Heat Transfer Heat Transfer Convection Convection ...
Heat Transfer •
•
Thermodynamics deals with the amount of heat transfer during a process from one equilibrium state to another Heat transfer theory deals with the rate of heat transfer between bodies
Convection – Forced convection: fluid motion resulting from external forces such as pumps, fans, or the wind – Natural convection: fluid motion caused by buoyancy forces induced by density differences due to temperature differences
– There must be a temperature difference – Heat flows from high-temperature medium to coldtemperature medium – Heat transfer stops when the two bodies reach the same temperature (thermal equilibrium)
Heat Transfer Three mechanisms of heat transfer: • Conduction: energy transfer between adjacent particles
Convection – Forced convection: fluid motion resulting from external forces such as pumps, fans, or the wind
(physical contact required)
• Convection: energy transfer by virtue of fluid movement • Radiation: energy transfer by electromagnetic waves (no medium required)
Convection •
Mechanism: Transfer of heat between a bulk fluid and a surface by mass movement – From a solid surface to a moving fluid or vice-versa
Convection •
Newton’s Law of Cooling •
Qconv = hAs (Ts − T∞ )
bulk fluid temperature far from the surface
– Fluid can be moving as a result of • the application of external forces • internal imbalanced forces
convection heat transfer coefficient
surface temperature surface area
1
Radiation •
Radiation
Mechanism: Thermal energy transferred by virtue of electromagnetic waves between bodies not in contact with each other
•
Radiant heating vs convection heating
– All bodies with T > absolute zero emit thermal radiation – No media required for propagation – Maximum radiation in a vacuum
Radiation •
Radiation
Stefan-Boltzmann Equation Convection Heating
– Thermal energy emitted by a surface
Ceiling T = 10OC
•
Qemit = ε σ As Ts emissivity
surface temperature
4
Radiant Heating Ceiling T = 18OC
Room Air T = 20OC
Room Air T = 20OC
surface area Wall T = 10OC
Wall T = 20OC
Floor T = 10OC
Floor T = 25OC
Stefan-Boltzmann constant
σ = 5.67 ×10 −8
W m2K 4
Radiation •
Conduction
Stefan-Boltzmann Equation – Radiation heat transfer between two surfaces •
Qrad = ε σ As (Ts ,1 − Ts , 2 ) emissivity
4
4
surface temperatures
•
Mechanism: Transfer of heat by molecular interaction or free electron movement – Occurs between parts of the same body or between two different bodies in physical contact
surface area
Stefan-Boltzmann constant
σ = 5.67 ×10 −8
W m2K 4
2
Conduction •
Conduction in Solids
Rate of heat transfer depends on
Metals
Non-metals
– Geometry
Heat transferred by
Heat transferred by
– Thickness
vibration of molecules
– Type of material
1) vibration of molecules 2) movement and collision of free electrons
– Temperature difference
Conduction •
Thermal Conductivities
Fourier’s Law of Heat Conduction •
dT dx
•
T1 − T2 L
Qcond = −kA Qcond = kA thermal conductivity
heat transfer area
temperature gradient
Conduction
Thermal Conductivities •
Solids: k for – Metals > non-metals – Crystalline materials > amorphous materials – Pure metals > alloys
•
Liquids and gases: k for – Small molecules > larger molecules
3
?
Heat Conduction •
What mechanism of heat transfer is involved in each scenario below?
