fundementals of heat and mass transfer kotandaraman (3rd edition): part 2

8
CONVECTIVE HEAT TRANSFER
Practical Correlations - Flow Over Surfaces
8.0

INTRODUCTION

In chapter 7 the basics of convection was discussed and the methods of analysis were enumerated,
correlations were obtained for laminar flow over flat plate at uniform temperature, starting from
basic principles and using the concept of boundary layer. The application of these correlations are
limited. However these equations provide a method of correlation of experimental results and
extension of these equations to practical situations of more complex nature. Though the basic
dimensionless numbers used remain the same, the constants and power indices are found to vary
with ranges of these parameters and geometries. In this chapter it is proposed to list the various
types of boundaries, ranges of parameters and the experimental correlations found suitable to
deal with these situations, as far as flow over surfaces like flat plates, cylinders, spheres and tube
banks are concerned.

Equations for heat transfer in laminar flow over flat plate were derived from basics in Chapter 7.
In this chapter additional practical correlations are introduced. Though several types of boundary
conditions may exist, these can be approximated to three basic types. These are (i) constant wall
temperature, (as may be obtained in evaporation, condensation etc., phase change at a specified
pressure) (ii) constant heat flux, as may be obtained by electrical strip type of heating and (iii) flow
with neither of these quantities remaining constant, as when two fluids may be flowing on either
side of the plate.
Distinct correlations are available for constant wall temperature and constant heat flux.
But for the third case it may be necessary to approximate to one of the above two cases.
8.1.1. Laminar flow: The condition is that the Reynolds number should be less than 5 × 105 or
as may be stated otherwise. For the condition that the plate temperature is constant the following
equations are valid with fluid property values taken at the film temperature.
Hydrodynamic boundary layer thickness (from Chapter 7)

δx = 5x/Rex0.5
...(8.1)
Thermal boundary layer thickness
δtx = δx Pr–0.333
...(8.2)
Displacement thickness and Momentum thickness are not directly used in heat
transfer calculations. However, it is desirable to be aware of these concepts.
334

Chapter 8

8.1 FLOW OVER FLAT PLATES


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

335

Displacement thickness is the difference between the boundary layer thickness
and the thickness with uniform velocity equal to free stream velocity in which the
flow will be the same as in the boundary layer. For laminar flow displacement thickness
is defined as

z FGH
δ

0

1−

IJ
K

u
dy
u∞

δd = δx/3
...(8.3)
Momentum thickness is the difference between the boundary layer thickness
and the layer thickness which at the free stream velocity will have the same
momentum as in the boundary layer.
Momentum thickness δm in the laminar region is defined by

z

δ

0

LM u F u I OP
MN u − GH u JK PQ dy
2

∞

∞

δm = δx/7

...(8.4)

ρ u∞
Friction coefficient defined as τ s/(ρ
Cfx = 0.664 Rex

2/2)

is given by

–0.5

...(8.5)

The average value of Cf in the laminar region for a length L from leading edge is given
by (Chapter 7)
CfL = 1.328 ReL–0.5

...(8.6)

The value of local Nusselt number is given by (Chapter 7)
Nux = 0.332 Rex0.5 Pr0.33
N uL = 2NuL = 0.664 ReL0.5 Pr1/3

...(8.7)
...(8.7 (a))

This is valid for Prandtl number range of 0.6 to 50.
For low values of Prandtl numbers as in the case of liquid metals, the local Nusselt
number is

Nux = 0.565 (Rex Pr)0.5

...(8.8)

This is valid for Prandtl number less than 0.05 (liquid metals) A more general expression
applicable for both low and high values of Prandtl number is given by
Nux =

0.3387 Rex 0.5 Pr 0.333

[1 + (0.0468 / Pr) 0.67 ]0.25

...(8.9)

This is valid for Pr < 0.05 and Pr > 50 and Rex Pr > 100. (liquid metals and silicones).

It may be seen that there is gap in the range of Prandtl number 0.6 to 0.1. If one goes through
property values of various fluids in practical application, it will be seen that no fluid is having Prandtl
numbers in this range.

8.1.2. Constant heat flux: The local Nusselt number is given by
Nux = 0.453 Rex0.5 Pr0.333

...(8.10)

Chapter 8

Note: The modification for very high values of Prandtl number is very little as can be seen in the
worked out problems.

336

FUNDAMENTALS OF HEAT AND MASS TRANSFER

This is also valid in the range of Prandtl numbers 0.6 to 50. In constant heat flux boundary
the plate temperature varies along the lengths. Hence the temperature difference between the
plate and the free stream varies continuously. The average difference in temperature
between the fluid and surface length x is given by
Twx – T∞ = (qx/k)/[0.6795 Rex0.5. Pr0.33]
...(8.11)
For low as well as high values of Prandtl numbers the relationship is (For Pr < 0.05 and
Pr > 50)
Nux =

0.453 Rex 0.5 Pr 0.333

[1 + (0.0207 / Pr) 0.67 ]0.25

...(8.12)

The property values are at film temperature.
In all cases, the average Nusselt number is given by
NuL = 2 NuL

This is applicable in all cases when Nu ∝
by

...(8.13)
Re0.5

Using the analogy between heat and momentum transfer the Stanton number is given

St Pr0.67 = Cf /2
...(8.14)
The equations (8.1) to (8.14) are applicable for laminar flow over flat plates. The choice
of the equation depends upon the values of Prandtl number and Reynolds numbers (laminar
flow).
Property values should be at the film temperature, (Ts + T∞)/2.
Eight examples follow, using different fluids at different conditions.
Example 8.1: In a process water at 30°C flows over a plate maintained at 10°C with a free
stream velocity of 0.3 m/s. Determine the hydrodynamic boundary layer thickness, thermal
boundary layer thickness, local and average values of friction coefficient, heat transfer coefficient
and refrigeration necessary to maintain the plate temperature. Also find the values of
displacement and momentum thicknesses. Consider a plate of 1 m × 1 m size.
Solution: The film temperature
= (30 + 10)/2 = 20°C
The property values are:
Kinematic viscosity
= 1.006 × 10–6 m2/s,
Thermal conductivity
= 0.5978 W/mK
Prandtl number
= 7.02, at 1m
u∞ x
0.3 × 1
=
= 2.982 × 105 ∴ laminar
ν
1.006 × 10 −6

δ x = 5x/Rex0.5 = 9.156 × 10–3 m = 9.156 mm
δ tx = δx. PR–0.33 = 9.156(7.02)–0.33 = 4.782 mm

Rex =
∴

Thermal boundary layer will be thinner if Pr > 1
Displacement thickness
δ d = δx/3 = 9.156/3 = 3.052 mm
Momentum thickness
δ m = δx/7 = 9.156/7 = 1.308 mm


337

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

Cfx = 0.664/Re0.5 = 0.664/(2.982 × 105)0.5 = 1.216 × 10–3
CfL = 2 × CfL = 2 × 1.216 × 10–3 = 2.432 × 10–3
Nux = 0.332 × Rex0.5 Pr0.33 = 0.332 × (2.982 × 105)0.5 × 7.020.33
= 347.15
hx = Nux

k
= 347.15 × 0.5978/1 = 207.52 W/m2K
L

h = 2 hL = 415.04 W/m2K
cooling required = hA ∆T = 415.04 × 1 × 1 × (30 – 10) = 8301 W or 8.3 kW.

