Tutorial 7 Summary of Week 7 lecture
The radiative properties of an object, i.e. 𝜀, 𝛼, 𝜌, 𝜏, are functions of the radiation wavelength 𝜆 and angle of the incidence, 𝜃.
In this unit, we shall make some assumptions about some objects that (a) remove the dependency of the properties on either 𝜆 and/or 𝜃; (b) set the value of some properties to 0 or 1.
Opaque body (wood, brick, metals)
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“Diffuse” reflector (as opposed to spectral
(or mirror-like) reflector); ie 𝜌 independent of 𝜃)
“Gray” (emissivity is not dependent on wavelength) “Diffuse and gray”
“Opaque, diffuse and gray”
WRITE THE ENERGY BALANCE EQUATION FOR RADIATIVE HEAT TRANSFER
In previous years I received queries from students on how to write up the energy balance equation for radiative heat transfer. Specifically, the query is about when to use eq (1) and when to use eq (2)
𝑞!”#,%&’ = 𝐴(𝐽 − 𝐺) (1) 0r
𝑞!”#,%&’ = 𝐴(𝐺 − 𝐽) (2)
Eq (1) is given in Cengel’s book (5th ed. P785, eq. 13-23) as the definition of the net rate of radiant heat transfer. qnet,rad, is defined as net radiant heat transfer from an object to the surroundings. However, the energy balance equation should be written with a proper drawing, which identify the directions of all processes of heat transfer.
Your lecture notes also offered a derivation, which shows that eq (1) can also be written in the form of
𝑞 = 𝐴𝜀𝜎(𝑇) − 𝑇) ) (3) !”#,%&’ ( (*%%
The symbols have their usual meanings. This equation has the instruction on how we should write the energy balance. If the surface temperature of the object is higher than the surroundings, heat will transfer from the object surface to the surroundings, we write the energy balance in the form of eq (2) (see Figure 1). If, however, the surface temperature of the object is lower than that of the surroundings, such as an esky in outdoor summer, heat will transfer from the surroundings to the surface of the object. Then we either use eq (1) (see Figure 2).
The Kirchhoff’s Law
We can also derive the Kirchhoff’s Law from the mini- energy balance approach
If we take the blackbody shell and a non-blackbody object enclosed in the shell, the object and the shell are at the same temperature. Therefore, even the blackbody shell and non- blackbody object are emitting and receiving radiation energy from each other, the net heat transfer rate to either of
the shell and the object is 0, since they are in thermal equilibrium.
Then if we look at the non-blackbody object, its heat transfer rate from the shell is:
0 = G – J, (where G is the irradiation from the blackbody shell, G = 𝜎𝑇). J has its normal meaning. Then
0 = G – (𝜌G – 𝜀𝜎𝑇) ) 0 = (1-𝜌)G – 𝜀𝜎𝑇) ,
assuming we do not consider 𝜏 here, therefore 1 – 𝜌 = 𝛼 0 = 𝛼G – 𝜀𝜎𝑇)
0=𝛼𝜎𝑇) -𝜀𝜎𝑇)
This leads to 𝛼 = 𝜀. This holds, even if 𝜏 of the non- blackbody is not zero.
ATMOSPHERIC AND SOLAR RADIATION
Gs is the total solar irradiance that reaches the out edge of the earth’s atmosphere, Gs = 1373 W/m2. Gs is also known as the solar constant, which represents the rate at which solar energy is incident on a surface normal to the sun’s rays at the outer edge of the atmosphere.
The direct and diffuse radiation incident on a horizontal Surface on earth’s surface.
𝐺(,-&% = 𝐺. ∙ 𝑐𝑜𝑠𝜃 + 𝐺’
The atmospheric emission is in the far infrared regions of the spectrum,
from 5 to 8 μm and above 13 μm. It is primarily due to the CO2 and H2O molecules.
it is found convenient in radiation calculations to treat the atmosphere as
a blackbody at some lower fictitious temperature that emits an equivalent amount of radiation energy. This fictitious temperature is called the effective sky temperature Tsky.
𝐺 =𝜎𝑇) (/0 (/0
The effective sky temperature does not deviate much
from the room temperature. We can apply the Kirchhoff’s law and take the absorptivity of a surface to be equal to its emissivity at room temperature, 𝛼 = 𝜀, if the effective sky temperature is not specifically given. Then the sky radiation absorbed by a surface can be expressed as
The net rate of radiation heat transfer to a surface exposed to solar and atmospheric radiation is determined from
an energy balance:
CALCULATION OF VIEW FACTORS
We take Q3 of our practical work as an example:
3. Consider a 5-m x 5-m x 5-m cubical furnace whose surfaces approximate black surfaces. The base, top and sides are maintained at 800K, 1500K and 500K, respectively. Determine:
a) View factors for the problem Basic relationships:
RADIATION HEAT TRANSFER – BLACK SURFACES
Blacksurfaces:𝛼= 𝜀=1; 𝜏=0; 𝜌=0
If we consider an enclosure consisting N black surfaces maintained at a specific temperature, the net radiation heat transfer from any surface i is determined by the sum of heat transfer from surface i to each of the surfaces in the enclosure.
RADIATION HEAT TRANSFER – DIFFUSE, GRAY SURFACES
Diffuse surface:
Gray surface:
First step: examine mini energy balance, particularly The radiosity J (total radiation energy leaving a surface per unit time and per unit area)
Forasurfaceithatisgrayandopaque(𝜀1 = 𝛼1 and 𝛼1 +𝜌1 =1), the radiosity can be expressed as
For a black surface
The net rate of radiation heat transfer FROM a surface i of surface area Ai is denoted by 𝑄̇ i and is expressed as
Thus, the direction of radiation is TO the surface if Ji > Ebi and FROM the surface if Ji < Ebi. Note, for the temperature to remain constant, there must be some other heat transfer mode operating (conduction, convection etc).
In an electrical analogy to Ohm’s law, this equation can be rearranged as
NET RADIATION HEAT TRANSFER BETWEEN ANY TWO SURFACES
Lecture recording has provided a good summary on this topic.
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