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

(bq) part 2 book fundementals of heat and mass transfer kotandaraman has contents: convective heat transfer—practical correlations—flow over surfaces, forced convection, natural convection, phase change processes—boiling, condensation freezing and melting, heat exchangers, thermal radia tion, mass transfer.

CONVECTIVE HEAT TRANSFER

Practical Correlations - Flow Over Surfaces

8

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 frombasic principles and using the concept of boundary layer The application of these correlations arelimited However these equations provide a method of correlation of experimental results andextension of these equations to practical situations of more complex nature Though the basicdimensionless numbers used remain the same, the constants and power indices are found to varywith ranges of these parameters and geometries In this chapter it is proposed to list the varioustypes of boundaries, ranges of parameters and the experimental correlations found suitable todeal with these situations, as far as flow over surfaces like flat plates, cylinders, spheres and tubebanks are concerned

8.1 FLOW OVER FLAT PLATES

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 boundaryconditions may exist, these can be approximated to three basic types These are (i) constant walltemperature, (as may be obtained in evaporation, condensation etc., phase change at a specifiedpressure) (ii) constant heat flux, as may be obtained by electrical strip type of heating and (iii) flowwith neither of these quantities remaining constant, as when two fluids may be flowing on eitherside 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 followingequations are valid with fluid property values taken at the film temperature

Hydrodynamic boundary layer thickness (from Chapter 7)

Chapter 8

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

Momentum thickness δm in the laminar region is defined by

u u

0 δ

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

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

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

worked out problems.

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

This is also valid in the range of Prandtl numbers 0.6 to 50 In constant heat flux boundarythe 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

.[ ( / ) ]

Pr

x

The property values are at film temperature.

In all cases, the average Nusselt number is given by

This is applicable in all cases when Nu ∝ Re0.5

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

by

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 (laminarflow)

Property values should be at the film temperature, (T s + 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:

Displacement thickness

δδδδδd = δx/3 = 9.156/3 = 3.052 mm

Momentum thickness

δδδδδm = δx/7 = 9.156/7 = 1.308 mm

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:

Nu x = 0 3387

1 0 0468

0.5 0.333 0.67 0.25

.[ ( / ) ]

Pr

x

+

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

Re x = u x∞ = × −

×ν

.[ ( / ) ]

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

4 1 2

15 06 10 6

= 3.187 × 105 ∴ laminar

For constant heat flux, the average temperature difference can be found by using

It does not make much of a difference

To determine the value of convection coefficient, equation (8.11) is used

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

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)

Nu x = 0.453 Re x0.5Pr0.333taking property values at 15.51°C

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

.[ ( / ) ]

Chapter 8

The average temperature difference:

∆T = h q= 1600000

11302 = 141.6° C

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

+

−

= 49.7 as compared to 49.83 Values are not very different.

Using equation (8.8), Nu x = 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/m 2 , 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

2 0 4

83 10 6

= 9639 ∴ laminarUsing equation (8.12)

Nu x = 0 453

1 0 0207

0.5 0.33 0.67 0.25

.[ ( / ) ]

at 75.73°C, ν = 46.82 × 10–6 m2/s, Pr = 609.6, k = 0.1389

Nu x = 0 453 2 0 4 46 82 10 609 6

1 0 0207 609 6

6 0.5 0.33 0.67 0.25

( / ) ( )[ ( / ) ]

To determine the plate temperature at the edge:

∆T = 10000174 3. = 57.4°C

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

MM x O Q PP−

At x o = 0, this will reduce to the regular expression given by equation (8.7) The average

value in this case will not be 2 Nu x and the above expression has to be integrated over thelength 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 h x 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 Re x = 0 3 1

1006 10 6

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

Distance x h x with heating h x with heating

cases is low and has to be specified In the turbulent region the velocity boundary layer thickness is given by

The average Nusselt number is given by

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

momentum and heat transfer

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 forconstant wall temperature

∴ Nu x (constant heat flux) = 1.04 Nu x (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

δd = 41.54/8 = 5.19 mm Momentum thickness

δδδδδm = 7

72 × 41.54 = 4.04 mm

As Re is in the border (< 108), we can calculate C fx using eqn (8.20) or (8.21)

C fx = 0.0592 × Re x–02 = 2.08 × 10 –3

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

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 ∴ TurbulentThe length at which flow turns tubulent:

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 × 10 5 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

A Re

Chapter 8

Example 8.11: Considering the data of Example 8.10, determine the average value of convection

coefficient and C f 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 m 2 /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 ∴ TurbulentAssuming Re cr = 5 × 105

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 ∴ turbulentCritical length: 0.664 m ∴ necessary to consider laminar region

Assuming turbulent flow throughout:

C f = 0 0592

0 8

Re L–0.2 = 3.94 × 10 –3

Taking laminar region into account

Nu L = Pr0.333 [0.037 × Re L0.8 – 871] = 3321

h = 26.17 W/m 2 K

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

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 alongthe 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 flowdiffers from the flow over a flat plate as shown in Fig 8.1

Flow separation

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 C D is defined by

Drag force = C D A f ρu∞2

2 Where A f is the frontal or projected area (for a cylinder of

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

Fig 8.3 Variation of Nusselt number with angular location.

