L1condCartesian2015-2
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Heat Transfer and Fluid FlowMMB801
Joanna Szmelter - Heat transfer
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CLASS TEST ARRANGEMENT 2014-2015The 45 minutes, open book, Class Test will take place on Friday in week6 13th March 2015 at 09:00 in rooms T003 (surnames beginning with A-J) and SMB014(surnames beginning with K-Z).
Students with alternative arrangements will be individually informed by e-mail about their room allocation.
The test cannot be moved. If for exceptional reasons you are unable to come to the test, please follow the impaired performance procedure.
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Heat TransferRecommended texts
Recommended reading.
C.A. Long “Essential Heat Transfer”, Longman 1999.
(Chapters 1,2 & 9)Y. Bayazitoglu, M. Nevati Ozisik, “Elements of Heat
Transfer”, McGraw-Hill Book Company, 1998. Y. A. Cengel “Heat Transfer. A Practical Approach”,
“Elements of Heat Transfer”, McGraw-Hill Book Company, 2003. (Chapters 1,2,3 &4)
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Heat Transfer in Design and Manufacturing
Maximize desired heat transfer:- heating of materials for processing - household applications (kettle, iron, etc.)- cooling of electrical components
http://www.appliancist.comhttp://www.imaging1.com/thermal/images/Nikon-P4-chip-thermal.jpg
http://www.gspsteelprofiles.com
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Heat Transfer in Design and Manufacturing
Minimize undesired heat transfer:- Maintaining a temperature difference
Thermos; Refrigerator; Windows
http://www.bbc.co.uk/schools/gcsebitesize/science/aqa_pre_2011/energy/heatrev3.shtml
http://www.appliancist.com
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Everyday Example: A thermos Flask
Energy transfer (Q)from teato surroundings
T(tea)= 80oCT(surroundings) = 20oC
Poorly insulated; cools rapidly
Tea(150 ml,
80oC)
Insulation reduces conduction (often a vacuum)
Stopper; prevents convection
Reflective surface reduces radiation(often stainless steel)
Vtea = 150 ml
How much energy is transferred?How long does it take?
ThermodynamicsHeat Transfer
Why this design? How does it work?
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What is Energy? Energy is a property of a substance
ALL substances have energy Various forms: kinetic (motion); potential; chemical
INTERNAL energy (U (J) or u (J/kg) where mass specific): Energy associated with substance’s constituent
molecules Increases with temperature
Energy is never created or destroyed, it is just changed from one form to another Or it is TRANSFERRED between substances
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What is Heat Transfer?
Definition: heat transfer is the energy transfer due to a temperature difference
Study the ability of materials and components to either transmit, absorb, or prevent the transfer of heat
A RATE of energy transfer (units: J/s = W)
Related to thermodynamics:1st law: Energy conservation Law: The energy transferred out of one object = energy transferred into another object2nd law: Entropy: heat is transferred from the hot object to the cold object
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Material Properties: Heat Capacity
Specific Heat (Cp): Amount of energy to increase the temperature of a given
mass of a substance (normally increasing 1 kg by 1 K)
Heat Capacity (C): Amount of energy to increase the temperature of a given
volume of a substance
Cp = Q/(mT) [J/kgK]
C= r*Cp [J/m3K]
Represents a substance’s capacity to store energy(Change in Specific Internal Energy =Du = CpDT)
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A Heat Capacity Example: The Mug v The Thermos
V = 150 mlTinitial = 90°CTsurr. = 20°CCp(tea) = 4.2 kJ/kgKrtea = 980 kg/m3
Find the change in internal energy of the tea after 1 hour if:
a) The tea in the mug has cooled to 25°C
b) The tea in the thermos has cooled to 85°C
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Answer: Part a): For the mug
Change in Specific Internal Energy =Du = CpDT Du = 4.2kJ/kgK * (90°C – 25°C) Du = 273kJ/kg
Change in internal energy of the tea in the mug = DU DU = m*Du = (r*V)*Du DU = 980 kg/m3 * 1.5 x 10-4 m3 * 273 kJ/kg DU = 40.1kJ
Part b): For the Thermos As above: Du = 21kJ/kg so DU = 3.1 kJ
The change in Total internal energy (DU) is the amount of heat transferred from the drink to the surroundings
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What is heat transfer?
Study of the movement of heat energy from one place to another by:– Conduction – Convection – Radiation
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Heat transfer occurs in many engineering applications and usually is a combination of the three heat transfer methods is present e.g. Engines Refrigeration and heating devicesManufacturing processesMechanical processesAnd from a small to a large scale – from solar systems to microelectronics
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Examples: Finite element solution of conduction equations
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ExamplesUrban boundary layers
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numerical models: applications to atmospheric flows
across a range of scales and physics
ΔX O(10-2) m O(102) m O(104) m O(107) m
Cloud turbulence Gravity waves Global flows Solar convection
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Heat transfer by CONDUCTION
The transfer of heat energy, within a stationary medium
FROM a high temperature region (high molecular energy)
TO a low temperature region (low molecular energy)
Heat transfer has a direction as well as magnitude
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Mechanisms of Conduction
In solids – due to collision (vibrations) and motion of free electrons. This is why good conductors of electricity are good conductors of heat
In liquids and gases – due to molecular collision and molecular diffusion
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Material Properties: Thermal Conductivity
Ability of a material to conduct heat Represented by k (units are W / m*K)
Definition: The rate at which heat is transferred through a unit thickness of a material per unit area per unit temperature difference
Related to collisions between molecules (gases, liquids) and molecular vibration (solids)
k varies between 0.026 W/mK for air and 2300 W/mK for diamond (graphite is less).