Example 8.2: Sodium potassium alloy (25% + 75%) at 300°C flows over a 20 cm long plate
element at 500°C with a free stream velocity of 0.6 m/s. The width of plate element is 0.1 m.
Determine the hydrodynamic and thermal boundary layer thicknesses and also the displacement
and momentum thicknesses. Determine also the local and average value of coefficient of friction
and convection coefficient. Also find the heat transfer rate.
Solution: The film temperature is (300 + 500)/2 = 400°C
The property values are:
Kinematic viscosity
= 0.308 × 10–6 m2/s, Pr = 0.0108,
Thermal conductivity = 22.1 W/mK, at 0.2 m,
Rex = 0.6 × 0.2/0.308 × 10–6 = 3.9 × 105 ∴ laminar
∴
δ x = 5x/Rex0.5 = 1.6 mm
δ tx = δx . Pr–0.33 = 7.25 mm
This is larger by several times. So most of the thermal layer is outside the velocity
boundary layers.
Displacement thickness:
δ d = 1.6/3 = 0.53 mm
Momentum thickness
δ m = 1.6/7 = 0.229 mm
It can be seen that thermal effect is predominant
Cfx = 0.664/Re0.5 = 0.664/(3.9 × 105)0.5 = 1.064 × 10–3
CfL = 2.128 × 10–3

Using equation (8.8) as the Prandtl number is very low (less than 0.05)
Nux = 0.565 × (Rex Pr)0.5 = 36.65
k
= 36.65 × 22.1/0.2 = 4050 W/m2K
L

2
h = 2 × hL = 8100 W/m K
Heat flow = 8100 × 0.2 × 0.1 × (500 – 300) = 32,399 W

Alternately using equation (8.9)
Nux =

0.3387 Rex 0.5 Pr 0.333

[1 + (0.0468 / Pr) 0.67 ]0.25

or

32.4 kW

Chapter 8

hx = Nux


338

FUNDAMENTALS OF HEAT AND MASS TRANSFER

=
∴

hx =

0.3387 × (3.9 × 10 5 ) 0.5 (0.0108) 0.33

[1 + (0.0468 / 0.0108) 0.67 ]0.25

= 33.79

33.79 × 22.1
= 3734 W/m2K
0.2

h = 7468 W/m2K Q = 29.87 kW
If equation (8.7) had been used Q = 40.5 kW, an over estimate.

Example 8.3: Engine oil at 80°C flows over a flat surface at 40°C for cooling purpose, the flow
velocity being 2 m/s. Determine at a distance of 0.4 m from the leading edge the hydrodynamic
and thermal boundary layer thickness. Also determine the local and average values of friction
and convection coefficients.
Solution: The film temperature is (80 + 40)/2 = 60°C
The property values are read from tables at 60°C as kinetic viscosity = 83 × 10–6 m2/s, Pr
= 1050. Thermal conductivity = 0.1407 W/mK
Rex =
∴

u∞ x
2 × 0.4
=
= 9639, laminar
ν
83 × 10 −6

δx = 5x/Rex0.5 = 0.02037 m = 20.37 mm
δ tx = δxPr–1/3 = 20.37 × 1050–0.333 = 2 mm

Thermal boundary layer is very thin as different from liquid metal-viscous effect is
predominant.
Cfx = 0.664/Rex0.5 = 6.76 × 10–3
CfL = 0.0135 (rather large)

As the values of Prandtl number is very high equation (8.9) can be used
Nux =
=
hx =

0.3387 Rex 0.5 Pr 0.33

[1 + (0.0468 / Pr) 0.67 ]0.25
0.3387 × 9639 0.5 × 1050 0.33
[1 + (0.0468 / 1050) 0.67 ]0.25

= 337.97/1.0003 = 337.87

Nux k 337.87 × 0.1407
=
= 118.85 W/m2K
x
0.4

h = 2hx = 118.85 × 2 = 237.69 W/m2K
For 1 m width the heat flow is given by

little.

Q = 237.69 × 0.4 × 1 (80 – 40) = 3803 W or 3.803 kW
If equation (8.7) is used Nu = 331.3 and h = 233.01 W/m2K. The difference is very

Example 8.4: Air at 20°C flows over a flat plate having a uniform heat flux of 800 W/m2.
The flow velocity is 4m/s and the length of the plate is 1.2 m. Determine the value of heat
transfer coefficient and also the temperature of the plate as the air leaves the plate.
Solution: As the plate temperature varies, the value of film temperature cannot be determined.
For the first trial, the properties of air at 20°C are used.


339

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

ν = 15.06 × 10–6, k = 0.02593 W/mK, Pr = 0.703
First, a check for laminar flow:
Re =

u∞ L
4 × 1.2
=
= 3.187 × 105
ν
15.06 × 10 −6

∴ laminar

For constant heat flux, the average temperature difference can be found by using
equation (8.11)
Tx − T∞ = (qL/k)/[0.6795 Rex0.5 Pr0.33] = 108.54°C

Now properties may be found at (108.54 + 20)/2 = 64.27°C
ν, m2/s

T°C

k, W/mK

Pr

60

18.97 ×

10–6

0.02896

0.696

70

20.02 × 10–6

0.02966

0.694

10–6

0.02926

0.695

64.27

19.42 ×

Using the equation again
Tw – T∞ =

800 × 1.2
1
.
0.02926 0.6795 (4 × 1.2 / 19.42 × 10 −6 ) 0.5 (0.695) 0.333

= 109.644°C
∴ Film temperature = 64.82°C
It does not make much of a difference.
To determine the value of convection coefficient, equation (8.11) is used.
Nux = 0.453 [Rex Pr]0.5 = 0.453
∴

hx =

LM 4 × 1.2 × 0.695 OP
N 19.42 × 10 Q
−6

0.5

= 187.75

187.75
× 0.02926 = 4.58 W/m2K
1.2

h = 9.16 W/m2K

To find the temperature at the trailing edge the basic heat flow equation is used:
(Tw – T∞) =
=

800 × 1.2
= 174.75°C
0.02926 × 187.75

Tw = 194.75°C.