This also shows that averaging out the convection coefficient is difficult The experimental

values measured by various researchers plotted using common parameters Re D and Nu D (loglog plot) is shown in Fig 8.4 It can be seen that scatter is high at certain regions and severalseparate straight line correlations are possible for various ranges Some researchers havelimited their correlations for specific ranges and specific fluids Thus a number of correlationsare available and are listed below

A very widely used correlation is of the form (1958)

Scatter Band Nu

(log)

D

(log) ReD

Fig 8.4 Variation of Nu D with Re D 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 Nu D

The properties are to be evaluated at the film temperature

0.7 < Pr < 500; 1 < Re D < 106 and

with n = 0.36 for Pr < 10 and n = 0.37 for Pr > 10

The values of C and m are tabulated below

The properties for Re and Pr should be at free stream temperature.

A two range (1972) correlation is given below: (f-film temp.)

Nu D = [0.43 + 0.50 Re D0.5] Pr0.38 Pr

Pr

f w

F

HG I KJ

0.25

(8.35(a))

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)

10–1 < Re D < 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

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

Nu D = 0.3 + 0 62

1 0 4

1282000

0 5 0 333

0 67 0 75

0 625 0 8

F

L N

for 100 < Re D < 107, Pe = Re D Pr > 0.2

The properties are to be evaluated at film temperature.

A modification of this equation for limited range of Reynolds number is

Nu D = 0.3 + 0 62

1 0 4

1282000

0.5 0.333 0.67 0.25

0.5

F

L N

2 × 104 < Re D < 4 × 105, Pr > 0.2

This equation use properties at film temperature and is applicable for all fluids Finally

for liquid metals another correlation (1975) is obtained as

Other correlations for liquid metals over cylinder are (1979)

1 ≤ Re D Pr ≤ 100.

Analytical results are also available for constant wall temperature.

Nu D = 1.015 (Re D Pr)0.5

For constant heat flux Nu D = 1.145 (Re D Pr)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 intodifferent ranges of parameters However a common warning is that most of these may giveresults varying as much as 25% from experimental results A single correlation applicable forvarious ranges will be easier to use in computer application (say 8.38)

In actual application one has to weight carefully the parameter ranges before choosingthe 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)

Square along face, side

Square along face, side

Plane, perpendicular, width

hexagon perpendicular to flats,

0.222 0.224 0.267 0.160 0.092 0.205 0.144 0.035 0.138 0.085

0.246 0.250 0.292 0.178 0.102 0.228 0.160 0.039 0.153 0.095

0.588 0.612 0.624 0.699 0.675 0.731 0.638 0.782 0.638 0.804

Chapter 8

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 T f and T w

0 687

0.25

0.5 0.333 0.67 0.25

0.5

Re Pr Pr

Re

+F

HG I KJ

L N

F

L N

0.5

+F

HG I KJ

L N

F

L N

= 167.36 ∴ h = 25.5 W/m 2 K

In this example all the various equation provide answers within a small band This isonly fortitious and not necessarily so in all cases The parameters are not in the extremerange

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∞, T f and T w

0.5 0.333 0.67 0.75

0.625 0.8

F

L N

0.67 0.25

6 0.625 0.8 ( ) ( )

×+F

L N

= 159.16 ∴ h = 203236 W/m2 K

Chapter 8

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.

Equation (8.38 (b)), Properties at film temperature:

Nu = 0.3 + 0 62

1 0 4

1282000

0.5 0.333 0.67 0.25

0.5

Re Pr Pr

Re

+F

HG I KJ

L N

F

L N

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

×

= 21.54 W/m 2 K.

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 Variouscorrelations have been obtained from experimental measurements and these are listed in thefollowing paras

The following three relations are useful for air with Pr = 0.71 (1954)

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)

100 < Re < 3 × 105For still higher values (1978)

∞

F

HG I KJ < 3.2For 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)

3.56 × 104 < Re < 1.525 × 105 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

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.

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 T f

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/m2 K

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.

