heat chap04 026

8/12/2019 Heat Chap04 026

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Chapter 4 Transient Heat Conduction

Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres

4-26C A cylinder whose diameter is small relative to its length can be treated as an infinitely longcylinder. When the diameter and length of the cylinder are comparable, it is not proper to treat thecylinder as being infinitely long. It is also not proper to use this model when finding the temperatures nearthe bottom or top surfaces of a cylinder since heat transfer at those locations can be two-dimensional.

4-27C Yes. A plane wall whose one side is insulated is equivalent to a plane wall that is twice as thickand is eposed to convection from both sides. !he midplane in the latter case will behave like an insulatedsurface because of thermal symmetry.

4-28C !he solution for determination of the one-dimensional transient temperature distribution involvesmany variables that make the graphical representation of the results impractical. In order to reduce thenumber of parameters, some variables are grouped into dimensionless quantities.

4-29C !he "ourier number is a measure of heat conducted through a body relative to the heat stored. !husa large value of "ourier number indicates faster propagation of heat through body. #ince "ourier number

is proportional to time, doubling the time will also double the "ourier number.

4-30C !his case can be handled by setting the heat transfer coefficient h to infinity ∞ since thetemperature of the surrounding medium in this case becomes equivalent to the surface temperature.

4-31C !he maimum possible amount of heat transfer will occur when the temperature of the body

reaches the temperature of the medium, and can be determined from Q mC T T p ima $ %= −∞ .

4-32C When the &iot number is less than '.(, the temperature of the sphere will be nearly uniform at alltimes. !herefore, it is more convenient to use the lumped system analysis in this case.

4-33 A student calculates the total heat transfer from a spherical copper ball. It is to be determinedwhether his)her result is reasonable.

Assumptions !he thermal properties of the copper ball are constant at room temperature.

Properties !he density and specific heat of the copper ball are ρ * + kg)m, and C p * '.+ k/)kg.°0

$!able A-%.

Analysis !he mass of the copper ball and the maimum amount of heat transfer from the copper ball are

k/('120%33''%$0k/)kg.-+.'%$kg4,.($56

kg4,.(1

m%(.'$%kg)m+,--$

1

ma

--

-

=°−°=−=

=

=

==

∞T T mC Q

DV m

i p

π π ρ ρ

Discussion !he student7s result of 23' k/ is not reasonable since it isgreater than the maimum possible amount of heat transfer.

2-(4

Copperball, 200 C

Q

8/12/2019 Heat Chap04 026



4-34 An egg is dropped into boiling water. !he cooking time of the egg is to be determined. √

Assumptions 1 !he egg is spherical in shape with a radius of r ' * 3.4 cm. 2 8eat conduction in the egg isone-dimensional because of symmetry about the midpoint. 3 !he thermal properties of the egg areconstant. 4 !he heat transfer coefficient is constant and uniform over the entire surface. 4 !he "ourier

number is τ 9 '.3 so that the one-term approimate solutions $or the transient temperature charts% are

applicable $this assumption will be verified%.

Properties !he thermal conductivity and diffusivity of the eggs are given to be k * '.1 W)m.°0 and α *'.(2×('-1 m3)s.

Analysis !he &iot number for this process is

Bihr

k

o= = °

° =

$ %$ . %

$ . %.

(2'' ' '34

' 112 3

W ) m . 0 m

W ) m. 0

3

!he constants λ ( (and A corresponding to this &iot number

are, from !able 2-(,

1.( and '+44. (( == Aλ

!hen the "ourier number becomes

3.'(+.'%1.($4+

44' 33( %'+44.-$

('

,' ≈= → =−

− → =

−

−=

−−

∞

∞τ θ

τ τ λ ee AT T

T T

i sph

!herefore, the one-term approimate solution $or the transient temperature charts% is applicable. !hen the

time required for the temperature of the center of the egg to reach 4'°0 is determined to be

min17.8==×

=α

τ=

−s('1+

)s%m('(2.'$

m%'34.'%$(+.'$31

33or t

2-(+

Water

97 C

Egg

T i = 8 C

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4-3

"!PROBLEM 4-35""GIVEN"D=0.055 "[m]"T_i=8 "[C]""T_o=0 [C] #$#m%&%$ &o '% (#$i%)"T_i*+i*i&,= "[C]"

=/400 "[1m2- C]"

"PROPERTIE"=0.6 "[1m-C]"#7#=0./4E-6 "[m21]"