Example 8.5: Water at 10°C flows over a flat plate with a uniform heat flux of 8.3 kW/m2. The
velocity of flow is 0.3 m/s. Determine the value of convective heat transfer coefficient and also
the temperature at a distance of 1 m from the leading edge.
Solution: As the film temperature cannot be specified the properties will be taken at 10°C for
the first trial
ν = (1.788 + 1.006) × 10–6/2 = 1.393 × 10–6 m2/s
Pr = (13.6 + 7.03)/2 = 10.31

Chapter 8

∴

qx
as (h = Nuk/x)
kNu x


340

FUNDAMENTALS OF HEAT AND MASS TRANSFER

k = (0.5524 + 0.5978)/2 = 0.5751 W/mK
at 1 m,
Rex = 0.3 × 1/1.393 × 10–6 = 2.154 × 105 ∴ laminar
The average temperature difference
=

q. L
1
.
k 0.6795 Re 0.5 Pr 0.333

=

8300 × 1
1
1
.
.
= 21.03°C
5 1/ 2

0.5751 0.6795 × (2.154 × 10 )
10.310.333

The property values can now be taken at 15.1°C and results refined.
The heat transfer coefficient can be determined using eqn. (8.10).
Nux = 0.453 Rex0.5Pr0.333
taking property values at 15.51°C
Nux = 465.9
∴
hx = 465.9 × 0.58762/1 = 273.8 W/m2K
Average value = 547.5 W/m2K (compare with example 8.1)
Temp. difference at 1 m:
h∆T = q
∴

h = Nu. k/x
∆T =

∴

∆T =

q
h

∴

∆T =

qx

Nu k

1
8300 × 1
qx
.
=
= 30.32°C
k Nux 0.58762 × 465.9

Example 8.6: Sodium postassium alloy (25% + 75%) at 300°C flows over a plate element
with free stream velocity of 0.6 m/s. The plate has a uniform heat generation rate of 1600
kW/m2. Determine the value of average convection coefficient for a length of 0.2 m. Also determine
the plate temperature at this point.
Solution: The Prandtl number has a value less than 0.05 and there is no equation to determine
the temperature difference. Equation (8.12) is used, starting with property values at 300°C
ν = 0.336 × 10–6 m2/s, Pr = 0.0134, k = 22.68
Rex = 0.6 × 0.2/0.366 × 10–6 = 3.279 × 105
∴ Laminar. Flor low value of Pr using equation (8.12).
Nux =
=
hx =

0.453 Rex 0.5 Pr 0.33

[1 + (0.0207 / Pr) 0.67 ]0.25
0.453 × 3.279 × 10 5 (0.0134) 0.333
[1 + (0.0207 / 0.0134) 0.67 ]0.25
49.83 × 22.68
= 5651 W/m2K

0.2

h = 11302.1 W/m2K

= 49.83


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

341

The average temperature difference:
q 1600000
=
= 141.6° C
h
11302
Compare with example 8.2. The results can be refined now taking property values at
300 + (141.6)/2 = 370.8°C (film temperature). Interpolating

∆T =

ν = – (0.366 – 0.308) × 0.708 + 0.366 = 0.325 × 10–6 m2/s
Pr = – (0.0134 – 0.0108) × 0.708 + 0.0134 = 0.0116
k = – (22.68 – 22.10) × 0.708 + 22.68 = 22.27 W/mK
Nux =

∴

0.453 × (0.6 × 0.2 / 0.325 × 10 −6 ) 0.5 (0.0116) 0.33

[1 + (0.0207 / 0.0116) 0.67 ]0.25

= 49.7 as compared to 49.83. Values are not very different.
Using equation (8.8), Nux = 0.565 (Re Pr)0.5 = 36.98, compared with 49.7.
Example 8.7: Engine oil at 60°C flows over a flat surface with a velocity of 2 m/s, the length of
the surface being 0.4m. If the plate has a uniform heat flux of 10 kW/m2, determine the value
of average convective heat transfer coefficient. Also find the temperature of the plate at the
trailing edge.
Solution: As the film temperature cannot be determined, the property values are taken at
free stream temperature of 60°C
Kinematic viscosity
= 83 × 10–6 m2/s, Pr = 1050, k = 0.1407 W/mK
Rex =

u∞ x
2 × 0.4
=
= 9639
ν
83 × 10 −6

∴ laminar

Using equation (8.12)
Nux =
hx =

0.453 Rex 0.5 Pr 0.33

[1 + (0.0207 / Pr) 0.67 ]0.25

=

0.453 . 9639 0.5 . 1050 0.333
[1 + (0.0207 / 1050) 0.67 ]0.25

= 451.95

451.95 × k 451.95 × 0.1407
=
= 158.97 W/m2K
x
0.4

h = hL × 2 = 317.94 W/m2K
The average temperature difference:
q 100000
=
= 31.45°C
h 317.94
Now the film temperature can be taken as

∆T =

at

ν

Pr

k

80°C

37 × 10–6

490

0.1384

60°C

10–6

1050

0.1407

83 ×

75.73°C, ν = 46.82 × 10–6 m2/s,

Pr = 609.6, k = 0.1389

Chapter 8

31.45
+ 60 = 75.73°C
2
Using property tables

342

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Nux =
hx =

0.453(2 × 0.4 / 46.82 × 10 −6 ) 0.5 (609.6) 0.33
[1 + (0.0207 / 609.6) 0.67 ]0.25

= 501.95

Nux k 501.95 × 0.1389
=
= 174.3 W/m2K
x
0.4

h = 348.6 W/m2K

∆T =

10000
= 28.7°C
348.6

The value can be refined further using new value of film temperature.
To determine the plate temperature at the edge:

10000
= 57.4°C
174.3
T = 60 + 57.4 = 117.4°C

∆T =
∴

Compare with example 8.3
8.1.3. Other Special Cases: Laminar constant wall temperature, with heating starting
at a distance x0 from the leading edge.
The correlation is obtained as below
Nux = 0.332. Rex

0.5 Pr0.33

LM1 − F x I
MN GH x JK
o

OP
PQ

0.75 −0.333

...(8.15)

At xo = 0, this will reduce to the regular expression given by equation (8.7). The average
value in this case will not be 2 Nux and the above expression has to be integrated over the
length to obtain the value.

