PROBLEM 3. KNON: Various geometric shapes involving two areas and. FIND: Shape factors, F and F, for each configuration. SSUMPTIONS: Surfaces are diffuse. NLYSIS: 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. 3.. 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, RL F = F =.0 = = 0. ( 3/) πrl 3π (b) Small sphere,, under concentric hemisphere,, where = Summation rule F + F + F3 = But F = F 3 by symmetry, hence F = 0.50 By reciprocity, F = F = 0.5 = 0.5. (c) Long duct (L): By inspection, F =.0 (d) Long inclined plates (L): By reciprocity, RL F = F =.0 = = 0.637 πrl π Summation rule, F = F = 0.6 = 0.363. Summation rule, F + F + F3 = But F = F 3 by symmetry, hence F = 0.50 0L By reciprocity, F = F = 0.5 = 0.707. / 0 L () (e) Sphere lying on infinite plane Summation rule, F + F + F 3 = But F = F 3 by symmetry, hence F = 0.5 By reciprocity, F = F 0 since. Continued..
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 3 is written as F3 + F3 + F33 =. Hence, F3 =.0. By reciprocity, F 3 3 = F3 πd π ( D/) D π F 3 = /.0 = 0.375. By reciprocity, π D πd F = F = /.0 0.5. = Summation rule for, F + F + F3 = or Note that by inspection you can deduce F = 0.5 (g) Long open channel (L): F = F F3 = 0.5 0.375 = 0.5. Summation rule for F + F + F3 = 0 but F = F 3 by symmetry, hence F = 0.50. By reciprocity, L F = F = = 0.50 = 0.637. ( π )/ 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.
PROBLEM 3.0 KNON: rrangement of perpendicular surfaces without a common edge. FIND: (a) relation for the view factor F and (b) The value of F for prescribed dimensions. SSUMPTIONS: () Diffuse surfaces. NLYSIS: (a) To determine F, it is convenient to define the hypothetical surfaces and 3. From Eq. 3.6, + F, 3, = F 3, + F 3, where F (,)(3,) and F (3,) may be obtained from Fig. 3.6. Substituting for F (3,) from Eq. 3.5 and combining expressions, find F3, = F3 + F F = ( ) F (,)( 3,) F 3 F ( 3,). + Substituting for F 3 from Eq. 3.6, which may be expressed as + F, 3 = F 3 + F 3. The desired relation is then F = ( + ) F (,)( 3,) + F 3 ( + ) F (,) 3 F ( 3,). (b) For the prescribed dimensions and using Fig. 3.6, find these view factors: Surfaces (,)(3,) L+ L L3 + L ( Y / X) = =, ( Z / X) = =.5, F(,)( 3,) = 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,) L L3 + L ( Y / X) = = 0.5, ( Z / X) = =.5, F3, = 0.3 Using the relation above, find F = L + L ) 0. + ( L ) 0.8 ( L + L ) 0.0 ( L ) 0.3 ] L F = [ ( 0.) + ( 0.8) ( 0.0) ( 0.3) ] = 0.0.
PROBLEM 3.8 KNON: Surface temperature of a semi-circular drying oven. FIND: Drying rate per unit length of oven. SSUMPTIONS: () Blackbody behavior for furnace wall and water, () Convection effects are negligible and bottom is insulated. PROPERTIES: Table -6, ater (35 K): hfg NLYSIS: pplying a surface energy balance, q = qevap = mh fg =.378 0 6 J / kg. where it is assumed that the net radiation heat transfer to the water is balanced by the evaporative heat loss. From Eq. 3.3 q = F σ T T. From inspection and the reciprocity relation, Eq. 3.3, Hence or D L F = F = = 0.637. ( π D/) L ( T T ) m π D m = = F σ L hfg ( m) ( 00 K) ( 35 K) π m 0.637 5.67 0 8 = m K.378 0 6 J / kg m = 0.09 kg / s m. COMMENTS: ir flow through the oven is needed to remove the water vapor. The water surface temperature, T, is determined by a balance between radiation heat transfer to the water and the convection of latent and sensible energy from the water.