"9N9L:I"Bi=;<$_o1$_o=D1">$om T#'7% 4-/ ?o$$%o*)i*@ &o &i Bi *Am'%$ % $%#)"7#m')#_/=/.6

9_/=3. 0863;T_o-T_i*+i*i&,1;T_i-T_i*+i*i&, =9_/<%;- 7#m')#_/ 2 <& #A

&im%=;&#A <$_o21#7#<Co*(%$&; mi*

!o "C# time "min#

' .+1 23.2

1' 2.31

1 2+.2

4' 3.+

4 4

+' 13.+3

+ 4'.1+

' +3.+

(((.(

50 55 60 65 0 5 80 85 -0 -5

30

40

50

60

0

80

-0

/00

//0

/0

To [C]

t i m e

[ m i n ]

2-(

8/12/2019 Heat Chap04 026



4-36 :arge brass plates are heated in an oven. !he surface temperature of the plates leaving the oven is to be determined.

Assumptions 1 8eat conduction in the plate is one-dimensional since the plate is large relative to itsthickness and there is thermal symmetry about the center plane. 3 !he thermal properties of the plate areconstant. 4 !he heat transfer coefficient is constant and uniform over the entire surface. !he "ourier



Properties !he properties of brass at room temperature are given to be k * ((' W)m.°0, α * .×('-1

m3)s

Analysis !he &iot number for this process is

BihL

k = =

°

° =

$+' %$ . %

$ %.

W ) m . 0 m

W ) m. 0

3''(

(('''('

!he constants λ ( (and A corresponding to this &iot

number are, from !able 2-(,

λ ( ('(' (''(+= =. . and A

!he "ourier number is

τ α

= = × ×

= >

−t

L3

1-- (' ('

''(' 2 ' 3

$ .

$ .. .

m ) s%$ min 1' s ) min%

m%

3

3

!herefore, the one-term approimate solution $or the transient temperature charts% is applicable. !hen thetemperature at the surface of the plates becomes

C44°= → =−

−

==λ=−

−=θ

−τλ−

∞

∞

%,$-4+.'4''3

4''%,$

-4+.'%('-.'cos$%''(+.($%)cos$%,$

%,$ %2.'$%('-.'$((

33(

t LT t LT

e L Le AT T

T t xT t L

iwall

Discussion !his problem can be solved easily using the lumped system analysis since &i ; '.(, and thusthe lumped system analysis is applicable. It gives

°+°=−+=→=−−

=°⋅⋅×

°⋅=====

°⋅⋅×=×

°⋅==→=

−−

∞∞

−

∞

∞ s1''%$s''(122.'$

1

3

1

31-

(-

0%4''-$304''%$%$ %$

''.'0%s)mW('32.m%$'(.'$

0W)m+'

%)$%$

0s)mW('32.)m('.

0W)m(('

eeT T T t T eT T

T t T

k L

h

LC

h

C LA

hA

VC

hAb

s

k C

C

k

bt i

bt

i

p p p

p p

α ρ ρ ρ

α ρ

ρ α

which is almost identical to the result obtained above.

2-3'

<lates

3°0

"urnace,

4''°0

8/12/2019 Heat Chap04 026



4-37 "!PROBLEM 4-3"

"GIVEN"L=0.031 "[m]"T_i=5 "[C]"T_i*+i*i&,= 00 "[C] #$#m%&%$ &o '% (#$i%)"&im%=/0 "[mi*] #$#m%&%$ &o '% (#$i%)"

=80 "[1m2- C]"

"PROPERTIE"=//0 "[1m-C]"#7#=33.E-6 "[m21]"

"9N9L:I"Bi=;<L1">$om T#'7% 4-/ ?o$$%o*)i*@ &o &i Bi *Am'%$ % $%#)"7#m')#_/=0./03

9_/=/. 00/8&#A=;#7#<&im%<Co*(%$&;mi* 1L2;T_L-T_i*+i*i&,1;T_i-T_i*+i*i&,=9_/<%;- 7#m')#_/ 2 <& #A <Co;7#m')#_/<L1L

! "C# !$ "C#

'' 3(.1

3 4.3

' 3.

4 1+.

1'' +2.(

13 .41' 2(.

14 2'.

4'' 221.

43 213.(

4' 244.+

44 2.2+'' '

+3 32.1+' 2'.3

+4 .+

'' 4(.2

time "min# !$ "C#

3 (21.42 322.+

1 3.+ (.

(' 221.

(3 2(.(2 3+.

(1 +.

(+ +.

3' 1'2.

