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7/28/2019 01. Introduction Overview of Mechanisms
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1. INTRODUCTION
Examples of Heat Transfer Problems
(1) Slide Projector,
(2) Ice Storage(3) Re-entry Aerodynamic Heating
(4) Pot Handle
(5) Under-window radiator of heating system
(6) Refrigerator
Focal Point in Heat Transfer
Determination of the temperature distribution in a region
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Modes of Heat Transfer
(ii) Convection: by mass motion
(iii)Radiation: by electromagnetic waves
Conduction: Fourier's LawS= wall area
L L = wall thickness
siT T1= inside surface temperature
soT T2= outside surface temperature
Is proportional toS? How?Q&
Is proportional toL ? How?x
Q&
Is proportional to (T1 - T2)? How?Q&
(i) Conduction: by molecular or atomic activity
Fig.1.1
x
xq
L
dx
siTsoT
A
0
Q&
S [m2]
T1 T2
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is called thermal conductivity, a property of material Equation (1.1) is valid for: (i) steady state, (ii) one-
dimensional conduction and (iii) constant .Reformulation of (1.1): Apply (1.1) to an element dx:
dx
T(x)dx)+T(xS=
dx
dx)+T(xT(x)S=Q
x
&
dxT(x),T1 dx),T(xT2 +
L
TTSQ 21
x
=& 1.1.[W]
Heat Transfer Rate (tepeln tok)
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(1.2) becomes
dxdTq
x=& (1.4)
For 3-D:
Equation (1.5) is known as Fourier's law
,x
Tq
x
=& ,
y
Tq
y
=&
z
Tq
z
=& (1.5)
Definition: Heat flux (mrn tepeln tok)x
q&
(1.2)dx
dTS=Q
x& [W]
S
Qq x
x
&
& = (1.3)[W/m2]
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Convection: Newton's Law of Cooling
Convection: Energy is transported by means of massmotion
Classification:
(1) Free convection
(2) Forced convection
Heat exchange between a surface and a fluid movingover it
TTq
ww&
where
wT = surface temperature
T = fluid temperature far away from the surface
wq& = surface (wall) flux [W/m2] (mrn tepeln tok)
T
sTw
Twq&
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Rewrite:
(1.6)= TTq ww
This isNewton's law of cooling.
NOTE:(1) is called heat transfer coefficient
(2) It is a defined quantity
(3) It depends on geometry, fluid properties and
motion
(4) To determine , the temperature distribution in the
fluid must be known
(5) Major objective in convection: Determination of
(souinitel pestupu tepla)
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Typical Values of Table 1.1Typical values of
5 - 30
20 - 1000
20 - 300
50 - 20,000
5,000 - 50,000
2,000 - 100,000
5,000 - 100,000
Free Convection
Gases
Liquids
Forced ConvectionGases
Liquids
Liquid metals
Phase change
Boiling liquids
Condensation
(W/m2 K)ProcessImportant:
problems.solve
tovaluestheseusenotDoonly.guide
aastablethisUse
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Radiation: Stefan-Boltzmann Law Transmission by electromagnetic waves
No medium is needed. Best in a vacuum
Maximum possible radiation: By an idealsurfacecalled blackbody.
Stefan-Boltzmann law for blackbody radiation flux:
T = surface temperature, measured in absolute degrees
(Kelvin)
=Stefan-Boltzmann constant
(1.7)4
0TE = [W]
= blackbody radiation flux (zivost)0E
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= 5.67 x 10-8
W/m2
-K4 (1.8)
Real surface:
E= radiation flux (zivost edho povrchu)
Emissivity, , a surface property defined as(1.9)
0E
E=
Combining (1.7) and (1.9)
4TE= (1.10)
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12Q& = Energy exchange between two surfaces
Absorptivity (absorptance) a : Fraction of radiationincident on a surface which is absorbed
Simplified model: Gray surface: = aSpecial case: A small gray surface enclosed by a much
larger surface
(1.11))T(TSQ 42
4
11112=&
( )1= small surface, ( )
2= large surface
Energy Exchange Between Two Bodies:
A Simplified Model
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Properties
Heat transfer depends on material and surface
properties as:
Conductivity Density
Viscosity
Specific heat Emissivity
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Example: Application of Problem SolvingMethodology
Square transistor is mounted
on a circuit board. Transistorsurface is cooled by convection.
Size, dissipated power, heat
transfer coefficient and ambienttemperature are known.
(1) Observations
(i) Schematic diagram
Determine surface
temperature.
T sT
board
transistor
1.3Fig.
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(ii) Dissipated electric energy is removed by
convection and radiation
(iii) Surface temperature is higher than ambient
temperature
(iv) Increasing dissipated power increases surface
temperature
(2) Problem DefinitionFind the relationship between power and surface
temperature
(3) Solution Plan
Apply Newton's law of cooling to the surface of the
transistor.
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(4) Plan Execution
(i) Assumptions(1) Steady state
(2) Dissipated electric energy leaves transistorsurface (no energy leaves from the back side)
(3) Negligible heat loss by radiation
(4) Uniform surface temperature
(5) Uniform heat transfer coefficient
(6) Constant ambient temperature
(ii) Analysis
Newton's law of cooling:
T sT
board
transistor
1.3Fig.
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= heat transfer coefficient = 8 K][W/m2
wq& = surface heat flux, 2W/m
wT = ? surface temperature, Co
T = ambient air temperature = 26 Co
Conservation of energy and assumptions (2) and (3):
surfacefromremovedHeatpowerDissipated =
wqSP &= (b)
S= surface area = 1 (cm) 1 (cm) = 1 = 0.0001 m2
(a)
= TTqww
&
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P= power dissipated in transistor = 0.04 W
(b) into (a)
)T(TS
Pw = (c)
Solving forw
T
S
TT
w
+=
(d)
(iii) Computations
(d) gives
C768.0,0001
0,0426
wT
o
=+=
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(iv) Checking
Dimensional check: Equation (d) is dimensionallycorrect. Each term has units of temperature.
Qualitative check: Equation (d) behaves correctly:
wT Decreasing the surface area increases
Decreasing increasesw
T
Limiting checks: Power off (P= 0): surface temperature = ambient
temperature
IncreasingPincreasesw
T
(d)S
PTT
w+=
If = 0 ( no heat is removed) then wT
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(5) Learning and Generalizing
< 85
o
CwT(i)
(ii) Examine assumptions (2) and (3):
Heat removed from the back side lowerscalculated
w
T
Radiation heat loss lowers calculatedw
T
(2) no energy leaves from the back side
(3) negligible heat loss by radiation
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)dt
dum
dt
dU(
dt
dEEEEE akakoutgin ====+&&&& [W]
Requirements for Energy Conservation
a) Balance for Energy flux
in every moment, [W].
b) Balance for amount
of heat, [J]. akoutginEEEE =+ [J]
Energy Conservation for a Control Volume
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Requirements for Energy Conservation
Energy Conservation for a Control Surface
0EEoutin
= &&