Fundementals of heat and mass transfer kotandaraman

13.6 Radiant Heat Exchange Between Black Surfaces 60413.7 Heat Exchange by Radiation Between Gray Surfaces 606 13.8 Effect of Radiation on Measurement of Temperature by a Bare Thermomete

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Professor Obert has observed in his famous treatise on Thermodynamics that concepts arebetter understood by their repeated applications to real life situations A firm conviction ofthis principle has prompted the author to arrange the text material in each chapter in thefollowing order.

In the first section after enunciating the basic concepts and laws mathematical modelsare developed leading to rate equations for heat transfer and determination of temperaturefield, simple and direct numerical examples are included to illustrate the basic laws Morestress is on the model development as compared to numerical problems

A section titled “SOLVED PROBLEMS” comes next In this section more involvedderivations and numerical problems of practical interest are solved The investigation of theeffect of influencing parameters for the complete spectrum of values is attempted here Problemsinvolving complex situations are shown solved in this section Two important ideas are stressed

in this section These are checking of dimensional homogeneity in the case of all equationsderived and the validation of numerical answers by cross checking This concept of validation

in professional practice is a must in all design situations

In the next section objective type questions are given These are very useful forunderstanding the basis and resolving misunderstandings

In the final section a large number of graded exercise problems involving simple tocomplex situations are included

In the first of the 14 chapters the basic laws for the three modes of heat transfer areintroduced and the corresponding rate equations are developed The use of electrical analogy

is introduced and applied to single and multimode heat transfer situations The need for iterativeworking is stressed in the solved problems

The second chapter deals with one dimensional steady state conduction Mathematicalmodels are developed by the three geometries namely Plate, Hollow Cylinder and Hollow Sphere.Multilayer insulation is also discussed The effect of variation of thermal conductivity on heattransfer and temperature field is clearly brought out Parallel flow systems are discussed.Examples on variation of area along the heat flow direction are included The use of electricalanalogy is included in all the worked examples The importance of calculating the temperaturegradient is stressed in many of the problems

In the third chapter models for conduction with heat generation are developed for threegeometric configurations namely plate, cylinder and sphere The effect of volume to surfacearea and the convection coefficient at the surface in maintaining lower material temperature

is illustrated Hollow cylindrical shape with different boundary conditions is discussed.Conduction with variable heat generation rate is also modelled

Fins/extended surface or conduction-convection situation is discussed in the fourthchapter Models for heat transfer and temperature variation are developed for four different

PREFACE TO THE THIRD EDITION

boundary conditions Optimisation of the shape of the fin of specified volume for maximumheat flow is discussed Circumferential fins and variable area fins are analysed The use ofnumerical method is illustrated Error in measurement of temperature using thermometer iswell discussed The possibility of measurement of thermal conductivity and convective heattransfer coefficient using fins is illustrated.

Two dimensional steady state conduction is discussed in the fifth chapter Exact analysis

is first developed for two types of boundary conditions The use of numerical method is illustrated

by developing nodal equations The concept and use of conduction shape factor is illustratedfor some practical situations

One dimensional transient (unsteady) heat conduction is discussed in Chapter 6 Threetypes of models arise in this case namely lumped heat capacity system, semi-infinite solid andinfinite solid Lumped heat capacity model for which there are a number of industrialapplications is analysed in great detail and problems of practical interest are shown solved.The condition under which semi-infinite solid model is applicable as compared to infinite solidmodel is clearly explained Three types of boundary conditions are analysed Infinite solidmodel for three geometric shapes is analysed next The complexity of the analytical solution isindicated Solution using charts is illustrated in great detail Real solids are of limiteddimensions and these models cannot be applied directly in these cases In these cases productsolution is applicable A number of problems of practical interest for these types of solids areworked out in this section In both cases a number of problems are solved using numericalmethods Periodic heat flow problems are also discussed

Concepts and mechanism of convection are discussed in the seventh chapter Afterdiscussing the boundary layer theory continuity, momentum and energy equations are derived.Next the different methods of solving these equations are discussed In addition to the exactanalysis approximate integral method, analogy method and dimensional analysis are alsodiscussed and their applicability is indicated General correlations for convective heat transfercoefficient in terms of dimensionless numbers are arrived at in this chapter

In Chapter 8, in addition to the correlations derived in the previous chapter, empiricalcorrelations arrived at from experimental results are listed and applied to flow over surfaceslike flat plate, cylinder, sphere and banks of tubes Both laminar and turbulent flows situationare discussed

Flow through ducts is discussed in Chapter 9 Empirical correlations for various situationsare listed Flow developing region, fully developed flow conditions, constant wall temperatureand constant wall heat flux are some of the conditions analysed Flow through non-circularpipes and annular flow are also discussed in this chapter

Natural convection is dealt with in Chapter 10 Various geometries including enclosedspace are discussed The choice of the appropriate correlation is illustrated through a number

of problems Combined natural and forced convection is also discussed

Chapter 11 deals with phase change processes Boiling, condensation, freezing andmelting are discussed Basic equations are derived in the case of freezing and melting andcondensation The applicable correlations in boiling are listed and their applicability isillustrated through numerical examples

Chapter 12 deals with heat exchangers, both recuperative and regenerative types TheLMTD and NTU-effectiveness methods are discussed in detail and the applicability of thesemethods is illustrated Various types of heat exchangers are compared for optimising the size

Thermal radiation is dealt with in Chapter 13 The convenience of the use of electricalanalogy for heat exchange among radiating surfaces is discussed in detail and is applied inalmost all the solved problems Gas radiation and multi-body enclosures are also discussed.Chapter 14 deals with basic ideas of mass transfer in both diffusion and convectionmodes A large number of problems with different fluid combinations are worked out in thischapter.

