact solution, calculate:

(i) The stream wise velocity component at the midpoint of the boundary layer, (ii) The maximum boundary layer thickness, and

(iii) The maximum value of the normal component of velocity at the trailing edge of the plare. Solution. Given t, = 20C, p = 1.205 kg/m3, p = 0.06533 kghm, L = 0.6 m, U = 1.8 m/s (i) The stream wise velocity component at the midpoint of the boundary layer, u: The boundary layer thickness by exact solution is given by

Obviously, the midpoint of the boundary layer y = - occurs at i 3 The stream wise velocity component is obtained form the Blasius solution in tabular form

(Refer table 7.1).

At U 11 = y $ = 2.5, we get - = 0.736 U

or u = 0.736 U = 0.736 x 1.8 = 1.325 m/s (Am.) (ii) The maximum boundary layer thickness, 6,: The maximum boundary layer thickness occurs at x = 0.6 m. Thus.

Re, = puL 1.205 x 1.8 x 0.6 - - p (0.06533/3600) = 71713, hence flow in laminar.

The boundary layer thickness at the trailing edge, '

5L 5 x 0.6 a,=---- 6-m- 0.01 12m or 11.2 mrn (Am.) -

(iii) The inaxilduni dalue of the normal component of velocity at the trailing edge, v : !% < $ $ t The maximum value of the normal cornfionent of velocity occurs at the outer edge of the

Forced Convection

u boundary layer where u = U. Hence for - = 1, we have U

" = 0.86 (Refer Table 7.1) U

7.2 LAMINAR TUBE FLOW 7.2.1. Development of Boundary Layer

In case of a pipe flow, the development of boundary layer proceeds in a fashion similar to that for flow along a flat plate. A fluid of uniform velocity entering a tube is retmded near the walls and a boundary layer begins to develop as shown in Fig. 7.12 by dotted lines. n e thickness of the boundary layer is limited to the pipe radius because of the flow being within a confined passage. Boundary layers from the pipe walls meet at the centre of the pipe and the entire flow acquires the characteristics of a boundary layer. Once the boundary layer thickness becomes equal to the radius of the tube there will not be any further change in the velocity distribution, this invariant velocity distribution is called fully developed v e l c p r o ie. , poiseulle flow (parabolic distribution).

Boundary Layer -1 ~ d l v developed

c-- k- Entrance length (L,) -I Fig. 7.12. The development of a laminar velarity profile in the intake region Le of a tube.

According to Langhar (1942), the entrance length (L,) is expwscd : 5 = 0.0575 Re where D represents the inside diameter of the pipe. 7.2.2 Velocity Distribution

Fig. 7.13 shows a horizontal circular pipe of radius R, having laminar flow of fluid through it. Consider a small concentric cylinder (fluid element) of radius r and length dx as a free body.

If z is the shear stress, the shear force F is given by F = z x 2 n r ~ d u

Let p be the intensity of pressure at left end and the intensity of pmsum at the dght end

' Thus the forces acting on the fluid element are: 1. The shear force, T x 2zr x dn on the surface of fluid element.

-

4m

2. The pressure force, p x n? on the left-end.

T


( 3. The pressure force, p + - . dr n? on the right end. ( E ] For steady flow, the net force on the cylinder must be zero.

Integrating the above equation w.r.t. 'r', we get --- -0-- @

1 Pipe 4 p dX Where C is the constant of integration and its value is obtained from the boundary condition:

Substituting this. value of C in Eqn. (7.97), we get

it--.------ x2 -------3: Fig. 7.13. Laminar flow through a circular pipe.

Equation 7.98 shows that the velocity distribution curye is a parabola (see Fig. 7.13). The maximum velocity occurs at the centre and is given by

From equation 7.98 and 7.99, we have a

II ' - Equation (7.95) shows that flow will occur only if pressure gradient exists in the direction - \ , - Equation 7.100 is the most commonly used equation for the velocity distribution for laminar

flow through pipes. This equation can be used to calculate the discharge as follows: The discharge through an elementary ring of thickness dr at radial distances r is given by

of flow. *. The negative sign shows that pressure decreases in the direction of flow.