8.5 FLOW OVER BLUFF BODIES

Prerpendicular to flat plate:

(i) Packing diameter D P = 6V

A where V is the volume and A is the surface area.

(ii) Void fraction ε = the empty volume/total volume of bed The equation to find the rate

of heat flow from gas to the packing is

h D k

ν1 −ε , where V s is the fluid velocity if the bed is empty

For heat flow from wall to gas (for cylinders)

8.6 FLOW ACROSS BANK OF TUBES

In most heat exhangers in use, tube bundles are used with one fluid flowing across tube bundles.First it is necessary to define certain terms before discussing heat transfer calculations Twotypes of tube arrangement are possible

(i) in line and (ii) staggered The distance between tube centres is known as pitch The pitch along the flow is known as (S n ) and the pitch in the perpendicular direction is called (S p).These are shown in Fig 8.5

Fig 8.5 Tube arrangements in tube banks.

Due to the obstruction caused by the tubes, the velocity near the tube increases and thisincreased value has to be used in the calculation of Reynolds number In the case of in line theactual velocity near the tubes

HG I KJ

L N

Reynolds number to be calculated based on V max The property values should be

at T f The value of C and n are tabulated below in Table 8.1 For larger values of S p /D, tubes

can be considered as individual tubes rather than tube bank

Property values at T∞ Re based on Vmax

The values of C and m are tabulated below.

Conditions C m

Inline, 10 3 < Re < 2 × 105 0.27 0.63 Inline, 2 × 10 5 < Re < 2 × 106 0.021 0.84 Staggered: 10 3 < Re < 2 × 105

S

p n

Liquid metals are now in use in heat exchangers

The available correlation for tube bank is

2000 < Re < 80000

For finned tubes correlations are more complicated but are available in hand books

Example 8.20: 20 mm OD copper tubes are arranged in line at 30 mm pitch perpendicular to

flow and 25 mm pitch along the flow The entry velocity of air is 1 m/s, and the air temperature

is 20°C The tube wall is at 40°C Determine the value of convection coefficient if the number of tubes along the flow is 6 (or Bank is 6 rows deep).

Solution:

From tables, C = 0.367, n = 0.586 (look carefully for S p /D and S n /D)

Property values at T f = 30°C are

Nu = 45.61 ∴ h = 61.0 W/m 2 K

But the bank is only 6 rows deep

(Value 0.95 is read from tables 8.2).

Example 8.21: Work out the value of h for staggered arrangement using data of example

8.20.

Solution: From tables for S p /D = 1.5 and S n /D = 1.25

C = 0.451, m = 0.568

Vmax = [S p /2(S D – D)] u∞= 3 m/s

Chapter 8

S D = Sn2 S p

2 0.52+F

HG I KJ

L N

SOLVED PROBLEMS

Problem 8.1: The local Nusselt number in the case of rough plate was correlated to give

Nu x = 0.04 Re x 0.9 Pr 1/3 Determine the average value upto a length L

k Pr u v L

Nu= 0 04Re L Pr Nu= Nu L

0 9

1

0 90.9 1/3

generally average will be (1/n) times Nu L where n is the index of x.

Problem 8.2: In testing a model, the following measurements were made

Average Nusselt No Reynolds number

Solution: The test is done using air at film temperature of 30°C If the results can be correlated

by an equation of the form Nu = C Re m Pr1/3 determine C and m For flow of air over a similar

surface of length 6 cm at 70°C with a velocity of 40 m/s, determine the average value of convectioncoefficient Air is at 110°C and 1 atm

Two readings are sufficient to determine the two unknowns, namely C and m However,

a check can be made to determine the average values of C and m

Both Nu and Re lie between data 2 and 3 as a check.

Problem 8.3: In flow over a wedge, the local Nusselt number is given by

11

1

4 3

32

,(m ) ( / )

Nu = 1.5 Nu L

β = 1.0, m = 1, 0 5 1 11 1

( + ) =

Problem 8.4: A plate 5 m long at 470°C has air flowing over it with a velocity of 2.5 m/s The

air is at 30°C Determine the heat to be supplied for every m length Assume unit width.