Example 8.8: Considering water at 30°C flowing over a flat plate 1 m × 1 m at 10°C with a free
stream velocity of 0.3 m/s, plot the variation of hx along the length if heating starts from 0.3 m
from the leading edge.
Solution: The film temperature = (30 + 10)/2 = 20°C
The property values are: ν = 1.006 × 10–6 m2/s, Pr = 7.02, k = 0.5978 W/mK
The maximum value of Rex =
∴

0.3 × 1
1.006 × 10 −6

= 2.98 × 105

Laminar flow exists all along.
Nux = 0.332

Rex0.5

Pr0.33

LM1 − F x I OP
MN GH x JK PQ
o

0.75 −0.333

hx = k.Nux/x
at x = 0.35:

F

GH

0.5978
0.3 × 0.35
× 0.332
hx =
0.35
1.006 × 10 −6

I
JK

0.5

(7.02) 0.333


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

LM1 − FG 0.3 IJ
MN H 0.35 K

OP
PQ

343

0.75 −0.333

= 733.93 W/m2K

Similarly for other values at 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1.0
Distance x
m

hx with heating
from x = 0
W/m2K

hx with heating
from x = 0.3 m
W/m2K

0.3

367.47

0

0.35

340.21

733.93

0.4

318.24

549.67

0.5

284.64

416.90

0.6

259.84

315.09

0.7

240.57

309.34

0.8

225.03

279.70

0.9

212.16

257.20

1.0

201.27

239.35

The average value over the heated length can be found only by integrating between
x = xo and x = L.

8.2 TURBULENT FLOW

to

107

The local friction coefficient defined as τw/(ρ u∞2/2) is given for the range Rex from 5 × 105
by
Cfx = 0.0592 Rex–0.2
...(8.20)
7
9
For higher values of Re in the range 10 to 10
Cfx = 0.37 [log10 Rex]–2.584
...(8.21)
The local Nusselt number is given by
Nux = 0.0296 Rex0.8Pr0.33
...(8.22)

Chapter 8

Rex > 5 × 105 or as specified. In flow over flat plate, the flow is initially laminar and after some
distance turns turbulent, the value of Reynolds number at this point being near 5 × 105. However,
there are circumstances under which the flow turns turbulent at a very short distance, due to
higher velocities or due to disturbances, roughness etc. The critical reynolds number in these
cases is low and has to be specified. In the turbulent region the velocity boundary layer
thickness is given by
δx = 0.381 x × Rex–0.2
...(8.16)
δt ≈ δx
...(8.17)
The displacement and momentum thickness are much thinner. The displacement
thickness is
δd = δx/8
...(8.18)
Momentum thickness is
δm = (7/72) δx
...(8.19)


344

FUNDAMENTALS OF HEAT AND MASS TRANSFER

The average Nusselt number is given by
Nu = 0.037 Re0.8 Pr0.33

...(8.23)

Nu = Nux/0.8 in this case as Nu is dependent on Rex0.8. Using analogy between

momentum and heat transfer

Nux =

(C fx / 2) Rex Pr
1 + 12.8 (C fx / 2) 0.5 ( Pr 0.68 − 1)

...(8.24)

To obtain the average value, this expression has to be integrated from x = 0 to x = L. But
this is more complex.
For constant heat flux, the Nusselt number is found to increase by 4% over the value for
constant wall temperature.
∴

Nux (constant heat flux) = 1.04 Nux(Constant wall temperature)...(8.25)

Example 8.9: Air at –10°C flows over a flat surface at 10°C with a free stream velocity of
80 m/s. The length of the plate is 3.1 m. Determine the location at which the flow turns
turbulent. Also determine the local and average value of convection coefficient assuming that
the flow is turbulent although. Compare the value of local heat transfer coefficient calculated
using the equation obtained by analogy. (8.24).
Solution: The film temperature is (–10 + 10)/2 = 0°C
The property values are
ν = 13.28 × 10–6 m2/s, Pr = 0.707, k = 0.02442 W/mK
Rex at = 3.1 m
Re = (80 × 3.10/13.28 × 10–6) = 1.8675 × 107
∴

turbulent flow exists

location at which

Re = 5 × 105 is

5 × 10 5 × 13.28 × 10 −6
= 0.083 m
80
This length is much shorter (2.7% of the total length) and so the assumption that flow is
turbulent all through is valid. Hydrodynamic boundary layer thickness

x=

δ x = 0.381 × 3.1 × (1.8675 × 107)–0.2 = 41.54 mm
Thermal boundary layer thickness is also
δ t = 41.54 mm
Displacement thickness
δd = 41.54/8 = 5.19 mm
Momentum thickness
7
× 41.54 = 4.04 mm
72
As Re is in the border (< 108), we can calculate Cfx using eqn (8.20) or (8.21).

δm =

Cfx = 0.0592 × Rex–02 = 2.08 × 10–3


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

345

Using 8.21
Cfx = 0.37 [log10 Rex ]–2.584 = 2.197 × 10–3
Local Nusselt number is given by eqn. 8.22
Nux = 0.0296 Rex0.8 Pr0.33
= 0.0296 × (1.8675 × 107)0.8 (0.707)0.33 = 17302.2
∴
Average value

hx =

Nu x k
= 136.3 W/m2K
L

Nu = 0.037 Rex0.8 × Pr0.33

∴
h = 170.4 W/m2K
If constant heat flux prevails, this value is increased by 4%. Using Analogy: using
Cfx by eqn. (8.20)
Nux =

(C fx / 2) Rex . Pr
1 + 12.8 (C fx / 2) 0.5 ( Pr 0.68 − 1)

= 15035

Example 8.10: Water at 30°C flows over a flat plate 1 m × 1 m at 10°C with a free stream
velocity of 4 m/s. Determine the thickness of boundary layers, displacement thickness, momentum
thickness, local and average value of drag coefficient and convection coefficient.
Solution: The film temperature = (30 + 10)/2 = 20°C. Property values at this temperature are
ν = 1.006 × 10–6 m2/s, Pr = 7.02, k = 0.5978 W/mK.
The maximum value of Reynolds number at 1 m is
= 4 × 1/1.006 × 10–6 = 3.976 × 106 ∴ Turbulent
The length at which flow turns tubulent:
(4 × x/1.006 × 10–6) = 5 × 105 ∴ x = 0.12575 m.
This is 12.5% of total length. As such the assumption that the flow is turbulent althrough
is not unacceptable.
Boundary layer thickness:
δ x = 0.381 × Rex–0.2 × 1 = 0.381 × (3.976 × 106)–0.2
= 0.1824 m = 18.24 mm
Thermal boundary layer thickness is also the same = 18.24 mm
Displacement thickness:
δ d = δx/8 = 2.28 mm
Momentum thickness
7
× δx = 1.773 mm
72
Cfx = 0.0592 × Rex–0.2 = 2.83 × 10–3

δm =

Chapter 8

By using Cfx for higher range:
Nux = 15922
These values are not very much different from the one using eqn. (8.22) (8%).

The values of convection coefficients calculated may be out by as much as 25% in certain
cases and as such these estimates are acceptable.


