TRANSMISSIVITY When radiation strikes a surface, part of it: Is absorbed Is reflected

FKKKSA Chem. Eng. Dept

THERMAL RADIATION

• Electromagnetic radiation emmited by a body due to large temperature difference

•  transmitted through space & vacuum

Mechanism of radiant heat transfer:

1. Thermal energy of a hot source at T1 is converted into energy of electromagnetic radiation waves

3. Electromagnetic waves are absored by the body and converted to thermal energy

•  medium not heated

2. Waves travel through intervening space in a straight lines & strike a cool object at T2


REFLECTIVITY, ABSORPTIVITY & TRANSMISSIVITY

ρ + α + τ = 1

Opaque body – no transmission (τ = 0)

ρ = reflectivity = fraction of irradiation reflected

α = adsorptivity = fraction absorbed

τ = transmissivity = fraction transmitted

ρ + α = 1

When radiation strikes a surface, part of it: Is absorbed Is reflected and the remaining part, if any, is transmitted


BLACK BODY

absorbs all incident radiation, regardless of wavelength and direction

emits radiation equally in all directions at the maximum rate for a given temperature and wavelength

serves as a standard against which all real surfaces are compared

closest approximation to a blackbody is a cavity with a small aperture whose inner surface is at a uniform temperature

a perfect absorber and emitter (no surface can emit more energy than a blackbody)


EMISSIVE POWER OF A BLACK BODY

– perfect absorber of radiation (absorb all radiation, α = 1)

– perfect emitter of radiation (radiates more energy)

where

Emissive power = total radiation emitted by a surface in all direction per unit surface area

σ = stefan-Boltzmann constant = 5.676 x 10-8 W/m2.K4

EB = blackbody emissive power (W/m2)

EB = σT4

T = surface temperature in absolute units (K)


EMISSIVITY

ε =

Emissivity represent emission characteristics of a body

Emissivity. ε = total emitted energy of the surface divided by the total emitted energy of a black body at the same temperature

No real surface can have an emissive power exceeding that of a black body at the same T


EMISSIVITY


INTENSITY

Total Intensity associated with blackbody emission

Total emissive power of blackbody

the rate at which radiation energy is emitted in the (θ,φ) direction per unit area normal to this direction and per unit solid angle about this direction


RADIOSITY

the rate at which radiation energy leaves a unit area of a surface in all directions

For a surface that is both a diffuse emitter and a diffuse reflector


THERMAL RADIATION

q = AσT4

Radiation from a gray body :

where

q = εAσT4

Radiation from a black body :

σ = stefan-Boltzmann constant = 5.676 x 10-8 W/m2.K4

ε = emissivity

A = surface area (m2)

Gray body: emit less than the blackbody (0< ε<1)

Black body: perfect absorber and emitter of thermal radiation (ε = 1)


KIRCHHOFF’S LAW

Kirchhoff’s law – The ability of a diffuse surface to emit radiation at a particular λ is equal to its ability to absorb radiation at λ at the same temperature T (at equilibrium),

– holds for any black or non black solid surface

ε, like α = low for polished metal surfaces

ε, like α = high for oxidized metal surfaces

The relation ε = α together with ρ = 1 - α allows the determination of all three properties of an opaque body


RADIATION TO A SMALL OBJECT FROM SURROUNDING

Small gray object at T1 in a large enclosure at T2 :

Net radiation from object 1:

q = A1ε1σT14 - A1α12σT2

4 = A1σ(ε1T14 - α12σT2

4 )

For engineering purpose q = εσA1 (T14 – T2

4)

A1ε1σT14

A1α12σT24

where ε = ε1 at T2


COMBINED RADIATION & CONVECTION

Total heat transfer

q = qconv.+ q rad. where

qconv. = hcA1(T1-T2)

q rad. = hrA1(T1-T2)

q = (hc + hr ) A1(T1-T2)


COMBINED RADIATION & CONVECTION

when ε = 1


RADIATION HEAT TRANSFER: GRAY SURFACES

q1 = - q2 = q12

Net radiation from surface 1 to surface 2 :


RADIATION RESISTANCE


VIEW FACTORS

View factor F12 – fraction of the radiation leaving surface 1 that is intercepted/reaching by surface 2

F12 Consider exchange dA1 and dA2 between differential areas

The view factor integral provides a general expression for


INFINITE PARALLEL PLANES


INFINITE PARALLEL BLACK PLANES

A1= A2 = A For black body, ε1 = ε2 = 1

q12 = A1F12σ(T14 - T2

4)

q21 = A2F21σ (T14 – T2

4)

for infinite parallel planes, F12 = F21 = 1

Net radiation from plane 2 to plane 1:

2 1

q12

q12 = A1σ(T14 - T2

4)


INFINITE PARALLEL GRAY PLANES

A1= A2 = A

for infinite parallel planes, F12 = F21 = 1

Net radiation from plane 1 to plane 2:

2 1

q12


EXAMPLE 4.11-6

A1<<< A2 , A1/A2 = 0 : q = ε1A1σ (T14 – T2

4)