33 13(.232 1.2

31 121.+

3+ 11.3

' 112

2-3(

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8/12/2019 Heat Chap04 026



4-39

"!PROBLEM 4-3"

"GIVEN"$_o=0.351 "[m]"T_i=400 "[C]"T_i*+i*i&,=/ 50 "[C]"

=60 "[1m2- C]""& im%= 0 [mi*] #$#m%&%$ &o '% (#$i%)"

"PROPERTIE"=/4. "[1m-C]"$o=00 "[@1m23]"C_=4 "[1@-C]"#7#=3.5E-6 "[m21]"

"9N9L:I"Bi=;<$_o1">$om T#'7% 4-/ ?o$$%o*)i*@ &o &i Bi *Am'%$ % $%#)"7#m')#_/=/.035

9_/=/. /558_/=0.40 ">$om T#'7% 4- ?o$$%o*)i*@ &o 7#m')#_/"&#A=;#7 #<&im %<Co*(%$&; mi* 1$_o2;T_o-T_i*+i*i&,1;T_i-T_i*+i*i&, =9_/<%;- 7#m')#_/ 2 <& #A

L=/ "[m] / m 7%*@& o+ &% ?,7i*)%$ i ?o*i)%$%)"V=i<$_o2<Lm=$o<V_m#=m<C_<;T_i-T_i*+i*i&,<Co*(%$&; 1_m#= /- <;T_o-T_i*+i*i&, 1;T_i-T_i*+i*i& ,<_/17#m')#_/

time [min] To [C] Q [kJ]

5 45. 44/

/0 4/3.4 8386

/5 40/.5 //05

0 30./ /5656

5 3.3 /046

30 368. 83

35 35 534

40 34.6 835

45 340.5 3//4

50 33/. 3383

55 33. 3640/

60 3/5.8 38853

2-32

8/12/2019 Heat Chap04 026



0 /0 0 30 40 50 60

300

30

340

360

380

400

40

440

0

5000

/0000

/5000

0000

5000

30000

35000

40000

time [min]

T o

[ C ]

Q

[ k J ]

temperature

heat

2-3

8/12/2019 Heat Chap04 026



4-40E :ong cylindrical steel rods are heat-treated in an oven. !heir centerline temperature when theyleave the oven is to be determined.

Assumptions 1 8eat conduction in the rods is one-dimensional since the rods are long and they havethermal symmetry about the center line. 2 !he thermal properties of the rod are constant. 3 !he heat

transfer coefficient is constant and uniform over the entire surface. 4 !he "ourier number is τ 9 '.3 so that

the one-term approimate solutions $or the transient temperature charts% are applicable $this assumptionwill be verified%.

Properties !he properties of AI#I stainless steel rods are given to be k * 4.42 &tu)h.ft.°", α * '.( ft3)h.

Analysis !he time the steel rods stays in the oven can be determined from

t = = =length

velocity

ft

ft ) min min * (+' s

'

('

!he &iot number is

2'4.'%"&tu)h.ft.42.4$

%ft(3)3%$".&tu)h.ft3'$ 3

=°

°==

k

hr Bi o

!he constants λ ( (and A corresponding to this &iot number are, from !able 2-(,

λ ( (' +4+2 ('= =. . and A


τ α = = =

t

r o3

'(

3 (3'32

$ .

$ ).

ft ) h%$ ) 1' h%

ft%

3

3

!hen the temperature at the center of the rods becomes

θ λ τ '

'(

' +4+2 ' 32(3 3

(' ' (3,$ . % $ . %

$ . % .cyl i

T T

T T A e e=

−

− = = =∞

∞

− −

)228°= → =−

−

''

(3.'(4''+

(4''T

T

2-31

*+en, 1700 )

'teel ro, 8 )

8/12/2019 Heat Chap04 026



4-41 #teaks are cooled by passing them through a refrigeration room. !he time of cooling is to bedetermined.