PROBLEM 3.3 KNON: ater-cooled heat flux gage exposed to radiant source, convection process and surroundings. FIND: (a) Net radiation exchange between heater and gage, (b) Net transfer of radiation to the gauge per unit area of the gage, (c) Net heat transfer to the gage per unit area of gage, (d) Heat flux indicated by gage described in Problem 3.98. SSUMPTIONS: () Heater and gauge are parallel, coaxial discs having blackbody behavior, () g h, (3) Surroundings are large compared to h and g. NLYSIS: (a) The net radiation exchange between the heater and the gage, both with blackbody behavior, is given by Eq. 3.3 having the form qh g = h Fhgσ( Th Tg ) = g Fghσ( Th T g). Note the use of reciprocity, Eq. 3.3, for the view factors. From Eq. 3.8, Fgh = D h / L + Dh = 0.m / 0.5 m + 0. m = 0.0385. ( π ) 8 qh g = 0.0 m / 0.0385 5.67 0 / m K 800 90 K = 69.0 m. (b) The net radiation to the gage per unit area will involve exchange with the heater and the surroundings. Using Eq. 3., q net,rad = q g /g = q hg /g + q surg / g. The net exchange with the surroundings is qsurg = surfsur g σ Tsur Tg = g Fg sur σ Tsur T g. 3 69.0 0 8 q net,rad = + 0.0385 5.67 0 / m K 300 90 K = 93.5 / m. π 0.0m / (c) The net heat transfer rate to the gage per unit area of the gage follows from the surface energy balance q net,in = q ) net,rad + q conv q net,in = 93.5 / m + 5 / m K( 300 90) K q net,in = 085 / m. (d) The heat flux gage described in Problem 3.98 would experience a net heat flux to the surface of 085 /m. The irradiation to the gage from the heater is G g = q h g / g = F gh σ T h = 89 /m. Since the gage responds to net heat flux, there would be a systematic error in sensing irradiation from the heater.
PROBLEM 3.3 KNON: Long V-groove machined in an isothermal block. FIND: Radiant flux leaving the groove to the surroundings and effective emissivity. SSUMPTIONS: () Groove surface is diffuse-gray, () Groove is infinitely long, (3) Block is isothermal. NLYSIS: Define the hypothetical surface with T = 0 K. The net radiation leaving, q, will pass to the surroundings. From the two surface enclosure analysis, Eq. 3.3, σ ( T ) T q = q = ε ε + + ε F ε Recognize that ε = and that from reciprocity, F = F where F =. Hence, q ( T ) T σ = ε + ε ith / = " tan0 /( " /cos0 ) = sin0, find 8 5.67 0 / m K 000 0 K q = = 6.7 k / m. ( 0.6) sin 0 + 0.6 The effective emissivity of the groove follows from the definition given in Problem 3. as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and at the same temperature as the cavity surface. For the present situation, 3 q q + 6.7 0 / m εe = = = = 0.8. E 8 b T σ T 5.67 0 / m K ( 000 K) COMMENTS: Note the use of the hypothetical surface defined as black at 0 K. This surface does not emit and absorbs all radiation on it; hence, is the radiant power to the surroundings.
PROBLEM 3.6 KNON: Heated tube with radiation shield whose exterior surface is exposed to convection and radiation processes. FIND: Operating temperature for the tube under the prescribed conditions. SSUMPTIONS: () Steady-state conditions, () No convection in space between tube and shield, (3) Surroundings are large compared to the shield and are isothermal, () Tube and shield are infinitely long, (5) Surfaces are diffuse-gray, (6) Shield is isothermal. NLYSIS: Perform an energy balance on the shield. E in E out = 0 q q conv qrad = 0 where q is the net radiation exchange between the tube and inner surface of the shield, which from Eq. 3.5 is, σ ( T T ) q = ε,i D + ε ε,i D Using appropriate rate equations for q conv and q rad, the energy balance is σ ( T T ) h( T T ) ε,oσ( T Tsur) 0 ε =,i D + ε,i D where ε =. Substituting numerical values, with / = D /D, and solving for T, ( 0 / 60) 5.67 0 8 / m K ( T 35 ) K 0 / m K( 35 300) K + 0.0/ 0.0 0 / 60 0. 5.67 0 8 / m K 35 90 K = 0 T = 75 K = 7 C. COMMENTS: Note that all temperatures are expressed in kelvins. This is a necessary practice when dealing with radiation and convection modes.
