PROBLEM cos 30. With A 1 = A 2 = A 3 = A 4 = 10-3 m 2 and the incident radiation rates q 1-j from the results of Example 12.

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1 PROBLEM. KNON: Rate at which radiation is intercepted by each of three surfaces (see (Example.). FIND: Irradiation, G[/m ], 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 is q j G j =. Aj ith A = A = A 3 = A 4 = 0-3 m and the incident radiation rates q -j from the results of Example., find. 0 3 G = =./m 0 3 m G 3 = = 8.0/m 0 3 m G 4 = = 9.8/m. 0 3 m COMMENTS: The irradiation could also be computed from Eq..5, which, for the present situation, takes the form Gj = Icosθjω j where I = I = 7000 /m sr and ω -j is the solid angle subtended by surface with respect to j. For example, G = Icosθω G = 7000/m sr 0 3 m cos60 cos 30 G =./m. ( 0.5m) Note that, since A is a diffuse radiator, the intensity I is independent of direction.

2 PROBLEM. KNON: A diffuse surface of area A = 0-4 m emits diffusely with total emissive power E = /m. FIND: (a) Rate this emission is intercepted by small surface of area A = m at a prescribed location and orientation, (b) Irradiation G on A, and (c) Compute and plot G as a function of the separation distance r for the range 0.5 r.0 m for zenith angles θ = 0, 30 and 60. ASSUMPTIONS: () Surface A emits diffusely, () A may be approximated as a differential surface area and that A r. ANALYSIS: (a) The rate at which emission from A is intercepted by A follows from Eq..5 written on a total rather than spectral basis. ( ) q = I e, θφ, Acosθdω. () Since the surface A is diffuse, it follows from Eq..3 that ( θφ) I e,, = Ie, = E π. () The solid angle subtended by A with respect to A is dω A cosθ r. (3) Substituting Eqs. () and (3) into Eq. () with numerical values gives 4 4 $ E A cosθ 5 0 m 5 0 m cos 30 q Acosθ = 0 m cos 60 sr (4) π r π sr ( 0.5m) q = 5,95 m sr 5 0 m.73 0 sr = ( ) = 4 $ ( ) (b) From section,.3, the irradiation is the rate at which radiation is incident upon the surface per unit surface area, 3 q G = = =.76 m (5) A m (c) Using the IHT workspace with the foregoing equations, the G was computed as a function of the separation distance for selected zenith angles. The results are plotted below. Continued...

3 PROBLEM. (Cont.) Irradiation, G (/m^) theta = 0 deg theta = 30 deg theta = 60 deg Separation distance, r (m) For all zenith angles, G decreases with increasing separation distance r. From Eq. (3), note that dω - and, hence G, vary inversely as the square of the separation distance. For any fixed separation distance, G is a maximum when θ = 0 and decreases with increasing θ, proportional to cos θ. COMMENTS: () For a diffuse surface, the intensity, I e, is independent of direction and related to the emissive power as I e = E/ π. Note that π has the units of [ sr ] in this relation. () Note that Eq..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.0, the irradiation on A may be expressed as Acosθ G = Ii,cosθ r Show that the result is G =.76 /m. Explain how this expression follows from Eq. (.5).

4 PROBLEM.3 KNON: Hot part, A p, located a distance x from an origin directly beneath a motion sensor at a distance L d = m. FIND: (a) Location x at which sensor signal S will be 75% that corresponding to x = 0, directly beneath the sensor, S o, and (b) Compute and plot the signal ratio, S/S o, as a function of the part position x for the range 0. S/S o for L d = 0.8,.0 and. m; compare the x-location for each value of L d at which S/S o = ASSUMPTIONS: () Hot part is diffuse emitter, () L d >> A p, A o. ANALYSIS: (a) The sensor signal, S, is proportional to the radiant power leaving A p and intercepted by A d, when Hence, S~qp d = Ip,e Apcosθp ωd p () Ld / cosθp = cosθd = = L d (Ld + x ) () R Ad cosθd 3/ ωd p = = A d L d (Ld + x ) (3) R L q d p d = Ip,e Ap Ad (Ld + x ) It follows that, with S o occurring when x= 0 and L d = m, S L d (Ld + x ) L = = d S o L d (Ld + 0 ) Ld + x so that when S/S o = 0.75, find, x = m (b) Using Eq. (5) in the IHT workspace, the signal ratio, S/S o, has been computed and plotted as a function of the part position x for selected L d values. (4) (5) Continued...