Assumptions 1 8eat conduction in the steaks is one-dimensional since the steaks are large relative to theirthickness and there is thermal symmetry about the center plane. 3 !he thermal properties of the steaks areconstant. 4 !he heat transfer coefficient is constant and uniform over the entire surface. !he "ourier



Properties !he properties of steaks are given to be k * '.2 W)m.°0 and α * '.(×('-4 m3)s

Analysis !he &iot number is

3''.'%0W)m.2.'$

%m'(.'%$0.W)m,$ 3

=°

°==

k

hL Bi

!he constants λ ( (and A corresponding to this

&iot number are, from !able 2-(,

λ ( (' 23+ ('((= =. . and A


3.'1'(.%2-3+.'cos$%'-((.($%(($3

%(($3

%)cos$%,$

3

3(

%2-3+.'$

((

>= → =−−

−−

=−−

−

−

∞

∞

τ

λ

τ

τ λ

e

L Le AT T T t LT

i

!herefore, the one-term approimate solution $or the transient temperature charts% is applicable. !hen thelength of time for the steaks to be kept in the refrigerator is determined to be

min102.6==×

==−

s1()s%m('(.'$

m%'(.'%$1'(.$34

33

α

τ Lt

2-34

#teaks

3°0

>efrigerated air

-((°0

8/12/2019 Heat Chap04 026



4-42 A long cylindrical wood log is eposed to hot gases in a fireplace. !he time for the ignition of thewood is to be determined.

Assumptions 1 8eat conduction in the wood is one-dimensional since it is long and it has thermalsymmetry about the center line. 2 !he thermal properties of the wood are constant. 3 !he heat transfer

coefficient is constant and uniform over the entire surface. 4 !he "ourier number is τ 9 '.3 so that the

one-term approimate solutions $or the transient temperature charts% are applicable $this assumption will be verified%.

Properties !he properties of wood are given to be k * '.(4 W)m.°0, α * (.3+×('-4 m3)s


''.2%0W)m.(4.'$

%m'.'%$0.W)m1.(-$ 3

=°

°==

k

hr Bi o



21+.( and '+(.( (( == Aλ

=nce the constant J ' is determined from !able 2-3

corresponding to the constant λ (*(.'+(, the "ourier number isdetermined to be

3(.'%344(.'$%21+.($''('

''23'

%)$%,$

3

3(

%'+(.($

('(

=τ → =−

−

λ=−

−

τ−

τλ−

∞

∞

e

r r J e AT T

T t r T oo

i

o

which is above the value of '.3. !herefore, the one-term approimate solution $or the transienttemperature charts% can be used. !hen the length of time before the log ignites is

min81.7==×

=α

τ=

−s2'2

)s%m('3+.($

m%'.'%$3(.'$

34

33or t

2-3+

(' cm Woo log, 10 C

8ot gases

4''°0

8/12/2019 Heat Chap04 026



4-43 A rib is roasted in an oven. !he heat transfer coefficient at the surface of the rib, the temperature ofthe outer surface of the rib and the amount of heat transfer when it is rare done are to be determined. !hetime it will take to roast this rib to medium level is also to be determined.

Assumptions 1 !he rib is a homogeneous spherical ob?ect. 2 8eat conduction in the rib is one-dimensional because of symmetry about the midpoint. 3 !he thermal properties of the rib are constant. 4 !he heat

transfer coefficient is constant and uniform over the entire surface. !he "ourier number is τ 9 '.3 so that

the one-term approimate solutions $or the transient temperature charts% are applicable $this assumption

will be verified%.

Properties !he properties of the rib are given to be k * '.2 W)m.°0, ρ * (3'' kg)m, C p * 2.( k/)kg.°0,

and α * '.(×('-4 m3)s.

Analysis $a% !he radius of the roast is determined to be

m'+1'-.'2

%m''3114.'$-

2

-

-

2

m''3114.'kg)m(3''

kg3.-

-

-

--

-

-

=== → =

=== → =

π π π

ρ ρ

V r r V

mV V m

oo


(3(4.'m%'+1'-.'$

1'%s2@-1'')s%$3m('(.'$3

34

3 =

×××==

−

or

t α τ

which is somewhat below the value of '.3. !herefore, the one-term approimate solution $or the transienttemperature charts% can still be used, with the understanding that the error involved will be a little morethan 3 percent. !hen the one-term solution can be written in the form

%(3(4.'$((

','

3(

3( 1.'

(1-.2

(1-1' λ τ λ θ −−

∞

∞==

−

− → =

−

−= e Ae A

T T

T T

i sph

It is determined from !able 2-( by trial and error that this equation is satisfied when Bi * ', which

corresponds to λ ( ('43 (++= =. . and A . !hen the heat transfer coefficient can be determined from

C.W-m16.92°=

°== → =

%m'+1'-.'$

%-'%$0W)m.2.'$

o

o

r

kBih

k

hr Bi

!his value seems to be larger than epected for problems of this kind. !his is probably due to the "ouriernumber being less than '.3.

$b% !he temperature at the surface of the rib is

C19. °= → =

−

−

==−

−=

−−

∞

∞

%,$'333.'