Application for a good conductor: •
Used in heat exchangers
•
Car radiators transfer heat from the coolant to the surrounding air
•
Made of brass, copper, or aluminum
•
Design maximizes rate of heat transfer: high surface area
A
B
t=0
Heat Conduction •
Heat-resistant tiles protect space shuttle from burning up during re-entry
Qcond = kA
•
Can withstand temperatures up to 23000F
Qcond =
•
Excellent insulators (poor thermal conductors)
•
Interior remains very hot
•
•
Composition of tiles: 99.8% silica – 10% amorphous fibers – 90% empty pore space
•
Low density
•
Scattering of lattice vibrations
T1 − T2 L
T1 − T2 R
R=
L kA
⎡ m oC 1 ⎤ ⎡ oC ⎤ R = ⎢m × × 2 ⎥⎢ ⎥ W m ⎦⎣ W ⎦ ⎣
Heat Conduction •
t = 30 min
Single-layer flat wall
•
Dissipate surface heat very quickly
B
Thermal Resistance
Application for a poor conductor:
•
A
Thermal Resistance •
R-factor – Manufacturers provide R-factors for various building materials
R=
L kA
R − factor =
L k
4
Thermal Resistance
Thermal Resistance •
•
Multi-layer flat wall (in series) •
•
Q A = QB = Q
Steady state
Multi-layer flat wall (in series) – Conduction resistance of each layer is connected in series
•
Wall A
• ⎡ L L ⎤ T1 − T3 = Q ⎢ A + B ⎥ k A k B A⎦ ⎣ A
Wall B
Wall A
Wall B
RA
RB
Analogy: Conservation of Mass •
•
•
•
mtotal = m1 = m2 = m3 = m4
•
Q = 1
2
3
4 RA
RB
T −T T1 − T3 = 1 3 ⎡ LA LB ⎤ [RA + RB ] ⎢ k A + k A⎥ B ⎦ ⎣ A
Thermal Resistance •
RB
RA
•
Thermal Resistance
Multi-layer flat wall (in series)
•
– Conduction resistance of each layer is connected in series
Multi-layer flat wall (in series) – Conduction resistance of each layer is connected in series – For a series of N layers in a flat wall
T −T QA = k A A 1 2 LA
T1 − T2 =
LA • QA kA A
T2 − T3 LB
T2 − T3 =
LB • QB kB A
•
•
QB = k B A
Wall A
Wall B
•
Q = •
Q = RA
Thermal Resistance •
Multi-layer flat wall (in series)
•
i
TH − TL n Li ∑ k i =1 i A
•
•
•
Q = QA + QB A
Wall A
Wall B
Analogy: Conservation of Mass •
•
•
•
B
•
mtotal = m1 + m2 + m3 + m4
L • L • T1 − T3 = A QA + B QB kA A kB A • ⎡ L L ⎤ T1 − T3 = Q ⎢ A + B ⎥ ⎣ k A A kB A ⎦
∑R
=
Multi-layer flat wall (in parallel)
T1 − T3 = (T1 − T2 ) + (T2 − T3 ) L • T2 − T3 = B QB kB A
n
Thermal Resistance
– Conduction resistance of each layer is connected in series
L • T1 − T2 = A QA kA A
TH − TL i =1
RB
TH − TL Req
RA
RB
1 2 3 4
5
Thermal Resistance •
Multi-layer flat wall (in parallel)
Thermal Resistance •
Multi-layer flat wall (in parallel)
– Conduction resistance of each layer is connected in parallel
– Conduction resistance of each layer is connected in parallel – For a series of N parallel layers in a flat wall A
•
Q A = k A A1
T1 − T2 LA
•
Q =
B
•
n
i =1
Thermal Resistance •
Thermal Resistance
Multi-layer flat wall (in parallel) •
– Conduction resistance of each layer is connected in parallel •
•
•
Q = Q A + QB •
Q=
Heat transfer through layers in SERIES •
Q =
A B
k A A1 (T1 − T2 ) + k B A2 (T1 − T2 ) LB LA
•
TH − TL Req
•
Thermal Resistance
Req = R1 + R2 + ... + RN
Heat transfer through layers in PARALLEL
Q =
• ⎡k A k A ⎤ Q = ⎢ A 1 + B 2 ⎥ (T1 − T2 ) LB ⎦ ⎣ LA
•
n kA 1 = (TH − TL ) ∑ i Ri i =1 Li
Q = (TH − TL ) ∑
T −T QB = k B A2 1 2 LB •
TH − TL Req
TH − TL Req
1 1 1 1 = + + ... + Req R1 R2 RN
Thermal Resistance
Multi-layer flat wall (in parallel) •
– Conduction resistance of each layer is connected in parallel •
•
•
Composite networks – Combined seriesparallel arrangement
Q = QA + QB A • ⎡k A k A ⎤ Q = ⎢ A 1 + B 2 ⎥ (T1 − T2 ) LB ⎦ ⎣ LA
B
• ⎡ 1 1 ⎤ Q = ⎢ + ⎥ (T1 − T2 ) R R B⎦ ⎣ A
6
Steady Heat Conduction in Flat Walls •
Thermal Resistance Concept
Thermal Contact Resistance •
In the previous analysis, we assumed perfect contact at the interface between each layer
•
BUT in reality, all surfaces are microscopically rough
– May also be applied to convection •
Qconv = hAs (Ts − T∞ ) •
Qconv =
(Ts − T∞ ) Rconv