346

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Cf = Cfx/0.8 = 3.54 × 10–3

Nux = 0.0298 Rex0.8Pr0.33 = 0.0296 × (3.976 × 106)0.8 × (7.02)0.3
= 10788.8
hx =

Nu x k 10788.8 × 0.5978
=
= 6449.6 W/m2K
x
1

Nu = 0.037 × Re0.8 × Pr0.33 = 13486

h = 8062 W/m2K
For constant heat flux, the average value is increased by 4%.

By using analogy
Nux =
∴

(C fx / 2)Re x . Pr

1 + 12.8 (C fx / 2) 0.5 ( Pr 0.68 − 1)

= 16967

hx = 10143 W/m2K

This is on the higher side.
8.2.1. The assumption that the flow is turbulent althrough (from start) may not be acceptable
in many situations. The average values are now found by integrating the local values
up to the location where Re = 5 × 105 using laminar flow relationship and then
integrating the local value beyond this point using the turbulent flow relationship and
then taking the average. This leads to the following relationship for constant wall temperature.
δx = 0.381x × Rex–0.2 – 10256x × Rex–1.0
...(8.26)
0.2
–1.0
CfL = 0.074 ReL – 1742 ReL
...(8.27)
5
This is for critical Reynolds number of 5 × 10 . A more general relationship can be used
for other values of critical Reynolds number.
CfL =

A
0.455
−
8.28
2.584
Re
(log 10 ReL )

L

Where A is given in the tabulation below
Recr
A
5
3 × 10
1050
5
5 × 10
1700
6
1 × 10
3300
3 × 106
8700
0.333
0.8
Nux = Pr
[0.037ReL – 871]
...(8.29)
5
for Recr = 5 × 10 , or more generally
NuL = Pr0.333 [0.037ReL0.8 – A]
...(8.30)
where
A = 0.037 Recr0.8 – 0.664 Recr0.5
...(8.31)
By analogy Stx. Pr0.666 = Cfx/2
...(8.32)

For large temperature differences, the estimates may be off the mark by as much as
25%. For low or high Prandtl numbers, these expressions are to be used with some reservations.
For constant heat flux, the value of h is to be increased by 4 percent.


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

347

Example 8.11: Considering the data of Example 8.10, determine the average value of convection
coefficient and Cf values taking into consideration the laminar region. Compare with problem
8.10.
Plate length 1 m, velocity = 4 m/s, plate temperature = 10°C, Water temperature = 30°C.
Film temperature = 20°C. The property values are ν = 1.006 × 10–6 m2/s, Pr = 7.02, k = 0.5978
W/mK.
Solution: The maximum value of Reynolds number
= 4 × 1/1.006 × 10–6 = 3.976 × 106 ∴ Turbulent
Assuming
Recr = 5 × 105
δ x = 0.381 x Rex–0.2 – 10256 x Rex–1.0
= 0.381 × 1 [4/1.006 × 10–6]–0.2 – 10256 × 1/[4/1.006 × 10–6]
= 0.01566 m or 15.66 mm (compared to 18.24 mm)
CfL = 0.074 ReL–0.2 – 1742 ReL–1.0
= 3.10 × 10–3 (compared to 3.54 × 10–3 in example 8.10)
Nu = Pr0.33 [0.037 ReL0.8 – 871]
= 11818 (compared to 13486).
Example 8.12: Air at –10°C flows over a flat plate at 10°C with a free stream velocity of
10 m/s, the length of the plate being 3.1 m. Determine the average value of friction coefficient
and convection coefficient taking into account the laminar length and compare the values with
those obtained assuming turbulent flow throughout. (example 8.9)

Solution: The film temperature = (– 10 + 10)/2 = 0°C
The property values are: ν = 13.28 × 10–6 m2/s, Pr = 0.707
k = 0.02442 W/mK
The maximum value of Reynolds number
= 3.1 × 10/(13.38 × 10–6) = 2.33 × 106 ∴ turbulent
Critical length: 0.664 m ∴ necessary to consider laminar region.
Assuming turbulent flow throughout:
0.0592
ReL–0.2 = 3.94 × 10–3
0.8
Taking laminar region into account
Cf =

Cf = 0.074 ReL–0.2 – 1742/ReL = 3.195 × 10–3

Heat transfer coefficient (turbulent all through)
NuL = 0.037 × ReL0.8 Pr0.33 = 4098
NuL × k
= 32.28 W/m2K
L
Considering laminar region

h=

NuL = Pr0.333 [0.037 × ReL0.8 – 871] = 3321
h = 26.17 W/m2K

Chapter 8

∴

348

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Note that at low velocities it will be better to consider the laminar region in taking
averages.

8.3. FLOW ACROSS CYLINDERS
The other type of flow over surfaces is flow across cylinders often met with in heat exchangers
and hot or cold pipe lines in the open. An important difference is the velocity distribution along
the flow. The obstruction by the cylinder causes a closing up of the streamlines and an increase
in pressure at the stagnation point. The velocity distribution at various locations in the flow
differs from the flow over a flat plate as shown in Fig. 8.1.

q

u¥

u¥

u¥

u¥

u¥

u¥

Flow
separation
1

2

3

4

5

Fig. 8.1. Velocity distribution at various angular locations in flow across cylinders.

As the flow pattern affects the heat transfer, it is found to be difficult to provide a
generalised analytical solution for the problem. The drag coefficient CD is defined by
Drag force = CD Af

ρu∞ 2
. Where Af is the frontal or projected area. (for a cylinder of
2

length of L it is equal to L.D). It is not based on the wetted area. The nature of variation of drag
coefficient for cylinder and sphere with Reynolds number is shown in Fig. 8.2. Reynolds number
should be calculated with diameter D as the length parameter and is some times referred as
ReD.
Thus a simple and single correlation for CD is difficult. The variation of local heat transfer
coefficient with angular location for two values of Reynolds number is shown in Fig. 8.3.
For angles upto 80°, the variation of Nusselt number can be represented by

LM FG θ IJ OP
MN H 90 K QP

hθ = 1.14 ReD0.5 Pr0.4 1 −

3

...(8.32(b))


CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

349

CD
100

Sphere

10

Cylinder

1

0.1
log Re

Fig. 8.2. Variation of CD with Reynolds number for flow over cylinders and spheres.

ReD

700
600

219,000
500
Nuq
400
300
70,800
200
100

0

40

80

120
q

160

This also shows that averaging out the convection coefficient is difficult. The experimental
values measured by various researchers plotted using common parameters ReD and NuD (log
log plot) is shown in Fig. 8.4. It can be seen that scatter is high at certain regions and several
separate straight line correlations are possible for various ranges. Some researchers have
limited their correlations for specific ranges and specific fluids. Thus a number of correlations

are available and are listed below.
A very widely used correlation is of the form (1958)
NuD = CRcDm Pr0.333

...(8.33)

Chapter 8

Fig. 8.3. Variation of Nusselt number with angular location.