- Equation (7.95) indicates that the shear stress varies linearly across the section (see Fig. 7.14). Its value is zero at the eentre of pipe ( r = 0 ) and maximum at the pipe wall given by I

I ... [7.95 (a)] Ffom Newton's law of viscosity,

1

I du 7 = p . - dv ...f i) Total discharge Q = ! ~ Q

--,

In this equation, the distance y is measured fmm the boundary. The radial distance r is related to distance y by the relation

Shear stress Velocity distribution distribution curve

y = R - r or d y = - d r The eqn. (i) becomes

" R2 - Q u,,

Average velocity of flow, u = - A = =- nR2 2

Comparing two values of T frm Eqns. 7.95 and 7.96, we have

Fig. 7.14. Shear stress and velocity distribution across a section. ..-.

402 Heat and Mass Transfer Forced Convection

Equation (7.101) shows that the average velocity is one-halfthe maximum velocity. Substituting the value of u,, from Eqn. (7.99), we have

ap The pressure gradient - is usually expressed in terms of a friction factor f, defined as ax

pU where - is dynamic pressure of the mean flow and D is the tube diameter. 2

From Eqn. (7.102) and (7.103), we get the friction factor as a simple function of Reynolds number,

which is valid for laminar tube flow, Re < 2300 Further, Eqn. (7.102) can be written as,

8p U -ap = -.ax

R2 The pressure difference between two sections 1 and 2 at distances, x, and x2 (see Fig. 7.12),

is given by

Obviously the head loss hL over a length of pipe varies directly as the first power of the rate of discharge Q and inversely as the fourth power of the pipe diameter. 7.2.3. Temperature Distribution

In order to estimate the distribution of temperature let us consider the flow of heat through an elementary ring of thickness dr and length dx as shown in Fig. 7.15. Considering the radial conduction (neglecting axial conduction) and axial enthalpy transport in the annular element, we have:

Heat conducted into the annular element,

dQr = - at k (271r.d~) - a r

Heat conducted out of the annular element,

, -

Net heat convected out of the annular element,

Elementary rine

~dnular Qr + dr element

Fig. 7.15. Analysis of energy in the tube flow.

Considering energy balance on the annular element, we obtain (Heat conducted in),,, = (Heat convected out),,,

dQr - dQr+dr = (dQmn,)ne, dt

= p (2nrdr)u c - dx ax at at

- k (2nr.h) - + k (271 (r + dr)} dx [$ + $ dr] = p (2nr.dr)u c - dx a r ax

at + k ( 2 n d r d x ) - a r

at

Neglecting second order terms, we get at

&.dr = p r u c --dx.dr ax 1 a

or ...( 7.106)

Inserting the value of u from eqn. (7.100), we get, ...( 7.107)

at or

...( 7.108)

at Let us consider the case of uniform heat flux along the wall, where we can take - as a ax

constant. Integrating Eqn. (7.108) we have

404 Heat and Mass TrAnsfer Forced. Convection 405 .

Integrating again, we have

(where C; and C2 a& the constants of integration). The boundary conditions are:

At r = 0, at - = 0 ar

At r = R, t = ts Applying the above boundary conditons, we get

Substituting the values of C, and C2 in Eqn. (7.109), we have

For determining the heat transfer coeficient for fully developed pipe flow, it is imperative to define a characteristic temperature of the fluid. It is the bulk temperature (tb) or the mixing up temperature of the fluid which is an average taken so as to yield the total energy carried by the fluid and is defined as the ratio of flux of enthalpy at a cross-section to the product of the mass flow rate and the specific heat of the fluid. Thus,

l R p ( 2 n r . d r ) u c p t tb =

O ...( 7.111) lR p (2 nr . dr) u cp

0

For an incompressible fluid having constant density and specific heat

l R u r r d r tb =

O ...( 7.112)