Rconv =
1 hAs
Steady Heat Conduction in Flat Walls •
Thermal Resistance Concept
Thermal Contact Resistance •
– Convective heat transfer viewed as another “layer” in series
Thermal resistance at an interface can be viewed as an additional layer of resistance to heat transfer
Req = ∑ Ri =
Steady Heat Conduction in Flat Walls •
L1 ⎛⎜ Leq, interface ⎞⎟ L2 + + k1 A ⎜⎝ keq, interface A ⎟⎠ k 2 A
Thermal Contact Resistance
Thermal Resistance Concept
Req = ∑ Ri =
– Various “layers” in series: convection and conduction
•
L1 ⎛⎜ Leq, interface ⎞⎟ L2 + + k1 A ⎜⎝ keq, interface A ⎟⎠ k 2 A
Thermal contact resistance – Is only significant for good heat conductors (metals)
•
Q =
Req = ∑ Ri =
T∞ ,1 − T∞ , 2 Req
L L 1 1 + 1 + 2 + h1 A k1 A k 2 A h2 A
– Can be ignored for poor heat conductors (insulators)
•
How to minimize – Apply a thermal grease between layers – Replace air at interface with a better conducting gas – Insert a soft metallic foil at interface
7
Steady Heat Conduction in Cylinders •
Fourier’s Law of Heat Conduction
Steady Heat Conduction in Cylinders •
– Expressed in cylindrical coordinates
•
•
Qcond
•
Qcond
2πLk T1 − T2 = r ln⎛⎜ 2 ⎞⎟ ⎝ r1 ⎠ •
Qcond =
Rcyl
Q =
ln⎛⎜ r2 ⎞⎟ r 1 1 + ⎝ 1⎠+ h1 A1 2πLk1 h2 A2
Req = ∑ Ri =
1
h1 (2π r1 L )
+
ln⎛⎜ r2 ⎞⎟ 1 ⎝ r1 ⎠ + 2πLk1 h2 (2π r2 L )
Steady Heat Conduction in Cylinders •
Multilayer hollow cylinder – Convection + conduction
r ln⎛⎜ 2 ⎞⎟ r1 ⎠ ⎝ = 2πLk
Cylinder subjected to convection at the inside and outside surfaces
•
Req
T1 − T2 Rcyl
Steady Heat Conduction in Cylinders •
T∞ ,1 − T∞ , 2
Req = ∑ Ri =
2πLk (T1 − T2 ) r ln⎛⎜ 2 ⎞⎟ ⎝ r1 ⎠
Steady Heat Conduction in Cylinders •
Q =
dT = −kA(r ) dr
Area normal to heat flow = circumference x length
Qcond =
Cylinder subjected to convection at the inside and outside surfaces
T∞ ,1 − T∞ , 2
Steady Heat Conduction in Cylinders •
Multilayer hollow cylinder
•
Q =
Req
T∞ ,1 − T∞ , 2 Req
r r ln⎛⎜ r2 ⎞⎟ ln⎛⎜ 3 ⎞⎟ ln⎛⎜ 4 ⎞⎟ r2 ⎠ r3 ⎠ r1 ⎠ 1 1 ⎝ ⎝ ⎝ + + + + Req = ∑ Ri = 2πLk1 2πLk 2 2πLk3 h1 A1 h2 A2 Based on inner area
Based on outer area
8
Critical Radius of Insulation •
Effect of adding insulation to a wall
Case Study: Enertia® House •
– Heat transfer area constant
T∞ ,1 − T∞ , 2
•
Q =
Req
Req = ∑ Ri =
1 1 L L + 1 + 2 + h1 A k1 A k 2 A h2 A
Critical Radius of Insulation •
Effect of adding insulation to a cylindrical pipe – 1. 2. –
Case Study: Enertia® House •
Two contradictory effects: Thermal conduction resistance increases Outer surface area for convection also increases Heat transfer rate may increase or decrease, depending on which effect dominates •
Q = •
Q=
Construction Material: Solid Wood – Solid wood house 5 times more massive than a stickframed house • Thermal inertia: the tendency of massive objects to maintain their temperature – Made of Southern yellow pine: highly resinous
T1 − T∞ Rins + Rconv
T1 − T∞ r ⎛ ln⎜ 2 ⎞⎟ 1 ⎝ r1 ⎠ + 2πLk1 h( 2π r2 L)
Critical Radius of Insulation •
Energy Efficient Design – Construction material – Outer envelope
– Always reduces heat transfer
Effect of adding insulation to a cylindrical pipe – Variation of heat transfer rate with outer radius of the insulation (r2)
rcr ,cylinder =
Case Study: Enertia® House •
Latent Heat of Fusion
hif = h f − hi
– Temperature remains constant during phase change
k h
9
Case Study: Enertia® House •
Southern yellow pine
•
Resin in the wood undergoes a phase change around 21oC Heat from sun melts resin into a liquid Resin stores energy of sun as latent heat of fusion Cooler temperatures cause resin to crystallize Latent heat of fusion released Temperature of wood remains constant at ~ 21oC during phase change Solid wood walls store energy during the day and release it during the night
• • • • • •
Case Study: Enertia® House •
Case Study: Enertia® House •
