Outline Unsteady Heat Transferlcaretto/me375/04-Unsteady.pdfUnsteady Conduction February 28 and...

Unsteady Conduction February 28 and March 7, 2007

ME 375 Heat Transfer 1

Unsteady Heat TransferUnsteady Heat Transfer

Larry CarettoMechanical Engineering 375

Heat TransferHeat Transfer

February 28 and March 7, 2007

2

Outline• Review material on fins• Lumped parameter model

– Basis for and derivation of model– Solving lumped-parameter problems

• Unsteady solutions using charts– Differential equation as basis for charts– Problem solving with charts – Semi-infinite solutions– Product solutions

3

Fin review• Adds surface to enhance heat transfer• Analysis for single fin linked to analysis of

surface with multiple fins• Equations for simple fins and charts for

fin efficiency and effectiveness• Rectangular fin equations: m = (hp/kAc)1/2

c

c

b mLxLm

TTTT

cosh)(cosh −

=−−

∞

∞ ccb

fin mLhpkATT

Qtanh=

− ∞

&

4

Lumped Parameter Model• Simplified model of unsteady heat

transfer for a particular problem– Solid object, with constant k and a uniform

initial temperature, Ti

– Placed in fluid environment with constant temperature, T∞, at zero time (t = 0)

– Convection to the solid with constant heat transfer coefficient, h

– Under certain conditions the temperature in the solid is assumed uniform

5

Parallel Resistance• Look at solid object initially at an initial

temperature, Ti, placed into a medium at another temperature T∞ with heat transfer coefficient h

• Convective resistance is 1/hA, conductive resistance is L/kA

hA1

kAL

Figure 3-6) in Çengel, Heat

and Mass Transfer

6

Uniform Temperature Basis

hA1

kAL

• T1 – T2 will be small compared to T∞1 – T1when hL/k is small

khL

hA

kAL

TTTT

kAL

TT

hA

TTQ ==−

−⇒

−=

−=

∞

∞11 11

212111&

Figure 3-6) in Çengel, Heat

and Mass Transfer



7

Application to Unsteady Case• Unsteady case: temperatures change• Special case: convection resistance is

much larger than conduction in solid• Result: temperature differences in the

solid are almost negligible• Idealization: Assume that solid is at

uniform temperature, T• Model: ρcpVdT/dt = hA(T∞ – T)

– V is volume8

Lumped Parameter Model• Basic model says that convection

energy into solid hA(T∞ – T) goes to increase uniform solid temperature, T, giving energy change ρcpVdT/dt

( ) ( )∞∞ −ρ

−=⇒−=ρ TTchA

dtdTTThA

dtdTc

pp VV

• Define characteristic length, Lc = V/A, and inverse time constant, b = hA/ρcpV= hA/ρcpLc

9

Lumped Parameter Solution

( ) ( ) 0==−−=−ρ

−= ∞∞ tatTTTTbTTchA

dtdT

ipV

• Solution is known to be exponential

• Have first-order differential equation with initial condition that T = Ti at t = 0

( ) ( ) ( ) ∞−

∞−

∞∞ +−=−=− TeTTToreTTTT bti

bti

• You can show that solution satisfies differential equation and initial condition

10

Lumped Parameter Analogy• Combine solution with b definition

( ) ( ) ( ) VpchAt

ibt

i eTTeTTTT ρ−

∞−

∞∞ −=−=−

• 1/hA is convection resistance and ρcpVis capacity of solid to absorb heat

thermalthermalpp CRt

chA

tchAt

=ρ

=ρ VV 1

• Result same as RC electrical circuit

11

When can we use this?• Saw that solid temperature becomes

closer to uniform as hL/k (known as Biotnumber, Bi) becomes smaller

• Criterion for application of lumped parameter solution is Bi = hLc/k < 0.1

• Problem: what is Lc for a cylinder with a diameter of 0.1 m and a length of 0.5 m?

m

mmDLDLD

LD

ALc 02273.0

1.04

5.02

142

1

42

42

2

=+

=+

=π+π

π

==V

12

When can we use this? II• So Lc = 0.02273 m; can we use lumped

parameter model if h = 80 W/m2·K and k = 240 W/m2·K?