350

FUNDAMENTALS OF HEAT AND MASS TRANSFER

NuD
(log)

Scatter
Band

(log)

ReD

Fig. 8.4. Variation of NuD with ReD for flow across cylinders.

Where C and m are tabulated below. The applicability of this correlation for very
low values of Prandtl number is doubtful. The length parameter in Nusselt number is diameter
D and Nusselt number is referred as NuD.

The properties are to be evaluated at the film temperature.
ReD

C

m

0.4–4.0

0.989

0.330

4–40.0

0.91

0.385

40–4000

0.683

0.466

4000–40000

0.193

0.618

40000–400000

0.0266

0.805

A more recent (1972) generalised form is
NuD = C ReD

mPrn

F Pr I
GH Pr JK

0.25

∞

...(8.34)

w

The validity for this correlation is for
0.7 < Pr < 500; 1 < ReD < 106
with n = 0.36 for Pr < 10 and n = 0.37 for Pr > 10
The values of C and m are tabulated below

and

ReD

C

m

1–4.0

0.75

0.4

40–103

0.51

0.5

0.26

0.6

0.076

0.7

103

–2×

105

2×

102

106

–

The properties for Re and Pr should be at free stream temperature.
A two range (1972) correlation is given below: (f-film temp.)
NuD = [0.43 + 0.50 ReD0.5] Pr0.38

F Pr I
GH Pr JK
f

w

0.25

...(8.35(a))


351

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

1 < ReD < 103

NuD = 0.25 ReD

0.6

Pr0.38

F Pr I
GH Pr JK
f

0.25

...(8.35(b))

w

103 < ReD < 2 × 105
This equation is applicable both for gases and liquids with the following conditions:
1. For gases the ratio of Pr numbers can be taken as unity.
2. For gases properties to be evaluated at film temperature.
3. For liquids properties to be evaluated at free stream temperature.
This is not suitable for very low and very high values of Prandtl numbers.
A correlation for liquids is given by (1965)
NuD = [0.35 + 0.56 ReD0.52] Pr0.33
...(8.36)
10–1 < ReD < 105 and properties at film temperature.
No indication is available for the applicability at low values of Pr. Another correlation
(1972) applicable over wider range is
NuD = [0.4 ReD0.5 + 0.06 ReD0.67] Pr0.4

LM µ OP
Nµ Q

0.25

∞

...(8.37)

w

10 < ReD < 105; 0.67 < Pr < 300
0.25 <

µ∞
< 5.2.
µw

The properties are to be evaluated at free stream temperature T∞ . Another set
of equations (1977) suitable for a wider range of parameters both Reynolds and Prandtl is
NuD

for

LM1 + F Re I
= 0.3 +
LM1 + F 0.4 I OP MN GH 282000 JK
MN GH Pr JK PQ
0.62 Re D 0.5 Pr 0.333
0.67

0.625

D

0.75

OP
PQ

0.8

...(8.38(a))

100 < ReD < 107, Pe = ReD Pr > 0.2
The properties are to be evaluated at film temperature.
A modification of this equation for limited range of Reynolds number is
NuD = 0.3 +

LM1 + F Re I
LM1 + FG 0.4 IJ OP MN GH 282000 JK
MN H Pr K PQ

0.62 ReD 0.5 Pr 0.333

0.67 0.25

D

0.5

OP
PQ

...(8.38(b))

This equation use properties at film temperature and is applicable for all fluids. Finally
for liquid metals another correlation. (1975) is obtained as
NuD = [0.8237 – ln (PeD0.5)]–1
...(8.39(a))
where
Pe = RePr

Chapter 8

2 × 104 < ReD < 4 × 105, Pr > 0.2


352

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Other correlations for liquid metals over cylinder are (1979)
NuD = 1.125 (ReD Pr)0.413

...(8.39(b))

1 ≤ ReD Pr ≤ 100.
Analytical results are also available for constant wall temperature.
NuD = 1.015 (ReDPr)0.5.

For constant heat flux NuD = 1.145 (ReDPr)0.5
This is applicable only for very low values of Pr and Pe. Nu will become negative for
higher values of Pe in eqn (8.39 (a)). Equations (8.33) to (8.39) are obtained from various
experimental results, the difference being that each one of these is dividing the spectrum into
different ranges of parameters. However a common warning is that most of these may give
results varying as much as 25% from experimental results. A single correlation applicable for
various ranges will be easier to use in computer application (say 8.38).
In actual application one has to weight carefully the parameter ranges before choosing
the equation to be used.
8.3.1. Flow Across non Circular Shapes: The general correlation used for gases, including
the Pr 0.333 in the constant is (1949)
Nu = C ReDn
Nu = C1ReD

...(8.40(a))

nPr0.33

...(8.40(b))

The values of C, C1 and n various shapes are tabulated below. The properties
are at film temperature.
Shape

ReD range

n

C

C1

Square along diagonal length
diagonal distance

5000–100000

0.588

0.222

0.246

Ellipse along major axis length
minor axis

2500–15000

0.612

0.224

0.250

Square along diagonal, length
diagonals distance

2500–7500

0.624

0.267

0.292

Square along face, side

2500–8000

0.699

0.160

0.178

Square along face, side

5000–100000

0.092

0.102

Plane, perpendicular, width

0.675

4000–15000

0.731

0.205

0.228

hexagon perpendicular to flats,
corner distance

5000–19500

0.638

0.144

0.160

hexagon perpendicular to flats,
corner distance

19500–100000

0.782

0.035

0.039

hexagon along flats, length,
between flats

5000–100000

0.638

0.138

0.153

ellipse along minor axis length,
major axis

3000–15000

0.804

0.085

0.095


353

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

Example 8.13: Air at 30°C flows across a steam pipe of 0.2 m dia at a surface temperature of
130°C, with a velocity of 6 m/s. Determine the value of convective heat transfer coefficient using
equations (8.33), (8.34), (8.35) and (8.37) and (8.38).
Solution: Property values are required both at T∞ and Tf and Tw
i.e.
30°C, (30 + 130)/2 = 80C and 130C

Pr
k
à
6
2
30C: 16 ì 10 m /s
0.701
0.02675
18.63 ì 10–6 Ns/m2
80°C: 21.09 × 10–6 m2/s
0.692
0.03047
21.08 × 10–6 kg/ms
–6
2
130°C: 26.625 × 10 m /s
0.687
0.034135
23.29 × 10–6 kg/ms
Equation (8.33), properties at film temp.:
Re = uD/ν = 6 × 0.2/(21.09 × 10–6) = 56,899
∴
Nu = CRem Pr0.33
From tables
C = 0.0266, m = 0.805
∴
Nu = 0.0266 (56899)0.805 (0.692)0.333 = 158.29
h = Nu ×