Envelope – Akin to the house’s “atmosphere” – Acts as a convection loop – Harnesses geothermal energy in subsurface
Envelope in Summer – Sun high – Roof well-insulated – Sun that does enter windows causes convection loop pulling cool air from basement
Case Study: Enertia® House •
Energy Efficient Design – Living space not in contact with outdoor temperatures, only air in envelope – Thermal inertia: small ΔT – R-factor becomes meaningless •
Qcond =
T1 − T2 R
– Heat not stored in air, but in walls – Can ventilate house without losing too much heat
Case Study: Enertia® House •
Envelope in Winter – Sun low – South-facing windows capture heat • Stored in walls during day • Released at night
10
•
Thermodynamics deals with the amount of heat transfer during a process from one equilibrium state to another Heat transfer theory deals with the rate of heat transfer between bodies
Convection – Forced convection: fluid motion resulting from external forces such as pumps, fans, or the wind – Natural convection: fluid motion caused by buoyancy forces induced by density differences due to temperature differences
– There must be a temperature difference – Heat flows from high-temperature medium to coldtemperature medium – Heat transfer stops when the two bodies reach the same temperature (thermal equilibrium)
Heat Transfer Three mechanisms of heat transfer: • Conduction: energy transfer between adjacent particles
Convection – Forced convection: fluid motion resulting from external forces such as pumps, fans, or the wind
(physical contact required)
• Convection: energy transfer by virtue of fluid movement • Radiation: energy transfer by electromagnetic waves (no medium required)
Convection •
Mechanism: Transfer of heat between a bulk fluid and a surface by mass movement – From a solid surface to a moving fluid or vice-versa
Convection •
Newton’s Law of Cooling •
Qconv = hAs (Ts − T∞ )
bulk fluid temperature far from the surface
– Fluid can be moving as a result of • the application of external forces • internal imbalanced forces
convection heat transfer coefficient
surface temperature surface area
1
Radiation •
Radiation
Mechanism: Thermal energy transferred by virtue of electromagnetic waves between bodies not in contact with each other
•
Radiant heating vs convection heating
– All bodies with T > absolute zero emit thermal radiation – No media required for propagation – Maximum radiation in a vacuum
Radiation •
Radiation
Stefan-Boltzmann Equation Convection Heating
– Thermal energy emitted by a surface
Ceiling T = 10OC
•
Qemit = ε σ As Ts emissivity
surface temperature
4
Radiant Heating Ceiling T = 18OC
Room Air T = 20OC
Room Air T = 20OC
surface area Wall T = 10OC
Wall T = 20OC
Floor T = 10OC
Floor T = 25OC
Stefan-Boltzmann constant
σ = 5.67 ×10 −8
W m2K 4
Radiation •
Conduction
Stefan-Boltzmann Equation – Radiation heat transfer between two surfaces •
Qrad = ε σ As (Ts ,1 − Ts , 2 ) emissivity
4
4
surface temperatures
•
Mechanism: Transfer of heat by molecular interaction or free electron movement – Occurs between parts of the same body or between two different bodies in physical contact
surface area
Stefan-Boltzmann constant
σ = 5.67 ×10 −8
W m2K 4
2
Conduction •
Conduction in Solids
Rate of heat transfer depends on
Metals
Non-metals
– Geometry
Heat transferred by
Heat transferred by
– Thickness
vibration of molecules
– Type of material
1) vibration of molecules 2) movement and collision of free electrons
– Temperature difference
Conduction •
Thermal Conductivities
Fourier’s Law of Heat Conduction •
dT dx
•
T1 − T2 L
Qcond = −kA Qcond = kA thermal conductivity
heat transfer area
temperature gradient
Conduction
Thermal Conductivities •
Solids: k for – Metals > non-metals – Crystalline materials > amorphous materials – Pure metals > alloys
•
Liquids and gases: k for – Small molecules > larger molecules
3
?