I R u r d r o

The averagetmean velocity (i) also known as the bulk mean velocity is calculated from the following definition:

- 2 R u = 3 In u r d r

"

Substituting this value of u in eqn. (7.1 12), we get

Substituting the value of u from eqns. 7.1 10 and 7.101 and that of t from eqn. (7.1 lo), we get

2 2 t b = s f o 2; [I - s ] b s - k a ' a x k { g - < + L } ] r d r 16 4 16R2

4

at 11 U * m R 2 - or tb = t s - - - 96 a ax The heat transfer coeflcient is calculated from the relation

From eqn. (7.1 lo), we have

where D is the diameter of the tube. The Nusselt number is given by

This shows that the Nusselt number for the fully developed laminar tube flow is constant and is independent of the Reynolds number and Prandtl number.

The first analytical solution for laminar flow for constant wall temperature was formulated at

by Graetz in 1885. Since - ax is not constant, therefore, the analysis of constant wall temperature is quite cumbersome. The final result comes out to be

Example 7.25. For laminar flow in a circular tube of 120 mm radius, the velocity and temperature distribution are given by the relations:


u = (2.7r - 3.2 ?) ; t = 85 ( I - 2.2r)OC where the distance r is measured from the tube surfnce. Calculate the following:

( i ) The average velocity and the mean bulk temperature of the fluid; (ii) The heat transfer coeficient based on the bulk mean temperature if the tube su$ace is

maintained at a constant miform temperature of 90C and there occurs a heat loss of 1000 kJ/h per metre length of the tube.

Solution. Given : u = (2.7 r - 3.2 3) ... Velocity distribution

t = 85 (1 - 2.2 r)OC ... Temperature distribution. ( i ) Average velocity ( Z ) and mean bulk temperature (t,) :

The average velocity is obtained by equating the volumetric flow to the integrated flow through an elementary ring of radius r and thickness dr.

= f R (2.7 r - 3.2 ?) rdr R2 0

Substituting R = 0.12 m, we have

. . . ( i )

Now,

u = 1.8 x 0.12 - 1.6 x 0.12' = 0.193 m/s (Am.) The mean bulk temperature is given by

lR u t r d r tb = 0 lR u r d r

0

j R u t r d r = J~ (2.7r - 3 . 2 q x 85(1-2 .2r)rdr 0 n .

... From eqn. (i)]

Forced Convection

Substituting, R = 0.12 m and ii = 0.193 mls, we get 170 (0.9 x 0.12 - 2.285 X 0.12' + 1.408 x 0.12)) = 68.290C (Ans.) tb = 0.193

(ii) Heat transfer coefficient, h: Q = hA (t, - tb) -

looo looo = 277.77 Jls ; t, = 90C ... Given

where Q = 1000 Mh per metre = 3600

277.77 h = = 16.97 W / m 2 OC (Am.) (271 x 0.12 x 1 ) (90 - 68.29)

Example 7.26. Lubricating oil at a temperature of 60C enters I cm diameter tube with a velocity of 3 d s . The tube su$ace is maintained at 40C. Assuming that the oil has the following average properties calculate the tube length required to cool the oil to 45OC.

p = 865 kg/&; k = 0.140 W/m K: cp = 1.78 kJ/kgC. Assume flow to be laminar (and fully developed)

-

Nu = 3.657 (AMIE Summer, 1997) Solution. Given : ti = 60C, to = 45'C; D = 1 cm = 0.01 m; U = 3 m/s, t, = 40C;

p = 865 kg/m3; k = 0.14 W/m K; cp = 1.78 kJ/kgC.

