Heat Conduction

Unsteady state In general, temperature is varying with direction and time

t,z,y,xfT

tTCq

zT

yT

xTk p2

2

2

2

2

2

Heat Conduction

Importance of External Versus Internal Resistance to Heat Transfer

Solid

Liquid

Heat transfer from surface to center of the solid will counter two resistance

1. External resistance

2. Internal resistance

Heat Conduction

Bi > 40 Finite internal and external resistance

Bi < 0.1 No internal and finite external resistance0.1 < Bi < 40 Finite internal resistance, no external resistance

Case 1Case 2Case 3

Internal resistance (L/k)External resistance (1/h)Bi =

Bi = hL k

L = characteristic length

Heat ConductionBodies with negligible internal temperature gradient

Major simplification: Assume that temperature variation within the body are negligible.

tfT In the case of material having very high thermal conductivity (k)

If initial temperature is Toh

Volume VSurface A

Tf

Tf

T0

Time

“Lump System”

Unsteady state of Heat Conduction

- At any instant of timeApply the first law of Thermodynamics

Rate at which heat transfer to the body

)TT(hAtTVC fp

Differential equation

)TT( fLet

hAt

VCp

Initial condition

0f0 )TT(

0TT,0tAt

Unsteady state of Heat ConductionBodies with Negligible Internal Temperature Gradients (cont)

VchAt

0f0

f peTTTT

Solution

Temperature ratio

Time

VchAt

0

pe

0

Tf

T0

Time

fTT Remaining temperature

Differential Equation of Heat Conduction

ExampleSteel ball

- R = 2.5 cm - k = 54 W/moC - p = 7833 kg/m3 - Cp = 0.465 kJ/kgoCFind ‘t’ to cool down from 850 oC to 250oC, if exposed to air at 50 oC (h = 100 W/moC)

tc.VA.hpeTR

25.05085050250

)TT()TT(TRf0

f

t)025.0

34)(10465.0)(7833(

)025.04)(100(.EXP25.033

2

s8.420t

Unsteady state of Heat ConductionExample ρ = 980 kg/m3

Rjacket= 0.5 m

hsteam= 5000 W/m2oC

Tinside jacket = 90 oC

Ttomato juice= 20 oC

Cp= 3.95 kJ/kgoCJacket hemispherical kettle

Tomato juice

Steam Find temperature of tomato juice after 6 min

Hemispherical kettle 2

2

r22r4A

33

r3/22r3/4V

r3

VA

Differential Equation of Heat Conduction

Jacket hemispherical kettle

Tomato juice

Steam

tc.VA.hpeTR

r3

)1095.3)(980()605)(5.0/3)(5000(.EXP

902090T

3

C16.83T o

Unsteady state of Heat ConductionBodies in which internal temperature gradients cannot be neglected

Unsteady state of Heat ConductionBodies in which internal temperature gradients cannot be neglected

Unsteady state in an finite slab

2b

tTCq

zT

yT

xTk p2

2

2

2

2

2

tTC

xTk p22

Unsteady state of Heat Conduction

tT

xT2

2

Define )TT( f

tx22

1. Initial condition

0f00 )TT(orTT,0t

2. Boundary conditions

0dxd

0x

bx

bx

hdxdk

1) 2)

Need sophisticate techniques

Unsteady state of Heat ConductionSolution obtained by “separation of variables” method in the form of a convergent series

1nn

t

nnn

n )xcos(ebcosbsinn

bsin22n

Where are called eigenvalues and are the roots of the equation

n21 ...,

)k/hb/()b(h/kbcot and

Unsteady state of Heat ConductionSolution in terms of dimensionless parameters

1nn

)b/t()b(

nnn

n

0

)bx,bcos(e

bcosbsinnbsin2

222n

Thus

bx,

bt,bf 2n

0

bx,

bt,

khbf 2

0

Unsteady state of Heat Conduction

khb

bx

Biot number (Ratio of thermal resistance of the solid to thermal resistance at the surface)

2bt Fourier number (Dimensionless of time)

Dimensionless measure of location in the slab

xb

Rx

In the case of sphere

Unsteady state of Heat ConductionPresent in graphic

0b/x0

0b/x

0

Note

bx,

bt,

khbf 2

0

Unsteady state of Heat ConductionPresent in graphic

0

0b/x

Midplane

Unsteady state of Heat ConductionPresent in graphic

At any point

0b/x

Unsteady state of Heat ConductionUnsteady state in a finite cylinder and in a sphereSimilarly, series solutions available for an infinitely long cylinder and a sphere, results are also presented graphically