A large number of short problems and fill in the blank type and true or false type

questions are provided to test the understanding of the basic principles

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1.0 Introduction 1

1.1 Heat Transfer 1

1.2 Modes of Heat Transfer 2

1.3 Combined Modes of Heat Transfer 8

1.4 Dimensions and Units 10

2.1 The General Model for Conduction Study 26

2.2 Steady Conduction in One Direction (One Dimensional) 30

2.3 Conduction in Other Shapes 41

2.4 One Dimensional Steady State Heat Conduction with Variable Heat

Conductivity or Variable Area Along the Section 42

2.5 Critical Thickness of Insulation 48

2.6 Mean Area Concept 50

4 HEAT TRANSFER WITH EXTENDED SURFACES (FINS) 128–175

6.1 A Wall Exposed to the Sun 202

6.2 Lumped Parameter Model 203

6.3 Semi Infinite Solid 207

6.4 Periodic Heat Conduction 213

6.5 Transient Heat Conduction in Large Slab of Limited Thickness, Long Cylinders

7 CONVECTION 285–333

7.0 Introduction 285

7.1 Mechanism of Convection 285

7.2 The Concept of Velocity Boundary Layer 287

7.3 Thermal Boundary Layer 289

7.4 Laminar and Turbulent Flow 291

7.5 Forced and Free Convection 292

7.6 Methods Used in Convection Studies 293

8.0 Introduction 334

8.1 Flow Over Flat Plates 334

8.2 Turbulent Flow 343

8.3 Flow Across Cylinders 348

8.4 Flow Across Spheres 356

8.5 Flow Over Bluff Bodies 359

8.6 Flow Across Bank of Tubes 360

9.1 Hydrodynamic Boundary Layer Development 386

9.2 Thermal Boundary Layer 387

9.3 Laminar Flow 388

9.4 Turbulent Flow 399

9.5 Liquid Metal Flow 402

9.6 Flow Through Non-circular Sections 404

9.7 The Variation of Temperature Along the Flow Direction 406

Solved Problems 408

Objective Questions 431

Exercise Problems 432

10.4 Correlations from Experimental Results 442

10.5 A More Recent Set of Correlations 446

10.6 Constant Heat Flux Condition—Vertical Surfaces 447

10.7 Free Convection from Inclined Surfaces 451

10.8 Horizontal Cylinders 454

10.9 Other Geometries 455

10.10 Simplified Expressions for Air 456

10.11 Free Convection in Enclosed Spaces 458

10.12 Rotating Cylinders, Disks and Spheres 459

10.13 Combined Forced and Free Convection 460

Solved Problems 461

Objective Questions 477

Exercise Problems 477

12.1 Over All Heat Transfer Coefficient 521

12.2 Classification of Heat Exchangers 524

12.3 Mean Temperature Difference—Log Mean Temperature Difference 526

12.4 Regenerative Type 531

12.5 Determination of Area in Other Arrangements 531

12.6 Heat Exchanger Performance 535

12.7 Storage Type Heat Exchangers 547

12.8 Compact Heat Exchangers 550

13.6 Radiant Heat Exchange Between Black Surfaces 604

13.7 Heat Exchange by Radiation Between Gray Surfaces 606

13.8 Effect of Radiation on Measurement of Temperature by a Bare Thermometer 61313.9 Multisurface Enclosure 614

13.10 Surfaces Separated by an Absorbing and Transmitting Medium 617

14.2 Diffusion Mass Transfer 657

14.3 Fick’s Law of Diffusion 657

14.4 Equimolal Counter Diffusion 659

14.5 Stationary Media with Specified Surface Concentration 660

14.6 Diffusion of One Component into a Stationary Component or

Unidirectional Diffusion 661

14.7 Unsteady Diffusion 661

14.8 Convective Mass Transfer 662

14.9 Similarity Between Heat and Mass Transfer 664

Solved Problems 664

Exercise Problems 680

Fill in the Blanks 682

State True or False 699

Short Questions 702

Appendix 707

References 712

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Unit Conversion Constants

Pressure 1 N/m 2 = 1.4504 × 10 –4 lbf/in 2 1 lbf/in 2 = 6894.8 N/m 2

Pressure 1 bar = 14.504 lbf/in 2 1 lbf/in 2 = 0.06895 bar

Specific heat 1 kJ/kg°C = 0.23884 Btu/lb°F 1 Btu/lb°F = 4.1869 kJ/kg°C Surface tension 1 N/m = 0.068522 lbf/ft 1 lbf/ft = 14.5939 N/m

Thermal conductivity 1 W/m°C = 0.5778 Btu/hr ft°F 1 Btu/hrft°F = 1.7307 W/m°C Convection coefficient 1 W/m 2 °C = 0.1761 Btu/hrft 2 °F 1 Btu/hr ft 2 °F = 5.6783 W/m 2 °C Dynamic viscosity 1 kg/ms = 0.672 lb/fts 1 lb/fts = 1.4881 kg/ms

Quantity S.I to Metric Metric to S.I.