40C Fig. 7.16

Length required, L : Q = m cp (ti - to)

= (PA# cp 0, - to) (where U = average velocity, Af= flow area)

I Also, ~ = h A 0 , j where A = heat transfer area = nDL, and


... ( Given)

Now, Q=5441.7=51.2XmLx 10.82

. . L = 5441.7 = 312.7 m (Ans.) 51.2 x n x 0.01 x 10.82

Example 7.27. When 0.5 kg of water per minute is passed through a tube of 20 mm diameter, it is found to be heated from 20C to 50C. m e heating is accomplisted by condensing steam on the sulface of the tube and subsequently the surface temperature of the tube is maintained at 85OC.Detennine the length of the tube required for fully developed flow.

Take the thermo-physical properties of water at 60C as : p = 983.2 kg/m3, cp = 4.178 kJ/kgK, k = 0.659 W/mC, v = 0.478 x 1@ m2/s Solution. Given : m = 0.5 kglmin, D = 20 mm = 0.02 m, ti = 20C, to = 50C

Length of the tube required for fully developed flow, L:

'The mean film temperature, rj = f /8fi + 2o + 50) = 6OOC 2

, Let us first determine the type of the flow. /

Reynolds number, D . U 0.02 x 0.0269 R e = - - - = 1125.5 V 0.478 x lov6

Since Re < 2000, hence the flow is laminar. With constant wall temperature having fully developed flow,

Nu = - hD = 3.65 ...[ Eqn. (7.11 8)] k

The rate of heat transfer, Q = A, h (t, - t,) = m cp (to - ti)

Here, t, = 20 + 50 2 = 35C = t,

or L = -- - 2.76 rn (Ans.) 377.8

B. TURBULENT FLOW 7.3. INTRODUCTION

The flow in the boundary layer, in majority of practical applications in the convective heat transfer, is turbulent rather than laminar. In a turbulent flow the irregular velocity fluctuations are mainly responsible for heat as well as momentum transfer. As the mixing in the turbulent flow is on a macroscopic scale with groups of particles transported in a zig-zig path through the fluid, the exchange mechanism is many times more effective than in laminar flow. Consequently. in turbulent flow, the rates of heat and momentum transfer and the associated friction and heat transfer coefficients are several times larger than that in laminar. Since the nature of turbulent flow is complex, therefore, it is difficult to solve the problems relating turbulent flow analytically. The heat transfer data can best be calculated by laboratory experiments; the other method of study is the analogy between heat and momentum transfer. A. Forced Convection-Flow over a Flat Plate 7.3.1. Turbulent Boundary Layer

Refer Fig. 7.17. As compared to laminar boundary layers, the turbulent boundary layers are thicker. Further in a turbulent boundary layer the velocity distribution is much more uniform, than in a laminar boundary layer, due to intermingling of fluid particles between different layers of the fluid. The velocity distribution in a turbulent boundary layer follows a logarithmic law i.e. u - log y, which can also be represented by a power

u - log y

0.99 U law of the type 19)

1 o7 Laminar sublayer

$ = ~ 1 " ...( 7.120) . . Smooth

flat plate

1 where n = - (approx) for Re < 7 but > 5 x 105

This is known as one-seventh Fig. 7.17. Turbulent boundary layer. power law.

The Eqn. (7.120). however, cannot be applied at the boundary itself because at y = 0, - = - ~ 6 - l ' ~ y4'7 = OJ. This difficulty is circumvented by considering the velocity in the viscous [El ; \-' / laminar sublayer to be linear and tangential to the seventh-root profile at the point, where the laminar sublayer merges with the turbulent part of the boundary layer.

Blasius suggested the following relation for viscous shear stress:

(for Re ranging from 5 x 10' to lo7) Let us now find the values of 6, z,, Cfi, for the velocity distribution given by Eqn. (7.120)

in , i . , ) 1

41 0 Heat and Mass Transfer Forced Convection

(i) Boundary layer thickness, 6: U Substituting the value of in Von Karman integral eqn. [7.28 (a)], we have

[In the expression, above, the limits have been taken from 0 to 6 instead of 6' to 6 since the laminar sublayer (6') is very thin]

Now equating the eqns. (7.122) and (7.121), we have

1/4 1 or -- d6 - - 0.0225 [A] x - 72 dx P U @)'I4 (cancelling puZ on both sides).

