**UFBA**

# Transferencia de calor -7th-Kreith-Sol

(Parte **1** de 5)

An Instructor’s Solutions Manual to Accompany

PRINCIPLES OF HEAT TRANSFER, 7TH EDITION, SI

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SI EDITION PREPARED BY: SHALIGRAM TIWARI Indian Institute of Technology Madras http://librosysolucionarios.net

1 | 1 |

2 | 85 |

3 | 231 |

4 | 311 |

5 | 421 |

6 | 513 |

7 | 607 |

8 | 683 |

9 | 781 |

10 | 871 |

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Chapter1Basic Modes of Heat Transfer2

1.1The Relation of Heat Transfer to Thermodynamics3 1.2Dimensions and Units7 1.3Heat Conduction9 1.4 Convection 17 1.5 Radiation 21 1.6Combined Heat Transfer Systems23 1.7 Thermal Insulation 45 1.8Heat Transfer and the Law of Energy Conservation51

References 58 Problems 58 Design Problems68

Chapter 2 Heat Conduction 70

2.1 Introduction 71 2.2The Conduction Equation71 2.3Steady Heat Conduction in Simple Geometries78 2.4 Extended Surfaces 95 2.5* Multidimensional Steady Conduction 105 2.6Unsteady or Transient Heat Conduction116 2.7*Charts for Transient Heat Conduction134 2.8 Closing Remarks 150

References 150 Problems 151 Design Problems163

Chapter3Numerical Analysis of Heat Conduction166

3.1 Introduction 167 3.2One-Dimensional Steady Conduction168 3.3 One-Dimensional Unsteady Conduction 180

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3.4*Two-Dimensional Steady and Unsteady Conduction195 3.5* Cylindrical Coordinates 215 3.6* Irregular Boundaries 217 3.7 Closing Remarks 221

References 221 Problems 2 Design Problems228

Chapter4Analysis of Convection Heat Transfer230

4.1 Introduction 231 4.2Convection Heat Transfer231 4.3Boundary Layer Fundamentals233 4.4Conservation Equations of Mass, Momentum, and Energy for Laminar Flow Over a Flat Plate235 4.5Dimensionless Boundary Layer Equations and Similarity

Parameters 239 4.6Evaluation of Convection Heat Transfer Coefficients243 4.7 Dimensional Analysis 245 4.8*Analytic Solution for Laminar Boundary Layer Flow Over a Flat Plate252 4.9*Approximate Integral Boundary Layer Analysis261 4.10*Analogy Between Momentum and Heat Transfer in Turbulent Flow Over a Flat Surface267 4.11Reynolds Analogy for Turbulent Flow Over Plane Surfaces273 4.12Mixed Boundary Layer274 4.13*Special Boundary Conditions and High-Speed Flow277 4.14 Closing Remarks 282

References 283 Problems 284 Design Problems294

Chapter 5 Natural Convection 296

5.1 Introduction 297 5.2Similarity Parameters for Natural Convection299 5.3Empirical Correlation for Various Shapes308 5.4*Rotating Cylinders, Disks, and Spheres322 5.5Combined Forced and Natural Convection325 5.6* Finned Surfaces 328 x Contents 67706_00_FM_pi-xi.qxd 5/14/10 9:32 AM Page x

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5.7 Closing Remarks 3

References 338 Problems 340 Design Problems348

Chapter6Forced Convection Inside Tubes and Ducts350

6.1 Introduction 351 6.2*Analysis of Laminar Forced Convection in a Long Tube360 6.3Correlations for Laminar Forced Convection370 6.4*Analogy Between Heat and Momentum Transfer in Turbulent Flow382 6.5Empirical Correlations for Turbulent Forced Convection386 6.6Heat Transfer Enhancement and Electronic-Device Cooling395 6.7 Closing Remarks 406

References 408 Problems 411 Design Problems418

Chapter7Forced Convection Over Exterior Surfaces420

7.1Flow Over Bluff Bodies421 7.2Cylinders, Spheres, and Other Bluff Shapes422 7.3*Packed Beds440 7.4Tube Bundles in Cross-Flow444 7.5*Finned Tube Bundles in Cross-Flow458 7.6*Free Jets461 7.7 Closing Remarks 471

References 473 Problems 475 Design Problems482

Chapter 8 Heat Exchangers 484

8.1 Introduction 485 8.2Basic Types of Heat Exchangers485 8.3Overall Heat Transfer Coefficient494 8.4Log Mean Temperature Difference498 8.5Heat Exchanger Effectiveness506 8.6*Heat Transfer Enhancement516 8.7*Microscale Heat Exchangers524

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8.8 Closing Remarks 525

References 527 Problems 529 Design Problems539

Chapter9Heat Transfer by Radiation540

9.1 Thermal Radiation 541 9.2 Blackbody Radiation 543 9.3 Radiation Properties 5 9.4The Radiation Shape Factor571 9.5Enclosures with Black Surfaces581 9.6Enclosures with Gray Surfaces585 9.7* Matrix Inversion 591 9.8*Radiation Properties of Gases and Vapors602 9.9Radiation Combined with Convection and Conduction610 9.10 Closing Remarks 614

References 615 Problems 616 Design Problems623

Chapter10Heat Transfer with Phase Change624

10.1Introduction to Boiling625 10.2Pool Boiling625 10.3Boiling in Forced Convection647 10.4 Condensation 660 10.5* Condenser Design 670 10.6*Heat Pipes672 10.7*Freezing and Melting683

References 688 Problems 691 Design Problems696

Appendix 1The International System of UnitsA3

Appendix 2Data TablesA6

Properties of SolidsA7 Thermodynamic Properties of LiquidsA14 Heat Transfer FluidsA23 xii Contents 67706_00_FM_pi-xi.qxd 5/14/10 9:32 AM Page xii

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Chapter 1

PROBLEM 1.1

The outer surface of a 0.2m-thick concrete wall is kept at a temperature of –5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/(m K). Determine the heat loss through a wall 10 m long and 3 m high.

