hAs (T −T )dt =mC ( ) =− ρ ln ) ∞ =− ρ · 2009-11-11 · 6 Copyright © The McGraw-Hill...

41
1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 1 FIGURE 4-1 A small copper ball can be modeled as a lumped system, but a roast beef cannot Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display 2 FIGURE 4-2 The geometry and parameters involved in the lumped system analysis. = dt during body the of energy the in increase The dt during body the o transfer Heat int ( ) dT mC dt T T hA p s = ( ) dt VC hA T T T T d p s ρ = t VC hA T T T t T p s i ρ = ) ( ln bt i e T T T t T = ) ( p s VC hA b ρ =

Transcript of hAs (T −T )dt =mC ( ) =− ρ ln ) ∞ =− ρ · 2009-11-11 · 6 Copyright © The McGraw-Hill...

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FIGURE 4-1

A small copper ball can be modeled as a lumped system, but a roast beef cannot

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FIGURE 4-2

The geometry and parameters involved in the lumped system analysis.

⎟⎟⎟

⎜⎜⎜

⎛=⎟⎟

⎞⎜⎜⎝

dtduringbodytheofenergy

theinincreaseThe

dtduringbodytheotransferHeat int

( ) dTmCdtTThA ps =−∞

( )dt

VChA

TTTTd

p

s

ρ−=

−−

tVChA

TTTtT

p

s

i ρ−=

−−

∞)(ln

bt

i

eTTTtT −

∞ =−−)(

p

s

VChA

=

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FIGURE 4-3

The temperature of a lumped system approaches the environment temperature as time gets larger.

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FIGURE 4-4

Heat transfer to or from a body reaches its maximum value when the body reaches the environment temperature.

[ ]∞−= TtThAtQ s )()(&

[ ]ip TtTmCQ −= )(

( )ip TTmCQ −= ∞max

sc A

VL =

khLBi c=

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FIGURE 4-5

bodythewithinConductionbodytheofsurfacetheatConvection

TT

LkhBi

c

=ΔΔ

=/

bodytheofsurfacetheatceresisConvectionbodythewithinceresisConduction

hkLBi c

tantan

1==

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FIGURE 4-6

Small bodies with high thermal conductivities and low convection coefficients are most likely to satisfy the criterion lumped system analysis.

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FIGURE 4-7

Analogy between heat transfer to a solid and passenger traffic to an island.

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FIGURE 4-8

When the convection coefficient h is high and k is low, large temperature difference occur between the inner and outer regions of a large solid.

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FIGURE 4-9

Schematic for Example 4-1.

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Assumption•The junction is spherical in shape with a diameter of D = 0.001 m.•Thermal properties of the junction and the heat transfer coefficient are constant.•Radiation effects are negligible

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( ) mmDD

D

AVL

sc

42

3

1067.1001.061

616

1−×=====

π

π

( )( ) 1.0001.0/35

1067.1/210 42

<=•

ו==

CmWmCmW

khLBi O

Oc

Analysis

In order to read 99 percent of the initial temperature difference

between the junction and the gas, we must have

01.0)(=

−−

TTTtT

i

∞− TTi

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Analysis

The value of the exponent b is

( )( )( )1

43

2

462.01067.1/320/8500

/210 −− =

ו•

=== smCkgJmkg

CmWLC

hVC

hAb O

O

cpp

s

ρρ

tsbt

i

eeTTTtT )462.0( 1

01.0)( −−−

∞ =→=−−

st 10=

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FIGURE 4-10

Schematic for Example 4-2.

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Assumption

.The body can be modeled as a 30-cm-diameter, 1.70-m-long cylinder..The thermal properties of the body and the heat transfer coefficient are constant..The radiation effects are negligible..The person was healthy (!) when he or she died with a body temperature of 37 0C.

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Analysis

( ) ( )( )( ) ( )

mmmm

mmrLr

LrAVL

sc 0689.0

15.027.115.027.115.0

22 2

2

200

20 =

+=

+==

πππ

πππ

( )( ) 1.089.0/617.0

0689.0/8 2

>=•

•==

CmWmCmW

khLBi O

Oc

( )( )( )15

3

2

1079.20689.0/4178/996

/8 −−×=•

•=== s

mCkgJmkgCmW

LCh

VChAb O

O

cpp

s

ρρ

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Analysis

tsbt

i

eeTTTtT )1079.2( 15

20372025)( −−×−−

∞ =−−

→=−−

hst 2.12860,43 ==

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FIGURE 4-11

Schematic of the simple geometries in which heat transfer is one-dimensional.

