슬라이드 1

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Transcript 슬라이드 1

INTRODUCTION
• What is Heat Transfer ?
• Continuum Hypothesis
• Local Thermodynamic Equilibrium
• Conduction
• Radiation
• Convection
• Energy Conservation
WHAT IS HEAT ?
In a solid body
Crystal : a three-dimensional periodic array of atoms
Oscillation of atoms about their various positions of
equilibrium (lattice vibration): The body possesses heat.
Conductors: free electrons ↔ Dielectics
Vibration of crystals with an atom
Longitudinal polarization vs. Transverse polarization
us-1
us-1
s-1
us+1
us
s
s+1
s+2
us+2
us+3
s+3
The energy of the oscillatory motions:
the heat-energy of the body
More vigorous oscillations:
the increase in temperature of the body
us
us+1
us+2
In a gas
The storage of thermal energy:
molecular translation, vibration and rotation
change in the electronic state
intermolecular bond energy
Energy
average kinetic energy
electronic
state 2
dissociation
energy for state 2
vibrational state
electronic
state 1
dissociation
energy for state 1
rotational state
Internuclear separation distance
(diatomic molecule)
1
3
2
Eu  mum  k BT
2
2
kB = 1.3807 × 10-23 J/K
at T = 300 K,
air M = 28.97 kg/kmol
2 1/ 2
um = 468.0 m/s
HEAT TRANSFER
Heat transfer is the study of thermal energy
transport within a medium or among neighboring
media by
• Molecular interaction: conduction
• Fluid motion: convection
• Electromagnetic wave: radiation
resulting from a spatial variation in temperature.
Energy carriers: molecule, atom, electron, ion,
phonon (lattice vibration), photon (electromagnetic wave)
CONTINUUM HYPOTHESIS
m
lim
Ex) density    V  V
V
 m microscopic
uncertainty
V
0
macroscopic
uncertainty
local value of
density
 V0  10 mm
9
3
V
(3×107 molecules at sea level, 15°C, 1atm)
• microscopic uncertainty
due to molecular random motion
• macroscopic uncertainty
due to the variation associated with
spatial distribution of density
In continuum, velocity and temperature
vary smoothly. → differentiable
Mean free path of air at STP (20°C, 1atm)
2 1/ 2
lm = 66 nm, um
 468.0 m/s
bulk motion vs molecular random motion
LOCAL THERMODYNAMIC EQUILIBRIUM
hot wall at Th
L
gas
cold wall at Tc
a) lm << L : normal pressure
b) lm ~ L : rarefied pressure
c) lm >> L
CONDUCTION
Gases and Liquids
• Due to interactions of
atomic or molecular
activities
• Net transfer of energy by random
molecular motion
• Molecular random motion→ diffusion
• Transfer by collision of random
molecular motion
Solids
• Due to lattice waves induced by
atomic motion
• In non-conductors (dielectrics):
exclusively by lattice waves
• In conductors:
translational motion of free electrons
as well
Fourier’s Law
Th
Qx
T  Th  Tc
T
Qx 
  t  A [J]
x
 x Tc
T
T
Qx

 k
heat flux qx 
x
x
A  t
A
[J/(m2s) = W/m2]
k: thermal conductivity [W/m·K]
T
As x → 0, qx   k
x
Notation
Q : amount of heat transfer [J]
Q
q : heat transfer rate [W], q 
t
q : heat transfer rate per unit area [W/m2]
Q
q 
A  t
q : heat transfer rate per unit length [W/m]
Q
q 
L  t
q  qA  qL
Heat Flux
z
qz
qx
x
vector quantity
q
qy
y
T
ˆ
qy  q  j   k
,
y
q  qx iˆ  qy ˆj  qz kˆ
T
ˆ
qx  q  i   k
x
T
ˆ
qz  q  k   k
z
 T ˆ  T ˆ  T ˆ 
j
q   k 
i
k   kT
y
z 
 x
Ex) T ( x , y )  x  y (0  x  1, 0  y  1)
y
1
q
T = constant line or
surface: isothermal
T  1.5 lines or surfaces
(isotherms)
T 1
T  0.5 1 x
 T ˆ T
i
q   kT   k 
y
 x
 qiˆ  q ˆj   kiˆ  kjˆ
x
y
qx   k , qy   k

