Solution Manual for Fundamentals of Heat and Mass Transfer 8th Edition by Bergman, Lavine, Incropera & DeWitt – Complete Chapter Solutions

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SOLUTION MANUAL
Fundamentals of Heat and Mass Transfer, 8th Edition
Authors: Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera & David P. DeWitt

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Table of Contents

1. Introduction
2. Introduction to Conduction
3. One-Dimensional, Steady-State Conduction
4. Two-Dimensional, Steady-State Conduction
5. Transient Conduction
6. Introduction to Convection
7. External Flow
8. Internal Flow
9. Free Convection
10. Boiling and Condensation
11. Heat Exchangers
12. Radiation: Processes and Properties
13. Radiation Exchange Between Surfaces
14. Diffusion Mass Transfer

Appendices

Appendix A: Thermophysical Properties of Matter
Appendix B: Mathematical Relations and Functions
Appendix C: Thermal Conditions Associated with Uniform Energy Generation in One-Dimensional, Steady-State
Systems
Appendix D: The Gauss–Seidel Method
Appendix E: The Convection Transfer Equations
Appendix F: Boundary Layer Equations for Turbulent Flow
Appendix G: An Integral Laminar Boundary Layer Solution for Parallel Flow over a Flat Plate

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PROBLEM 1.1


KNOWN: Temperature distribution in wall of Example 1.1.

FIND: Heat fluxes and heat rates at x = 0 and x = L.

SCHEMATIC:




ASSUMPTIONS: (1) One-dimensional conduction through the wall, (2) constant thermal conductivity,
(3) no internal thermal energy generation within the wall.

PROPERTIES: Thermal conductivity of wall (given): k = 1.7 W/m·K.

ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
dT
q = −k (1)
x
dx
Since the temperature distribution is T(x) = a + bx, the temperature gradient is

dT
=b (2)
dx

Hence, the heat flux is constant throughout the wall, and is

dT
q = −k = −kb = −1.7 W/m K (−1000 K/m) = 1700 W/m2 <
x
dx

Since the cross-sectional area through which heat is conducted is constant, the heat rate is constant and is

qx = qx (W  H ) = 1700 W/m2 (1.2 m × 0.5 m) = 1020 W <

Because the heat rate into the wall is equal to the heat rate out of the wall, steady-state conditions exist. <

COMMENTS: (1) If the heat rates were not equal, the internal energy of the wall would be changing
with time. (2) The temperatures of the wall surfaces are T1 = 1400 K and T2 = 1250 K.

, PROBLEM 1.2

KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid
extruded insulation.

FIND: (a) The heat flux through a 3 m  3 m sheet of the insulation, (b) the heat rate through
the sheet, and (c) the thermal conduction resistance of the sheet.

SCHEMATIC:
m222
A = 49m
9m



k = 0.029
qcond

T1 – T2 = 112
102˚˚CC
C


T1 T2


2205mm
L = 25 mm
x

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.

ANALYSIS: (a) From Equation 1.2 the heat flux is
dT T1 - T2 W 12 K W
q = -k =k = 0.029 × = 13.9 <
x
dx L mK 0.025 m m2

(b) The heat rate is
W
q = q  A = 13.9 × 9 m2 = 125 W <
x x 2
m

(c) From Eq. 1.11, the thermal resistance is

Rt,cond = T / qx = 12 K / 125 W = 0.096 K/W <
COMMENTS: (1) Be sure to keep in mind the important distinction between the heat flux
(W/m2) and the heat rate (W). (2) The direction of heat flow is from hot to cold. (3) Note that
a temperature difference may be expressed in kelvins or degrees Celsius. (4) The conduction
thermal resistance for a plane wall could equivalently be calculated from Rt,cond = L/kA.