Pressure 1 N/m 2 = 10.19 × 10 –6 kgf/cm 2 1 kgr/cm 2 = 98135 N/m 2

(heat, work) 1 Nm (= 1 J) = 0.1019 kgf m 1 kgf m = 9.81 Nm (J)

Energy

Specific heat 1 kJ/kg°C = 0.2389 kcal/kg°C 1 kcal/kg°C = 4.186 kJ/kg°C Surface tension 1 N/m = 0.1019 kgf/m 1 kgf/m = 9.81 N/m

Thermal conductivity 1 W/m°C = 0.86 kcal/hrm°C 1 kcal/hrm°C = 1.163 W/m°C Convection coefficient 1 W/m 2 °C = 0.86 kcal/hrm 2 °C 1 kcal/hrm 2 °C = 1.163 W/m 2 °C Dynamic viscosity 1 kg/ms (Ns/m 2 ) = 0.1 Poise 1 poise = 10 kg/ms (Ns/m 2 ) Kinematic viscosity 1 m 2 /s = 3600 m 2 /hr 1 m 2 /hr = 2.778 × 10 –4 m 2 /s

1 Stoke = cm 2 /s = 0.36 m 2 /hr = 10 –4 m 2 /s Universal gas const 8314.41 J/kg mol K = 847.54 m kgf/kg mol K

= 1.986 kcal/kg mol K Gas constant in air (SI) = 287 J/kg K

Stefan Boltzmann const 5.67 × 10 –8 W/m 2 K 4 = 4.876 × 10 –8 kcal/hr m 2 K 4

AN OVERVIEW OF HEAT TRANSFER

The driving potential or the force which causes the transfer of energy as heat is thedifference in temperature between systems Other such transport processes are the transfer ofmomentum, mass and electrical energy In addition to the temperature difference, physicalparameters like geometry, material properties like conductivity, flow parameters like flowvelocity also influence the rate of heat transfer

The aim of this text is to introduce the various rate equations and methods ofdetermination of the rate of heat transfer across system boundaries under different situations

locations and time i.e T (x, y, z, τ) and dT/dx, dT/dy, dT/dz, dT/dτ etc These two are interrelated,

one being dependent on the other However explicit solutions may be generally required forone or the other

The basic laws governing heat transfer and their application are as below:

1 First law of thermodynamics postulating the energy conservation principle: This

law provides the relation between the heat flow, energy stored and energy generated in a

given system The relationship for a closed system is: The net heat flow across the system bondary + heat generated inside the system = change in the internal energy, of the system This will also apply for an open system with slight modifications.

The change in internal energy in a given volume is equal to the product of volume

density and specific heat ρcV and dT where the group ρcV is called the heat capacity of the

system The basic analysis in heat transfer always has to start with one of these relations

2 The second law of thermodynamics establishing the direction of energy transport

as heat The law postulates that the flow of energy as heat through a system boundary willalways be in the direction of lower temperature or along the negative temperature gradient

3 Newtons laws of motion used in the determination of fluid flow parameters.

4 Law of conservation of mass, used in the determination of flow parameters.

5 The rate equations as applicable to the particular mode of heat transfer.

1.2 MODES OF HEAT TRANSFER

1.2.1 Conduction: This is the mode of energy transfer as heat due to temperature difference within a body or between bodies in thermal contact without the involvement of mass flow and mixing This is the mode of heat transfer through solid

barriers and is encountered extensively in heat transfer equipment design as well as in heatingand cooling of various materials as in the case of heat treatment The rate equation in this

mode is based on Fourier’s law of heat conduction which states that the heat flow by conduction in any direction is proportional to the temperature gradient and area perpendicular to the flow direction and is in the direction of the negative gradient.

The proportionality constant obtained in the relation is known as thermal conductivity, k, of

the material The mathematical formulation is given in equation 1.1

The units used in the text for various parameters are:

Q – W, (Watt), A – m2, dT – °C or K (as this is only temperature interval, °C and K can

be used without any difficulty) x – m, k – W/mK.

For simple shapes and one directional steady conditions with constant value of thermalconductivity this law yields rate equations as below:

1 Conduction, Plane Wall (Fig 1.1), the integration of the equation 1.1 for a plane

wall of thickness, L between the two surfaces at T1 and T2 under steady condition leads toequation 1.2 The equation can be considered as the mathematical model for this problem

Example 1.1: Determine the heat flow across a plane wall of 10 cm thickness with a constant

thermal conductivity of 8.5 W/mK when the surface temperatures are steady at 100°C and 30°C The wall area is 3m 2 Also find the temperature gradient in the flow direction.

Solution: Refer to Fig 1.1 and equation 1.2:

Fig 1.2 Electrical analogy (a) conduction circuit (b) Electrical circuit.

The denominator in equation 1.2, namely L/kA can be considered as thermal resistance

for conduction An electrical analogy is useful as a concept in solving conduction problems

and in general heat transfer problems

1.2.2 Thermal Conductivity: It is the constant of proportionality in Fourier’s equation and

plays an important role in heat transfer The unit in SI system for conductivity is W/mK It is

a material property Its value is higher for good electrical conductors and single crystals likediamond Next in order or alloys of metals and non metals Liquids have conductivity less thanthese materials Gases have the least value for thermal conductivity

In solids heat is conducted in two modes 1 The flow of thermally activated electronsand 2 Lattice waves generated by thermally induced atomic activity In conductors thepredominant mode is by electron flow In alloys it is equal between the two modes In insulators,the lattice wave mode is the main one In liquids , conduction is by atomic or molecular diffusion

In gases conduction is by diffusion of molecules from higher energy level to the lower level