Liquids are typically in between: water ~0.61 W/mK Depends mainly on temperature but also pressure
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Thermal Conductivity in Solids
Electron flow – dominant for metals (free electrons) Lattice vibration – depends on lattice structure
TLTH Tequal
Free electrons transfer energy(move from TH to TL)
Material k (W/mK)Air 0.024Water 0.58Copper 401Nickel 91Copper/nickel alloy (55/45) 23Silver 429Diamond(depends on orientation)
900-2300
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Heat transfer by CONDUCTION
Q
Governed by FOURIER’S LAW, based on experimental observation
Where: is the rate of heat energy transfer (rate of heat flow) [W] or [J/s ] k is thermal conductivity [W/(K·m)], x is a linear distance [m]A is an area perpendicular to x [m²] T is temperature [oC or K].
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Heat transfer by CONDUCTION per unit area
q
Where: is the heat flux i.e. the rate of heat energy transfer per unit area (W/m2),
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Materials k, W/(m C) at 300 K
Copper 401
Nickel 91
Aluminium 237
Bronze 52
Steel (mild) 70
Insulation materials 0.05 – 1.0
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Conduction The transfer of energy within a substance More energetic ‘particles’ pass energy to adjacent lower-
energy ‘particles’ Applies to solids, liquids, and vapours
Rate depends on: Material (thermal conductivity, k), thickness, orientation,
temperature difference, surface area
Thigh
Tlow
xTkAQconduction D
DhighQ lowQ
lowhigh QQ lowhigh TT
xD
xTkAQconduction D
D
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Rate of Change of Internal
Energy due to Conduction
+ Rate of Energy Generation in the element
= Rate of increase of internal energy over
time dt
tTmc
tE
tEQQQ gxxx
D
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Rate of increase of internal energy over time dt
tTcxA
tTcV
tTmc
tE
D
rr
Where m ≡ mass (kg)r ≡ density of the material (kg/m3)V ≡ volume of the element (m3)c ≡ specific heat (J/kg K)T ≡ temperature (K)t ≡ time (s)
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Derivation of 1D heat conduction equation
Apply an energy balance for the rate of
energy transfer
Taylor expansion gives
.
xQx
xQx
xQx
xxxx
xxxx
xxxx
D
D
D
D
D
D
)(
)(
)(
tTcxA
tEQQQ gxxx
D
D r
...!3
)(!2
)()()()( 3
3
2
2
D
D
D
D
xx
xfxx
xfxxxfxfxxf
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Derivation of 1D heat conduction equation
xAxTk
xQQ xxx D
D )(
tTcxAQQQ gxxx
D D r
From Taylor expansion
From Fourier law
Note also that
Returning to an energy balance for the rate of energy
transfer
1D heat conduction equation
. tTcg
xTk
x
r)(
Volumetric heat generation term W/m³gg
xTkAQx
xQx
QQ xxxx D
D )(
xAgQg D
xAD
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Heat conduction equation in Cartesian coordinates
1D conduction equation
3D conduction equation
tTcg
xTk
x
r)(
tTcg
zTk
zyTk
yxTk
x
r)()()(
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1D Heat conduction equation in Cartesian coordinates – SPECIAL CASES
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Kelvin temperature conversion formulae
from Kelvin to Kelvin Celsius
[ ] = [℃ K] − 273.15
[K] = [ ] + 273.15℃
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Problems - Cartesian heat conductionExample 1.
The ends of a steel bar with thermal conductivity 60 W/K m are maintained at 100 oC and 500 oC. The length of the bar is 1 m. Find the temperature distribution.
T1 = 100 oC
T2 = 500 oC tTcg
xTk
x
r)(
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Subject to boundary conditions:
400
xT
02
2
xT
1CxT
21 CxCT
and temperature distribution:
4001001500
1000100
112
2211
CCT
CCCT
mxatCTT
mxatCTT
1500
0100
2
1
T1 = 100 oC, x=0
T2 = 500 oC
100400 xT
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x
T
0 1
100
500
slope=400
100400 xT400
xT
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Subject to boundary conditions
400
xT
02
2
xT
1CxT
21 CxCT
Temperature distribution:
4005001100
5000500
1211
2212
CCT
CCCT
mxatCTT
mxatCTT
0500
1100
2
1
T1 = 100 oC,
T2 = 500 oC x=0
Reverse the direction of the coordinate system
500400 xT