Solution: The values to be determined are the heat transfer in the first, second and subsequent

m lengths This can be done by calculating the average heat transfer coefficient for 1 m, 2 m,

3 m 4 m and 5 m lengths and then finding the heat transfer in each of the case and then takingthe difference

The film temperature = (470 + 30)/2 = 250°C

The property values are: ν = 40.61 × 10–6 m2/s, Pr = 0.677, k = 0.04268 W/mK

The flow is completely in the laminar region:

∴ Nu = 0.664 Re0.5Pr0.33 The values and h are tabulated above

The heat flow: (width 1 m) Q = hA (T w – T∞), T w = 470°C, T∞ = 30°C

The heat transfer in the extended length decreases as it should

Problem 8.5: A motor cycle travels at 100 kmph On the engine head a fin of 0.16 m

length and 0.04 m width is exposed to convection on both sides The fin surface is at 300°C and air is at 20°C Determine the rate of heat removal from the fin assuming turbulent flow prevails all through.

Solution: In this case it is assumed that due to disturbances turbulent flow starts even at low

Reynolds numbers

The film temperature = (300 + 20)/2 = 160°C

The property values are ν = 30.09 × 10–6 m2/s, Pr = 0.682, k = 0.0364 W/mK

The heat transfer is only about 50% of that assuming turbulent conditions

Problem 6: An aircraft travelling at 300 kmph has a wing span of 2 m and is at an altitude

where the pressure is 0.7 bar and temperature is –10°C The wing absorbs solar radiation at

800 W/m 2 Determine the wing surface temperature under this condition.

Solution: This problem may be modelled as uniform heat flux model The film temperature is

not known Assuming 0°C, the property values are

This is in the turbulent region (Fully turbulent condition is assumed as L cr = 0.11 m)

The local Nusselt number is calculated using

Nu x = 0.0296 Re x0.8 Pr0.33 = 9476

h x = 115.7 W/m2K

Chapter 8

For constant heat flux = h x = h x × 1.04 = 120.3 W/m 2 K

∆T at this location:

800 = 120.2 × 1 × ∆T = 6.65°C

∴ Wing temp = – 3.35°C at the trailing edge.

The assumption of 0°C as T f is in error The new value can be now used and the resultsrefined

At a distance of 1 m from leading edge

Re = (8.8/2) × 106, fully turbulent condition is assumed

Nu x = 0.0296 Re x0.8 Pr0.33 = 5442

h x = 132.9 W/m2KConstant heat flux

h x = 1.04 × 132.9 = 138.2 W/m2K

The minimum wing surface temperature is –10°C and the maximum about – 3.34°C.

Problem 8.7: A surface 1 m × 1m size has one half very rough and the other half smooth The

surface is at 100°C Air at 72 kmph and 20°C flows over the surface If the flow direction is reversed is there a possibility of change in the average value of convection coefficient ? If initially the rough area is at the leading edge, determine the change in the value.

Solution: If the rough surface is at the leading edge, then the flow is turbulent all through.

However, if the smooth surface is at the leading edge, turbulence will begin only at Re = 5 × 105

or at the beginning of the rough surface So, the convection coefficient will depend on thedirection of flow

The film temperature is (100 + 20)/2 = 60°C

The property values are: ν = 18.97 × 10–6 m2/s, Pr = 0.696, k = 0.02896 W/mK Initially the rough surface is at the leading edge So the flow is turbulent all through Re x = 1.054 × 106(calculated)

L0 037 Re L Pr

0.8 0.333 ( )

When the smooth surface is at leading edge,

The Reynolds number at mid location

It may so happen that rough surface may start even before the value of Re = 5 × 105 Insuch a case the critical Reynolds number should be taken as the Reynolds number at thelocation where the rough area begins Then the number 871 will be different Equation (8.00)and (8.31) should be used in such a case

i.e Nu = Pr0.333 (0.37 Re L0.8 – A) where

A = 0.037 Re cr0.8 – 0.664 Re cr0.5

Problem 8.8: Wind blows at 20 kmph parallel to the wall of adjacent rooms The first room

extends to 10 m and the next one to 5 m The wall is 3.2 m high The room inside is at 20°C and the ambient air is at 40°C The walls are 25 cm thick and the conductivity or the material is 1.2 W/mK On the inside convection coefficient has a value of 6 W/m 2 K Determine the heat gain through the walls of each room.

Solution: The film temperature is not known But it has to be between 40°C and 20°C A

choice is made as 35°C

The property values are: v = 16.48 × 10–6 m2/s, Pr = 0.70, k = 0.02716 W/mK

Reynolds numbers at 10 m and 15 m locations are:

∴ Average for last 5 m = (12.324 × 15 – 12.783 × 10)/5 = 11.406 W/m 2 K

Heat gain in the first room

12 783

0 2512

161

16

Chapter 8

Problem 8.9: Icebergs 1 km long by 0.8 km wide and 0.3 km thick at 0°C are proposed to be

towed to arid regions for obtaining supply of fresh water If the average water temperature is 10°C and if the iceberg is to travel at 1.2 km/hour, determine the thickness of ice melted per hour The latent heat of ice is 334 kJ/kg Assume that the iceberg is towed along the 1 km direction.