Thermal conductivity is formed to vary with temperature In good conductors, thermal conductivity decreases with temperature due to impedance to electron flow of higher electron densities In insulators, as temperature increases, thermal atomic activity also increases and hence thermal conductivity increases with temperature In the case of gases, thermal conductivity increases with temperature due to increased random activity

of atoms and molecules Thermal conductivity of some materials is given in table 1.1

Table 1.1 Thermal conductivity of some materials at 293 K

Material Thermal conductivity, W/mK

The variation of thermal conductivity of various materials with temperature is shown

in Fig 1.3

Silver(99.9%) Aluminum(pure) Magnesium(pure)

Lead Mercury

Uo (dense) 2

Magnesite Fireclay brick(burned 1330°C)

Carbon(amorphous)

Water

Hydrogen

Asbestos sheets (40 laminations/m) Engine oil

Copper(pure)

Air

CO2

Solids Liquids Gases(at atm press)

Fig 1.3 Effect of temperature on thermal conductivity of materials.

1.2.3 Thermal Insulation: In many situations to conserve heat energy, equipments have to

be insulated Thermal insulation materials should have a low thermal conductivity This isachieved in solids by trapping air or a gas in small cavities inside the material It may also beachieved by loose filling of solid particles The insulating property depends on the material aswell as transport property of the gases filling the void spaces There are essentially three types

of insulating materials:

1 Fibrous: Small diameter particles or filaments are loosely filled in the gap between

surfaces to be insulated Mineral wool is one such material, for temperatures below 700°C.Fibre glass insulation is used below 200°C For higher temperatures refractory fibres likeAlumina (Al2O3) or silica (S1O2) are useful

2 Cellular: These are available in the form of boards or formed parts These contain

voids with air trapped in them Examples are polyurethane and expanded polystyrene foams

3 Granular: These are of small grains or flakes of inorganic materials and used in

preformed shapes or as powders

The effective thermal conductivity of these materials is in the range of 0.02 to 0.04W/mK

Chapter 1

1.2.4 Contact Resistance: When two

different layers of conducting materials are

placed in thermal contact, a thermal

resistance develops at the interface This

is termed as contact resistance A

significant temperature drop develops at

the interface and this has to be taken into

account in heat transfer calculation The

contact resistance depends on the surface

roughness to a great extent The pressure

holding the two surfaces together also

influences the contact resistance When the

surfaces are brought together the contact

is partial and air may be trapped between

the other points as shown in Fig 1.4

Some values of contact resistance for

different surfaces is given in table 1.2

Table 1.2.

Surface type Roughness µm Temp Pressure atm R, m 2 °C/W × 10 4

1.2.5 Convection: This mode of heat transfer is met with in situations where energy is

transferred as heat to a flowing fluid at the surface over which the flow occurs This mode isbasically conduction in a very thin fluid layer at the surface and then mixing caused by theflow The energy transfer is by combined molecular diffusion and bulk flow The heat flow isindependent of the properties of the material of the surface and depends only on the fluidproperties However the shape and nature of the surface will influence the flow and hence theheat transfer Convection is not a pure mode as conduction or radiation and hence involves

several parameters If the flow is caused by external means like a fan or pump, then the mode is known as forced convection If the flow is due to the buoyant forces caused by temperature difference in the fluid body, then the mode is known as free or natural convection.

In most applications heat is transferred from one fluid to another separated by a solid surface

So heat is transferred from the hot fluid to the surface and then from the surface to the coldfluid by convection In the design process thus convection mode becomes the most importantone in the point of view of application The rate equation is due to Newton who clubbed all the

parameters into a single one called convective heat transfer coefficient (h) as given in equation 1.3 The physical configuration is shown in Fig 1.5 (a).

Fig 1.4 Contact resistance temperature drop

T2

Tc1

Tc2

T1T

x 0

Insulated

Solid A Solid B Gap between solids

Fig 1.5 Electrical analogy for convection heat transfer

Example 1.2: Determine the heat transfer by convection over a surface of 0.5 m 2 area if the surface is at 160°C and fluid is at 40°C The value of convective heat transfer coefficient is 25 W/m 2 K Also estimate the temperature gradient at the surface given k = 1 W/mK.

Solution: Refer to Fig 1.5a and equation 1.3

Q = hA (T1 – T2) = 25 × 0.5 × (160 – 40) W = 1500 W or 1.5 kW

The resistance = 1/hA = 1/25 × 0.5 = 0.08°C/W.

The fluid has a conductivity of 1 W/mK, then the temperature gradient at the surface

1.2.6 Radiation: Thermal radiation is part of the electromagnetic spectrum in the limited

wave length range of 0.1 to 10 µm and is emitted at all surfaces, irrespective of the temperature.Such radiation incident on surfaces is absorbed and thus radiation heat transfer takes placebetween surfaces at different temperatures No medium is required for radiative transfer butthe surfaces should be in visual contact for direct radiation transfer The rate equation is due

to Stefan-Boltzmann law which states that heat radiated is proportional to the fourth power

of the absolute temperature of the surface and heat transfer rate between surfaces is given in

equation 1.4 The situation is represented in Fig 1.6 (a).

where, F—a factor depending on geometry and surface properties,

σ—Stefan Boltzmann constant 5.67 × 10–8 W/m2K4 (SI units)

A—m2, T1, T2 → K (only absolute unit of temperature to be used).

Fig 1.6 Electrical analogy-radiation heat transfer.

Example 1.3: A surface is at 200°C and has an area of 2m 2 It exchanges heat with another surface B at 30°C by radiation The value of factor due to the geometric location and emissivity

is 0.46 Determine the heat exchange Also find the value of thermal resistance and equivalent convection coefficient.