or d6 = 0.0225 x

Integrating both sides, we have 4 - 6'14 = 0.23 14 - 5 (;J + c (where C = constant of integration)

Let boundary layer be assumed to be turbulent over the entire length of plate. \Hence, at x = 0, 6 = 0 :. C = 0

1/5 0.37 1~ = 0.0371 [&) x x = - (Re,)'"

6 0.371 or - - --

X (Re,)'" (ii) Shear stress, 70 :

Substituting the value of 6 from eqn. (7.123), we get P

= 0.0225 pu2 [ 371i p ux- (Re,) 'I5 b

...[ Eqn. (7.12111

(iii)'~ocal skin friction (drag) coefficient, CB: . c We know

Also

[Eqn. (7.124)l

1 r0 = cfX x 1. p d [Eqn. (7.30)l

Now equating the eqns. (7.124) and (7.30), we have

(iv) Average value of skin friction (drag) coefficient, Cf :


This is valid for 5 x lo5 < Re, < lo7. -

For Reynolds number between lo7 and lo9 the following relationship suggested by Prandtl and Schlichting holds good,

7.3.2. Total Drag Due to Laminar and Turbulent Boundary Layers When the leading edge is not very rough, the turbulent boundary layer does not begin at

the leading edge, it is usually preceded by the laminar boundary layer. The point of transition from laminar to turbulent layer depends upon the intensity of turbulence. The distance xc (Fig. 7.18) of the transition from the leading edge can be obtained from critical Reynolds number which normally ranges from 3 x lo5 to 3 x lo6. - I

___+

u, t, + I

pi- ~ a m i y layer I Leading edge Turbulent layer b Fig. 7.18. Drage due to laminar and turbulent boundary layers.

Drag force (FD = F) for the turbulent boundary layer can be estimated from the following relation:

Fturb. = (Fturb.)total - (Fturb.) x, where (Fturb,)total = the drag which would occur if a turbulent boundary extends along

the entire length of the plate, and (Ftu,.)xc = the drag due to fictitious turbulent.bundary layer from the leading

edge to a distance x,. Let us assume that the pikeis long enough so that Reynolds number is greater than lo7,

then the turbulent drag is given by 0.455 pu2 0072 pu2

x - x ( L x B ) - - Fturb. = Ooglo ~ e ~ ) ~ ~ ~ ~ 2 115 x 2 x (xc x B) . . . ( i ) (Re,)

where L = length of the plate, B = width of the plate, and U = free stream velocity.

The laminar boundary layer prevails within the length xc and its contribution to drag force is given by

Forced Cohvection 41 3

...( ii)

1.328 x, 0.455 L + ( loglo ~"4 . '~ - (Re,)"'

Assuming that transition occurs at Re, = 5 x 10';

1 - LBpu2 Also, Ftotal = c x - ~ A U ~ = C f x - f 2 2

[where cf = average value of skin friction (drag) coefficient] Equating the above two equations, we have

1670 0.455 ...( 7.129) cf = (loglo ~ e , ) ' . ~ ~ Re,

The above relation in general may be rewritten as

Fig. 7.19. Skin friction (drag) coefficient for smooth flat plates.


where the value of A depends upon the value of critical Reynolds number Re, (at which the laminar boundary layer transforms to turbulent boundary layer).

The values of A for Re,, lo5, 5 x lo5 and 10, are 360, 1670, 3300 respectively.

Fig. 7.19 shows a log-log plot of cf versus Reh Example 7.28. A flat plate 5 m long and 0.75 m wide is kept parallel to the flow of water

which is flowing at a velocity of 5 d s . If the average drdg coeficient for turbulent flow past this flat plate is expressed as: cf = 0'455 2,58 find the drag force on both sides of the plate.