−−

=TT

TtxTtxi

),(),(θ

LxX =

khLBi =

2Ltατ =

Dimensionless temperature

Dimensionless distance from the center

Dimensionless heat transfer coefficient

Dimensionless time

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FIGURE 4-12

Transient temperature profiles in a plane wall exposed to convection from its surfaces for Ti > T∞.

2.0,)cos(),(),( 11

21 >=

−−

= −

∞ τλθ τλ LxeATT

TtxTtxi

wall

2.0,)(),(),( 0101

21 >=

−−

= −

∞ τλθ τλ rrJeATT

TtrTtri

cyl

2.0,)sin(),(),(01

011

21 >=

−−

= −

∞ τλλθ τλ

rrrreA

TTTtrTtr

isph

Plane wall

Cylinder

Sphere

τλθ21

10

,0−

∞ =−−

= eATTTT

iwall

τλθ21

10

,0−

∞ =−−

= eATTTT

icyl

τλθ21

10

,0−

∞ =−−

= eATTTT

isph

Center of plane wall (x=0)

Center of cylinder (r=0)Center of sphere (r=0)

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FIGURE 4-13

Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature Ti subjected to convection from both sides to an environment at temperature T∞with a convection coefficient of h.

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FIGURE 4-14

Transient temperature and heat transfer charts for a long cylinder of radius r0 initially at a uniform temperature Ti subjected to convection from all sides to an environment at temperature T∞ with a convection coefficient of h.

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FIGURE 4-15

Transient temperature and heat transfer charts for a sphere of radius r0 initially at a uniform temperature Ti subjected to convection from all sides to an environment at temperature T∞ with a convection coefficient of h.

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FIGURE 4-16

The specified surface temperature corresponds to the case of convection to an environment at T∞ with a convection coefficient h that is infinite.

( )( )ip

ip

TTVC

TTmCQ

−=

−=

ρmax

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FIGURE 4-17

The fraction of total heat transfer Q/Qmax up to a specified time t is determined using the Gröber charts.

1

1,0

max

sin1λλθ wall

wallQ

Q−=⎟⎟

⎞⎜⎜⎝

( )1

11,0

max

21λλθ J

QQ

cylcyl

−=⎟⎟⎠

⎞⎜⎜⎝

31

111,0

max

cossin31λ

λλλθ −−=⎟⎟

⎞⎜⎜⎝

⎛sph

sphQ

Q

Plane wall

Cylinder

Sphere

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FIGURE 4-18

Fourier number at time t can be viewed as the ratio of the rate of heat conducted to the rate of heat stored at that time.

( )TT

tLCLkL

Lt

p ΔΔ

== 3

2

2

ατ .

3

3

LvulumeofbodyainstoredisheatwhichatrateThe

LvolumeofbodyaofLacrossconductedisheatwhichatrateThe

=

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FIGURE 4-19

Schematic for Example 4-3.

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Assumption

•.The egg is spherical in shape with a radius of r0=2.5cm.•.Heat conduction in the egg is one-dimension because of thermal symmetry about the midpoint.•.The thermal properties of egg and the heat transfer coefficient are constant.•The Fourier number is τ>0.2 so that the one term approximate solutions are applicable.

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Analysis

( )( ) 1.08.47/627.0

025.0/1200 20 >=

••

==CmW

mCmWk

hrBi O

O

The coefficients λ1 and A1 for a sphere corresponding to this Bi are, from table 4-1,λ1=3.0753A1=1.9958

2.0209.09958.19559570 22

1 )0753.3(1

0 >=→=−−

→=−− −−

∞ τττλ eeATTTT

i

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Analysis

( )( ) min4.14865/10151.0

025.0209.026

220 ≈=

×== − s

smmrt

ατ

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FIGURE 4-20

Schematic for Example 4-4.

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Assumption

•. Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetryabout the center plane.•The thermal properties of the plate and the heat transfer coefficient are constant.•The Fourier number is τ>0.2 so that the one-term approximate solutions are applicable.

