Fundamentals of Heat and Mass Transfer - CH012

Fundamentals of Heat and Mass Transfer - CH012

(Parte 1 de 11)

PROBLEM 12.1 KNOWN: Rate at which radiation is intercepted by each of three surfaces (see (Example 12.1).

FIND: Irradiation, G[W/m2 ], at each of the three surfaces.

ANALYSIS: The irradiation at a surface is the rate at which radiation is incident on a surface per unit area of the surface. The irradiation at surface j due to emission from surface 1 is

1j j j

With A1 = A2 = A3 = A4 = 10-3 m2 and the incident radiation rates q1-j from the results of Example

12.1, find

COMMENTS: The irradiation could also be computed from Eq. 12.15, which, for the present situation, takes the form where I1 = I = 7000 W/m2×sr and w1-j is the solid angle subtended by surface 1 with respect to j. For example,

10m cos60 cos 30

22G 12.1W/ m.= Note that, since A1 is a diffuse radiator, the intensity I is independent of direction.


KNOWN: A diffuse surface of area A1 = 10-4m2 emits diffusely with total emissive power E = 5 × 104

FIND: (a) Rate this emission is intercepted by small surface of area A2 = 5 × 10-4 m2 at a prescribed location and orientation, (b) Irradiation G2 on A2, and (c) Compute and plot G2 as a function of the

ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) A1 may be approximated as a differential surface

ANALYSIS: (a) The rate at which emission from A1 is intercepted by A2 follows from Eq. 12.5 written on a total rather than spectral basis.

()e,1 e,1 1I, I Eθφ π==(2)

The solid angle subtended by A2 with respect to A1 is

Substituting Eqs. (2) and (3) into Eq. (1) with numerical values gives

EA cos qA cos sr 0.5mπ

(b) From section 12, 2.3, the irradiation is the rate at which radiation is incident upon the surface per unit surface area,

(c) Using the IHT workspace with the foregoing equations, the G2 was computed as a function of the separation distance for selected zenith angles. The results are plotted below.


PROBLEM 12.2 (Cont.)

Irradiation, G2 (W/m^2)

theta2 = 0 deg theta2 = 30 deg theta2 = 60 deg

For all zenith angles, G2 decreases with increasing separation distance r2From Eq. (3), note that dω2-1

and, hence G2, vary inversely as the square of the separation distance. For any fixed separation distance,

G2 is a maximum when θ2 = 0° and decreases with increasing θ2, proportional to cos θ2.

COMMENTS: (1) For a diffuse surface, the intensity, Ie, is independent of direction and related to the emissive power as Ie = E/ π. Note that π has the units of []sr in this relation.

(2) Note that Eq. 12.5 is an important relation for determining the radiant power leaving a surface in a prescribed manner. It has been used here on a total rather than spectral basis.

(3) Returning to part (b) and referring to Figure 12.10, the irradiation on A2 may be expressed as

Ac osGI cos r θθ=

Show that the result is G2 = 2.76 W/m2. Explain how this expression follows from Eq. (12.15).

PROBLEM 12.003

KNOWN: Intensity and area of a diffuse emitter. Area and rotational frequency of a second surface, as well as its distance from and orientation relative to the diffuse emitter.

FIND: Energy intercepted by the second surface during a complete rotation. SCHEMATIC:

ASSUMPTIONS: (1) A1 and A2 may be approximated as differentially small surfaces, (2) A1 is a diffuse emitter.

ANALYSIS: From Eq. 12.5, the rate at which radiation emitted by A1 is intercepted by A2 is

where θ1 = 0 and θ2 changes continuously with time. The amount of energy intercepted by both sides of A2 during one rotation, ∆E, may be grouped into four equivalent parcels, each corresponding to rotation over an angular domain of 0 ≤ θ2 < π/2. Hence, with dt = dθ2/2,θ the radiant energy intercepted over the period T of one revolution is

r ππ θθ θθθ

2r ad/s 0.50m

COMMENTS: The maximum rate at which A2 intercepts radiation corresponds to θ2 = 0 and is qmax = Ie A1 A2/r2 = 4 × 10-6 W. The period of rotation is T = 2π/2θ = 3.14 s.

PROBLEM 12.004

KNOWN: Furnace with prescribed aperture and emissive power.

FIND: (a) Position of gauge such that irradiation is G = 1000 W/m2, (b) Irradiation when gauge is tilted θd = 20o, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for the range 100 ≤ L ≤ 300 m and tilt angles of θd = 0, 20, and 60o .

ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad << L2 .

ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per unit area of the surface. That is

fddGqA→= fdefdfqIAcosfθω→−=(1,2)

where fdq→ is the radiant power which leaves Af and is intercepted by Ad. From Eqs. 12.2 and 12.5, dfω − is the solid angle subtended by surface Ad with respect to Af,


df d dA cos Lωθ− Noting that since the aperture emits diffusely, Ie = E/π (see Eq. 12.14), and hence


Solving for L2 and substituting for the condition θf = 0o and θd = 0o ,


(b) When θd = 20o, qf→d will be reduced by a factor of cos θd since ωd-f is reduced by a factor cos θd. Hence,

G = 1000 W/m2 × cos θd = 1000W/m2 × cos 20o = 940 W/m2 .<

(c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected θd. Note that G decreases inversely as L2. As expected, G decreases with increasing θd and in the limit, approaches zero as θd approaches 90o .

(Parte 1 de 11)