• Criterion for application of lumped parameter solution is Bi = hLc/k < 0.1

( )1.00076.0240

02273.0802

<=

⋅

⋅==

KmW

mKm

W

khLBi c

• So lumped parameter model is valid



13

Application of the Model• If the cylinder analyzed previously has

cp = 900 J/kg·K, ρ = 2700 kg/m3, an initial temperature of 350OC and an environmental temperature of 30oC, how long will it take to reach 50oC?

• Given: Lc = 0.02273 m, h = 80 W/m2·K, cp = 900 J/kg·K, ρ = 2700 kg/m3, Ti = 350OC, T∞ = 30oC, T = 50oC

• Find: time, t, to reach T = 50oC14

Application of the Model II• Given: h = 80 W/m2·K, cp = 900 J/kg·K,

ρ = 2700 kg/m3, Ti = 350OC, T∞ = 30oC, T = 50oC; Find: t

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=⇒−=−∞

∞−∞∞ TT

TTb

teTTTTi

bti ln1

( ) smKkgJ

mkg

sWJ

KmW

Lch

chAb

cpp

001449.0

02273.09002700

180

3

2=

⋅

⋅⋅=ρ

=ρ

=V

15

Application of the Model IIIs

CCCCs

TTTT

bt oo

oo

i1914

303503050ln

001449.0ln1

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=∞

∞

• Lumped parameter approach easy to use and does not require application of more complex calculations

• Applicable to general geometry• Requires Bi = hLc/k < 0.1• Next consider unsteady problems with

spatial variation16

Review Conduction Equation

genp ezTk

zyTk

yxTk

xtTc &+

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

ρ

genp ezTk

zyTk

yxTk

xtTc &+

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

ρ

0 for no heat generation

0 for one dimensional heat transfer

For constant k, bring k outside the derivative

2

2

2

2

xT

ck

tT

xTk

tTc

pp

∂∂

ρ=

∂∂

⇒∂∂

=∂∂

ρ

17

1D, constant k, • Define the thermal diffusivity, α = k/ρcp

0egen =&

TL

EM

ML

LTE 23

=Θ⋅

Θ⋅⋅• Typical units are m2/s or ft2/s

• Dimensions:

2

2

2

2

xT

tT

xT

ck

tT

p ∂∂

α=∂∂

⇒∂∂

ρ=

∂∂

• Solution requires initial (t=0, all x) and boundary conditions

18

What is 1D?• In theory, one dimensional

heat transfer implies that the slab is infinite (or insulated) in the y and z directions

• Practically this means that the y and z dimensions are so large that end effects are not importantFigure 4-11(a) in

Çengel, Heat and Mass Transfer



19

Specific Problem• Problem: at t = 0, a large

slab initially at Ti is placed in a medium at temperature T∞ with a heat transfer coefficient, h

• Coordinates: Choose x = 0 as center of slab (which runs from –L to L) for this symmetric problemFigure 4-11(a) in

Çengel, Heat and Mass Transfer

20

Specific Problem II• Initial condition: at t = 0, T

= Ti for all x• Boundary conditions: At

x = 0, ∂T/∂x = 0 for symmetry. At x = L an energy balance gives –k∂T/∂x = h(T – T∞)

• Dimensionless form: shows important combi-nations of variables

Figure 4-11(a) in Çengel, Heat and Mass

Transfer

21

Specific Problem Conditions• Differential equation with boundary and

initial condition for T(x,t)

( )[ ]∞=

=

−=∂∂

−

=∂∂

=∂∂

α=∂∂

TtLThxTk

xTTxT

xT

tT

Lx

xi

,

0)0,(0

2

2

• Define dimensionless distance, ξ = x/L and dimensionless time, τ = αt/L2, called the Fourier number

22

Dimensionless Equation Form• Define dimensionless

temperature ratio

( ) ( )τΘ=τΘ=ξ∂Θ∂

−

=ξ∂Θ∂

=ξΘξ∂Θ∂

=τ∂Θ∂

=ξ

=ξ

,1,1

01)0,(

1

02

2

BikhL

• Get dimensionless differential equation and initial/boundary conditions for Θ(ξ,τ)

∞

∞

−−

=ΘTTTT

i

23

Dimensionless Result

• We started with the following variables to solve for T: Ti, T∞, x, L, t, α, h, k