Solution: Refer to equation 1.4 and 1.5 and Fig 1.6.

The denominator in the resistance terms is also denoted as h r A where h r = Fσ (T1 + T2)

(T12 + T22) and is often used due to convenience approximately h r = Fσ FT1+T2

HG 2 I KJ

3 The

determination of F is rather involved and values are available for simple configurations in the form of charts and tables For simple cases of black surface enclosed by the other surface F = 1 and for non black enclosed surfaces F = emissivity (defined as ratio of heat radiated by a

surface to that of an ideal surface)

In this chapter only simple cases will be dealt with and the determination of F will be taken up in the chapter on radiation The concept of h r is convenient, though difficult to arrive

at if temperature is not specified The value also increases rapidly with temperature

1.3 COMBINED MODES OF HEAT TRANSFER

Previous sections treated each mode of heat transfer separately But in practice all the threemodes of heat transfer can occur simultaneously Additionally heat generation within the solidmay also be involved Most of the time conduction and convection modes occur simultaneouslywhen heat from a hot fluid is transferred to a cold fluid through an intervening barrier Considerthe following example A wall receives heat by convection and radiation on one side Afterconduction to the next surface heat is transferred to the surroundings by convection andradiation This situation is shown in Fig 1.7

Fig 1.7 Combined modes of heat transfer.

The heat flow is given by equation 1.6

Q A

where h r1 and h r2 are radiation coefficients and h1 and h2 are convection coefficients

Example 1.4: A slab 0.2 m thick with thermal conductivity of 45 W/mK receives heat from a

furnace at 500 K both by convection and radiation The convection coefficient has a value of

50 W/m 2 K The surface temperature is 400 K on this side The heat is transferred to surroundings

at T∞2both by convection and radiation The convection coefficient on this side being 60 W/m 2 K Determine the surrounding temperature.

Assume F = 1 for radiation.

Solution: Refer Fig 1.7 Consider 1 m2 area Steady state condition

= 7092.2 W.

U V|

1.3.1 Overall Heat Transfer Coefficient: Often when several resistances for heat flow is

involved, it is found convenient to express the heat flow

where U is termed as overall heat transfer coefficient having the same unit as convective

heat transfer coefficient, h The value of U can be obtained for a given area A by equation 1.8.

where R1, R2, R3, are the resistances in series calculated based on the reas A1, A2, A3 etc

1.3.2 Energy Balance With Heat Transfer: There are situations when a body receives heat

by convection and radiation and transfer part of it to the surroundings and stores the remaining

in the body by means of increase in temperature In such a situation, the rate of temperaturechange can be obtained by the equation 1.9 Heat generation may also be included

and specific heat of the body

When equilibrium is reached, dT

dτ = 0, So

Example 1.5: In a cylindrical shaped body of 30 cm diameter and 30 cm length heat is generated

at a rate 1.5 × 10 6 W/m 3 The surface temperature is 400°C The convection coefficient is 200 W/m 2 K Heat is convected and radiated to the surroundings at 100°C The radiation factor is one The solid has a density of 19000 kg/m 3 and a specific heat of 0.118 kJ/kgK Determine the rate of change of temperature of the body at that instant in °C/s.

Solution: Refer equation 1.4 and Fig 1.8

(q – Q R – Q C ) dτ = ρVC dT

Possible simplifications are.

(i) no heat generation, (ii) no radiation or, (iii) steady state etc, which will reduce one of

the terms to be zero

1.4 DIMENSIONS AND UNITS

For numerical estimation of heat transfer rate units of various parameters become necessary.All equations should be dimensionally homogeneous Dimensions are universal and there is nodifference from country to country But the systems of unit varies from country to country

Three popular systems are (1) FPS (foot, pound, second, °F) (2) MKS (metre, kilogram, second

°C) and (3) SI (metre, kilogram, second, K) system of units In this text SI system of units is

adopted The units used for various quantities is listed in table 1.2 and conversion factors aregiven separately

Table 1.3 Units adopted for various quantities

0.3 m 0.3 m

Fig 1.8

(Contd )

Chapter 1

The units for other parameters will be defined as and when these are used

In solving numerical problems, consistent sets of units should be used Otherwise theanswer will be meaningless

Example 1.6: Convert the following units into their equivalent SI units :

(i) BTU/hr ft°F, (ii) BTU/hr ft 2 °F From published tables the following are read 1J = 9.4787 × 10 –4 BTU, 1m = 39.370 inches, kg = 2.2046 lb, °C = 9/5°F.

Solution: (i) Therefore, 1 BTU = 1/9.4787 × 10–4 J = 1054.997 J, ft = (12/39.37) m

A series of steps listed below will be useful in analysing and estimating heat transfer

1 List the available data for the problem situation Then look for additional data fromother sources, like property listings

2 Sketch a schematic diagram for the system involved and identify the basic processesinvolved (Physical model)

3 List the simplifying assumptions that are reasonable This should be checked later

4 Apply the rate equations and conservation laws to the situation (Mathematical model)

5 Try to validate the results obtained This is an important step, which is often overlookedwith disastrous results

SOLVED PROBLEMSCombined Convection and Radiation

Problem 1: A surface is at 200°C and is exposed to surroundings at 60°C and convects and

radiates heat to the surroundings The convection coefficient is 80W/m 2 K The radiation factor

is one If the heat is conducted to the surface through a solid of conductivity 12 W/mK, determine the temperature gradient at the surface in the solid.