5 PROBLEM.3 (Cont.) 0.8 Signal ratio, S/So Part position, x (m) Sensor position, Ld = 0.8 m Ld = m Ld =. m hen the part is directly under the sensor, x = 0, S/S o = for all values of L d. ith increasing x, S/S o decreases most rapidly with the smallest L d. From the IHT model we found the part position x corresponding to S/S o = 0.75 as follows. S/S o L d (m) x (m) If the sensor system is set so that when S/S o reaches 0.75 a process is initiated, the technician can use the above plot and table to determine at what position the part will begin to experience the treatment process.

6 PROBLEM 3. KNON: Various geometric shapes involving two areas A and A. FIND: Shape factors, F and F, for each configuration. ASSUMPTIONS: Surfaces are diffuse. ANALYSIS: The analysis is not to make use of tables or charts. The approach involves use of the reciprocity relation, Eq. 3.3, and summation rule, Eq Recognize that reciprocity applies to two surfaces; summation applies to an enclosure. Certain shape factors will be identified by inspection. Note L is the length normal to page. (a) Long duct (L): By inspection, F =.0 By reciprocity, A RL 4 F = F =.0 = = 0.44 A ( 3/4) πrl 3π (b) Small sphere, A, under concentric hemisphere, A, where A = A Summation rule F + F + F3 = But F = F 3 by symmetry, hence F = 0.50 A A By reciprocity, F = F = 0.5 = 0.5. A A (c) Long duct (L): By inspection, F =.0 (d) Long inclined plates (L): By reciprocity, A RL F = F =.0 = = A πrl π Summation rule, F = F = 0.64 = Summation rule, F + F + F3 = But F = F 3 by symmetry, hence F = 0.50 A 0L By reciprocity, F = F = 0.5 = A / 0 L () (e) Sphere lying on infinite plane Summation rule, F + F + F 3 = But F = F 3 by symmetry, hence F = 0.5 A By reciprocity, F = F 0 since A. A Continued..

7 PROBLEM 3. (Cont.) (f) Hemisphere over a disc of diameter D/; find also F and F 3. By inspection, F =.0 Summation rule for surface A 3 is written as F3 + F3 + F33 =. Hence, F3 =.0. By reciprocity, A F 3 3 = F3 A πd π ( D/) D π F 3 = /.0 = By reciprocity, A π D πd F = F = / A 4 = Summation rule for A, F + F + F3 = or Note that by inspection you can deduce F = 0.5 (g) Long open channel (L): F = F F3 = = 0.5. Summation rule for A F + F + F3 = 0 but F = F 3 by symmetry, hence F = By reciprocity, A L 4 F = F = = 0.50 = A ( π )/4 L π COMMENTS: () Note that the summation rule is applied to an enclosure. To complete the enclosure, it was necessary in several cases to define a third surface which was shown by dashed lines. () Recognize that the solutions follow a systematic procedure; in many instances it is possible to deduce a shape factor by inspection.

8 PROBLEM 3. KNON: Geometry of semi-circular, rectangular and V grooves. FIND: (a) View factors of grooves with respect to surroundings, (b) View factor for sides of V groove, (c) View factor for sides of rectangular groove. ASSUMPTIONS: () Diffuse surfaces, () Negligible end effects, long grooves. ANALYSIS: (a) Consider a unit length of each groove and represent the surroundings by a hypothetical surface (dashed line). Semi-Circular Groove: A F = ; F = F = A ( π /) F = / π. Rectangular Groove: A4 F4,,3 ( ) = ; F(,,34 ) = F4,,3 ( ) = A+ A + A3 H+ + H F,,3 4 = / + H. ( ) ( ) V Groove: A F 3 3, ( ) = ; F(,3 ) = F A 3, ( ) = + A (b) From Eqs. 3.3 and 3.4, F = sin θ. (,) 3 From Symmetry, F3 = /. Hence, ( ) A F 3 = F3 = F 3. A / / + sinθ sinθ F = or F = sin θ. / /sinθ (c) From Fig. 3.4, with X/L = H/ = and Y/L, F 0.6. COMMENTS: () Note that for the V groove, F 3 = F 3 = F (,)3 = sinθ, () In part (c), Fig. 3.4 could also be used with Y/L = and X/L =. However, obtaining the limit of F ij as X/L from the figure is somewhat uncertain.