• We now see that T = (Ti – T∞)Θ + T∞, where Θ depends on ξ, τ, and Bi

( ) ( )τΘ=τΘ=ξ∂Θ∂

−

=ξ∂Θ∂

=ξΘξ∂Θ∂

=τ∂Θ∂

=ξ

=ξ

,1,1

01)0,(

1

02

2

BikhL

Lx

LtTTTT

i

=ξ

α=τ

−−

=Θ∞

∞

2

24

Equation Solution• Solution is infinite series with an infinite

set of dimensionless parameters λn that depends on Bi

∑∞

=

τλ− ξλλ+λ

λ=Θ

1cos

2sin2sin4 2

nn

nn

n ne

• The values of λn are the roots of the equation λn/Bi = cot λn– Next chart shows first seven λn values for

Bi = 1



25

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 2 4 6 8 10 12 14 16 18 20

λ

Func

tions Line

CotangentIntersections

1

cot

=

λ=λ

BiBi

Finding λ

λ7 = 18.9023λ6 = 15.7713λ5 = 12.6453λ4 = 9.52933λ3 = 6.43730λ2 = 3.42561λ1 = 0.86033

26

Dimensionless Temperature Difference in a Slab with hL/k = 0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Dimensionless distance x/L

tau = 0.01tau = 0.05tau = 0.1tau = 0.2tau = 0.3tau = 0.4tau = 0.5tau = 0.6tau = 0.8tau = 1tau = 1.5tau = 2tau = 3tau = 4tau = 5

∞

∞

−−

TTTT

i

27

Figure 6. Convection Boundary Condition, hL/k = 1

tau = 0.02tau = 0.05tau = 0.1tau = 0.2

tau = 0.5

tau = 1

tau = 2

tau = 3

tau = 4

tau = 0.75

tau = 1.5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x/L

Dim

ensi

onle

ss T

empe

ratu

re

∞

∞

−−

TTTT

i

28

Figure 7. Convection Boundary Condition, hL/k = 10

tau = 0.003

tau = 0.005

tau = 0.01

tau = 0.02tau = 0.05tau = 0.2

tau = 0.75

tau = 1

tau = 1.5

tau = 0.1

tau = 0.5

tau = 0.35

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x/L

Dim

ensi

onle

ss T

empe

ratu

re

∞

∞

−−

TTTT

i

29

Center-line Temperature, T0• Rewrite general result to introduce An

∑∑∞

=

τλ−∞

=

τλ−

∞

∞ ξλ=ξλλ+λ

λ−−

=Θ11

coscos2sin2

sin4 22

nnn

nn

nn

n

i

nn eAeTTTT

• Result for center-line (x = 0)

∑∑∞

=

τλ−∞

=

τλ−

∞

∞ =λ=−−

=Θ11

00

220cos

nn

nnn

i

nn eAeATTTT

• Heisler charts show Θ0 as a function of τ = αt/L2 with 1/Bi = k/Lh as a parameter– An and λn depend on Bi

30

Slab Center-line (x = 0) Temperature Chart Figure 4-15(a) in Çengel, Heat and Mass Transfer



31

Problem• Find time required to cool the centerline

temperature of 0.3 m thick plate to 50oC if k = 50 W/m·K, α = 15x10-6 m2/s and initial temperature is 400oC. The heat transfer coefficient is 80 W/m2·K and the environmental temperature = 20oC.

• Given: T0 = 50oC, Ti = 400oC, T∞ = 20oC, L = 0.3/2 = 0.15 m, k = 50 W/m·K, α = 15x10-6 m2/s, h = 80 W/m2·K Find: t

