Page 1 1 Tutorial -6 CH-206: Transfer Processes-1 Department of Chemical Engineering, IIT Roorkee 1. 2. Determine the view factor F12 between the rectangular surfaces shown in Figure: 3. A hole of area dA = 2 cm 2 is opened on the surface of a large spherical cavity whose inside is maintained at T = 800 K. (a) Calculate the radiation energy streaming through the hole in all directions into space. (b) Find the radiation energy streaming per unit solid angle in the direction making a 60 o angle with the normal to the surface of the opening. 4. Determine the view factors F12 and F21 for the very long ducts shown in Figure without using any view factor tables or charts. Neglect end effects. Consider a 20-cm-diameter spherical ball at 800 o C suspended in air as shown in figure: Assuming the ball closely approximates a blackbody; determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 µm. Page 2 1 Tutorial -6 CH-206: Transfer Processes-1 Department of Chemical Engineering, IIT Roorkee 1. 2. Determine the view factor F12 between the rectangular surfaces shown in Figure: 3. A hole of area dA = 2 cm 2 is opened on the surface of a large spherical cavity whose inside is maintained at T = 800 K. (a) Calculate the radiation energy streaming through the hole in all directions into space. (b) Find the radiation energy streaming per unit solid angle in the direction making a 60 o angle with the normal to the surface of the opening. 4. Determine the view factors F12 and F21 for the very long ducts shown in Figure without using any view factor tables or charts. Neglect end effects. Consider a 20-cm-diameter spherical ball at 800 o C suspended in air as shown in figure: Assuming the ball closely approximates a blackbody; determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 µm. 2 Page 3 1 Tutorial -6 CH-206: Transfer Processes-1 Department of Chemical Engineering, IIT Roorkee 1. 2. Determine the view factor F12 between the rectangular surfaces shown in Figure: 3. A hole of area dA = 2 cm 2 is opened on the surface of a large spherical cavity whose inside is maintained at T = 800 K. (a) Calculate the radiation energy streaming through the hole in all directions into space. (b) Find the radiation energy streaming per unit solid angle in the direction making a 60 o angle with the normal to the surface of the opening. 4. Determine the view factors F12 and F21 for the very long ducts shown in Figure without using any view factor tables or charts. Neglect end effects. Consider a 20-cm-diameter spherical ball at 800 o C suspended in air as shown in figure: Assuming the ball closely approximates a blackbody; determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 µm. 2 rn.AJ fl-rY /'s ek I- effVvV~ ' To~ b/~b0c2t ~'S0've- ? ~ ~fL~_ B4~ )LJ f b ~ IT TV =c (:Gl ~/DgfJ,r..V,k't) (mK)~ :=- 2. 3. 2 'X ID 3 vJj?n Y :::'23.2 ~yJ/~v. ,(>2 ~ boJ.J.- ~h :(3.2 ~J1~ e~ /,-, ~ ~ of e-/~~vw. ..h'L 'l!