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Heat Transfer. Conduction. Direction. From hot to cold. Heat Transfer by Conduction. Conduction is the process by which heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move. Heat Transfer by Convection. - PowerPoint PPT Presentation
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Heat Transfer
Heat Transfer by Conduction
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 Convection
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 currents.convection currents.
Convection is significantly affected Convection is significantly affected by geometry of heated surfaces. (wall, by geometry of heated surfaces. (wall, ceiling, floor)ceiling, floor)
Heat Transfer by Radiation
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 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
ConvectioConvectionnRadiationRadiation
Heat 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: J/s, cal/s, and Btu/hTypical units are: J/s, cal/s, and Btu/h
H = Heat current (J/s)
A = Surface area (m2)t = Temperature differenceL = Thickness of material
Thermal Conductivity
t1 t2
t = t2 - t1
The thermal conductivity The thermal conductivity k of a material is a k 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 Conductivity
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 SI units, typically small measures for In SI units, typically small measures for length L and area A must be converted to length L and area A must be converted to meters and square meters, respectively, meters and square meters, respectively, before substitution into formulas.before substitution into formulas.
In SI units, typically small measures for In SI units, typically small measures for length L and area A must be converted to length L and area A 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 Conductivity
Taken literally, this means that for a 1-in. Taken literally, this means that for a 1-in. thick plate of glass whose area is 1 ftthick plate of glass whose area is 1 ft22 and and whose sides differ in temperature by 1 Fwhose sides differ in temperature by 1 F00, , heat will be conducted at the rate of 5.6 heat will be conducted at the rate of 5.6 Btu/h.Btu/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 ftarea in ft22 time in hours, time in hours, length in seconds, and length in seconds, and quantity of heat in Btu’s.quantity of heat in Btu’s.
Glass k = 5.6 Btu in./ftGlass k = 5.6 Btu in./ft22h Fh F00
Thermal 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 Conductivity
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: A large glass window measures 2 m wide and 6 m high. The
inside surface is at 200C and the outside surface is at 120C. How many joules of heat pass through this window in one
hour? Assume L = 1.5 cm and that k = 0.8 J/s m C0. 200C 120C
t = t2 - t1 = 8 C0
0.015 m
AQ = ?
= 1 hA = (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: The wall of a freezing plant is composed of 8 cm of corkboard and
12 cm of solid concrete. The inside surface is at -200C and the outside surface is +250C. What is the interface
temperature ti? 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.): Finding the interface temperature for a composite 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.): Simplifying, we 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 rate of allows us to determine the rate of
heat flow per unit of area, H/A.heat flow per unit of area, H/A.
The quantity H/A is same for cork or concrete:The quantity H/A 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.): Constant steady state 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 H/A is constant so Over time H/A is constant so different k’s cause different different k’s cause different t’st’s
Cork:Cork: t = 21.9t = 21.900C - (-20C - (-2000C) = 41.9 CC) = 41.9 C00
Concrete:Concrete: t = 25t = 2500C - 21.9C - 21.900C = 3.1 C = 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.): Constant steady state 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) = 41.9 CC) = 41.9 C00
Concrete:Concrete: t = 25t = 2500C - 21.9C - 21.900C = 3.1 C = 3.1 CC00
2 20.7 W/mH
A
2 20.7 W/mH
A
Note that 20.7 Joules of heat per Note that 20.7 Joules of heat per second pass through the composite second 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 13.5 times as large as for the cork is 13.5 times 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
Summary: Heat Transfer
Convection is the process Convection 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: Heat energy is Conduction: Heat 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 Conductivity
H = Heat current (J/s)
A = Surface area (m2)t = Temperature differenceL = Thickness of material
t1 t2
t = t2 - t1
The thermal conductivity The thermal conductivity k of a material is a k 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 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
