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