9 PROBLEM 3.0 KNON: Arrangement of perpendicular surfaces without a common edge. FIND: (a) A relation for the view factor F 4 and (b) The value of F 4 for prescribed dimensions. ASSUMPTIONS: () Diffuse surfaces. ANALYSIS: (a) To determine F 4, it is convenient to define the hypothetical surfaces A and A 3. From Eq. 3.6, A + A F, 3,4 = A F 3,4 + A F 3,4 ( ) ( )( ) ( ) ( ) where F (,)(3,4) and F (3,4) may be obtained from Fig Substituting for A F (3,4) from Eq. 3.5 and combining expressions, find A F3,4 = A F3 + A F4 ( ) F 4 = ( A A ) F (,)( 3,4) A F 3 A F ( 3,4). A + Substituting for A F 3 from Eq. 3.6, which may be expressed as A + A F, 3 = A F 3 + A F 3. ( ) ( ) The desired relation is then F 4 = ( A + A ) F (,)( 3,4) + A F 3 ( A + A ) F (,) 3 A F ( 3,4). A (b) For the prescribed dimensions and using Fig. 3.6, find these view factors: Surfaces (,)(3,4) L+ L L3 + L4 ( Y / X) = =, ( Z / X) = =.45, F(,)( 3,4) = 0. Surfaces 3 L L3 ( Y / X) = = 0.5, ( Z / X) = =, F3 = 0.8 Surfaces (,)3 L+ L L3 ( Y / X) = =, ( Z / X) = =, F(,) 3 = 0.0 Surfaces (3,4) L L3 + L4 ( Y / X) = = 0.5, ( Z / X) = =.5, F3,4 ( ) = 0.3 Using the relation above, find F4 = ( ) L + L ) 0. + ( L ) 0.8 ( L + L ) 0.0 ( L ) 0.3 ] L F4 = [ ( 0.) + ( 0.8) ( 0.0) ( 0.3) ] = 0.0.

10 PROBLEM 3. KNON: Coaxial, parallel black plates with surroundings. Lower plate (A ) maintained at prescribed temperature T while electrical power is supplied to upper plate (A ). FIND: Temperature of the upper plate T. ASSUMPTIONS: () Plates are black surfaces of uniform temperature, and () Backside of heater on A insulated. ANALYSIS: The net radiation heat rate leaving A i is N Pe = qij = AFσ T T + AF3σ T T3 j= ( ) ( ) ( ) ( ) Pe = Aσ F T T + F3 T T sur From Fig. 3.5 for coaxial disks (see Table 3.), R = r / L = 0.0 m / 0.0 m = 0.5 R = r / L = 0.0 m / 0.0 m =.0 + R + S = + = + = 9.0 R ( 0.5) () / / F = S S 4( r / r ) = 9 9 4( 0. / 0.) = From the summation rule for the enclosure A, A and A 3 where the last area represents the surroundings with T 3 = T sur, F + F3 = F3 = F = = Substituting numerical values into Eq. (), with A = π D / 4 = m, = m / m K T 500 K ( ) ( ) ( ) T 300 K = T T 300 ( ) find by trial-and-error that T = 456 K. COMMENTS: Note that if the upper plate were adiabatic, T = 47 K.

11 PROBLEM 3.53 KNON: Emissivities, diameters and temperatures of concentric spheres. FIND: (a) Radiation transfer rate for black surfaces. (b) Radiation transfer rate for diffuse-gray surfaces, (c) Effects of increasing the diameter and assuming blackbody behavior for the outer sphere. (d) Effect of emissivities on net radiation exchange. ASSUMPTIONS: () Blackbody or diffuse-gray surface behavior. ANALYSIS: (a) Assuming blackbody behavior, it follows from Eq ( ) π( ) ( ) ( ) ( ) 4 4 q = AF σ T T = 0.8 m / m K 400 K 300 K = 995. (b) For diffuse-gray surface behavior, it follows from Eq. 3.6 ( ) π( ) σ A T 4 T / m K m K 4 q = = = 9. ε r ε ε r (c) ith D = 0 m, it follows from Eq / m Kπ ( 0.8m) ( 400 K) ( 300 K ) q = = ith ε =, instead of 0.05, Eq. 3.6 reduces to Eq. 3.7 and ( ) ( ) ( ) ( ) 4 4 q = σaε T T = / m K π 0.8 m K 300 K = 998. Continued..

12 PROBLEM 3.53 (Cont.) (d) Using the IHT Radiation Tool Pad, the following results were obtained Net radiation exchange increases with ε and ε, and the trends are due to increases in emission from and absorption by surfaces and, respectively. COMMENTS: From part (c) it is evident that the actual surface emissivity of a large enclosure has a small effect on radiation exchange with small surfaces in the enclosure. orking with ε =.0 instead of ε = 0.05, the value of q is increased by only ( )/983 =.5%. In contrast, from the results of (d) it is evident that the surface emissivity ε of a small enclosure has a large effect on radiation exchange with interior objects, which increases with increasing ε.