> y~'ct..j.,oh.- ?&Y .1~, pVy ~v. at.-fk .5.M~~ Wr~ -or r/k bv-D . G) tAL..f-vW ~ ~of YAh'al-ro,.,. e-r-,"T7;}- ~'~ ~..ft..tL .e.vvIi'r<2- bcJ) in .;- ~'h ' /~' c£.. ..f124 "v,./~ b;r ~ f f ctr >;} ../k b!<=t<(,01; hvci~s..; ve.- pcniJoff o1,l-cV~ tJJ,a-y<t- tcJ ~ ~ SM""-~ ~flI0 1~ bt'..11 ~ ~ (t<'vu, , h\~ ;~vj, A:s ;;: 7r:P'"V = IT (0. 2 ~ ~ 4£ ~ 5 h-\iYl G-o £ 30V S . e/)/VI./ ~ S' (VIZ... PCJvJ f2..r ' /~ ~.jfl!..,rhAl~ ~ kJ . (c) ik Sp~ I,..J hi ~ bO~ rJ: l\.- ~tJvNfZ ~~ ~ 3~ P/PvrvK 'f d.J~ -hn'1:, ~ ~ (0vJ . .. ) Page 4 1 Tutorial -6 CH-206: Transfer Processes-1 Department of Chemical Engineering, IIT Roorkee 1. 2. Determine the view factor F12 between the rectangular surfaces shown in Figure: 3. A hole of area dA = 2 cm 2 is opened on the surface of a large spherical cavity whose inside is maintained at T = 800 K. (a) Calculate the radiation energy streaming through the hole in all directions into space. (b) Find the radiation energy streaming per unit solid angle in the direction making a 60 o angle with the normal to the surface of the opening. 4. Determine the view factors F12 and F21 for the very long ducts shown in Figure without using any view factor tables or charts. Neglect end effects. Consider a 20-cm-diameter spherical ball at 800 o C suspended in air as shown in figure: Assuming the ball closely approximates a blackbody; determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 µm. 2 rn.AJ fl-rY /'s ek I- effVvV~ ' To~ b/~b0c2t ~'S0've- ? ~ ~fL~_ B4~ )LJ f b ~ IT TV =c (:Gl ~/DgfJ,r..V,k't) (mK)~ :=- 2. 3. 2 'X ID 3 vJj?n Y :::'23.2 ~yJ/~v. ,(>2 ~ boJ.J.- ~h :(3.2 ~J1~ e~ /,-, ~ ~ of e-/~~vw. ..h'L 'l!> y~'ct..j.,oh.- ?&Y .1~, pVy ~v. at.-fk .5.M~~ Wr~ -or r/k bv-D . G) tAL..f-vW ~ ~of YAh'al-ro,.,. e-r-,"T7;}- ~'~ ~..ft..tL .e.vvIi'r<2- bcJ) in .;- ~'h ' /~' c£.. ..f124 "v,./~ b;r ~ f f ctr >;} ../k b!<=t<(,01; hvci~s..; ve.- pcniJoff o1,l-cV~ tJJ,a-y<t- tcJ ~ ~ SM""-~ ~flI0 1~ bt'..11 ~ ~ (t<'vu, , h\~ ;~vj, A:s ;;: 7r:P'"V = IT (0. 2 ~ ~ 4£ ~ 5 h-\iYl G-o £ 30V S . e/)/VI./ ~ S' (VIZ... PCJvJ f2..r ' /~ ~.jfl!..,rhAl~ ~ kJ . (c) ik Sp~ I,..J hi ~ bO~ rJ: l\.- ~tJvNfZ ~~ ~ 3~ P/PvrvK 'f d.J~ -hn'1:, ~ ~ (0vJ . .. ) G " ~, [ b><f(C;T)-I] .5.7'73 )(I~ vJ . /A w,j'h1 v . Prj 5;[4f( /·ltE8-7- XID~.J() -/7 3~ C~K) :J - 3P1tcfJ vJ/'Mv.r . Page 5 1 Tutorial -6 CH-206: Transfer Processes-1 Department of Chemical Engineering, IIT Roorkee 1. 2. Determine the view factor F12 between the rectangular surfaces shown in Figure: 3. A hole of area dA = 2 cm 2 is opened on the surface of a large spherical cavity whose inside is maintained at T = 800 K. (a) Calculate the radiation energy streaming through the hole in all directions into space. (b) Find the radiation energy streaming per unit solid angle in the direction making a 60 o angle with the normal to the surface of the opening. 