Solution: The temperature of the ice is taken as 0°C So the film temperature = 5°C.

Properties of water are : (using 0°C and 20°C values)

∴ mass of ice melted = 10.03 × 106/3.34 × 105 = 30.03 kg/hr

The thickness melted= 30.03 mm/hr or 3 cm/hr.

Problem 8.10: Water flows over a flat plate having a uniform heat generation rate The plate

is 15 cm × 15 cm side Water is at 20°C and the flow velocity is 3 m/s Determine the heat that may be carried away by the water if the maximum temperature of the plate is not to exceed 80°C.

Solution: The property values can be evaluated at a mean temperature i.e 50°C in this case

∴ The flow is turbulent as Re > 5 × 105

This assumes an average value of h But the maximum temperature is to be at edge So

the local Nusselt number should be used for better estimate

As the heat generation at every location is the same this method may be used

These calculations can serve only as a first estimate, and refinements are necessary toget at more accurate values

Problem 8.11: Glycerine at 30°C flows past a 30 cm square flat plate at a velocity of 1.5 m/

s The drag force measured was 8.9 N Determine the value of convection coefficient for such a system.

Solution: This problem has to use the analogy method C f can be calculated using the dragforce The film temperature is taken as 30°C

density = 1258 kg/m3, ν = 501 × 10–6 m2/s, Pr = 5380, k = 0.2861 W/mK

Force on 1 m2 = 8 9

0 3 0 3

× N

1258 152

Problem 8.12: Helium at a pressure of 0.15 atm and 30°C flows over a flat plate at 70°C at a

velocity of 50 m/s The plate is 1 m long Calculate the value of convection coefficient.

Solution: The film temperature is (70 + 30)/2 = 50°C

The property values are : density = (0.178 + 0.130)/2 kg/m3

ν = ((105 + 176)/2) × 10–6 m2/s, Pr = (0.684 + 0.667)/2

k = (0.14304 + 0.1791)/2

Chapter 8

As the pressure is not 1 atm., the value of

µ = νρ only remains constant ∴ ν1ρ1 = ν2ρ2

Problem 8.13: Liquid ammonia at – 20°C flows with a velocity of 5 m/s over a plate 0.45 m

length at 20°C Determine the value of average convection coefficient.

Solution: The film temperature is (– 20 + 20)/2 = 0°C

The property values are

ρ = 640 kg/m3, ν = 0.373 × 10–6 m2/s

Pr = 2.050, k = 0.5396 W/mK

Re = 5 × 0.45 /0.373 × 10–6 = 6.03 × 106Turbulent flow prevails

Nu = (0.037 Re0.8 – 871) Pr0.333 = 11381.7

The value is high as it is liquid flow at a high velocity

Problem 8.14: Calculate the value of convection coefficient for flow of the following fluids at

10°C across a pipe 20 mm dia at 30°C, the flow velocity being 5 m/s

(a) Air (b) Water (c) engine oil (d) liquid ammonia.

Solution: (a) The property values for air are

(c) Engine oil at 20°C

ν = 901 × 10–6 m2/s, Pr = 10400, k = 0.1454 W/mK

The Prandtl number value is very high One suitable correlation can be

Note that for liquids higher value (of one or two orders of magnitude) of convectioncoefficient is obtained for the same velocity of flow

Problem 8.15: A wire 0.5 mm dia is at 40°C in a cross flow of air at 20°C while dissipating 35

W/m Determine the velocity of the air stream.

The convective heat transfer coefficient can be determined from the data.

Nu = C Re m Pr0.333

h D k

= 1114 08 0 0005

0 02675

Chapter 8

∴ The velocity is 63 m/s (nearest) R e = 1968

Check by another equation

∴ The estimate is acceptable

Problem 8.16: Air at 20°C flows over a cylinder 10 cm dia at 60°C with a velocity of 10 m/s.

Compare the value of convective heat transfer coefficient with a plate of length πD/2 with other parameters remaining the same.

Solution: The properties at T f = (20 + 60)/2 = 40°C are

The curved surface provides greater turbulence and higher convection coefficient

Problem 8.17: Approximating a human body as a cylinder of 0.3 m dia and 1.75 m long, at

surface temperature of 30°C exposed to winds at 15 kmph at 10°C, determine the rate of heat loss.

Solution: This problem may give an idea about the chilling in cold winds.

The film temperature = (30 + 10)/2 = 20°C

Property values are ν = 15.06 × 10–6 m2/s