ˆj    k iˆ  ˆj



• temperature : driving potential of
heat flow
• heat flux : normal to isotherms
along the surface of T(x, y, z) = constant
ds
T(x, y, z) = constant
T ˆ T ˆ T ˆ
T 
i
j
k
x
y
z
T
T
T
dx 
dy 
dz  0
dT 
x
y
z
q   kT
 T  ds  0  q  ds  0
ds  dxiˆ  dyjˆ  dzkˆ
Steady-State One Dimensional Conduction
T = T(x) only
qx   qx
qx
x
x
x  x
steady-state
qx  qx  qx  qx  0
dT
qx   k
dx
dqx
2

 x  O  x 
qx   qx  qx 


dx
dT d  dT 
2
  x  
 k
  k

x

O



dx dx 
dx 
d  dT 
2

0
qx    k

x

O

x





dx 
dx 
or
d  dT 
k
 O  x  0


dx 
dx 
d  dT 
k
0
As x → 0,


dx  dx 
d 2T
When k = const.,
0
2
dx
RADIATION
Thermal Radiation
10-2 m
10-1 m
1m
0.4 0.7
ultra violet
visible
10 m
infrared
thermal radiation
102 m
103 m
Characteristics of Thermal Radiation
1. Independence of existence and
temperature of medium
Ex) ice lens
ice lens
black carbon paper
2. Acting at a distance
Ex) sky radiation
• electromagnetic wave or photon
• photon mean free path
• ballistic transport
diffusion
• volume or integral phenomena
conduction
• fluid: molecular random motion
• solid: lattice vibration (phonon)
free electron
diffusion or differential phenomena
as long as continuum holds
3. Spectral and Directional Dependence
• quanta • history of path
Blackbody spectral
emissive power
surface emission
Two Points of View
1. Electromagnetic wave
• Maxwell’s electromagnetic theory
• Useful for interaction between
radiation and matter
2. Photons
• Planck’s quantum theory
• Useful for the prediction of spectral
properties of absorbing, emitting
medium
Radiating Medium
• Transparent medium
ex: air
• Participating medium
emitting, absorbing and scattering
ex: CO2, H2O
• Opaque material
Stefan-Boltzmann’s law
• Blackbody: a perfect absorber
• Blackbody emissive power
Eb  qb,e   T 4 [W/m2 ]
  5.6696  108 W/m2  K 4
Stefan by experiment (1879): Eb ~ T 4
Boltzmann by theory (1884): Eb   T 4
Planck’s law
(The Theory of Heat Radiation, Max Planck, 1901)
spectral distribution of hemispherical
emissive power of a blackbody in vacuum
2 C1
E l b  5 C / lT
l e 2 1


C1  hC , C 2  hC0 / k
2
0
C0: speed of light in vacuum: 2.9979×108 m/s
h: Planck constant: 6.6260755×10-34 J•s
k: Boltzmann constant: 1.380658×10-23 J/K


El,b (W/m2.mm)
Eb 

0
E l ,b d l

0
l
5

2 C1
dl
C 2 / lT
e
1

 T4
For a real surface,
Wavelength, l (mm)
Blackbody spectral
emissive power
E   T 4
 : emissivity
Surface Radiation
Ray-tracing method vs Net-radiation method
G
J
G: irradiation [W/m2]
J: radiosity [W/m2]
J   T   G ,  : reflectivity
4
q 
diffuse-gray
surface at T
q  J  G
4

q   T   G  G
  T  1    G
4
    1,  : absorptivity
4


q



T
  G
Kirchihoff’s law   
q  J  G
J   T 4   G
q  J 

1

1

 J   T
4

 T   J  
4

 J  J   T 


1

1
4
4

T
  J
G T4
q   T 4   G
  T 4   G
   T  G 
4
q 
diffuse-gray
surface at T
J   T   G
1
G   J   T 4 
4


4

q   T   J   T 



4
4
4
  T 
J


T


T
 J



1
1
4
Ex) a body in an enclosure
q1
q1 
T2, 2, A2
T1, 1, A1

[W]

A1  1

 1

 1 A2   2

1

A1
4
4

1,
q1  1 A1 T1  T2
when
A2
Tsur
q
Ts, , , A

A1 T14  T24
4
 Tsur
q   AT  AT

4
q   A Ts4  Tsur

4
s
4
sur
Surrounding can be regarded as
a blackbody.