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Analysis

( )( ) 8.4502.0120

10012 =•

•==

mCmWCmW

hLk

Bi o

o

( )( )( )

6.3502.0

607/109.332

26

2 =××

==−

mssm

Ltατ

46.00 =−−

TTTT

i

8.451==

hLk

Bi

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Analysis

1==LL

Lx

99.00

=−−

TTTT

455.099.046.00

0

=×=−−

−−

=−−

TTTT

TTTT

TTTT

ii

( )( ) ( )( )( )

13

2

00185.002.0/380/8530

/1202

2 −=•

•==== s

mCkgJmkgCmW

LCh

LACAh

VChAb o

o

ppp

s

ρρρ

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Analysis

( )ssbt

i

etTeTTTtT 420)00185.0( 1

50020500)()( −−−

∞ =−−

→=−−

CtT o279)( =

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FIGURE 4-21

Schematic for Example 4-5.

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Assumption

•Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the centerline.•The thermal properties of the shaft and the heat transfer coefficient are constant.•The Fourier number is τ>0.2 so that the one-term approximate solutions are applicable.

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Analysis

( )( ) 86.11.080

9.1412

0

=•

•==

mCmWCmW

hrk

Bi o

o

( )( )( )

07.11.0

6045/1095.32

26

20

=××

==−

mssm

rtατ

40.00 =−−

TTTT

i

( ) ( ) CTTTT oi 3602006004.02004.00 =−+=−+= ∞∞

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Analysis

( ) ( ) ( ) kgmmmkgLrVm 2.24811.0/7900 2320 ==== πρπρ

( ) ( )( )( ) kJCCkgkJkgTTmCQ ooip 354,47200600/477.02.248max =−•=−= ∞

537.086.11

11

===Bi

Bi

( ) ( ) 309.007.1537.0 222

2

=== τα Bik

th

62.0max

=Q

Q

( ) kJkJQQ 360,29354,4762.062.0 max =×==

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Analysis Alternative solution

( )( ) 537.0/9.14

1.0/80 20 =

••

==CmW

mCmWk

hrBi O

O

970.01 =λ 122.11 =A

( ) ( ) 41.0122.1 07.1970.01

00

221 ===

−−

= −−

∞ eeATTTT

i

τλθ

( ) ( ) CTTTT oi 36420060041.020041.00 =−+=−+= ∞∞

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20

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Analysis Alternative solution

( )( ) 636.0

970.0430.041.02121

1

110

max

=×−=−=λλθ J

QQ

( ) kJkJQQ 120,30354,47636.0636.0 max =×==

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FIGURE 4-22

Schematic for a semi-infinite body.

i

i

i TTTtxT

TTTtxTtx

−−

=−−

−=−∞∞

∞ ),(),(1),(1 θ

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+−⎟⎟

⎞⎜⎜⎝

⎛=

−−

∞ kth

txerfc

kth

khx

txerfc

TTTtxT

i

i αα

αα 2

exp2

),(2

2

dueerfc u∫ −−=ξ

πξ

0

221)(

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α2),(

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21

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FIGURE 4-23

Variation of temperature with position and time in a semi-infinite solid initially at Ti subjected to convection to an environment at T∞with a convection heat transfer coefficient of h (from P.J. Schneider, Ref. 10).

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FIGURE 4-24

Schematic for Example 4-6.

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Assumption

•The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be asemi-infinite medium with a specified surface temperature of –100C.•The thermal properties of the soil are constant.

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Analysis

∞=k

th α

( ) ( )( ) 6.0

10151001,1 =−−−−

−=−−

−∞

TTTtxT

i

36.02

==t

ξ

( )( )( ) shsdayhdayst 61078.7/3600/2490 ×==

( )( ) mssmtx 77.01078.7/1015.036.022 626 =×××== −αξ

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23

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Analysis alternative solution

( ) 60.02

,=⎟⎟

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α

( )( ) mssmtx 80.01078.7/1015.037.022 626 =×××== −αξ

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display46

FIGURE 4-25

The temperature in a short cylinder exposed to convection from all surfaces varies in both the radial and axial directions, and thus heat is transferred in both directions.

cylinderinitei

wallplanei

cylindershorti TT

TtrTTT

TtxTTT

TtxrTinf

),(),(),,(⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎞⎜⎜⎝

⎛−

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24

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FIGURE 4-26

A short cylinder of radius r0 and height a is the intersection of along cylinder of radius r0 and plane wall of thickness a.