Solution: Refer equation 1.10

Heat convected + heat radiated = heat conducted considering 1m2,

Problem 2: Heat is conducted through a material with a temperature gradient of – 9000 °C/m.

The conductivity of the material is 25W/mK If this heat is convected to surroundings at 30°C with a convection coefficient of 345W/m 2 K, determine the surface temperature.

If the heat is radiated to the surroundings at 30°C determine the surface temperature.

Solution: In this case only convection and conduction are involved.

– kAdT/dx = hA(T1 – T2) Considering unit area,

– 25 × 1 × (– 9000) = 345 × 1 (T1 – 30)

Therefore, T 1 = 682.17°C

In this case conduction and radiation are involved

Heat conducted = Heat radiated

– 25 × 1 × (– 9000) = 5.67 [(T1/100)4 – (303/100)4]

Therefore, T 1 = 1412.14K = 1139°C.

Problem 3: There is a heat flux through a wall of 2250W/m 2 The same is dissipated to the surroundings by convection and radiation The surroundings is at 30°C The convection coefficient has a value of 75W/m 2 K For radiation F = 1 Determine the wall surface temperature.

Solution: For the specified condition, Consider unit area.

The heat conducted = heat convected + heat radiated

Using the rate equations, with absolute temperature

2250 = T2 303

1 75 1

−

×/ + 5.67 × 1[(T2/100)4 – (303/100)4]

Chapter 1

This equation can be solved only by trial It may be noted that the contribution of

(T2/100)4 is small and so the first choice of T2 can be a little less than 4489/13.227 = 340K The

values of the reminder for T2 = 300, 310, 320, 330 are given below:

Problem 4: The outside surface of a cylindrical cryogenic container is at – 10°C The outside

radius is 8 cm There is a heat flow of 65.5 W/m, which is dissipated to the surroundings both

by radiation and convection The convection coefficient has a value of 4.35 W/m 2 K The radiation factor F = 1 Determine the surrounding temperature.

In this case, heat conducted = heat convected + heat radiated

Temperature should be in Kelvin consider unit length:

65.5 = 2 × π × 0.08 [4.35 {T s – 263} + 5.67 {(T s/100)4 – (263/100)4}]

This reduces to (T s/100)4 + 0.767 T s – 272.6 = 0

This equation has to be solved by trial

The first trial value can be chosen near 272.6/0.767 = 355.4 K

The surrounding temperature is 277.75K or 4.75°C.

Check: Q = hA(T s – T1) + σA[(T s/100)4 – (263/100)4]

= 4.35 × π × 0.08 × 2 (277.75 – 263) + 5.67 × 2 × π × 0.08 × 1[2.77754 – 2.634]

= 32.25 + 33.26 = 65.51 W checks to a very reasonable value

Problem 5: A spherical reactor vessel of outside radius 0.48 m has its outside temperature as

123.4°C The heat flow out of the vessel by convection and radiation is 450 W Determine the surrounding temperature.

Solution: In this case Temp should be in K,

Radiation

T = Tµ s

r = 0.48 m2

T = 123.4°C21.5 W/m K2

Problem 6: A solid receives heat by radiation over its surfaces at 4kW and the heat convection

rate over the surface of the solid to the surroundings is 5.2 kW, and heat is generated at a rate

of 1.7 kW over the volume of the solid, determine the heat capacity of the solid if the time rate of change of the average temperature of the solid is 0.5°C/s.

Solution: The energy balance yields: Heat received by radiation – heat convected + heat

generated = heat stored

But, heat stored = heat capacity × change in temperature

Q r dτ – Q c dτ + qdτ = ρVC dT

dT d

Solution: The energy equation yields:

Heat received + heat generated = heat stored

Heat stored = Volume × density × specific heat × temp rise

Q dτ + qV dτ = ρVC dT.

dT d

Q qV VC

τ = ρ+  , Q = 240J/s, q = 100000J/m3/s

2500 0.23 = 0.1°C/s

Time rate of temperature change = 0.1°C/s.

Problem 8: A spherical mass 1m diameter receives heat from a source at 160°C by radiation

and convects heat to the surroundings at 30°C, the convection coefficient being 45 W/m 2 K Determine the steady state temperature of the solid.

Assume F = 1 for radiation.

Solution: Using energy balance,

dτ = 0,

heat received by radiation = heat convected

σA (T s4 – T4) = hA(T – T∞)

It is to be noted that the temperature values

should be in absolute units cancelling A on both

sides and substituting the values

5.67 160 273100 100

4

4+

L

R S|

T|

U V|

1 m

T

h = 45 W/m K230°C

Fig 1.12

Temp, K 330 331 332 332.1

Therefore the equilibrium temperature is 332K or 59°C.

Check: heat convected 45 [332 – (273 + 30)] = 1305 W/m2

heat received = 5.67 (4.334 – 3.324) = 1304.3 W/m2

Checks within reasonable limits

Problem 9: A person sits in a room with surrounding air at 26°C and convection coefficient

over the body surface is 6 W/m 2 K The walls in the room are at 5°C as the outside temperature

is below freezing If the body temperature is 37°C, determine the heat losses by convection and radiation Assume F = 1.0 for radiation Consider a surface area of 0.6 m 2

Solution: Heat loss by convection: hA (T1 – T2) = 6 × 0.6 (37 – 26) W = 39.6 W

Heat loss by radiation: σA (T14 – T24) Note that T should be in K.

