Solution Manual For Fundamentals of Heat and Mass Transfer 8th Edition by Theodore L. Bergman| All Chapters Covered| Latest Upadate 2026

SOLUTION MANUAL FOR
Fundamentals of Heat and Mass Transfer
by Theodore L. Bergman, Adrienne S. Lavine
8th Edition

,PROBLEM 1.2
KNOWN: Heat Rate, Q, Through One-Dimensional Wall Of Area A, Thickness L, Thermal
Conductivity K And Inner Temperature, T1.

FIND: The Outer Temperature Of The Wall,

T2. SCHEMATIC:




ASSUMPTIONS: (1) One-Dimensional Conduction In The X-Direction, (2) Steady-State Conditions,
(3) Constant Properties.

ANALYSIS: The Rate Equation For Conduction Through The Wall Is Given By Fourier’s Law,
Dt T1 − T2
Q = Q = Q A = -K  A = Ka .
Cond X X
Dx L

Solving For T2 Gives
Qcond L
T =T − .
2 1
Ka

Substituting Numerical Values, Find

3000W  0.025m
T2 = 415 C -$




0.2W / M K  10m2

T2 = 415 C - 37.5 C
$ $




T2 = 378 C. $

<
COMMENTS: Note Direction Of Heat Flow And Fact That T2 Must Be Less Than T1.

, PROBLEM 1.2
KNOWN: Inner Surface Temperature And Thermal Conductivity Of A Concrete Wall.
FIND: Heat Loss By Conduction Through The Wall As A Function Of Ambient Air Temperatures Ranging From
-15 To 38C.
SCHEMATIC:




ASSUMPTIONS: (1) One-Dimensional Conduction In The X-Direction, (2) Steady-State Conditions,
(3) Constant Properties, (4) Outside Wall Temperature Is That Of The Ambient Air.
ANALYSIS: From Fourier’s Law, It Is Evident That The Gradient, Dt Dx = − QX K , Is A Constant, And
Hence The Temperature Distribution Is Linear, If QX And K Are Each Constant. The Heat Flux Must
Be Constant Under One-Dimensional, Steady-State Conditions; And K Is Approximately Constant If It
Depends

( )
Only Weakly On Temperature. The Heat Flux And Heat Rate When The Outside Wall Temperature Is T2
= -15C Are Dt T −T
QX = −K = K 1 2 = 1W M K = 133.3 W M2
25 C − −15 C $ $



. (1)
Dx L 0.30 M

Qx = QX  A = 133.3 W M2  20 M2 = 2667 W . (2) <
Combining Eqs. (1) And (2), The Heat Rate Qx Can Be Determined For The Range Of Ambient Temperature, -15
 T2  38C, With Different Wall Thermal Conductivities, K.

3500


2500
Heat loss, qx (W)




1500


500


-500


-1500
-20 -10 0 10 20 30 40

Ambient Air Temperature, T2 (C)

Wall Thermal Conductivity, K = 1.25
W/M.K K = 1 W/M.K, Concrete Wall
K = 0.75 W/M.K


For The Concrete Wall, K = 1 W/MK, The Heat Loss Varies Linearily From +2667 W To -867 W
And Is Zero When The Inside And Ambient Temperatures Are The Same. The Magnitude Of The
Heat Rate Increases With Increasing Thermal Conductivity.
COMMENTS: Without Steady-State Conditions And Constant K, The Temperature Distribution In A
Plane Wall Would Not Be Linear.

, PROBLEM 1.3
KNOWN: Dimensions, Thermal Conductivity And Surface Temperatures Of A Concrete Slab.
Efficiency Of Gas Furnace And Cost Of Natural Gas.
FIND: Daily Cost Of Heat Loss.
SCHEMATIC:




ASSUMPTIONS: (1) Steady State, (2) One-Dimensional Conduction, (3) Constant Properties.
ANALYSIS: The Rate Of Heat Loss By Conduction Through The Slab Is

T − T2 7C
Q = K (LW) 1 = 1.4 W / M K (11m 8 M ) = 4312 W <
T 0.20 M

The Daily Cost Of Natural Gas That Must Be Combusted To Compensate For The Heat Loss Is
Q Cg 4312 W $0.01/ MJ
C = ( T ) = (24 H / D 3600 S / H ) = $4.14 / D <
D
F 0.9 106 J / MJ
COMMENTS: The Loss Could Be Reduced By Installing A Floor Covering With A Layer Of
Insulation Between It And The Concrete.