( )wallplanei

wall TTTtxTtx ⎟⎟

⎞⎜⎜⎝

⎛−

−=

∞),(,θ

cylinderinitei

cyl TTTtrTtr

inf

),(),( ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=∞

∞θ

( )solid

initesemiisemi TT

TtxTtxinf

inf),(,

−∞

∞− ⎟⎟

⎞⎜⎜⎝

⎛−

−=θ

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display48

FIGURE 4-27

A long solid bar of rectangular profile a X b is intersection of two plane wall of thicknesses a and b.

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛=⎟⎟

⎞⎜⎜⎝

1max2max1max2,max

1Q

QQ

QQ

QQ

Q

Dtotal

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛=⎟⎟

⎞⎜⎜⎝

2max1max3max1max2max1max3,max

111Q

QQ

QQ

QQ

QQ

QQ

QQ

Q

Dtotal

),(),(),,(tan

tytxTT

TtyxTwallwall

bargularreci

θθ=⎟⎟⎠

⎞⎜⎜⎝

⎛−

rectangular bar

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25

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FIGURE 4-28

Schematic for Example 4-7.

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FIGURE 4-28

Assumption

•Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. • The thermal properties of the cylinder and the heat transfer coefficient are constant.•The Fourier number is τ > 0.2 so that the one-term approximate solutions are applicable.

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26

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Analysis (a)

( )( )( )

48.806.0

900/1039.32

25

2 =×

==−

mssm

Ltατ

( ) ( ) 8.0,0,0 =⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=∞

TTTtTt

iwallθ

( )( ) 63006060

11012 .

m.Cm/WCm/W

hLk

Bi O

O

=⋅

•==

( )( )( )

2.1205.0

900/1039.32

25

2 =×

==−

mssm

rtατ

( ) ( ) 5.0,0,0 =⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=∞

TTTtTt

icylθ

( )( ) 73605060

11012 .

m.Cm/WCm/W

hrk

Bi O

O

o

=⋅

•==

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Analysis (a)

( ) ( ) ( ) 4.05.08.0,0,0,0,0=×=×=⎟⎟

⎞⎜⎜⎝

⎛−

∞ ttTT

TtTcylwall

cylindershorti

θθ

( ) ( )∞∞ −+= TTTtT i4.0,0,0

( ) C°=−+= 63251204.025

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27

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Analysis (b)

106.006.0

==mm

Lx

( ) 98.0,=⎟⎟

⎞⎜⎜⎝

⎛−−

TTTtLT

o

( )( ) 6.3006.0/60

/11012 =⋅

•==

mCmWCmW

hLk

Bi O

O

( ) ( ) ( ) 784.08.098.0,,, =×=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=−−

=∞

TTTT

TTTtLT

TTTtLTtL

i

o

oiwallθ

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Analysis (b)

( ) ( ) ( ) 392.05.0784.0,0,,0,=×=×=⎟⎟

⎞⎜⎜⎝

⎛−

∞ ttLTT

TtLTcylwall

cylindershorti

θθ

( ) ( )∞∞ −+= TTTtLT i392.0,0,

( ) C°=−+= 2.6225120392.025

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28

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FIGURE 4-29

Schematic for Example 4-9.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display56

FIGURE 4-28

Assumption

•Heat conduction in the semi-infinite cylinder is two dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. • The thermal properties of the cylinder and the heat transfer coefficient are constant.•The Fourier number is τ > 0.2 so that the one-term approximate solutions are applicable.

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29

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Analysis

( )( )( )