Heat Conduction •
What mechanism of heat transfer is involved in each scenario below?
Application for a good conductor: •
Used in heat exchangers
•
Car radiators transfer heat from the coolant to the surrounding air
•
Made of brass, copper, or aluminum
•
Design maximizes rate of heat transfer: high surface area
A
B
t=0
Heat Conduction •
Heat-resistant tiles protect space shuttle from burning up during re-entry
Qcond = kA
•
Can withstand temperatures up to 23000F
Qcond =
•
Excellent insulators (poor thermal conductors)
•
Interior remains very hot
•
•
Composition of tiles: 99.8% silica – 10% amorphous fibers – 90% empty pore space
•
Low density
•
Scattering of lattice vibrations
T1 − T2 L
T1 − T2 R
R=
L kA
⎡ m oC 1 ⎤ ⎡ oC ⎤ R = ⎢m × × 2 ⎥⎢ ⎥ W m ⎦⎣ W ⎦ ⎣
Heat Conduction •
t = 30 min
Single-layer flat wall
•
Dissipate surface heat very quickly
B
Thermal Resistance
Application for a poor conductor:
•
A
Thermal Resistance •
R-factor – Manufacturers provide R-factors for various building materials
R=
L kA
R − factor =
L k
4
Thermal Resistance
Thermal Resistance •
•
Multi-layer flat wall (in series) •
•
Q A = QB = Q
Steady state
Multi-layer flat wall (in series) – Conduction resistance of each layer is connected in series
•
Wall A
• ⎡ L L ⎤ T1 − T3 = Q ⎢ A + B ⎥ k A k B A⎦ ⎣ A
Wall B
Wall A
Wall B
RA
RB
Analogy: Conservation of Mass •
•
•
•
mtotal = m1 = m2 = m3 = m4
•
Q = 1
2
3
4 RA
RB
T −T T1 − T3 = 1 3 ⎡ LA LB ⎤ [RA + RB ] ⎢ k A + k A⎥ B ⎦ ⎣ A
Thermal Resistance •
RB
RA
•
Thermal Resistance
Multi-layer flat wall (in series)
•
– Conduction resistance of each layer is connected in series
Multi-layer flat wall (in series) – Conduction resistance of each layer is connected in series – For a series of N layers in a flat wall
T −T QA = k A A 1 2 LA
T1 − T2 =
LA • QA kA A
T2 − T3 LB
T2 − T3 =
LB • QB kB A
•
•
QB = k B A
Wall A
Wall B
•
Q = •
Q = RA
Thermal Resistance •
Multi-layer flat wall (in series)
•
i
TH − TL n Li ∑ k i =1 i A
•
•
•
Q = QA + QB A
Wall A
Wall B
Analogy: Conservation of Mass •
•
•
•
B
•
mtotal = m1 + m2 + m3 + m4
L • L • T1 − T3 = A QA + B QB kA A kB A • ⎡ L L ⎤ T1 − T3 = Q ⎢ A + B ⎥ ⎣ k A A kB A ⎦
∑R
=
Multi-layer flat wall (in parallel)
T1 − T3 = (T1 − T2 ) + (T2 − T3 ) L • T2 − T3 = B QB kB A
n
Thermal Resistance
– Conduction resistance of each layer is connected in series
L • T1 − T2 = A QA kA A
TH − TL i =1
RB
TH − TL Req
RA
RB
1 2 3 4
5
Thermal Resistance •
Multi-layer flat wall (in parallel)
Thermal Resistance •
Multi-layer flat wall (in parallel)
– Conduction resistance of each layer is connected in parallel
– Conduction resistance of each layer is connected in parallel – For a series of N parallel layers in a flat wall A
•
Q A = k A A1
T1 − T2 LA
•
Q =
B
•
n
i =1
Thermal Resistance •
Thermal Resistance
Multi-layer flat wall (in parallel) •