PROBLEM 3.69 KNON: Blackbody simulator design consisting of a heated circular plate with an opening over a well insulated hemispherical cavity. FIND: (a) Radiant power leaving the opening (aperture), D a = r o /, (b) Effective emissivity of the cavity, ε e, defined as the ratio of the radiant power leaving the cavity to the rate at which the circular plate would emit radiation if it were black, (c) Temperature of hemispherical surface, T hc, and (d) Compute and plot ε e and T hc as a function of the opening aperture in the circular plate, D a, for the range r o /8 D a r o /, for plate emissivities of ε p = 0.5, 0.7 and 0.9. SSUMPTIONS: () Plate and hemispherical surface are diffuse-gray, () Uniform radiosity over these same surfaces. NLYSIS: (a) The simulator can be treated as a three-surface enclosure with one re-radiating surface ( ) and the opening ( 3 ) as totally absorbing with no emission into the cavity (T 3 = 300 K). The radiation leaving the cavity is the net radiation leaving, q which is equal to q 3. Using Eq. 3.30, σ ( T ) T q 3 cav = q = q3 = () ( ε) / ε + F3 + [( /F ) + ( /3F3 )] + ( ε3) / ε3 3 Using the summation rule and reciprocity, evaluate the required view factors: F + F + F3 = F3 = 0 F = F3 + F3 + F33 = F3 =. Substituting numerical values with ε 3 =, T 3 = 300 K, π 0 - m, 3 = r ( r /) = ro r o / = 5πr o /6 =.95 π a = π o =.963 0-3 m and / 3 = 5, and multiplying numerator and denominator by, σ ( T T3 ) qcav = q = () ( ε) / ε + F3 + [( /F ) + ( /3F3 )] + 0 { } Continued..
PROBLEM 3.69 (Cont.) { } [ ] 8.95 0 m 5.67 0 / m K 600 300 K qcav = q = =.6 0.9 / 0.9 + 0 + + 5 / + 0 (b) The effective emissivity is the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the opening and temperature of the inner surface of the cavity. That is, qcav.6 εe = = 0.873 3 8 3σ T.963 0 m 5.67 0 / m K ( 600 K) = (c) From a radiation balance on, find J, (3) Eb J σ 600 J q =.6 = = J = 730 / m ( ε) / ε ( 09 )/ 0.9 () From a radiation balance on with J 3 = E b3 = σt3 = 59.9 / m and J = σt, find J J J J3 J 730 / m J 59.9 + = + = 0 ( / F ) ( / 3F3) 3 ( /.95 0 m ) ( /.963 0 m ) (5) J = 6873 / m T = 590 K. (d) Using the foregoing equations in the IHT workspace, ε e and T were computed and plotted as a function of the opening, D a, for selected plate emissivities, ε p. From the upper-left graph, ε e decreases with increasing opening, D a, as expected. In the limit as D a 0, ε 3 since the cavity becomes a complete enclosure. From the upper-right graph, T hc, the temperature of the re-radiating hemispherical surface decreases as D a increases. In the limit as D a 0, T will approach the plate temperature, T p = 600 K. The effect of decreasing the plate emissivity is to decrease ε e and decrease T. hy is this so? Continued..
PROBLEM 3.69 (Cont.) COMMENTS: The IHT Radiation, Tool, Radiation Tool, Radiation Exchange nalysis, Three- Surface Enclosure with Re-radiating Surface, is especially convenient to perform the parametric analysis of part (c). copy of the IHT workspace that can generate the above graphs is shown below. // Radiation Tool Radiation Exchange nalyses, Reradiating Surface /* For the three-surface enclosure, 3 and the reradiating surface, the net rate of radiation transfer from the surface to surface 3 is */ q = (Eb Eb3) / ( ( eps)/(eps * ) + /( * F3 + /(/( * F) + /(3 * F3))) + ( eps3)/(eps3 * 3)) // Eq 3.30 /* The net rate of radiation transfer from surface 3 to surface is */ q3 = q /* From a radiation energy balance on, */ (J J) / (/( * F)) + (J J3)/(/( * F3) ) = 0 // Eq 3.3 /* where the radiosities J and J3 are determined from the radiation rate equations expressed in terms of the surface resistances, Eq 3. */ q = (Eb J) / (( eps) / (eps * )) q3 = (Eb3 J3) / (( eps3) / (eps3 * 3)) // The blackbody emissive powers