A1
T
4
sur
4
q1   Tsur
A1
T
4
sur
q1  q2
A2
4
q2   Tsur
A2
Why is the irradiation on the small object the
4
same as  Tsur
?
q1
q2
4
 Tsur
A1
A2
4
q1   Tsur
A1
4
q2   Tsur
A1 F12
F: view factor
F11  F12  1
F21  F22  1
F22  0  F21  1
A2
Reciprocity: A1F12  A2 F21  A2  F12 
A1
A2
4
4
4
q2   Tsur A1F12   Tsur A1
  Tsur
A2
A1
CONVECTION
energy transfer due to bulk or macroscopic motion of fluid
bulk motion: large number of molecules
moving collectively
• convection: random molecular motion
+ bulk motion
• advection: bulk motion only
U ,T
y
u
T
Ts
x
solid wall
• hydrodynamic (or velocity)
boundary layer
• thermal (or temperature)
boundary layer
at y = 0, velocity is zero: heat transfer only
by molecular random motion
U ,T
u
y
T
kf
ks
Ts


solid wall
When radiation is negligible,
nˆ
kf
ks
T 

n  
T 

n  
T 

T


qs   k f
 ks


n  
n  
 h Ts  T 
h : convection heat transfer
coefficient [W/m2.K]
Newton’s Law of Cooling
x
u ,T
qconv
Ts
qcond
qcond  qconv
u ,T
T
T
Ts
T 
qs   k f
 h  Ts  T 

n  
Ts
Convection Heat Transfer Coefficient
ks
T 
T 
h



Ts  T  n   Ts  T  n  
kf
not a property: depends on geometry and
fluid dynamics
• forced convection
• free (natural) convection
• external flow
• Internal flow
• laminar flow
• turbulent flow
ENERGY CONSERVATION
First law of thermodynamics
• control volume (open system)
• material volume (closed system)
control volume
Ein
Eg , Est
Eout
Ein  Eg  Eout  Est
In a time interval t: Ein  Eg  Eout   Est
steady-state: Est  0  Ein  Eg  Eout  0
Surface Energy Balance
Ein
sur.
Ex)
Ts
Tsur
T0

qcond,s
Ein  Eout
Eout
 f
qcond,

nˆ
 nd,s  qcond , f  qrad

qco
  qrad

 qconv
T 
T 

ks
 k f
 qrad


n  
n  

 qrad
T
4
 h Ts  T    Ts4  Tsur

Example 1.2
air
T  25 C
h  15 W/m 2  K
q
E
L
Ts = 200 C
 = 0.8
D  70mm
G
Tsur = 25 C
Find:
1) Surface emissive power E and irradiation G
2) Pipe heat loss per unit length, q 
Assumptions:
1) Steady-state conditions
2) Radiation exchange between the pipe and the room is
between a small surface in a much larger enclosure.
3) Surface emissivity = absorptivity
air
q
T  25 C
h  15 W/m 2  K
E
L
Ts = 200 C
 = 0.8
D  70mm
Tsur = 25 C
G
1. Surface emissive power and irradiation
E   Ts4  0.8(5.67  108 W/m 2  K 4 )(473 K)4  2, 270 W/m 2
4
 5.67  108 W/m 2  K 4 (298 K)4  447 W/m 2
G   Tsur
air
T  25 C
h  15 W/m 2  K
q
E
L
Ts = 200 C
 = 0.8
D  70mm
Tsur = 25 C
G
2. Heat loss from the pipe

4
qloss  qconv  qrad  hA Ts  T    A Ts4  Tsur

4
 h  DL Ts  T     DL   Ts4  Tsur
q 


A   DL
qloss
 15 W/m 2  K   0.07 m  200  25  C
L
+ 0.8   0.07m  5.67  10-8 W/m 2  K 4 4734  2984 K 4

 577 W/m  421 W/m  998 W/m

Example 1.2
air
T  25 C
h  15 W/m 2  K
q
E
L
Ts = 200 C
 = 0.8
D  70mm
Tsur = 25 C
G
qloss  qconv  qrad
Q: Why not qloss  qconv  qrad  qcond ?
Conduction does not take place ?
Ts = 200 C
T
Ts
 = 0.8
air

qcond,s
T  25 C
h  15 W/m 2  K
T
 r
qcond,f

qrad
Tsur = 25 C
  qcond,f
  qcond,s
  qrad

qloss
dT 
dT 
4
4
 ks
 k f
   Ts  Tsur 


dr  s
dr  f
4
 h  Ts  T     Ts4  Tsur

  qra
 d ,
qloss  qconv
  qconv

qcond,f
Example 1.4
Hydrogen-air Proton Exchange Membrane (PEM) fuel cell
Three-layer membrane electrode
assembly (MEA)
Anode: (exothermic) 2H2  4H+  4e 
Tc  Tsat