32

Solution• Chart relates (T0 – T∞)/(Ti – T∞), k/hL,

and αt/L2.• From given data we can find

167.415.01

80

50

0789.020400

20500 =

⋅

⋅==−

−=

−−

∞

∞

mKmWKmW

hLk

CCCC

TTTT

oo

oo

i

• Find τ = αt/L2 = 11.9 (see next chart)( ) hsx

smxmLt 96.41079.1

101515.09.11 4

26

22===

ατ

= −−

Input Θ0= 0.079Input k/hL = 4.167

Find τ = 11.9

Figure 4-15(a) in Çengel, Heat

and Mass Transfer 34

Chart II• Can find T at any

x/L from this chart once T at x = 0 is found from previous chart

• See basis for this chart on the next page

∞

∞

−−

=ΘΘ

TTTT

00

Figure 4-15(b) in Çengel, Heat and Mass Transfer

35

Temperature Approximations

• For “large” τ (> 0.2) series in numerator and denominator converge in one term

L

L

++

+ξλ+ξλ=

−−

=ΘΘ

τλ−τλ−

τλ−τλ−

∞

∞22

21

22

21

21

2211

00

coscos

eAeA

eAeATTTT

• Previous equation for Θ/Θ0

ξλ≈++

+ξλ+ξλ=

−−

=ΘΘ

τλ−τλ−

τλ−τλ−

∞

∞1

21

2211

00coscoscos

22

21

22

21

L

L

eAeA

eAeATTTT

36

Temperature Approximations II

• Depends only on ξ = x/L and λ1 which depends on Bi = hL/k

• Must first determine Θ0 as in previous example to get Θ

• What is Θ at x = L in that example?

• For τ > 0.2

ξλ≈−−

=ΘΘ

∞

∞1

00cos

TTTT



37

Answer• Example had T0 =

50oC, T∞ = 20oC,and k/hL = 4.167

• This chart gives (T – T∞)/(T0 – T∞) = 0.91 for x/L = 1

• So T = (0.91)·(50oC – 20oC) + 20oC = 47.3oC

∞

∞

−−

=ΘΘ

TTTT

00


38

Alternative Approximation• Return to series for “large” τ (> 0.2) that

converges in one term

ξλ≈ξλ=−−

=Θ τλ−∞

=

τλ−

∞

∞ ∑ 111

coscos21

2eAeA

TTTT

nnn

i

n

• Can use this approximation to find Θ and can also solve for τ and ξ when τ > 0.2

⎟⎟⎠

⎞⎜⎜⎝

⎛ξλ−

−λ

−≈τ∞

∞

1121 cos

1ln1ATT

TT

i ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−−

λ≈ξ

τλ−∞

∞−21

1

1

1

1cos1

eATTTT

i

39

Table Extract• Use Table 4-2 in text

to find λ1 and A1 for given Bi

• Find T at any ξ = x/L and τ = αt/L2 from

ξλ≈−−

=Θ τλ−

∞

∞11 cos

21eA

TTTT

i

( ) ∞τλ−

∞ +⎟⎠⎞⎜

⎝⎛ ξλ−≈ TeATTT i 11 cos

21

40

Approximate Equations• Repeat example with T0 = 50oC, Ti =

400oC, L = 0.3/2 = 0.15 m, T∞ = 20oC, k = 50 W/m·K, α = 15x10-6 m2/s, h = 80 W/m2·K to find t for T0 = 50oC and T at surface for this time

• Recall 1/Bi = 4.167 so Bi = 0.24• Interpolate in table to find λ1 = 0.4684

and A1 = 1.0367 for Bi = 0.24– Last two slides have interpolation details

• First find time for T0 = 50oC

41

Approximate Equations II• Data: T0 = 50oC, Ti = 400oC, L =0.15 m, T∞

= 20oC, k = 50 W/m·K, α = 15x10-6 m2/s, h = 80 W/m2·K, λ1 = 0.4476 and A1 = 1.0334

( ) 74.110cos0367.1

1204002050ln

4684.01

cos1ln1

211

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

•−=⎟⎟⎠

⎞⎜⎜⎝

⎛ξλ−

−λ

−≈τ∞

∞

CCCC

ATTTT

oo

ooi

( ) hsx

smxmLt 83.41076.1

101515.074.11 4

26

22===

ατ

= −42

Approximate Equations III• Find surface temperature for τ = 11.74

( ) ( )

( ) ( ) ( )( )[ ]{ } CeCC

CeATTTTTTT

ooo

oii

8.4614684.0cos0334.120400

20cos

74.114684.0

112

21

=−=

+=⎟⎠⎞⎜

⎝⎛ ξλ−+≈Θ−+=

−

τλ−∞∞∞∞

• Results similar to charts– 4.83 h vs. 4.96 h to reach T0 = 50oC– Surface temperature 46.8oC vs. 47.3oC

