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Chapter 18. Heat Chapter 18. Heat TransferTransfer
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics
Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007
TRANSFER OF HEAT is minimized by multiple layers of beta cloth. These and other insulating materials protect spacecraft from hostile environmental conditions. (NASA)
Objectives: After finishing Objectives: After finishing this unit, you should be this unit, you should be able to:able to:
• Demonstrate your understanding Demonstrate your understanding of of conductionconduction, , convectionconvection, and, and radiationradiation, and give examples., and give examples.
• Solve Solve thermal conductivitythermal conductivity problems based on quantity of problems based on quantity of heat, length of path, temperature, heat, length of path, temperature, area, and time.area, and time.
• Solve problems involving the Solve problems involving the rate rate of radiationof radiation and and emissivityemissivity of of surfaces.surfaces.
Heat Transfer by Heat Transfer by ConductionConduction
Conduction is the process by which heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move.
Conduction Direction
From hot to cold.
Heat Transfer by Heat Transfer by ConvectionConvection
Convection is the process by which heat energy is transferred by the actual mass motion of a heated fluid.
ConvectionHeated fluid rises and is then Heated fluid rises and is then replaced by cooler fluid, producing replaced by cooler fluid, producing convection currentsconvection currents..
Convection is significantly affected Convection is significantly affected by by geometry geometry of heated surfaces. (wall, of heated surfaces. (wall, ceiling, floor)ceiling, floor)
Heat Transfer by Heat Transfer by RadiationRadiation
Radiation
Sun
Radiation is the process by which heat energy is transferred by electromagnetic waves.
Atomic
No medium is required !No medium is required !
Kinds of Heat TransferKinds of Heat Transfer
Consider the operation of a typical coffee Consider the operation of a typical coffee maker:maker:
Think about how heat is Think about how heat is transferred by:transferred by:
ConductionConduction??ConvectioConvection?n?RadiationRadiation??
Heat CurrentHeat Current
SteamSteam IceIce
( / )Q
H J s
The The heat currentheat current HH is defined as the is defined as the quantity of heat quantity of heat QQ transferred per unit of transferred per unit of time time in the direction from high in the direction from high temperature to low temperature.temperature to low temperature.
Typical units are: Typical units are: J/s, cal/s, and Btu/hJ/s, cal/s, and Btu/h
H = Heat current (J/s)
A = Surface area (m2)t = Temperature differenceL = Thickness of material
Thermal ConductivityThermal Conductivity
t1 t2
t = t2 - t1
The The thermal conductivity thermal conductivity kk of a material is a of a material is a measure of its ability to measure of its ability to conduct heat.conduct heat.
QLk
A t
QLk
A t
Q kA t
HL
Q kA t
HL
0
J
s m CUnits
The SI Units for The SI Units for ConductivityConductivity
HotHot ColdCold QLk
A t
QLk
A t
For Copper: k = 385 J/s m C0 For Copper: k = 385 J/s m C0
Taken literally, this means that for a Taken literally, this means that for a 1-m1-m length of copper whose cross section is length of copper whose cross section is 1 1 mm22 and whose end points differ in and whose end points differ in temperature by temperature by 1 C1 C00, heat will be , heat will be conducted at the rate of conducted at the rate of 1 J/s1 J/s..
In In SI unitsSI units, typically small measures for , typically small measures for lengthlength LL and and area Aarea A must be converted to must be converted to meters and square meters, respectively, meters and square meters, respectively, before substitution into formulas.before substitution into formulas.
In In SI unitsSI units, typically small measures for , typically small measures for lengthlength LL and and area Aarea A must be converted to must be converted to meters and square meters, respectively, meters and square meters, respectively, before substitution into formulas.before substitution into formulas.
Older Units for ConductivityOlder Units for Conductivity
Taken literally, this means that for a Taken literally, this means that for a 1-in.1-in. thick plate of glass whose area is thick plate of glass whose area is 1 ft1 ft22 and and whose sides differ in temperature by whose sides differ in temperature by 1 F1 F00, , heat will be conducted at the rate of heat will be conducted at the rate of 5.6 5.6 Btu/hBtu/h..
t = 1 F0
L = 1 in.
A=1 ft2
Q=1 Btu
h
Older units, still active, use Older units, still active, use common measurements for common measurements for area in area in ftft22 time in time in hourshours, , length in length in secondsseconds, and , and quantity of heat in quantity of heat in Btu’sBtu’s..
