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College of Engineering and Computer ScienceMechanical Engineering Department

Mechanical Engineering 375 Heat Transfer

Spring 2007 Number 17629 Instructor: Larry Caretto

Solutions to Exercise Six More Unsteady Heat Transfer

1. The soil temperature in the upper layers of the earth varies with the variations in theatmospheric conditions. Before a cold front moves in, the earth at a location is initially ata uniform temperature of 1 oC. Then the area is su!"ected to a temperature of #1 oC andhigh winds that resulted in a convection heat transfer coefficient of $ %&m ' (oC on theearth)s surface for a period of 1 h. Ta*ing the properties of the soil at that location to !e *+ . %&m( oC and + 1.- 1 /0 m ' &s, determine the soil temperature at distances , 1 , ' ,and 0 cm from the earth)s surface at the end of this 1 /h period.

We can assume that the solution or a semi!in inite me"ium applies here# In particular thetemperature$ %$ at any "istance$ &$ an" time$ t$ 'hen there is a con(ection boun"ary con"itionstarting at t ) 0 *'hen the semi!in inite region has an initial temperature$ % i$ is gi(en by theollo'ing e+uation#

+

=

+

k t h

t x

erfcet

xerfc

T T T T k

t hk hx

i

i

44

2

2

%he terms in this e+uation using the heat trans er coe icient can be oun" as ollo's#

( ) 11287.337.33360010106.19.0

40 22

225

2 ====

k t h

h s

h s

m xW C m

C mW

k t h o

o

,or & ) 0$ the parameter &-*. t/ 1-2 ) 0 or & ) 10 cm ) 0$1 m$ this parameter becomes

( )06588.0

360010

106.14

1.04 25

==

h s

h s

m x

mt

x

,or "epths o 20 an" 0 cm &-*. t/ 1-2 'ill be t'o an" i(e times this (alue so that &-*. t/ 1-2 )0#1 13 or & ) 20 cm an" 0# 29. or & ) 0 cm# 4pplying the parameters or & ) 0 gi(es

( ) ( ) ( )7.3300 11280 += +

erfceerfc

T T T T

i

i

I 'as not able to compute the secon" term# 5oth the argument o the e&ponential an" theargument o the complimentary error unction ga(e me a numerical error# When I compute" the

pro"uct or an increasing argument I got a limit o ero or this pro"uct# 5ecause ot this I too thepro"uct to be ero not only in the case o & ) 0$ but in the case o all other & (alues since they 'illha(e e(en larger arguments than the ones or & ) 0# Setting this term to ero gi(es the ollo'ingresults#

( ) ( ) ( )

=

+=

+=

t x

erfcC C t

xerfcC C C

t x

erfcT T T T oooooii 42010

4101010

4

,or & ) 0$ the parameter &-*. t/ 1-2 ) 0 an" 'e get

8acaran"a * ngineering/ ail Co"e ;hone: 313#677#6..3!mail: lcaretto

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( ) ( ) ( ) ( )( ) ===

= 1201002010

42010 C C erfcC C

t x

erfcC C T oooooo

10 oC #

,or & ) 10 cm$ the parameter &-*. t/ 1-2 ) 0#06 33 an" 'e get

( ) ( ) ( ) ( )( ) ===

= 9257.0201006588.020104

2010 C C erfcC C t

xerfcC C T oooooo

8.5 oC #

,or & ) 20 cm$ the parameter &-*. t/ 1-2 ) 0#1 13 an" 'e get

( ) ( ) ( ) ( )( ) ===

= 8519.020101318.02010

42010 C C erfcC C

t

xerfcC C T oooooo

7.0 oC #

,or & ) 0 cm$ the parameter &-*. t/ 1-2 ) 0# 29. an" 'e get

( ) ( ) ( ) ( )( ) ===

= 6418.020103294.02010

42010 C C erfcC C

t

xerfcC C T oooooo

2.8 oC #

'. short !rass cylinder 2 + 304 *g&m4

, cp + .43 *5&*g(o

C, * 6 11 %&m(o

C, and +4.4 1 /0 m ' &s7 of diameter D + 3 cm and height 8 + 10 cm is initially at a uniformtemperature of T i + 10 oC. The cylinder is now placed in atmospheric air at ' oC, whereheat transfer ta*es place !y convection with a heat transfer coefficient of h + $ %&m ' (oC.Calculate 2a7 the center temperature of the cylinder9 2!7 the center temperature of the topsurface of the cylinder9 and 2c7 temperature of the top outer radius of the cylinder, and 2d7the total heat transfer from the cylinder 10 min after the start of the cooling.