10 m long, 3 m high, and 0.2 m thick concrete wall Thermal conductivity of the concrete (k) = 1.2 W/(m K)

Temperature of the inner surface (Ti) = 20°C Temperature of the outer surface (To) = –5°C

The heat loss through the wall (qk) ASSUMPTIONS

One dimensional heat flow The system has reached steady state

SKETCH L = 0.2 m

=1 0m H =3m

SOLUTION The rate of heat loss through the wall is given by Equation (1.2) qk = AKL (ΔT)

(20°C – (–5°C)) qk = 4500 W COMMENTS

Since the inside surface temperature is higher than the outside temperature heat is transferred from the inside of the wall to the outside of the wall.

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PROBLEM 1.2

The weight of the insulation in a spacecraft may be more important than the space required. Show analytically that the lightest insulation for a plane wall with a specified thermal resistance is that insulation which has the smallest product of density times thermal conductivity.

GIVEN Insulating a plane wall, the weight of insulation is most significant

Show that lightest insulation for a given thermal resistance is that insulation which has the smallest product of density (ρ) times thermal conductivity (k)

One dimensional heat transfer through the wall Steady state conditions

The resistance of the wall (Rk), from Equation (1.13) is

Rk = LAk where

L = the thickness of the wall A = the area of the wall The weight of the wall (w) is w = ρ A L Solving this for L

L = w Aρ

Substituting this expression for L into the equation for the resistance

Rk = 2 w kAρ

∴ w = ρ k Rk A2 Therefore, when the product of ρ k for a given resistance is smallest, the weight is also smallest.

Since ρ and k are physical properties of the insulation material they cannot be varied individually. Hence in this type of design different materials must be tried to minimize the weight.

PROBLEM 1.3

A furnace wall is to be constructed of brick having standard dimensions 2.5 cm × 1 cm

× 7.5 cm. Two kinds of material are available. One has a maximum usable temperature of 1040°C and a thermal conductivity of 1.7 W/(m K), and the other has a maximum temperature limit of 870°C and a thermal conductivity of 0.85 W/(m K). The bricks cost the same and can be laid in any manner, but we wish to design the most economical wall for a furnace with a temperature on the hot side of 1040°C and on the cold side of 200°C. If the maximum amount of heat transfer permissible is 950 W/m2 for each square foot of area, determine the most economical arrangements for the available bricks.

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Furnace wall made of 2.5 cm × 1 cm × 7.5 cm bricks of two types

Type 1 bricks Maximum useful temperature (T1, max) = 1040°C

Thermal conductivity (k1) = 1.7 W/(m K)

Type 2 bricks Maximum useful temperature (T2, max) = 870°C

Thermal conductivity (k2) = 0.85 W/(m K) Bricks cost the same

Wall hot side temperature (Thot) = 1040°C and wall cold side temperature (Tcold) = 200°C Maximum permissible heat transfer (qmax/A) = 950 W/m2

The most economical arrangement for the bricks

One-dimensional, steady state heat transfer conditions Constant thermal conductivities The contact resistance between the bricks is negligible

SKETCH Type 2 Bricks

Type 1 Bricks

Since the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, the most economical wall would use as few type 1 bricks as possible. However, there should be thick enough layer of type 1 bricks to keep the type 2 bricks at 870°C or less.

For one-dimensional conduction through type 1 bricks (from Equation 1.2) qk = kAL ΔT maxq

(Thot – T12) where L1 is the minimum thickness of the type 1 bricks. Solving for L1

L1 = 1 maxk

(Thot – T12)

950 W/m (1040 – 870)K = 0.3042 m = 30.42 cm http://librosysolucionarios.net

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This thickness can be achieved by using 4 layers of type 1 bricks using the 7.5 cm dimension. Similarly, for one-dimensional conduction through type 2 bricks

L2 = 2 maxk

(T12 – Tcold)

950 W/m (870 – 200)K = 0.6 m = 60 cm

This thickness can be achieved with 8 layers of type 2 bricks using the 7.5 cm dimension.

Therefore, the most economical wall would be built using 4 layers of type 1 bricks and 8 layers of type 2 bricks with the three inch dimension of the bricks used as the thickness.

PROBLEM 1.4

To measure thermal conductivity, two similar 1-cm-thick specimens are placed in an apparatus shown in the accompanying sketch. Electric current is supplied to the 6-cm by 6-cm guarded heater, and a wattmeter shows that the power dissipation is 10 watts (W). Thermocouples attached to the warmer and to the cooler surfaces show temperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of the material at the mean temperature in W/(m K).

Thermal conductivity measurement apparatus with two samples as shown Sample thickness (L) = 1 cm = 0.01 cm

(Parte **1** de 5)