Glass k = 5.6 Btu in./ftGlass k = 5.6 Btu in./ft22h Fh F00
Thermal ConductivitiesThermal ConductivitiesExamples of the two systems of units used for thermal
conductivities of materials are given below:
Copper:Copper:
Concrete or Concrete or Glass:Glass:
Corkboard:Corkboard:
385385 26602660
0.8000.800 5.65.6
0.0400.040 0.300.30
MaterialMaterialoJ/s m C 2 0Btu in/ft h F
Examples of Thermal Examples of Thermal ConductivityConductivity
Aluminum:Aluminum:
Comparison of Heat Currents for Similar Conditions: L = 1 cm (0.39 in.); A = 1 m2 (10.8 ft2); t = 100
C0
Copper:Copper:
Concrete or Concrete or Glass:Glass:
Corkboard:Corkboard:
2050 kJ/s2050 kJ/s 4980 Btu/h4980 Btu/h
3850 kJ/s3850 kJ/s 9360 Btu/h9360 Btu/h
8.00 kJ/s8.00 kJ/s 19.4 Btu/h19.4 Btu/h
0.400 kJ/s0.400 kJ/s 9.72 Btu/h9.72 Btu/h
Example 1:Example 1: A large glass window A large glass window measures measures 2 m2 m wide and wide and 6 m6 m high. The high. The inside surface is at inside surface is at 202000CC and the and the outside surface is at outside surface is at 121200CC. How many . How many joules of heat pass through this window joules of heat pass through this window in in one hourone hour? Assume ? Assume L =L = 1.5 cm1.5 cm and and that that k = 0.8 J/s m Ck = 0.8 J/s m C00. .
200C 120C
t = t2 - t1 = 8 C0
0.015 m
AQ = ?
= 1 h
A = (2 m)(6 m) = 12 mA = (2 m)(6 m) = 12 m22
; Q kA t kA t
H QL L
0 2 0(0.8 J/m s C )(12 m )(8 C )(3600 s)
0.0150 mQ
Q = 18.4 MJQ = 18.4 MJ
Example 2:Example 2: The wall of a freezing The wall of a freezing plant is composed of plant is composed of 8 cm8 cm of of corkboard and corkboard and 12 cm12 cm of solid of solid concrete. The inside surface is at concrete. The inside surface is at --202000CC and the outside surface is and the outside surface is +25+2500CC. What is the interface . What is the interface temperature temperature ttii?? ttii 252500CC-20-2000CC
HHAA
8 cm 12 cm8 cm 12 cm
SteadSteady y
FlowFlow
Note:Note:Cork Concrete
H H
A A
0 01 2
1 2
( 20 C) 25 C -
L Li ik t k t
0 01 2
1 2
( 20 C) (25 C - )
L Li ik t k t
Example 2 (Cont.):Example 2 (Cont.): Finding the Finding the interface temperature for a composite interface temperature for a composite wall.wall.
ttii 252500CC-20-2000CC
HHAA
8 cm 12 cm8 cm 12 cm
SteadSteady y
FlowFlow
0 01 2
1 2
( 20 C) (25 C - )
L Li ik t k t
Rearranging factors gives:Rearranging factors gives:
0 01 2
2 1
L( 20 C) (25 C - )
L i i
kt t
k
01 2
02 1
L (0.04 W/m C )(0.12 m)0.075
L (0.8 W/m C )(0.08 m)
k
k
Example 2 (Cont.):Example 2 (Cont.): Simplifying, we Simplifying, we obtain:obtain:
ttii 252500CC-20-2000CC
HHAA
8 cm 12 cm8 cm 12 cm
SteadSteady y
FlowFlow
0 0(0.075)( 20 C) (25 C - )i it t
0.0750.075ttii + + 1.51.500C = 25C = 2500C - C - ttii
From which:From which: ti = 21.90Cti = 21.90C
Knowing the interface temperature Knowing the interface temperature ttii allows us to determine the allows us to determine the rate of rate of
heat flowheat flow per unit of area, H/Aper unit of area, H/A..