2.091.21.0

605/1071.92

25

2 >=××

==−

mssm

Ltατ

( )( ) 05.0/273

1.0/120 20 =

•⋅

==CmW

mCmWk

hrBi O

O

( )⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+−⎟

⎞⎜⎝

⎛=− − kth

txerfc

kth

khx

txerfctxsemi

αα

αα

θ2

exp2

,1 2

2

inf

( ) ( ) ( ) 762.00124.1,0 91.23126.010

221 ==== −− eeAtcylτλθθ

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Analysis

( ) 0074.0086.0 22

2

2

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

kth

kth αα

( )( )44.0

605/1071.9215.0

2 25=

××==

− ssmm

txα

ξ

( ) ( )( )086.0

/237300/1071.9/120 252

=°⋅

×°⋅=

CmWssmCmW

kth α

( )( ) 0759.0/237

15.0/1202

2

=°⋅

°⋅=

CmWmCmW

khx

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30

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Analysis

( ) ( ) ( ) 73407620963000 ...t,t,xTT

Tt,,xTcyleintinfsemi

cylindereintinfsemii

=×=×=⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

−∞

∞ θθ

( ) ( )∞∞ −+= TT.Tt,,xT i73400

( ) C. °=−+= 15115200734015

( ) 457008330533801 ..exp. ×+−=

( ) ( ) ( ) ( )086.044.00074.00759.0exp44.01,inf +++−=− erfcerfctxsemoθ

9630.=

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FIGURE 4-30

Schematic for Example 4-10.

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FIGURE 4-28

Assumption

•Heat conduction through the steaks is one-dimensionalsince the steaks from a large layer relative to their thickness and there is thermal symmetry about the center plane. • The thermal properties of the steaks and the heat transfer coefficient are constant.•The Fourier number is τ > 0.2 so that the one-term approximate solutions are applicable.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display62

Analysis

( ) ( )( ) 770

157152

0

.TT

Tt,LT=

−−−−

=−−

( ) Ccm/Wm/.

Ccm/W.Lk

.h O

O

⋅=⋅

== 22

30100151

45051

1

111

==cmcm

Lx

511 .hLk

Bi==

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FIGURE 4-31

Typical growth curve of microorganisms.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display64

FIGURE 4-32

The factors that affect the rate of growth of microorganisms.

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FIGURE 4-33

The rate of growth of microorganisms in a food product increases exponentially with increasing environmental temperature.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display66

FIGURE 4-34

Freezing may stop the growth of microorganisms, but it may not necessarily kill them.

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FIGURE 4-35

Recommended refrigeration and freezing temperatures for most perishable foods.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display68

FIGURE 4-36

Typical freezing curve of food item.

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FIGURE 4-37

Typical cooling curve of a beef carcass in the chilling and holding rooms at an average temperature of 0oC (from ASHRAE, Handbook: Refrigeration, Ref. 3, Chap. 11, Fig. 2).

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display70

FIGURE 4-38

Various cuts of beef (from National Livestock and Meat Board).

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FIGURE 4-39

Variation of tenderness of meat stored at 2oC with time after slaughter.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display72

FIGURE 4-40

The freezing time of meat can be reduced considerably by using low temperature air at high velocity.

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FIGURE 4-41

A refrigerated truck dock for loading frozen items to a refrigerated truck.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display74

FIGURE 4-42

Air chilling causes dehydration and thus weight loss for poultry, whereas immersion chilling causes a weight gain as a result of water absorption.

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FIGURE 4-43

The storage life of fresh poultry decreases exponentially with increasing storage temperature.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display76

FIGURE 4-44

The variation of freezing time of poultry with air temperature (from van der Berg and Lentz, Ref. 11).

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FIGURE 4-45

The variation of temperature of the breast of 6.8-kg turkeys initially at 1oC with depth during immersion cooling at -29oC (from van der Berg and Lentz, Ref. 11).

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display78

FIGURE 4-46

Schematic for Example 4-5.

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FIGURE 4-28

Assumption

•Water evaporates at a rate of 0.080 kg/s. •All the moisture in the air freezes in the evaporator.

Properties

( ) Ckg/kJ....contentwater..Cp °⋅=×+=×+= 143580512681512681

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display80

Analysis (a)

( )( ) ( ) s/kg.s/carcass/kgcarcasses 563360010285450 =×=

gainheatlightsfanbeefchillroom,total QQQQQ &&&&& +++=

( ) ( )timeCooling/cooledmassbeefTotalmbeef =

( ) ( )( )( ) kWCCkg/kJ.s/kg.TmCQ beefbeef 2351536143563 =°−°=Δ=&

kW27713326235 =+++=

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Analysis (b)

s/m.m/kg.s/kgmV

air

airair

33 978

2921102

===ρ&&

( ) ( )( ) kWkg/kJs/kg.mhQwaterfgeevaporativ,beef 19924900800 ===&

( ) ( ) ( )[ ] s/kg.C.Ckg/kJ.

kWTC

Qmairp

airair 0102

2700061277

=°−−°⋅

=&

&