• For full solution T0 = 50oC in 4.83 h and surface temperature = 46.7oC at that time



43

Cylinder and Sphere

Figure 4-11 in Çengel, Heat and Mass

Transfer

Same problem has similar chart solutions44

1D Cylinder, constant k,

gen

p

ezTk

z

Tkrr

Tkrrr

tTc

&+∂∂

∂∂

+

φ∂∂

φ∂∂

+∂∂

∂∂

=∂∂

ρ

211

Figure 2-3 from Çengel, Heat and Mass Transfer

rTr

rrtT

rTr

rrk

tTcp ∂

∂∂∂α

=∂∂

∂∂

∂∂

=∂∂

ρ

0egen =&

45

Same Problem for Cylinder• Constant initial temperature, Ti, and

convection at outer radius r0

( )[ ]∞=

=

−=∂∂

−

=∂∂

=∂∂

∂∂α

=∂∂

TtrThrTk

rTTrT

rTr

rrtT

rr

ri

,

0)0,(

0

0

0

• Define dimensionless distance, ξ = r/r0and dimensionless time, τ = αt/r0

2,

46

Cylinder Center (r = 0) Temperature Chart Figure 4-16(a) in Çengel, Heat and Mass Transfer

47

gen

p

eTkr

Tkr

rTkr

rr

tTc

&+φ∂

∂φ∂

∂θ

+θ∂

∂θ

θ∂∂

θ

+∂∂

∂∂

=∂∂

ρ

22

2

22

sin1

sinsin1

1

Figure 2-3 from Çengel, Heat and Mass Transfer

1D Sphere, constant k, 0egen =&

rTr

rrtT

rTr

rrk

tTcp ∂

∂∂∂α

=∂∂

∂∂

∂∂

=∂∂

ρ 22

22

48

Same Problem for Sphere• Constant initial temperature, Ti, and

convection at outer radius r0

( )[ ]∞=

=

−=∂∂

−

=∂∂

=∂∂

∂∂α

=∂∂

TtrThrTk

rTTrT

rTr

rrtT

rr

ri

,

0)0,(

0

0

22

0

• Define dimensionless distance, ξ = r/r0and dimensionless time, τ = αt/r0

2,



49

Sphere Center (r = 0) Temperature Chart Figure 4-17(a) in Çengel, Heat and Mass Transfer

50

(T – T∞)/(T0 – T∞) Charts

Figures 4-16(b) and 4-17(b) in Çengel, Heat and Mass

Transfer

51

More Approximate Solutions• Cylinder and sphere also have

approximate solutions for τ > 0.2– Values of A1 and λ1 still depend on Bi and

are different for each geometry

⎟⎟⎠

⎞⎜⎜⎝

⎛λ=

−−

=Θ τλ−

∞

∞

0101

21

rrJeA

TTTT

i

⎟⎟⎠

⎞⎜⎜⎝

⎛λ

λ=

−−

=Θ τλ−

∞

∞

01

1

01 sin

21

rr

rreA

TTTT

i

• Cylinder

• Sphere

52

More Approximate Solutions II

Table 4-2 in Çengel, Heat and Mass Transfer

53

More Approximate Solutions III• The function J0(x) is the zero order

Bessel function– Tables in text or get value from Excel

function besselj(x,0) or Matlab function besselj(0,x)

• Excel function requires analysis tool pack add-in to be installed

• Here Bi = hr0/k for cylinder and sphere is different from Bi = hV/(kA) for lumped parameter solution

54

Semi-Infinite Solids• Plane that

extends to infinity in all directions

• Practical applications: large area for short times– Example: earth

surface locallyFigure 4-24 in Çengel, Heat and Mass Transfer



55

Two Semi-infinite Solid Results• Initial temp Ti,

surface temperature set to Ts at t = 0

⎟⎠

⎞⎜⎝

⎛α

=−−

txerf

TTTT

si

s

4

• Initial temp Ti, convection starts at t = 0 to T∞ with heat transfer coefficient, h

⎟⎟⎠

⎞⎜⎜⎝

⎛ α+

α−⎟

⎠

⎞⎜⎝

⎛α

=−− ⎟

⎟⎠

⎞⎜⎜⎝

⎛ α+

∞ kth

txerfce

txerf

TTTT k

thkhx

i

i

44

2

2

56

Error Functions• Defined integrals erf(x) is called error

function and erfc(x) = 1 – erf(x) is the complementary error function– Excel/ Matlab functions erf(x) and erfc(x)