(log,, Re,) Take v = 0.011 x I@ m2/s Solution: Given : L = 5 m, B = 0.75 m, U = 5 d s , v = 0.01 1 x 10'' m2/s Drag force on both sides of the plate: The Reynolds number at the end of the plate is given by

This confirms that the nature of flow is turbulent. The average drag coefficient is expressed as

c - 0.455 - [log,, ~ e d ~ . ~ ~

:. The drag force on both sides of the plate,

FD = 2 x c f x [ ;~AU' ) [where A = area of one side] 1

= 2 x 2.642 x x IT x 1000 x (5 x 0.75) x 5'1 = 247.68 N (Ans.) L J

Example 7.29. A submarine can be assumed to have cylindrical shape with rounded nose. Assuming its length to be 50 m and diameter 5.0 m, determine the total Dower reauired to overcome boundary friction if it cruises at 8 rn/s velocity in sei water at 20C (p = 1030 kg/m3), v = I x lo4 m2/s.

Solution. Length of submarine, L = 5 0 m Diameter of submarine, D = 50 m Velocity of submarine, U = 8 d s Density of sea water, p = 1030 kg/m3 Kinematic viscosity of sea water, v = 1 x lo6 m2/s

Total power required to overcome boundary friction, P :

Reynolds number.

The length over which boundary layer wilf be laminar is given by Ux - = 5 x 1 0 5 0 r x = 5 x 1 6 ~ ~ v ' U

Forced Convection 41 5

This being very small, contribution to total drag from laminar boundary layer is negligible;

hence cf is given by

Area, A = ?rDL = n x 5 x k = 785.4 m2 :. Drag force,

1 I FD = cf x 5 p~d = 0.001765 x x 1030 x 785.4 x 8'

Hence total power required to overcome boundary friction,

Example 7-30. Find the ratio of friction drag on the front half and rear half of the f2ot plate kept at zero incidence in a stream of uniform velocity, i f the boundary layer is turbulent over the whole plate.

Solution. The average coefficient of drag (cf) for turbulent boundary layer is given by UL

For the entire plate, ReL = - v

Ux UL For the first half of the plate, Re, = - = - v 2v Drag force per unit width for the entire plate is,

- pu2 FD = Cf x - x area per unit width 2

\ / Similarly the drag force per unit width for the front halfportion of the plate is

FD, -

:. Drag force for the rear half portion of the plate is

Hence,

1 FD,

?x(2)l" -

- - - -

0S74 = 1.347 (Ans.)

1 FD2 1 - - 1 5 - 0.574 2 Example 7.31. A stream-lined train is ZOO m long with a typical cross-section having a

perimeter of 9 m above the wheels. If the kinematic viscosity of air at the prevailing tempemture is 1.5 x m2/s and density 1.24 kg/m3, determine the su$ace drag (friction drag) of the train when running at 90 km/h.

41 6 Heat and Mass Transfer

Make allowance for the fact that boundary layer changes from laminar to turbulent on the train surface.

Solution. Length of the train, L = 200 m Perimeter of cross-section of the train above wheels, P = 9 m :. Surface area, A = L x P = 2 0 0 x 9 = 1800m2 Kinematic viscosity of air, v = 1.5 x lo-' m2/s Density of air, p = 1.24 kg/m3

Free stream velocity, U = 90 kmh = 9ox 1000 = 25 ds 3600 Friction drag, FD : The Reynolds number with length of the train as the characteristic length,

uL. 25 200 = 3.333 x 108 - Re, = - - v 1.5x10-~

Obviously the boundary layer is turbulent. Assuming that the abrupt transition from laminar to turbulent flow occurs at a Reynolds

number of 5 x lo5, the average coefficient of drag,

- - 0.455 1670

[log,, (3.333 x 108)12." - 3.333 x lo8 = 0.001807 - 5.01 x = 0.0018

The approximate friction drag over the train surface, - 1 1 FD = Cf x 2 ~ A V ~ = 0.0018 x - X 1.24 x 1800 x 25'= 1255.5 N (Ans.)

2 Example 7.32. A

heat and mass transfer rk rajput

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