∴

k
= 21.11 W/m2K
D

F Pr I
GH Pr JK

0.25

∞

Nu = CRem Prn

Equation (8.34)

w

Properties at free stream temp. : at 30°C
Re = (6 × 0.2/16 × 10–6) = 75000 = 0.75 × 105
From tables,
C = 0.26, m = 0.6 as Pr < 10, n = 0.36,
Nu = 0.26 (75000)0.6 (0.701)0.36

∴

= 192.8
Equation (8.35 b), is applicable
Nu = 0.25

∴

Re0.6

FG 0.701IJ
H 0.687 K

0.25

h = 25.79 W/m2K (k at 30°C, 0.02675)

Pr0.38

F Pr I
GH Pr JK
f

0.25

w

= 1. for gases properties at film temp. at 80°C
Nu = 0.25 (56899)0.6 (0.692)0.38
= 154.97

h=

∴

Equation (8.37) Properties at T∞, 30°C
Nu = [0.4

Re0.5

+ 0.06

Re0.67]

Pr0.4

154.9 × 0.03047
= 23.61 W/m2K
0.2

LM à OP
Nà Q

0.25



w

L F 6 ì 0.2 I
= M0.4 × G
MN H 16 × 10 JK
−6

0.5

F 6 × 0.2 I OP 0.701 LM 18.63 × 10 OP
+ 0.06 G
H 16 × 10 JK PQ
N 23.29 × 10 Q
0.67

−6

0.4

−6 0.25
−6

Chapter 8

for gases: (Prf /Prw

)0.25


354

FUNDAMENTALS OF HEAT AND MASS TRANSFER

= 180.76 ∴ h = 24.18 W/m2K, (k = 0.02675)
Equation (8.38 (b)) properties at film temperature

LM1 + FG Re IJ OP
LM1 + FG 0.4 IJ OP MN H 282000 K PQ

MN H Pr K PQ
LM1 + FG 56899 IJ OP
0.62 (56899) (0.701)
= 0.3 +
MN H 282000 K PQ
LM1 + FG 0.4 IJ OP
NM H 0.701K QP
0.5

0.62 Re0.5 Pr 0.333

Nu = 0.3 +

0.67 0.25

0.5

0.5

0.333

0.67 0.25

= 167.36 ∴ h = 25.5 W/m2K
In this example all the various equation provide answers within a small band. This is
only fortitious and not necessarily so in all cases. The parameters are not in the extreme
range.
Example 8.14: Liquid sodium at 300°C flows across a tube 0.05 m outside dia at 500°C with
a velocity of 8 m/s. Determine the value of convective heat transfer coefficient using suitable
correlations.

Solution: Property values may be required at all the three temperature T∞, Tf and Tw.
T°C

ν, m2/s

Pr

k, W/mK

µ = νρ, kg/ms

300

0.394 ×

10–6

0.0063

70.94

878 × 10–6

400

0.330 × 10–6

0.0056

63.97

854 × 10–6

500

10–6

0.0053

56.99

829 × 10–6

0.289 ×

The possible correlations are only 8.33 and 8.38 (a). Equation (8.33) properties at film
temp., 400°C, Nu = CRem Pr0.333
Re = 8 × 0.05/0.330 × 10–6 = 1212121, (1.212 × 106) values of C and m are only up to
400,000 C = 0.0266, m = 0.805
∴
Nu = 0.0266 (1.212 × 106)0.805 × 0.00560.333 = 372.85,
∴
equation (8.38(a))

h=

372.85 × 63.97
= 477022
0.05

or

0.477 × 106 W/m2K

LM1 + FG Re IJ
Nu = 0.3 +
LM1 + FG 0.4 IJ OP MN H 282000 K
MN H Pr K PQ
0.62 Re0.5 Pr 0.333

0.67 0.75

= 0.3 +

0.62 (1.212 × 10 6 ) 0.5 (0.0056) 0.333

= 159.16

LM1 + FG 0.4 IJ OP
MN H 0.0056 K PQ

0.67 0.25

∴

h = 203236 W/m2K

OP
PQ

0.625 0.8

LM F 1.212 × 10 I
MN1 + GH 282000 JK
6

OP
PQ

0.625 0.8


355

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

The correlation 8.33 is an older one and hence the values obtained from the more recent
correlation (8.38 (a)) has to be taken as more reliable.
Example 8.15: Water at 30°C flows across a pipe 10 cm OD at 50°C with a velocity of 0.6 m/s.
Determine the value of convection coefficient using applicable correlations.
v, m2/s

Pr

k, W/mK

ρ, kg/m3

T∞, 30°C

0.8315 × 10–6

5.68

0.6129

997.5

Tf, 40°C

0.657 × 10–6

4.34

0.628

995

Tw, 50°C

0.5675 × 10–6

3.68

0.63965

990

Solution: Equation (8.33): Properties at Tf
Re = 0.1 × 0.6/0.657 × 10–6 = 91,324, (9.13 × 104)

Nu = C. Rem Pr0.333
From tables
C = 0.0266, m = 0.805
Nu k
= 2683 W/m2K
D
Equation (8.34): properties at free stream temperature, T∞

∴

Nu = 427.22

∴h=

Re = 0.1 × 0.6/0.8315 × 10–6 = 72159
Nu = CRem Prn (Prf /Prw)0.25
= 0.26 (72159)0.6 (5.68)0.37 (5.68/3.68)0.25
= 453.1, h = 2777 W/m2K
Equation (8.36): Properties at film temperature
Nu = [0.35 + 0.56 Re0.52] Pr0.333
Re < 105, so applicable
∴
Nu = [0.35 + 0.56 (91324)0.52] [4.34]0.333
= 347.5, h = 2182 W/m2K
Equation (8.35 (b)) (properties at free stream temp.)
Re = 72159 Applicable
Nu = 0.25 Re0.6 Pr0.38 (Prf /Prw)0.25
= 0.25 (72159)0.6 (5.68)0.38

FG 5.68 IJ

H 3.68 K

0.25

= 443.3 ∴ h = 2717 W/m2K
(note the similarity between equation (8.34) and this)
Equation (8.37) Properties at T∞
+ 0.06

Re0.67)

Pr0.4

LM µ OP
Nµ Q
∞

w

Re = 72159
Nu = 475.6 ∴

h = 2915 W/m2K

0.25

Chapter 8

Nu = (0.4

Re0.5


356

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Equation (8.38 (b)), Properties at film temperature:
Nu = 0.3 +