• Excel requires analysis took pack add-in

• See plots on next chart

∫∞

−

π=−=

x

texerfxerfc22)(1)(∫ −

π=

xtexerf

0

22)(

57

Error Function and Complement

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x

Func

tions

∫ −

π=

xtexerf

0

22)(

∫∞

−

π=−=

x

texerfxerfc22)(1)(

58Figure 4-29 in Çengel, Heat and Mass Transfer

Semi-Infinite Medium Results

59

Problem• How deep should water pipes be buried

in a soil that is initially at 20oC to avoid freezing if the surface is at –15oC for 60 days? (Assume soil α = 1.4x10-7 m2/s)

• Given: Soil with α = 1.4x10-7 m2/s and Ti = 20oC must not freeze for t = 60 days if Ts= -15oC. Find: depth required

• Assumption: Required depth will set T = 0oC at t = 60 days

60

Solution• Use equation for semi-infinite medium

with fixed surface temperature( )( ) ⎟

⎠

⎞⎜⎝

⎛α

==−−

−−=

−−

txerf

CCCC

TTTT

oo

oo

si

s

442857.0

1520150

• From tables or erfinv function of Matlabfind erf-1(.42857) = 0.40019 t4x α=

( )d

sds

mxtx 8640060104.140019.040019.027−

=α=

x = 0.682 m



61

Multidimensional Solutions• Can get multidimensional solutions as

product of one dimensional solutions– All one-dimensional solutions have initial

temperature, Ti, with convection coefficient, h, and environmental temperature, T∞, starting at t = 0

– General rule: ΘtwoD = ΘoneΘtwo where Θoneand Θtwo are solutions from charts for plane, cylinder or sphere

62

Multidimensional Example•Solution for finite cylinder is product of solution for infinite cylinder and infinite slab

•Slab solutions have thickness 2L so we use L = a/2 in this case

Figure 4-35 in Çengel, Heat

and Mass Transfer

63

Multidimensional Example II

( )

( )

( )

slabinfinitei

cylinderinfinitei

cylinderfinitei

TTTtxT

TTTtrT

TTTtxrT

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

•⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞

,

,

,,

x = a/2

x = -a/2Figure 4-35 in Çengel, Heat

and Mass Transfer

64

Multidimensional Problem• A cylinder is 0.3 m high, with a radius of

0.1 m, k = 50 W/m·K, α = 15x10-6 m2/s, and initial temperature of 400oC. The heat transfer coefficient is 80 W/m2·K and the environmental temperature is 20oC. Find temperature in the centre of the cylinder and at its corner after one hour.

65

Multidimensional Solution• We can apply the following equation:

( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtxTTT

TtrTTT

TtxrT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,,,,

• First get the temperatures at the center ( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtTTT

TtTTT

TtT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,0,0,0,0

• We find infinite quantities from charts 66

Multidimensional Solution II• We have separate Biot and Fourier

numbers for the two infinite geometries

( )

( )40.2

15.0

36001015

167.415.01

80

50

2

26

2 ==α

=τ=

⋅

⋅=

−

m

ss

mx

Lt

mKmWKmW

hLk

( )

( )40.5

1.0

36001015

25.61.01

80

50

2

26

200

==α

=τ=

⋅

⋅=

−

m

ss

mx

rt

mKmWKmW

hrk



67

τ = 5.4

Θ0 =

0.22Enter cylinder chart with values ofτ = 5.4 and k/hr0 = 6.25 to find Θ0 = 0.22

Figure 4-16(a) in Çengel, Heat and Mass Transfer68

τ = 2.4

Slab chart uses τ= 2.4 and k/hL = 4.167 to find Θ0 = 0.62

Figure 4-15(a) in Çengel, Heat and Mass Transfer

69

Multidimensional Solution III• Use Θ0 values just found( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtTTT

TtTTT

TtT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,0,0,0,0

( ) ( )( ) 1364.062.022.03600,0,0==⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

cylinderfinitei TT

TsT

( )( )( )( ) CCCC

TTTsTToooo

i

721364.020400201364.0)3600,0,0(0

=−+=−+== ∞∞

70

Multidimensional Solution IV• Edge temperature (x = L = 0.15 m and r

= r0 = 0.1 m) found as follows( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtLTTT

TtrTTT

TtrLT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,,,, 00

• Each term on right is product of center solution times ratio of point to center– Must find this for each one dimensional

solution separately • Use auxiliary charts for this ratio

71

Slab• For k/hL = 4.167

and x/L = 1, Θ/Θ0= 0.875

• Previously found that Θ0 = 0.62

• So slab component of product solution is (0.62)(0.875) = 0.543

∞

∞

−−

=ΘΘ

TTTT

00


k/hL = 4.167

72

Cylinder• For k/hr0 = 6.25

and r/r0 = 1, Θ/Θ0= 0.93

• Previously found that Θ0 = 0.22

• So cylinder component of product solution is (0.22)(0.93) = 0.205

∞

∞

−−

=ΘΘ

TTTT

00


k/hr0 = 6.25



73

Multidimensional Solution V( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtLTTT

TtrTTT

TtrLT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,,,, 00

( ) ( )( ) 111.0543.0205.0,, 0 ==⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

cylinderfinitei TT

TtrLT

( )( )( )( ) CCCC

TTTsrLTToooo

iedge

62111.02040020111.0)3600,,( 0

=−+

=−+== ∞∞

74

Repeat Using One-term Values( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtLTTT

TtrTTT

TtrLT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,,,, 00

( )⎟⎟⎠

⎞⎜⎜⎝

⎛λ=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

− τλ−

∞

∞

0101

21,

rrJeA

TTTtrT

cylinderinfinitei

( )LxeA

TTTtLT

slabinfinitei

11 cos, 21 λ=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

− τλ−

∞

∞

75

Repeat Using One-term Values II

( )LxeA

TTTtLT

slabinfinitei

11 cos, 21 λ=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

− τλ−

∞

∞

For infinite slab, τ = 2.40 and Bi = 0.24; interpolation in Table 4-2 for Bi = 0.24 gives λ1= 0.4684 and A1 = 1.0367

546.015.015.04684.0cos0367.1 )40.2()4684.0( 2

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

mme

76

Repeat Using One-term Values IIIFor infinite cylinder, τ = 5.40 and Bi = 0.16; interpolation in Table 4-2 for Bi = 0.24 gives λ1= 0.5469 and A1 = 1.0388

191.01.01.05469.00388.1 0

)40.5()5469.0( 2=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

mmJe

( )⎟⎟⎠

⎞⎜⎜⎝

⎛λ=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

− τλ−

∞

∞

0101

21,

rrJeA

TTTtrT

cylinderinfinitei

77

Repeat Using One-term Values IVNow have individual terms in product solution

( )( ) 1045.0191.0546.0 ==

( ) ( ) ( )

slabinfinitei

cylinderinfinitei

cylinderfinitei TT

TtLTTT

TtrTTT

TtrLT⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

∞

∞

∞

∞

∞

∞ ,,,, 00

( )( )( )( ) CCCC

TTTsrLTToooo

iedge

601045.020400201045.0)3600,,( 0

=−+

=−+== ∞∞

78

Interpolation• Given table with two data pairs (x1,y1)

and (x2,y2) we want to find y for some value of x between x1 and x2

( ) ( )1212

111

12

121 yy

xxxxyxx

xxyyyy −

−−

+=−−−

+=

• Here known x is Bi and we are trying to find y = λ1 so interpolation formula is

( )112

1,12,11,11 BiBi

BiBi−

−

λ−λ+λ=λ



79

Interpolation II• Find λ1 for Bi = 0.24

– Bi = 0.24 is between Bi1 = 0.2 (λ1,1 = .4328) and Bi2 = 0.3 (λ1,2 = .5218)

( )112

1,12,11,11 BiBi

BiBi−

−λ−λ

+λ=λ

( )2.24.2.3.4328.5218.4328. −

−−

+=

4684.=

Outline Unsteady Heat Transferlcaretto/me375/04-Unsteady.pdfUnsteady Conduction February 28 and...

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Transcript of Outline Unsteady Heat Transferlcaretto/me375/04-Unsteady.pdfUnsteady Conduction February 28 and...