T1, ε1, A1

T2, ε2, A2

F12 = 1


RADIATION HEAT TRANSFER


RADIATION SHIELD

Radiation shields – planes with low ε (high ρ) used to reduce the net radiation transfer between two surfaces


RADIATION SHIELD

Noting that F13=F23=1 and A1=A2=A3=A for infinite parallel plates,

q 12

The radiation heat transfer through large parallel plates separated by N radiation shields

q 12


RADIATION SHIELDS

q12 = Aσ(T14 T2

4) = Aσ(T14 T2

4)

q12 = Aσ(T14 T2

4)

(A1= A2 = A , F13 = F32 = 1)

If ε1 = ε2 = ε31= ε31 = ε

Without shield:


RADIATION SHIELDS

With 1 shield:

q12= Aσ(T14 T2

4)

q12 = Aσ(T14 T2

4)

For N shields (all emissivities are equal)

q12 = Aσ(T14 T2

4) = Aσ(T14 T2

4)


RADIATION SHIELDS

q12 = Aσ(T14 T2

4)

For N shields (all emissivities are equal)

Heat flux will be reduced to


EXAMPLE

A thin aluminum sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 800K andT2 = 500K and have emissivities ε1 = 0.2 and ε2 =0 .7,respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result to that without the shield. Answers: 806 W/m2 , 3625 W/m2


VIEW FACTOR RELATIONS

A1F12= A2 F21

General equation:

F12 = 1 Surface 1 can only see surface 2 :

Reciprocity relation :

ΣF12 = F11 + F12 = 1 Summation rule :

F11 = 0

Surface 1 cannot see itself (flat or convex):



F12 = 1



Superposition rule :



Symmetry rule :

13

2



N(N – 1)/2 view factors—reciprocity relation

N2 view factors for N surfaces

[N2 – N – N(N – 1)/2] = N(N – 1)/2 determined directly

Consider a 2 surface enclosure N = 2 surfaces

N2 = 4 view factors

N(N – 1)/2 = 1 reciprocity relations & determined directly


EXAMPLE 4.11-3

dA2

dx

A2

A1

x a

Since dA1 is a differential area



Determine F21 & F22 :

Reciprocity

Summation

Results F12 = 1

By inspection : F12 = 1 all radiation leaving inner surface is intercepted by outer surface (determined directly)


VIEW FACTOR FOR ADJACENT PERPENDICULAR RECTANGLES


VIEW FACTOR BETWEEN PARALLEL PLANES DIRECTLY OPPOSED


VIEW FACTOR BETWEEN TWO ALIGNED PARALLEL RECTANGLES OF EQUAL SIZE


VIEW FACTOR BETWEEN TWO COAXIAL PARALLEL DISKS


VIEW FACTOR FOR TWO CONCENTRIC CYLINDERS OF FINITE LENGTH


EXAMPLE

Consider the 5-m x 5-m x 5-m cubical furnace, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Determine (a)   the net rate of radiation heat transfer between the base and the side surfaces, (b) the net rate of radiation heat transfer between the base and the top surface,

and (c) the net radiation heat transfer from the base surface.

Answers: (a)394 kW (b) -1319 kW (c) -925 kW


EXAMPLE

A furnace is of cylindrical shape with R = H = 2 m. The base, top, and side surfaces of the furnace are all black and are maintained at uniform temperatures of 500, 700, and 1200 K, respectively. Determine the net rate of radiation heat transfer to or from the top surface during steady operation. Answer: -1.538 x 106 W


EXAMPLE

A furnace is of cylindrical shape with D =2.5 cm and H = 10 cm. The emissivities of the base, top, and side surfaces are 0.85, 0.85 and 0.8 and are maintained at uniform temperatures of 400, 400, and 500 oC, respectively. Determine the net rate of radiation heat transfer to or from the side surface reaching to each of the base and top surfaces . Answer: 3.382 W


EXAMPLE

Determine the view factors F13 and F23 between the rectangular surfaces as shown. Answers :


EXAMPLE

Determine the view factors from the base of the pyramid shown below to each of its four side surfaces. The base of the pyramid is a square, and its side surfaces are isosceles triangles. Answer: F12 = 1/4


RERADIATING SURFACES

Surface that is well insulated on one side and the convection effects may be neglected on the opposite (radiating) side.

Surface must lose as much radiation energy as it gains (reradiate all the radiation enegy it receives)

Net heat transfer through it is zero



No reradiating wall:

q12 = 12A12σ(T14 - T2

4)

F12 = 12

A1 12= A2 21

Parallel black planes:



q12 = F 12A12σ(T14 - T2

4)

A1F 12= A2F 21

Parallel gray planes:

F


EXAMPLE

Consider a 4-m x 4-m x 4-m cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of 550 K and 1100 K, respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace. Answer: 747921 W


EXAMPLE

In a cubical oven, the top wall is maintained at T1 = 800K and has an emissivity ε1 = 0.8, the floor is at T2 = 600K and has an emissivity ε2 = 0.8, and the four lateral walls are reradiating surfaces. Calculate the net radiation heat flux leaving the top surface. Answer: 731069 W

TRANSMISSIVITY When radiation strikes a surface, part of it: Is absorbed Is reflected

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