The quantity The quantity H/AH/A is same for cork or concrete: is same for cork or concrete:
H;
A
Q kA t k tH
L L
H
; A
Q kA t k tH
L L
Example 2 (Cont.):Example 2 (Cont.): Constant steady state Constant steady state flow.flow.
ttii 252500CC-20-2000CC
HHAA
8 cm 12 cm8 cm 12 cm
SteadSteady y
FlowFlowH
; A
Q kA t k tH
L L
H
; A
Q kA t k tH
L L
Over time Over time H/AH/A is constant so is constant so different different k’sk’s cause different cause different t’st’s
Cork:Cork: t = 21.9t = 21.900C - (-20C - (-2000C) = C) = 41.9 C41.9 C00
Concrete:Concrete: t = 25t = 2500C - 21.9C - 21.900C = C = 3.1 3.1 CC00
Since H/A is the same, let’s just choose concrete alone:Since H/A is the same, let’s just choose concrete alone:
0 0H (0.8 W/mC )(3.1 C )
A 0.12 m
k t
L
2 20.7 W/m
H
A
2 20.7 W/mH
A
Example 2 (Cont.):Example 2 (Cont.): Constant steady state Constant steady state flow.flow.
ttii 252500CC-20-2000CC
HHAA
8 cm 12 cm8 cm 12 cm
SteadSteady y
FlowFlow
Cork:Cork: t = 21.9t = 21.900C - (-20C - (-2000C) = C) = 41.9 C41.9 C00
Concrete:Concrete: t = 25t = 2500C - 21.9C - 21.900C = C = 3.1 3.1 CC00
2 20.7 W/mH
A
2 20.7 W/mH
A
Note that Note that 20.7 Joules20.7 Joules of heat per of heat per secondsecond pass through the composite pass through the composite wall. However, the temperature wall. However, the temperature interval between the faces of the interval between the faces of the cork is cork is 13.5 times13.5 times as large as for the as large as for the concrete faces.concrete faces.
If A = 10 m2, the heat flow in 1 h would be ______?
If A = 10 m2, the heat flow in 1 h would be ______?745 kW745 kW
RadiationRadiationThe The rate of radiationrate of radiation R R is the energy is the energy emitted per unit area per unit time (power emitted per unit area per unit time (power per unit area).per unit area).
Q PR
A A
Q PR
A A Rate of RadiationRate of Radiation
(W/m(W/m22):):
Emissivity, e : 0 > e > 1Emissivity, e : 0 > e > 1
Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4
Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4
4PR e T
A
4PR e T
A
Example 3:Example 3: A spherical A spherical surface surface 12 cm12 cm in radius is in radius is heated to heated to 62762700CC. The . The emissivity is emissivity is 0.120.12. What . What power is radiated?power is radiated?
2 24 4 (0.12 m)A R
A = 0.181 mA = 0.181 m22
T = 627 + 273; T = 627 + 273; T = T = 900 K900 K
4P e AT 4P e AT-8 4 2 4(0.12)(5.67 x 10 W/mK )(0.181 m )(900 K)P
P = 808 WP = 808 WPower Radiated from Surface:Power Radiated from Surface:
A
6270C
Find Power Radiated
Summary: Heat TransferSummary: Heat Transfer
ConvectionConvection is the process is the process by which heat energy is by which heat energy is transferred by the actual transferred by the actual mass motion of a heated mass motion of a heated fluid.fluid.
Conduction: HeatConduction: Heat energy is energy is transferred by adjacent transferred by adjacent molecular collisions inside a molecular collisions inside a material. The medium itself material. The medium itself does not move.does not move.
Radiation is the process by which heat energy is transferred by electromagnetic waves.
Summary of Thermal Summary of Thermal ConductivityConductivity
H = Heat current (J/s)
A = Surface area (m2)t = Temperature differenceL = Thickness of material
t1 t2
t = t2 - t1
The The thermal conductivity thermal conductivity kk of a material is a of a material is a measure of its ability to measure of its ability to conduct heat.conduct heat.
QLk
A t
QLk
A t
Q kA t
HL
Q kA t
HL
0
J
s m CUnits
Summary of RadiationSummary of Radiation
Rate of RadiationRate of Radiation (W/m(W/m22):):
The The rate of radiationrate of radiation R R is the energy is the energy emitted per unit area per unit time (power emitted per unit area per unit time (power per unit area).per unit area).
Q PR
A A
Q PR
A A
Emissivity, e : 0 > e > 1Emissivity, e : 0 > e > 1
Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4
Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4
4PR e T
A
4PR e T
A
R
Summary of FormulasSummary of Formulas
QLk
A t
QLk
A t
Q kA t
HL
Q kA t
HL
0
J
s m CUnits
H;
A
Q kA t k tH
L L
H
; A
Q kA t k tH
L L
Q PR
A A
Q PR
A A 4P
R e TA
4P
R e TA
4P e AT 4P e AT
CONCLUSION: Chapter 18CONCLUSION: Chapter 18Transfer of HeatTransfer of Heat