(1-.2

(1-%,$

'-43.-

%rad'-43.-sin$%++.($

)

%)sin$%,$%,$

%(3(4.'$%'-43.-$

(

((

33(

t r T t r T

er r

r r e A

T T

T t r T t r

oo

oo

oo

i

o spho

λ

λ θ

τ λ

2-3

=ven

(1°0

ib,

4. C

8/12/2019 Heat Chap04 026



$c% !he maimum possible heat transfer is

k/3'+'0%.2(1%$0k/)kg.(.2%$kg3.$%$ma =°−°=−= ∞ i p T T mC Q

!hen the actual amount of heat transfer becomes

Q

Q

Q Q

o sphma

,

ma

sin$ % cos$ %$ . %

sin$ . % $ . % cos$ . %

$ . %.

. $ . %$

= − −

= − −

=

= = =

( ( ' 1'43 '43 '43

'43'4+

' 4+ ' 4+ 3'+'

( ( (

(

θ λ λ λ

λ

k/% 1629 %&

$d % !he cooking time for medium-done rib is determined to be

hr3≅==×

==

= → =−

− → =

−

−=

−

−−

∞

∞

min(+(s+11,(')s%m('(.'$

m%'+1'-.'%$(--1.'$

(--1.'%++.($(1-.2

(1-4(

34

33

%'-43.-$(

','

33(

α

τ

τ θ τ τ λ

o

i sph

r t

ee AT T

T T

!his result is close to the listed value of hours and 3' minutes. !he difference between the two results isdue to the "ourier number being less than '.3 and thus the error in the one-term approimation.

Discussion !he temperature of the outer parts of the rib is greater than that of the inner parts of the ribafter it is taken out of the oven. !herefore, there will be a heat transfer from outer parts of the rib to theinner parts as a result of this temperature difference. !he recommendation is logical.

4-44 A rib is roasted in an oven. !he heat transfer coefficient at the surface of the rib, the temperature ofthe outer surface of the rib and the amount of heat transfer when it is well-done are to be determined. !hetime it will take to roast this rib to medium level is also to be determined.

Assumptions 1 !he rib is a homogeneous spherical ob?ect. 2 8eat conduction in the rib is one-dimensional

because of symmetry about the midpoint. 3 !he thermal properties of the rib are constant. 4 !he heattransfer coefficient is constant and uniform over the entire surface. !he "ourier number is τ 9 '.3 so that

the one-term approimate solutions $or the transient temperature charts% are applicable $this assumptionwill be verified%.

Properties !he properties of the rib are given to be k * '.2 W)m.°0, ρ * (3'' kg)m, C p * 2.( k/)kg.°0,

and α * '.(×('-4 m3)s

Analysis $a% !he radius of the rib is determined to be

m'+1'-.'

2

%m''314.'$-

2

-

-

2

m''314.'kg)m(3''

kg3.-

-

-

--

-

-

=== → =

=== → =

π π

π

ρ ρ

V r r V

mV V m

oo


(++(.'m%'+1'-.'$

1'%s(@-1'')s%$2m('(.'$3

34

3 =

×××==

−

or

t α τ

which is somewhat below the value of '.3. !herefore, the one-term approimate solution $or the transienttemperature charts% can still be used, with the understanding that the error involved will be a little morethan 3 percent. !hen the one-term solution formulation can be written in the form

2-'

=ven

(1°0

ib,

4. C

8/12/2019 Heat Chap04 026



%(++(.'$((

','

3(

3( 2-.'

(1-.2

(1-44 λ τ λ θ −−

∞

∞==

−

− → =

−

−= e Ae A

T T

T T

i

sph

It is determined from !able 2-( by trial and error that this equation is satisfied when Bi * 2., which

corresponds to 42'3.(and2''.3 (( == Aλ . !hen the heat transfer coefficient can be

determined from.

C.W-m22. 2 °=°== → =%m'+1'-.'$

%-.2%$0W)m.2.'$o

o

r kBih

k hr Bi

$b% !he temperature at the surface of the rib is

C142.1 °= → =−

−

==−

−=

−−

∞

∞

%,$(-3.'(1-.2

(1-%,$

2.3

%2.3sin$%42'3.($

)

%)sin$%,$%,$

%(++(.'$%2.3$

(

((

33(

t r T t r T

er r

r r e A

T T

T t r T t r

oo

oo

oo

i

o spho

λ

λ θ

τ λ

$c% !he maimum possible heat transfer is

k/3'+'0%.2(1%$0k/)kg.(.2%$kg3.$%$ma =°−°=−=

∞ i p T T mC Q

!hen the actual amount of heat transfer becomes

%&112===

=−

−=−

−=

k/%3'+'%$434.'$434.'