4. Determine the view factors F12 and F21 for the very long ducts shown in Figure without using any view factor tables or charts. Neglect end effects. Consider a 20-cm-diameter spherical ball at 800 o C suspended in air as shown in figure: Assuming the ball closely approximates a blackbody; determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 µm. 2 rn.AJ fl-rY /'s ek I- effVvV~ ' To~ b/~b0c2t ~'S0've- ? ~ ~fL~_ B4~ )LJ f b ~ IT TV =c (:Gl ~/DgfJ,r..V,k't) (mK)~ :=- 2. 3. 2 'X ID 3 vJj?n Y :::'23.2 ~yJ/~v. ,(>2 ~ boJ.J.- ~h :(3.2 ~J1~ e~ /,-, ~ ~ of e-/~~vw. ..h'L 'l!> y~'ct..j.,oh.- ?&Y .1~, pVy ~v. at.-fk .5.M~~ Wr~ -or r/k bv-D . G) tAL..f-vW ~ ~of YAh'al-ro,.,. e-r-,"T7;}- ~'~ ~..ft..tL .e.vvIi'r<2- bcJ) in .;- ~'h ' /~' c£.. ..f124 "v,./~ b;r ~ f f ctr >;} ../k b!<=t<(,01; hvci~s..; ve.- pcniJoff o1,l-cV~ tJJ,a-y<t- tcJ ~ ~ SM""-~ ~flI0 1~ bt'..11 ~ ~ (t<'vu, , h\~ ;~vj, A:s ;;: 7r:P'"V = IT (0. 2 ~ ~ 4£ ~ 5 h-\iYl G-o £ 30V S . e/)/VI./ ~ S' (VIZ... PCJvJ f2..r ' /~ ~.jfl!..,rhAl~ ~ kJ . (c) ik Sp~ I,..J hi ~ bO~ rJ: l\.- ~tJvNfZ ~~ ~ 3~ P/PvrvK 'f d.J~ -hn'1:, ~ ~ (0vJ . .. ) G " ~, [ b><f(C;T)-I] .5.7'73 )(I~ vJ . /A w,j'h1 v . Prj 5;[4f( /·ltE8-7- XID~.J() -/7 3~ C~K) :J - 3P1tcfJ vJ/'Mv.r . 9fC1-) ~lIrial :£ (4) (2) (3) (1) 3m 1m 1m 1m 1m Solution (1): Theview factors between the rectangular surfaces shown in the figure are tobe determined. Assumptions Thesurfaces are diffuse emitters and reflectors. Analysis We designate the different surfaces asfollows: shaded part ofperpendicular surface by (1), bottom part ofperpendicular surface by (3), shaded part ofhorizontal surface by (2), and front part ofhorizontal surface by (4). (a) From Fig.12-6(V ~ = ~}F23 = 0.25 and i = %)F2~(1+3) = 0.32 L., 1 L 1 I -=- -=- W 3 W 3 superposition rule: F2~('+3) = F21 +F23 ~ F21 = F2~(1+3) - F23 = 0.32 - 0.25 = 0.07 . reciprocity rule: AI =A 2 ~AIFI2 =A 2 F 2I ~F12 =F 21 =0.07 (b) From Fig.12-6;GJ L2 1 LI 2} -=- and -=- F.(4+2)~3 =0.15 W 3 W 3 and ~=~ W 3 L, 2} and - = - F(4+2)~(1+3) = 0.22 W 3 superposition rule: F(4+2;~(1+3) = F(4+2)~1 +F(4+2)~3 ~F(4+2)~1 = 0.22-0.15 = 0.07 reciprocity rule: A(4+2) F(4+2)~1= AIFI~(4+2) A(4+2) 6 ~Fl~(4+2) =--F(4+2)~1 =-(0.07)=0.14 1m AI 3 superposition rule: FI~(4+2) = F I4 +F I2 1m 3m (1) (3) 2m ••••••••••••••••••• y•••••••••••••••••••••••••••••••••••••••••••••••••• ~F14 =0.14-0.07 =0.07 since F)2 = 0.07(frompart a). Note that F\4 in part (b) is equivalent to F)2 inpart (a). (c) Wedesignate shaded part oftop surfaceby (1), remaining part oftop surface by (3), remaining part ofbottom surface by (4), and shaded part ofbottom surface by (2). From Fig.~,Q) L 2 2} L 2 2) -=- -=- D 2 D 2 F(2+4)~(1+3)= 0.20 and FI4 = 0.12 !2=~ !2=.~ D 2 D 2 superposition rule: F(2+4)~(1+3) = F(2+4)~1 +F(2+4)~3 symmetryrule: F(2+4HI = F(2+4)~3 f\ '. 1m······· .... 1m'"" 2m (1) (3) (4) (2) }\,I m ........ 1m ··················1··············::1 "', : '\'Read More

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