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

Solutions Manual for Fundamentals of Heat and Mass Transfer, 8e
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| Bergman, Lavine, Incropera, DeWitt (All Chapters Download link at th
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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 | = |qx|(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:
| 9m
A |= 4 m
9|m
|m
222




k |= |0.029
qcond

| T1 |– |T2 1102˚C
10˚C
12 C
|C



| =


T1 T2


L |= |2205|m
20
25 m
|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
2
|(W/m ) |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.

, PROBLEM |1.3

KNOWN: | Thickness |and |thermal |conductivity |of |a |wall. | Heat |flux |applied |to |one |face |and
|temperatures |of |both |surfaces.


FIND: | Whether |steady-state |conditions |exist.
SCHEMATIC:

L |= |10 |mm

T2 |= |30C

q” |= |20 |W/m2
qcond


T1 |= |50C k |= |12 |W/m∙K




ASSUMPTIONS: | (1) |One-dimensional |conduction, |(2) |Constant |properties, |(3) |No |internal |energy
|generation.



ANALYSIS: | Under |steady-state |conditions |an |energy |balance |on |the |control |volume |shown |is

qin | = |qout | = |qcond | = |k(T1 |−|T2|)|/|L |=|12 |W/m||K(50C|−|30C)|/|0.01 |m |= |24,000 |W/m2


Since |the |heat |flux |in |at |the |left |face |is |only |20 |W/m2, |the |conditions |are |not |steady |state. <

COMMENTS: | If |the |same |heat |flux |is |maintained |until |steady-state |conditions |are |reached, |the
|steady-state |temperature |difference |across |the |wall |will |be




T |= | qL|/|k | = |20 |W/m2 ||0.01 |m|/12 |W/m||K |= |0.0167 |K

which |is |much |smaller |than |the |specified |temperature |difference |of |20C.

, PROBLEM |1.4
KNOWN: | Inner |surface |temperature |and |thermal |conductivity |of |a |concrete |wall.
FIND: | Heat |loss |by |conduction |through |the |wall |as |a |function |of |outer |surface |temperatures |ranging |from
|-15 |to |38C.


SCHEMATIC:




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



ANALYSIS: | From |Fourier’s |law, |if | qx | and |k |are |each |constant |it |is |evident |that |the |gradient,
dT dx |= |−qx | k |, |is |a |constant, |and |hence |the |temperature |distribution |is |linear. | 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 25∘C |− | −15∘C ( )
qx | =|−k = |k | 1 2 | =|1W | m||K =|133.3|W | m2 |. (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 |outer |surface |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

OAmbient
m
sidbe
utsi die
e
sn |air
rtfa
acire |temperature, | T2 | (C)
u|surface

| 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 |linearly |from |+2667 |W |to |-867 |W |and |is |zero
|when |the |inside |and |outer |surface |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.