%he cylin"er can be sol(e" as the pro"uct o the solution or the in inite cylin"er an" that o thein inite slab# %he temperature at any point in the cylin"er 'ill be the pro"uct o t'o solutions# Ithe ,ourier number is greater than 0#2 'e can use the appro&imate solutions sho'n belo'#

( ) ( ) ( )c s

i

icyl slab

i r r

J e A L x

e At r t r T T T T

t r x

===

0

1011121

21 cos,,,,

%he subscripts s or slab an" c or cylin"er in"icate that the (alues o 4 1 an" 1 'ill be "i erent orthe "i erent solutions# In a""ition$ the ,ourier numbers 'ill be "i erent or the slab *'ith a hal'i"th$ L ) *1 cm/-2 ) 7# cm ) 0#07 m$ an" the cylin"er 'ith a ra"ius o *3 cm/-2 ) . cm ) 0#0.m# =sing the total time o 1 minutes ) 900 s$ the ,ourier number or the cylin"er is

( )( )

2.0069.1904.0

9001039.32

25

20

>===

m

s s

m x

r

t cyl

%he ,ourier number or the slab is

( )( )

2.0424.504.0

9001039.32

25

2 >===

m

s s

m x

L

t

Since both ,ourier numbers are greater than 0#2 the use o the appro&imate solution is >usti ie"#

Ne&t 'e compute the 5iot numbers to "etermine the (alues o 4 1 an" 1 or each solution#

( ) 02727.0110

075.040

2 =

==

W C m

mC m

W k

hL Bi

o

o slab


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( ) 01455.0110

04.040

2 =

==

W C m

mC m

W k

hr Bi

o

oo

cyl

,or 5i ) 0#. %able .!2 sho's that that 4 1 ) 1#00. an" 1 ) 0#1620 or the slab an" 4 1 ) 1#00 6an" 1 ) 0#1667 or the cylin"er# We can no' compute the center temperature o the cylin"er'here & ) r ) 0# *Note that cos*0/ ) 8 0*0/ ) 1#/

[ ] [ ]c sc si

e Ae Ar r

J e A L x

e AT T T T 21212121

110

10111 cos

=

=

Substituting numerical (alues or $ 41 an" 1 or each geometry gi(es#

[ ] [ ] ( ) ( )[ ] ( ) ( )[ ] ( )( ) 511.0587.0871.000361.10045.1 069.191667.0424.51620.011 222121 ====

eee Ae AT T T T

c si

We can then in" the temperature at the center o the cylin"er#

( ) ( ( ) =+=+= 511.02015020511.0 C C C T T T T oooi 86.4 oC #

%o in" the temperature at the top center o the cylin"er$ the ra"ial solution 'ill still ha(e the (alueo 0# 37$ but the rectangular solution has to be e(aluate" at & ) L *e+ui(alent to ) &-L ) 1/# %hissolution is

( ) ( ) ( ) ( ) 860.01620.0cos0045.1cos, 424.51620.01122

1 ==

= e

L x

e At L si

i slab

So the pro"uct at the top center o the cylin"er )s *0#360/*0# 37/ ) 0# 0 # With this 'e in"the "esire" temperature as

( ) ( )( ) =+=+= 505.02015020505.0 C C C T T T T oooi 85.6 oC #%o compute the temperature at the top outer ra"ius 'e can use the a&ial solution that 'e >ust

obtaine"$ but 'e not ha(e to "etermine the ra"ial solution at the point 'here r-r o ) 1# %hisre+uires that 'e multiply the e&ponential solution or r ) 0 by the 5essel unction 8 o*1r-r o/ ) 8 o*1/) 8 o*0#1667/ ) 0#99 1# %his gi(es a ra"ial solution at the outer ra"ius o *0#99 1/*0# 37/ ) 0# 3 #ultiplying this by the solution slab ) 0#360 or the rectangular problem oun" abo(e gi(es apro"uct solution ) 0# 01# Sol(ing or the temperature gi(es

( ) ( ( ) =+=+= 501.02015020501.0 C C C T T T T oooi 85.2 oC #Note that this problem has a lo' 5iot number an" the temperature throughout the soli" is close touni orm#

%he heat trans er can be compute" rom the e+uations or ?-? ma& or the one!"imensionalsolutions gi(en in e+uations *.! / an" *.! ./ on page 2 o the te&t#

( )1

11,0

max1

1,0

max

21sin1

J

QQ

QQ

cyl

cylinder

wall

slab

=

=

%he pro"uct solution or a t'o!"imensional case is gi(en by the ollo'ing mo"i ie" pro"uctsolution in e+uation 9.! / on page 2 1# @ere the 2A solution is 'ritten in terms o the irst an"secon" "imension#


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+

=

first ond first D Q

QQ

QQ

QQ

Q

maxsecmaxmax2max

1

We can compute the (alues or the cylin"er an" the slab as the irst an" secon" "imensions"irectly# We no' 0$slab )0#371 or the slab an" 0$cyl )0# 37 or the cylin"er rom the solution orthe center temperatures$ so 'e can substitute these (alues as 'ell as the appropriate (alues or1 to "etermine the ?-? ma& ratio or each "imension#

133.01620.0

1620.0sin871.01

sin1

1

1,0

max

===

wall

slabQ

Q

Similarly$ or the in inite cylin"er solution 'e get

( ) ( ) ( ) 415.01677.0

0835.0587.021

1677.0

1677.0587.021 1

max

===

J Q

Q

cylinder

No' 'e can compute the (alue o ?-? ma& or the short clin"er#

( ) 493.0133.01415.0133.0

1maxsecmaxmax2max

=+= + = first ond first D Q

QQ

QQ

QQ

Q

%o compute ?$ 'e ha(e to irst in" ? ma %his is the heat trans er i the entire soli" changestemperature rom % i to %B#

( ) ( )

( )( ) ( ) kJ C C C kg kJ

mmm

kg T T c L DT T mcQ

ooo

i pi p

32520150389.0

15.08.8350

)2(

3

max

=

===

We thus in" ? ) ? ma&*?-? ma&/ ) * 2 8/*0#.9 / or Q 160 !" #


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