– Conduction resistance of each layer is connected in parallel •
•
•
Q = Q A + QB •
Q=
Heat transfer through layers in SERIES •
Q =
A B
k A A1 (T1 − T2 ) + k B A2 (T1 − T2 ) LB LA
•
TH − TL Req
•
Thermal Resistance
Req = R1 + R2 + ... + RN
Heat transfer through layers in PARALLEL
Q =
• ⎡k A k A ⎤ Q = ⎢ A 1 + B 2 ⎥ (T1 − T2 ) LB ⎦ ⎣ LA
•
n kA 1 = (TH − TL ) ∑ i Ri i =1 Li
Q = (TH − TL ) ∑
T −T QB = k B A2 1 2 LB •
TH − TL Req
TH − TL Req
1 1 1 1 = + + ... + Req R1 R2 RN
Thermal Resistance
Multi-layer flat wall (in parallel) •
– Conduction resistance of each layer is connected in parallel •
•
•
Composite networks – Combined seriesparallel arrangement
Q = QA + QB A • ⎡k A k A ⎤ Q = ⎢ A 1 + B 2 ⎥ (T1 − T2 ) LB ⎦ ⎣ LA
B
• ⎡ 1 1 ⎤ Q = ⎢ + ⎥ (T1 − T2 ) R R B⎦ ⎣ A
6
Steady Heat Conduction in Flat Walls •
Thermal Resistance Concept
Thermal Contact Resistance •
In the previous analysis, we assumed perfect contact at the interface between each layer
•
BUT in reality, all surfaces are microscopically rough
– May also be applied to convection •
Qconv = hAs (Ts − T∞ ) •
Qconv =
(Ts − T∞ ) Rconv
Rconv =
1 hAs
Steady Heat Conduction in Flat Walls •
Thermal Resistance Concept
Thermal Contact Resistance •
– Convective heat transfer viewed as another “layer” in series
Thermal resistance at an interface can be viewed as an additional layer of resistance to heat transfer
Req = ∑ Ri =
Steady Heat Conduction in Flat Walls •
L1 ⎛⎜ Leq, interface ⎞⎟ L2 + + k1 A ⎜⎝ keq, interface A ⎟⎠ k 2 A
Thermal Contact Resistance
Thermal Resistance Concept
Req = ∑ Ri =
– Various “layers” in series: convection and conduction
•
L1 ⎛⎜ Leq, interface ⎞⎟ L2 + + k1 A ⎜⎝ keq, interface A ⎟⎠ k 2 A
Thermal contact resistance – Is only significant for good heat conductors (metals)
•
Q =
Req = ∑ Ri =
T∞ ,1 − T∞ , 2 Req
L L 1 1 + 1 + 2 + h1 A k1 A k 2 A h2 A
– Can be ignored for poor heat conductors (insulators)
•
How to minimize – Apply a thermal grease between layers – Replace air at interface with a better conducting gas – Insert a soft metallic foil at interface
7
Steady Heat Conduction in Cylinders •
Fourier’s Law of Heat Conduction
Steady Heat Conduction in Cylinders •
– Expressed in cylindrical coordinates
•
•
Qcond
•
Qcond
2πLk T1 − T2 = r ln⎛⎜ 2 ⎞⎟ ⎝ r1 ⎠ •
Qcond =
Rcyl
Q =
ln⎛⎜ r2 ⎞⎟ r 1 1 + ⎝ 1⎠+ h1 A1 2πLk1 h2 A2
Req = ∑ Ri =
1
h1 (2π r1 L )
+
ln⎛⎜ r2 ⎞⎟ 1 ⎝ r1 ⎠ + 2πLk1 h2 (2π r2 L )
Steady Heat Conduction in Cylinders •
Multilayer hollow cylinder – Convection + conduction
r ln⎛⎜ 2 ⎞⎟ r1 ⎠ ⎝ = 2πLk
Cylinder subjected to convection at the inside and outside surfaces
•
Req
T1 − T2 Rcyl
Steady Heat Conduction in Cylinders •
T∞ ,1 − T∞ , 2
Req = ∑ Ri =
2πLk (T1 − T2 ) r ln⎛⎜ 2 ⎞⎟ ⎝ r1 ⎠
Steady Heat Conduction in Cylinders •
Q =
dT = −kA(r ) dr
Area normal to heat flow = circumference x length