0b/r0

0b/r

0

Note

Unsteady state of Heat ConductionUnsteady state in an finite cylinder

Unsteady state of Heat ConductionPresent in graphic

At any point

Unsteady state of Heat ConductionUnsteady state in a sphere

Unsteady state of Heat ConductionPresent in graphic

At any point

Unsteady state of Heat Conduction

T0 = 200ºC

Problem

Tf = 20ºCTf = 20ºC

Tx = 0 = ?Tx = b = ?

at t = 5 min find

1 cm

Unsteady state of Heat ConductionSignificant of Biot Number

The value of Biot number decides whether internal temperature gradient are importantUsual criterion

Bi < 0.1 Neglect internal temperature gradients

For a slab (hb/k) < 0.1For a cylinder/sphere (hR/k) < 0.1

Unsteady state of Heat ConductionTwo and three dimensional unsteady state heat conductionLong rectangular bar

2a

2b

- Initially at a temp. T0- Immerged at t=0 in

surroundings at a temp. Tf- Surface heat transfer is h

T = f(x, y, t)

Unsteady state of Heat ConductionTwo and three dimensional unsteady state heat conductionDifferential equation

2a

2b

tTCq

zT

yT

xTk p2

2

2

2

2

2

tT1

yT

xT

2

2

2

2

t1

yx 22

2

2

)TT( fwhere

Unsteady state of Heat ConductionDifferential equation

t1

yx 22

2

2

Initial conditions t = 0, T = T00f0 )TT(

Boundary conditions

0x 0x

0y 0y

2a

x

y

axax

hx

k

byby

hx

k

2b

Unsteady state of Heat ConductionSolution

y0x00

Product solution

x0

is the solution for dimensionless temperature in an

infinite slab of width 2a

y0

is the solution for dimensionless temperature in an

infinite slab of width 2b

They both are derived from the charts

Unsteady state of Heat ConductionRectangular box (2a x 2b x 2c)

z0y0x00

In a similar way

How to calculate HT in multiple dimensional case? (Finite objects)

1) TRoverall= TR1 x TR2 x TR3

2) TRoverall= TR3

3) TRoverall= TRslab x TRcylinder

ExampleD = 6.8 cm L = 10.3 cm

T0= 29.5 oC Tstream = 115.5 oC

h = 4542 W/m2oC k = 0.83 W/moCα = 2.007 x 10-7 m2/s

จงหาวาอุณหภูมิที่ตรงกลาง เทาไรเม่ือเวลาครบ 45 min.- วางซอนกัน- วางช้ันเดียว

Applying one-dimension HT for the case of a long cylinder

R = 0.034 m

Case 1 Stacked over each other

1/Bi = k/hR = 0.005375Fo= αt/R2 = 0.4638

0.13

TR = (T – Tf)/(T0-Tf)

T = 104.3 oC= 0.13

Case 2

CylinderTR = 0.13

Plane wall1/Bi = k/hR = 0.0036

Fo= αt/R2 = 0.209TR = 0.8

0.8 wallplanecylinderoverall TRTRTR

104.08.013.0TRoverall

C6.106T ocenter

Unsteady state of Heat ConductionA food product of cylindrical shape (length = 10 cm, diameter = 10 cm) is initially at30oC. What will be its center temperature after exposure to heating medium at 100oCfor 124 minutes assuming no thermal resistance at the surface? Given: k = 0.571W/(m.oC), = 1,055 kg/m3 and C = 4.015 kJ/(kg.oC).

If this food product was produced in cubical and spherical shapes having same volume,what will be the temperatures at the center under identical heat conditions?

Unsteady state of Heat Conductionผลิตภัณฑทางการเกษตรรูปทรงกระบอก (cylinder) ขนาด 7 cm ยาว 5 cm มีอุณหภูมิเร่ิมตนที่ 30 ºCถูกนําไปใสใน autoclave ท่ีอุณหภูมิ 100 ºC จงหาวาอุณหภูมิที่ตรงกลางของผลิตภัณฑน้ีจะเปนเทาไรเม่ือเวลาผานไป 60 นาที กําหนดให k = 0.571 W/m ºC, = 1,055 kg/m3, Cp = 4.015 kJ/kg ºCกําหนดใหสมการหาอุณหภูมิท่ีจดุศูนยกลางของผลิตภัณฑท่ีรูปทรงตางๆ ดังน้ี Infinite plane ln TR = -(0.413 + 0.70Fo) Infinite cylinder ln TR = -(0.319 + 1.56Fo) Sphere ln TR = -(0.159 + 2.50Fo) ถาผลิตภัณฑเปลี่ยนเปนทรงกลม (sphere) และรูปทรงลูกเตา (cubic shape) ท่ีปริมาตรเทาเดิม จงเปรียบเทียบอุณหภูมิของผลิตภัณฑแตละชนิด