LM1 + FG Re IJ OP
LM1 + FG 0.4 IJ OP MN H 282000 K PQ
MN H Pr K PQ
0.62 Re0.5 Pr 0.333

0.5

0.67 0.25

Re = 91324 Pr = 4.34, k = 0.628
Nu = 458.4 ∴ h = 2877 W/m2K
In this case also, the values by various correlations fall in a small band varying from
2181-2915 with fewer lower than 2700. For water and air, the correlations give less scatter.
Example 8.16: Air flows across an elliptical tube 0.1 m by 0.15 m perpendicular to the minor
axis with a velocity of 2.4 m/s. Air is at 20°C and the tube surface is at 40°C. Determine the
value of convection coefficient.
Solution: The properties are required at the film temperature i.e. 30°C
ν = 16 × 10–6 m2/s, Pr = 0.701, k = 0.02675
Re = 0.1 × 2.4/16 × 10–6 = 15000
Nu = C. Ren, From tables C = 0.224, n = 0.612

∴
Nu = 0.224 × 150000.612 = 80.54
h=

80.54 × 0.02675
= 21.54 W/m2K.
0.1

8.4 FLOW ACROSS SPHERES
There are a number of applications for flow over spheres in industrial processes. As in the case
of flow across cylinders, the flow development has a great influence on heat transfer. Various
correlations have been obtained from experimental measurements and these are listed in the
following paras.
The following three relations are useful for air with Pr = 0.71 (1954)
Nu = 0.37 Re0.6 17 < Re < 7000
...(8.41)
With Properties evaluated at film temperature.
The next correlation can be used for higher values of Re (1978)
Nu = 2 + [0.25 Re + 3 × 10–4 Re1.6]0.5
...(8.42)
5
100 < Re < 3 × 10
For still higher values (1978)
Nu = 430 + 5 × 10–3 Re + 0.025 × 10–9 Re2 – 3.1 × 10–17 Re3 ...(8.43)
3 × 105 < Re < 5 × 106
The next correlation is for liquids (1946)
...(8.44)
NuPr–0.3 = 0.97 + 0.68 Re0.5
1 < Re < 2000 with properties at Tf
For oils and water and for higher values of Re (1961)

NuPr–0.3

FG µ IJ
Hµ K
w
∞

0.25

= 1.2 + 0.53 Re0.54

...(8.45)


357

CONVECTIVE HEAT TRANSFER-PRACTICAL CORRELATIONS-FLOW OVER SURFACES

1 < Re < 200000 with properties at T∞
A more recent and a general equation is (1972)
Nu = 2 + (0.4

Re0.5

Re0.67)

+ 0.06

Pr0.4

FG µ IJ
Hµ K

0.25

w

...(8.46)

∞

3.5 < Re < 8 × 104, 0.7 < Pr < 380 and properties at T∞
1<

FG µ IJ < 3.2
Hµ K
w
∞

For a sphere falling in a fluid like quenching in hot bath,
Nu = 2 + 0.6 Re0.5 Pr0.333 [25 (x/D)]–0.7
...(8.47)
For low values of Pr (liquid metals)
Nu = 2 + 0.386 (Re Pr)0.5
...(8.48)
4
5
3.56 × 10 < Re < 1.525 × 10 with properties at film temperature.
These relations also provide values in the range of ± 25%.
Example 8.17: Air at 30°C flows over a sphere of 0.1 m dia with a velocity of 8 m/s. The solid

surface is at 50°C. Determine the value of convection coefficient.
Solution: The property values are
°C

ν × 106 m2/s

Pr

k, W/mK

à ì 106, kg/ms

T

30

16

0.701

0.02675

18.63

Tf

40

16.96

0.699

0.02756

19.12

Tw

50

17.95

0.698

0.02820

19.61

At film temp.:
Re = 0.1 ì 8/16.96 × 10–6 = 47170
∴ Equation (8.41) can be used
Nu = 0.37. Re0.6 = 235.72, ∴ h = 64.96 W/m2K
Using eqn. (8.42)
Nu = 2 + (0.25 Re + 3 × 10–4 Re1.6]0.5 = 146.25
∴
h = 40.3 W/m2K
Another possible equation is (8.46) (properties at T∞)
Nu = 2 + (0.4 Re0.5 + 0.06 Re0.67) Pr0.4

Fà I

GH à JK

0.25



w

6

0.5

L 0.1 ì 8 OP OP (0.701) L 18.63 O
+ 0.06 M
MN 19.61 PQ
N 16 × 10 Q PQ
0.67

0.4

0.25

−6

= 151.88
∴
h = 40.62 W/m2K
The equation (8.46) being the latest correlation, it is safer to consider the value of 40.62
W/m2K for convection coefficient.

Chapter 8

L L 0.1 × 8 O
= 2 + M0.4 M
MN N 16 × 10 PQ


358

FUNDAMENTALS OF HEAT AND MASS TRANSFER

Example 8.18: Engine oil flows over a sphere of 4 cm dia with a velocity of 0.31 m/s. The oil
is at 40°C and the ball is at 80°C. Determine the value of convection coefficient.
Solution: Two possible correlations are 8.45 and 8.46.
Eqn. (8.45),
Nu

Pr–0.3

FG µ IJ
Hµ K

0.25

w

= 1.2 + 0.53 Re0.54

∞

with properties at T∞
The property values are

T∞
Tf

Tw

°C

ν, m2/s

Pr

k,W/mK,

ρ kg/m3

40

241 × 10–6

2870

0.1442

876

60

83 ×

10–6

1050

0.1407

864

37 ×

10–6

490

0.1384

852

80

Re =

Nu. (2870)–0.3

0.31 × 0.04
241 × 10 −6

LM 37 × 10

N 241 × 10

= 51.45 ∴ The equation (8.45) is valid

−6

× 852
−6
× 876

OP
Q

0.25

L 0.31 × 0.04 OP
= 1.2 + 0.53 M
N 241 × 10 Q

0.54

−6

Nu × 0.057 = 1.2 + 4.45 = 5.65
∴
Nu = 99.07
∴
h = 357.15 W/m2K
The other correlation is given by equation (8.46)
Nu = 2 + (0.4 Re0.5 + 0.06 Re0.67) Pr0.4

LM µ OP
Nµ Q

0.25

∞

w

with properties at T∞
∴

Nu = 2 + 144.2 = 146.2

∴

h = 527.12 W/m2K

Here one of the conditions µ∞/µw < 3.2 is not satisfied and the ratio is about 6.5.
The other possible correlation is 8.44 with properties at Tf
Nu Pr–0.3 = 0.97 + 0.68 Re0.5
Nu × 0.124 = 0.97 + 0.68 (0.31 × 0.04/83 × 10–6)0.5
= 0.97 + 0.68 (149.4)0.5
Nu = 74.81 ∴ h = 263.15 W/m2K
Note that the scatter is a little more than 25% between 263.15 and 357.15.
Example 8.19: Liquid sodium at 200°C flows over a sphere at 400°C, the diameter of the
sphere being 4 cm. The velocity of flow is 0.8 m/s. Determine the value of convection coefficient.