434.'%2.3$

%2.3cos$%2.3$%2.3sin$%2-.'$-(

%cos$%sin$-(

ma

--(

(((,

ma

QQ

Q

Q spho

λ

λ λ λ θ

$d % !he cooking time for medium-done rib is determined to be

hr4===×

=α

τ=

=τ → =−

− → =

−

−=θ

−

τ−τλ−

∞

∞

min'.32's2'-,(2)s%m('(.'$

m%'+1'-.'%$(44.'$

(44.'%42'3.($(1-.2

(1-4(

34

33

%2.3$(

','

33(

o

i

sph

r t

ee AT T

T T

!his result is close to the listed value of 2 hours and ( minutes. !he difference between the two results is probably due to the "ourier number being less than '.3 and thus the error in the one-term approimation.

Discussion !he temperature of the outer parts of the rib is greater than that of the inner parts of the ribafter it is taken out of the oven. !herefore, there will be a heat transfer from outer parts of the rib to theinner parts as a result of this temperature difference. !he recommendation is logical.

2-(

8/12/2019 Heat Chap04 026



4-4 An egg is dropped into boiling water. !he cooking time of the egg is to be determined.

Assumptions 1 !he egg is spherical in shape with a radius of r ' * 3.4 cm. 2 8eat conduction in the egg isone-dimensional because of symmetry about the midpoint. 3 !he thermal properties of the egg areconstant. 4 !he heat transfer coefficient is constant and uniform over the entire surface. !he "ourier



Properties !he thermal conductivity and diffusivity of the eggs can be approimated by those of water at

room temperature to be k * '.1'4 W)m.°0, α = k C p) ρ * '.(21×('-1 m3)s $!able A-%.


3.1%0W)m.1'4.'$

%m'34.'%$0.W)m+''$ 3

=°

°==

k

hr Bi o



3.(and'--.- (( == Aλ

!hen the "ourier number and the time period become

(1--.'%3.($(''+

(''1' 33( %'--.-$

('

,' = → =−

− → =

−

−=

−−

∞

∞τ θ

τ τ λ ee AT T

T T

i sph

which is somewhat below the value of '.3. !herefore, the one-term approimate solution $or the transienttemperature charts% can still be used, with the understanding that the error involved will be a little morethan 3 percent. !hen the length of time for the egg to be kept in boiling water is determined to be

min14.1==×

==−

s+21)s%m('(21.'$

m%'34.'%$(1--.'$

31

33

α

τ or t

2-3

Water

100 C

Egg

T i = 8 C

8/12/2019 Heat Chap04 026



4-46 An egg is cooked in boiling water. !he cooking time of the egg is to be determined for a location at(1('-m elevation.

Assumptions 1 !he egg is spherical in shape with a radius of r ' * 3.4 cm. 2 8eat conduction in the egg isone-dimensional because of symmetry about the midpoint. 3 !he thermal properties of the egg and heattransfer coefficient are constant. 4 !he heat transfer coefficient is constant and uniform over the entire

surface. !he "ourier number is τ 9 '.3 so that the one-term approimate solutions $or the transient

temperature charts% are applicable $this assumption will be verified%.

Properties !he thermal conductivity and diffusivity of the eggs can be approimated by those of water at

room temperature to be k * '.1'4 W)m.°0, α = k C p) ρ * '.(21×('-1 m3)s $!able A-%.


3.1%0W)m.1'4.'$

%m'34.'%$0.W)m+''$ 3

=°

°==

k

hr Bi o



3.(and'--.- (( == Aλ

!hen the "ourier number and the time period become

(434.'%3.($2.2+

2.21' 33( %'--.-$

('

,' = → =−

− → =

−

−=

−−

∞

∞τ θ

τ τ λ ee AT T

T T

i sph

which is somewhat below the value of '.3. !herefore, the one-term approimate solution $or the transienttemperature charts% can still be used, with the understanding that the error involved will be a little morethan 3 percent. !hen the length of time for the egg to be kept in boiling water is determined to be

min14.9==×

==−

s+)s%m('(21.'$

m%'34.'%$(434.'$31

33

α

τ or t

2-

Water

94.4 C

Egg

T i = 8 C

heat chap04 026

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