+
Cathode: O2  4e  4H  2H2O
 56.4 C
T  Tsur
 25 C
  0.88
P  I  Ec  15 [A]  0.6 [V]=9 [W]
2H2  O2  2H2O
Role of electrolytic membrane
1. transfer hydrogen ions
2. serve as a barrier to electron
transfer
Membrane needs a moist state
to conduct ions.
Liquid water in cathode: block
oxygen from reaching cathode
reaction site → need to control Tc
The convection heat coefficient, h
h  10.9 W  s0.8 / m 2.8  K  V 0.8
Find: The required cooling air velocity, V,
needed to maintain steady state
operation at Tc = 56.4ºC.
h  10.9 W  s0.8 / m2.8  K V 0.8
Eg  11.25 W
Assumptions:
1) Steady-state conditions
2) Negligible temperature variations
within the fuel cell
3) Large surroundings
4) Insulated edge of fuel cell
5) Negligible energy flux by the gas
or liquid flows
Energy balance on the fuel cell
Ein  Eg  Eout  Est
Eg  Eout  qconv  qrad
Eg  11.25 W  qconv  qrad
qconv  hA  Tc  T 
qconv  Eg  qrad

4
qrad   A Tc4  Tsur


4
hA  Tc  T   E g   A Tc4  Tsur
h  10.9 W  s0.8 / m2.8  K V 0.8


4
Eg   A Tc4  Tsur
A  Tc  T 
V  9.4 m/s


Example 1.5
section A-A
A
A
k
cubical cavity
ice of mass M at the fusion
temperature Tf  0 C
T1  Tf
L
W
T1  Tf
Tf  0 C
Find:
Expression for time needed to melt the ice, tm
Assumptions:
1) Inner surface of wall is at Tf through the process.
2) Constant properties
3) Steady-states, 1-D conduction through each wall
4) Conduction area of one wall = W 2 ( L  W )
section A-A
Tf
Ein  Eg  Eout  Est
k
M
Ein  Est
Est
E in
Ein  qcond  t m
T1  Tf
Est  Mhsf
L
qcond  kA
hsf : latent heat of fusion
T1  T f
L
T  Tf

2 1
 6kW
L

tm 

 k 6W

 t m  Mhsf

Mhsf L
6kW 2 T1  T f 
2

T1  T f
L
Example 1.7
Tsur  30 C
Glamp  2000 W/m 2
air
T  20 C
2  h  200 W/m 2  K
T ?
Coating to
be cured
  0.8,   0.5
k  1.2 W/ m  K
Find:
1) Cure temperature T for h  15W/m2  K
2) Effect of air flow on the cure temperature for 2  h  200 W/m2  K
Value of h for which the cure temperature is 50°C.
Assumptions:
1) Steady-state conditions
2) Negligible heat loss from back surface of plate
3) Plate is very thin and a small object in large surroundings,
coating absorptivity     0.5 w.r.t. irradiation from the
surroundings
Tsur  30 C
air
T  20 C
2  h  200 W/m 2  K

qconv
Glamp  2000 W/m 2

qrad
 Glamp
coating
  0.8,   0.5
T ?

qcond
k  1.2 W/ m  K
Ein  Eg  Eout  Est  Ein  Eout
Ein  Glamp

4
  qrad
  qcond
  h T  T    T 4  Tsur
Eout  qconv
4
 Glamp  h T  T    T 4  Tsur

h  15 W/m 2  K
T  377 K  104 C

2  h  200 W/m 2  K
h(T  50 C) = 51.0 W/m 2
Tsur  30 C
y T
air
T  20 C
2  h  200 W/m 2  K
T ?
Glamp  2000 W/m 2
Coating to
be cured
  0.8,   0.5
Tk  1.2 W/ m  K
Ein  Eout , Ein  Glamp

dT
4
4
  qrad
   k f
Eout  qcond,f
   T  Tsur 

dy  f
dT 
4
4
 h  T  T     T  Tsur 
   k s
qcond,s

dy  s
  qrad


q
conv
 0?
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