for and 3 are Eb = sigma * T^ Eb3 = sigma * T3^ // For the reradiating surface, J = Eb Eb = sigma * T^ // Stefan-Boltzmann constant, /m^ K^ sigma = 5.67E-8 // Effective emissivity: epseff = q / (3 * Eb) // Eq (3) // reas: = pi * ( ro^ ra^) = 0.5 * pi * ( * ro)^ 3 = pi * ra^ // Hemisphere, s = 0.5 * pi * D^ // ssigned Variables T = 600 // Plate temperature, K eps = 0.9 // Plate emissivity T3 = 300 // Opening temperature, K; Tsur eps3 = 0.9999 // Opening emissivity; not zero to avoid divide-by-zero error ro = 0. // Hemisphere radius, m Da = 0.05 // Opening diameter; range ro/8 to ro/; 0.05 to 0.050 Da_mm = Da * 000 // Scaling for plot Ra = Da / // Opening radius
PROBLEM 3.77 KNON: Dimensions, surface radiative properties, and operating conditions of an electrical furnace. FIND: (a) Equivalent radiation circuit, (b) Furnace power requirement and temperature of a heated plate. SSUMPTIONS: () Steady-state conditions, () Opaque, diffuse-gray surfaces, (3) Negligible plate temperature gradients, () Back surfaces of heater are adiabatic, (5) Convection effects are negligible. NLYSIS: (a) Since there is symmetry about the plate, only one-half (top or bottom) of the system need be considered. Moreover, the plate must be adiabatic, thereby playing the role of a re-radiating surface. (b) Note that = 3 = m and = (0.5 m m) = m. From Fig. 3., with X/L = Y/L =, F 3 = 0.6. Hence F = F3 = 0.38, and F3 = F = 0.38. It follows that lso, F = 0.38 =.5m 3 F = ( 0.6) =.8 m, ( ε) / ε= 0./ 3.6 m = 0.078 m 33 F = ( 0.38) =.5 m, ( ε) / ε = 0.7 /. m = 0.583m. Eb = σ T = 5.67 0 8 / m K 800 K = 3, / m, Eb = σ T 8 = 5.67 0 / m K 00 K = 5 / m. The system forms a three-surface enclosure, with one surface re-radiating. Hence the net radiation transfer from a single heater is, from Eq. 3.30, Eb E q b = ε ε + + ε F + [ /3 F + /33 F ] ε Continued..
( 3, 5) / m ( + + ) PROBLEM 3.77 (Cont.) q =. k. 0.078 0.06 0.583 m = The furnace power requirement is therefore q elec = q = 3.8 k, with Eb J q =. ( ε) / ε where ε J = Eb q = 3, / m,00 0.078m ε lso, J =,679 / m. ε J = Eb q =, 5 / m (,00 ) 0.583m ε J = 3,98 / m. From Eq. 3.3, JJ3 J3J = /3 F /33 F J J3 3F3.5 = = = 0.63 J3 J F3.8.63J 3 = J+ 0.63J =, 69 + 8537 = 3,66 / m J3 Since J 3 = E b3, = 9,3 / m / ( σ 8 ) / T3 = E b3/ = 9,3/ 5.67 0 = 76 K. COMMENTS: () To reduce q elec, the sidewall temperature T, should be increased by insulating it from the surroundings. () The problem must be solved by simultaneously determining J, J and J 3 from the radiation balances of the form Eb J = F ( J J) + F 3( JJ3) ( ε) / ε Eb J / ε ( ε ) = F J J + F3 JJ3 0= 3 F J3 J + F3 J3J.
PROBLEM 3.78 KNON: Geometry and surface temperatures and emissivities of a solar collector. FIND: Net rate of radiation transfer to cover plate due to exchange with the absorber plates. SSUMPTIONS: () Isothermal surfaces with uniform radiosity, () bsorber plates behave as blackbodies, (3) Cover plate is diffuse-gray and opaque to thermal radiation exchange with absorber plates, () Duct end effects are negligible. NLYSIS: pplying Eq. 3. to the cover plate, it follows that N ε J Jj ε E b J = = F ( J J ) F 3 ( J J 3). ε ε j + = ( F i ij) From symmetry, F = F 3 = 0.5. lso, J = E b and J 3 = E b3. Hence Eb J = 0.0556 J Eb Eb3 or with Eb = σt,.j= Eb+ 0.0556 Eb+ Eb3.J 8 8 = 5.67 0 ( 98) / m + 0.0556( 5.67 0 ) ( 333) + ( 33) / m J = 76.6 / m From Eq. 3.9 the net rate of radiation transfer from the cover plate is then 8 b 5.67 0 E J ( 98) 76.6 q = = = ( 65.5" ). ( ε) / ε ( 0.9 )/ 0.9( ") The net rate of radiation transfer to the cover plate per unit length is then q = ( q / ") = 66 / m. COMMENTS: Solar radiation effects are not relevant to the foregoing problem. ll such radiation transmitted by the cover plate is completely absorbed by the absorber plate.