Total heat loss = 81.88 W.

Problem 10: A person stands in front of a fire at 650 C in a room where air is at 5°C Assuming

the body temperature to be 37°C and a connection coefficient of 6 W/m 2 K, the area exposed to convection as 0.6m 2 , determine the net heat flow from the body The fraction of radiation from the fire of 1m 2 are reaching the person is 0.01.

Solution: Heat loss by convection = hA(T1 – T2) = 6 × 0.6(37 – 5) = 115.2 W

Substituting the values, heat gain by radiation = σA(T14 – T24)

Net heat gain = 406.3 – 115.2 = 291.1 W.

This shows that sudden exposure to the high temperature warms up a person quickly

Problem 11: A electric room heater (radiator) element is 25 cm long and 4 cm in diameter The

element dissipates heat to the surroundings at 1500 W mainly by radiation, the surrounding temperature being 15°C Determine the equilibrium temperature of the element surface.

Chapter 1

Solution: At equilibrium, neglecting convection,

Q = σA(T14 – T24)Using absolute units of temperature,

Solution: The energy balance yields, (Fig 1.14)

The incident heat rate = convection on the front side + convection on the back side

Substituting the values, and considering 1m2

800 = 15 (T – 30) + 10(T – 30)

Check: 15(62 – 30) + 10(62 – 30) = 800 W

Problem 13: A thin plate receives radiation on one side from a source at 650°C and radiates on

the other face to a surface at 150°C Determine the temperature of the plate Take F = 1 Neglect convection heat flow.

Solution: The energy conservation leads to (Fig 1.15)] radiation received by the surface =

radiation from the surface

σA(T14 – T4] = σA[T4 – T24]Remembering to use Kelvin scale,

650 273

150 273100

T1

1500 W

4+

F

L N

= 39336.66 which is 2 × Q 1

Problem 14: Air at 120°C flows over a plate 20 mm thick and the temperatures in the middle

10mm layer of the plate was measured using thermo couples and were found to be 42°C and 30°C The thermal conductivity of the material is known to be 22.5 W/mK Determine the average convection coefficient over the plate.

Solution: The surface temperature T s and Q

can provide the means for the determination

of the convection coefficient

Using the rate equation,

Q = hA(T s – T∞)

Using the temperature drop and the

thermal conductivity of the wall material, Q

can be determined using

having constant thermal conductivity The drop in temperature over a 10mm layer is, 42 – 30

= 12°C Hence, over 5mm, the drop will be 6°C Hence the surface temperature = 42 + 6 = 48°C.

Fig 1.16

5 mm 42°C

10 mm

Chapter 1

Substituting, 27000 = h × 1 (120 – 48)

Problem 15: In a solar flat plate heater some of the heat is absorbed by a fluid while the

remaining heat is lost over the surface by convection the bottom being well insulated The fraction absorbed is known as the efficiency of the collector If the flux incident has a value of 800 W/m 2 and if the collection temperature is 60°C while the outside air is at 32°C with a convection coefficient of 15 W/m 2 K, determine the collection efficiency Also find the collection efficiency if collection temperature is 45°C.

Solution: The heat lost by convection = Q = hA(T1 – T2)

Fig 1.17

Assuming unit area , Q = 15 × 1(60 – 32) = 420 W

Therefore efficiency of the collector = 800 420

Solar heat flux: 800 W/m2, h = 15 W/m2K Ambient temp = 32°C

Problem 16: A glass plate at 40°C is heated by passing hot air over it with a convection coefficient

of 18 W/m 2 K If the temperature change over 1mm thickness is not to exceed 5°C to avoid distortion damage, determine the maximum allowable temperature of the air Thermal conductivity of the plate material is 1.4 W/mK.

TairHot air,

Fig 1.19

Solution: The heat flow by conduction = heat flow

by convection

The conduction heat flow is found using the

allowable temperature drop over 1mm thickness

Problem 17: A surface at 260°C convects heat at steady

state to Air at 60°C with a convection coefficient of 30

W/m 2 K If this heat has to be conducted through wall with

thermal conductivity of 9.5 W/mK, determine the

temperature gradient in the solid.

Solution: Energy balance yields the relation, heat

conducted = heat convected

Assuming Unit area

= – kA(dT/dx) = hA(T2 – T∞)

Therefore dT/dx = (–h/k) (T2 – T∞)

(30/9.5) (260 – 60) = – 631.5°C/m

or, – 6.315°C/cm.

Problem 18: A thin metal sheet receives heat on one side from a fluid at 80°C with a convection

coefficient of 100 W/m 2 K while on the other side it radiates to another metal sheet parallel to it The second sheet loses heat on its other side by convection to a fluid at 20°C with a convection coefficient of 15 W/m 2 K Determine the steady state temperature of the sheets The two sheets exchange heat only by radiation and may be considered to be black and fairly large in size.

Solution: The energy balance provides (Fig 1.19) heat received convection by

sheet 1 = heat radiation exchange between sheet 1 and 2

= heat convected by sheet 2

h1A(T∞1 – T1) = σA(T14 – T24) = h2A(T2 – T∞2)Substituting the values: considering unit area

100 × 1(353 – T1) = 15 × 1(T2 – 293)

Problem 19: Heat is conducted at steady state through a solid with temperature gradient of

– 5°C/cm, the thermal conductivity of the solid being 22.5 W/mK If the heat is exchanged by radiation from the surface to the surroundings at 30°C, determine the surface temperature.