Qcond =
Cylinder subjected to convection at the inside and outside surfaces
T∞ ,1 − T∞ , 2
Steady Heat Conduction in Cylinders •
Multilayer hollow cylinder
•
Q =
Req
T∞ ,1 − T∞ , 2 Req
r r ln⎛⎜ r2 ⎞⎟ ln⎛⎜ 3 ⎞⎟ ln⎛⎜ 4 ⎞⎟ r2 ⎠ r3 ⎠ r1 ⎠ 1 1 ⎝ ⎝ ⎝ + + + + Req = ∑ Ri = 2πLk1 2πLk 2 2πLk3 h1 A1 h2 A2 Based on inner area
Based on outer area
8
Critical Radius of Insulation •
Effect of adding insulation to a wall
Case Study: Enertia® House •
– Heat transfer area constant
T∞ ,1 − T∞ , 2
•
Q =
Req
Req = ∑ Ri =
1 1 L L + 1 + 2 + h1 A k1 A k 2 A h2 A
Critical Radius of Insulation •
Effect of adding insulation to a cylindrical pipe – 1. 2. –
Case Study: Enertia® House •
Two contradictory effects: Thermal conduction resistance increases Outer surface area for convection also increases Heat transfer rate may increase or decrease, depending on which effect dominates •
Q = •
Q=
Construction Material: Solid Wood – Solid wood house 5 times more massive than a stickframed house • Thermal inertia: the tendency of massive objects to maintain their temperature – Made of Southern yellow pine: highly resinous
T1 − T∞ Rins + Rconv
T1 − T∞ r ⎛ ln⎜ 2 ⎞⎟ 1 ⎝ r1 ⎠ + 2πLk1 h( 2π r2 L)
Critical Radius of Insulation •
Energy Efficient Design – Construction material – Outer envelope
– Always reduces heat transfer
Effect of adding insulation to a cylindrical pipe – Variation of heat transfer rate with outer radius of the insulation (r2)
rcr ,cylinder =
Case Study: Enertia® House •
Latent Heat of Fusion
hif = h f − hi
– Temperature remains constant during phase change
k h
9
Case Study: Enertia® House •
Southern yellow pine
•
Resin in the wood undergoes a phase change around 21oC Heat from sun melts resin into a liquid Resin stores energy of sun as latent heat of fusion Cooler temperatures cause resin to crystallize Latent heat of fusion released Temperature of wood remains constant at ~ 21oC during phase change Solid wood walls store energy during the day and release it during the night
• • • • • •
Case Study: Enertia® House •
Case Study: Enertia® House •
Envelope – Akin to the house’s “atmosphere” – Acts as a convection loop – Harnesses geothermal energy in subsurface
Envelope in Summer – Sun high – Roof well-insulated – Sun that does enter windows causes convection loop pulling cool air from basement
Case Study: Enertia® House •
Energy Efficient Design – Living space not in contact with outdoor temperatures, only air in envelope – Thermal inertia: small ΔT – R-factor becomes meaningless •
Qcond =
T1 − T2 R
– Heat not stored in air, but in walls – Can ventilate house without losing too much heat
Case Study: Enertia® House •
Envelope in Winter – Sun low – South-facing windows capture heat • Stored in walls during day • Released at night
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