Radiation and Heat Transfer
Radiation
C
ontents:
•
Basic Concept
•
Example
•
Whiteboards
Radiation
Hot objects radiate photons (black body radiation)
Heating up a poker
Demo - Show red hot paperclip
Demo – Start silver/black demo
Bonfire/barn fire/stoves
The sun
Frost
Preheat the oven/drying tennis balls
Radiation
ΔQ
/
t
= Heat flow rate in J/s or Watts
e = emissivity (0
<
e
<
1)
(Temp?)
σ
= 5.67x10
-8
Wm
-2
K
-4
(Stefan Boltzmann constant)
A = Area (m
2
)
T = High temperature (K)
Net heat transfer
(emitted) - (absorbed)
T
1
= object
T
2
= surroundings
e = e?
Q/t = (0.75)(1.72 m
2
)(5.67E-8 Wm
-2
K
-4
)((273.15+33 K)
4
-(273.15+18 K)
4
) = 116.97 W
120 W
Example: A person has a surface area of 1.72 m
2
and a skin
temperature of 33
o
C. They are standing in a room whose
walls are at 18
o
C, and their skin has an emissivity of 0.75.
What is the net rate of radiative heat flow out of their body?
Demo – check temperatures
on radiation demo
Silver lined blankets/bags reflect radiation back to you
Radiation
1-3
Q/t = (0.92)(0.0096 m
2
)(5.67E-8 Wm
-2
K
-4
)((273.15 + 20.50)
4
-(273.15+20.00 K)
4
) = 0.025296… W
0.025 W, 25 mW
A power converter is a cube 4.0 cm on a side, and is made of
black plastic with an emissivity of 0.92. It maintains a
temperature of 20.50
o
C in a room whose walls are at 20.00
o
C. What is its net rate of radiative heat transfer?
34 = e(4
(0.03 m)
2
)(5.67E-8 Wm
-2
K
-4
)((273.15 + 250.0)
4
-(273.15+19.0 K)
4
)
0.78
A 3.0 cm radius (sperical) light bulb radiates 34 Watts of
energy. If it has a surface temperature of 250.0
o
C in a room
whose walls are at 19.0
o
C, what is its emissivity?
1000 = (0.95)(6.00 m
2
)(5.67E-8 Wm
-2
K
-4
)(T
4
-(273.15+18.00 K)
4
), T = 318.417
T = 45
o
C
318 K, 45
o
C
If you put a 1000. Watt hair dryer into a 1.00 m cube box with
an emissivity of 0.95, it would heat up the box’s surface until
energy flowed out (assume radiatively) as fast as it flowed in.
What would be the surface temperature of the box in Celsius if
the room it was in had walls at 18.0
o
C?
Concepts of radiation and heat transfer through examples like heating objects, net heat transfer equations, and radiative heat flow calculations. Discover how emissivity, temperature differentials, and surface characteristics impact the rate of heat transfer and energy flow in various scenarios. Learn about practical applications such as using silver-lined blankets to reflect radiation and understanding the emissivity of different materials.
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Presentation Transcript
Radiation Contents: Basic Concept Example Whiteboards
Radiation Hot objects radiate photons (black body radiation) Heating up a poker Demo - Show red hot paperclip Demo Start silver/black demo Bonfire/barn fire/stoves The sun Frost Preheat the oven/drying tennis balls
Radiation Q = 4 e AT t Q/t = Heat flow rate in J/s or Watts e = emissivity (0 < e < 1) (Temp?) = 5.67x10-8 Wm-2K-4(Stefan Boltzmann constant) A = Area (m2) T = High temperature (K)
Net heat transfer Q 4 4 4 4 = = e AT e AT ( ) e A T T 1 2 1 2 t (emitted) - (absorbed) T1 = object T2 = surroundings e = e?
Example: A person has a surface area of 1.72 m2 and a skin temperature of 33oC. They are standing in a room whose walls are at 18 oC, and their skin has an emissivity of 0.75. What is the net rate of radiative heat flow out of their body? Q/t = (0.75)(1.72 m2)(5.67E-8 Wm-2K-4)((273.15+33 K)4-(273.15+18 K)4) = 116.97 W 120 W
Silver lined blankets/bags reflect radiation back to you Demo check temperatures on radiation demo
Radiation 1-3
A power converter is a cube 4.0 cm on a side, and is made of black plastic with an emissivity of 0.92. It maintains a temperature of 20.50 oC in a room whose walls are at 20.00 oC. What is its net rate of radiative heat transfer? Q/t = (0.92)(0.0096 m2)(5.67E-8 Wm-2K-4)((273.15 + 20.50)4-(273.15+20.00 K)4) = 0.025296 W 0.025 W, 25 mW
A 3.0 cm radius (sperical) light bulb radiates 34 Watts of energy. If it has a surface temperature of 250.0 oC in a room whose walls are at 19.0 oC, what is its emissivity? 34 = e(4 (0.03 m)2)(5.67E-8 Wm-2K-4)((273.15 + 250.0)4-(273.15+19.0 K)4) 0.78
If you put a 1000. Watt hair dryer into a 1.00 m cube box with an emissivity of 0.95, it would heat up the box s surface until energy flowed out (assume radiatively) as fast as it flowed in. What would be the surface temperature of the box in Celsius if the room it was in had walls at 18.0 oC? 1000 = (0.95)(6.00 m2)(5.67E-8 Wm-2K-4)(T4-(273.15+18.00 K)4), T = 318.417 T = 45 oC 318 K, 45 oC