Solution: Energy balance yields the

relation (Fig 1.21)

Heat conducted = heat radiated

– kA.dT/dx = σA(T24 – T s4)

Considering unit area and

substituting the values

– 22.5 × – 5 × 100 = 5.67 [(T2/100)4

– (303/100)4](The gradient should be converted

to °C/m by multiplying by 100)

Therefore T 2 = 674.4K or 401.4°C.

Problem 20: A satellite in space is of 2m dia and internal heat generation is 2000 W If it is

protected from direct solar radiation by earths shadow determine its surface temperature.

Fig 1.21

T2Radiation

T = T = 303 K s

– 5°C/cm or – 500°C/m

dT dx

— =

Solution: In the absence of atmosphere the only possible way heat is dissipated is by radiation.

The temperature of the space may taken as 0K

Hence heat generated = heat radiated

T|

U V|

W|

Problem 21: A heat flux meter on the outside surface of a wall shows 10 W/m 2 The wall is 0.2

m thick and conductivity is 1.5 W/mK Determine the temperature drop through the wall.

R or ∆T = QR R = kA L , A = 1, Q = 10J/s.

∴ ∆T = 10 × 0.2/(1.5 × 1) = 1.33°C.

EXERCISE PROBLEMS1.1 Model the following heat transfer situations Specify heat flows and storages Try to write down the mathematical expressions.

(i) Solar heating of the road surface

(ii) A steam pipe passing through an open space between two buildings

(iii) Heat transfer from a person in a warm room in the cold season

(iv) Pressure cooker-warming up-cooling down

(v) Pressure cooker-steady conditions

(vi) A rod with one end in a furnace and the remaining surface in atmosphere

(vii) A wire carrying current, exposed to air

(viii) A water heater (electrical) with hot water being drawn out with cold water admission (ix) Cake being baked in an oven or a fruit placed in a refrigeration

(x) A frying pan placed on a stove.

1.2 Choose the correct statement in each question.

(i) A pipe carrying steam at about 300°C traverses a room, the air being still at 30°C The major fraction of the heat loss will be by (a) conduction to the still air (b) convection to the air (c) radiation to the surroundings (d) conduction and convection put together.

(ii) A satellite in space exchanges heat with its surroundings by (a) conduction (b) convection (c) radiation (d) conduction as well as convection.

(iii) For the same temperature drop in the temperature ranges of 300–400°C the heat flow rate will be highest by (a) conduction process (b) convection process (c) radiation process (d) other

factors should be known before any conclusion.

(iv) In the cold season a person would prefer to be near a fire because (a) the conduction from the fire will be better (b) the convection will be better if he is near the fire (c) direct unimpeded radiation will provide quick warmth (d) combined conduction and convection will be better (v) A finned tube hot water radiator with a fan blowing air over it is kept in rooms during winter.

The major portion of the heat transfer from the radiator to air is due to

(vi) For a specified heat input and a given volume which material will have the smallest ture rise (Use data book if necessary) (a) steel (b) aluminium (c) water (d) copper.

tempera-Chapter 1

(vii) When a hot metal piece is left to cool in air the time rate of cooling of the outer layer will be (a) slower at start and faster near the end

(b) faster at start and slower near the end

(c) both rates will be the same

(d) this will depend on the material.

(viii) A thin black plate at temperature T receives radiation from a surface at Temperature T1 and

radiates to a surface at T2 If all surfaces are black at steady state

(a) (T1 – T) > (T – T2) (b) (T1 – T) < (T – T2)

(c) (T1 – T) = (T – T2) (d) can be any one of a, b or c.

(ix) The temperature profile (in) a slab initially at a constant temperature and then allowed to

cool by convection for a short time will be as shown in Fig :

Tih

T

Tih

T

Fig 1.22

Answer to problem 1.2: (i) c, (ii) c, (iii) d, (iv) c, (v) b, (vi) c, (vii) b, (viii) b, (ix) c.

1.3 A wall is exposed on one side to a heat flux of 1.5 kW/m2 which is conducted through the wall For

the following combinations, determine the temp drop through the wall (a) thickness 0.16m and k

= 1.4, 15, 25, 45, 210 and 340 W/mK (ii) Thickness 0.25m and k as above Plot the temperature drop against the radio (L/k) and also (k/L).

1.4 The heat flux through a layer of material 40 mm thick conducting heat under steady state with

a temperature drop of 40°C, was measured as 106 W Determine the thermal conductivity of the material.

1.5 A glass pane is 8mm thick and the inside surface temperature was 25°C and outside surface

temperature was 33°C If k = 1.4 W/mK determine the heat flow through an area of 0.8m × 1m

size pane.

1.6 The surface temperature of a plate over which air flows was measured as 80°C The air

tempera-ture was 40°C In order to maintain the surface temperatempera-ture over an area of 0.1m 2 , the heater rating required was found to be 1.5 kW Determine the value of convection coefficient.

1.7 A strip heater of area 0.2m2 and rating of 1200W is fixed on a vertical wall and mostly convects the heat into the room air at 20°C Determine the value of convective heat transfer coefficient if the surface temperature of the heater is not to exceed 60°C Indicate whether such a value can be achieved by natural convection.

1.8 A strip heater with an area of 0.05m2 has to radiate at 600°C to surroundings at 30°C mine the rating assuming that convection is negligible.

Deter-1.9 The filament of an incandescent lamp of 60 W rating has a total surface area of 40mm2 If the surrounding is at 30°C and if 90% of the power is converted to heat and radiated, determine the temperature of the filament.