Question

Done Expert Q&A Consider the perpendicular rectangles shows ieally. -04 0 03 -0.2m r-0.5m Determine the shape factor F 02 A drying oven consists of a long semicincular dact of diameter D 1m. T-1200 -325K D-1 Materials to be dried cover the base of the oven while the wall is maintained at 1200 K. What is the drying rate per unit length of the oven Okpls-m) if a water coted layer of material is maintained at 323 K doring the drying process? Blackbody behavior may be assumed for the water surface and the Done Expert Q&A 03 A room is represented by the following enclosure where the ceiling () has an emissivity of 08 and is maintained at 40°C by embedded electrical heating elements. Heaters are also used to maintain the floor (2) of emissivity 0.9 at 50°C. The right wall (3) of emissivity 0.7 reaches a temperature of 15C on a cold winter day. The left wall (4) and end walls (5A) and (5B) are very well insulated. To simplify the analysis, treat the two end walls as a single sur- face (5). 4 m Assuming the surfaces are diffuse and gray. fiad the net radiation beat transfer from each surface

Answer 1

i IOm 6 m ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Surfaces are isothermal and irradiated uniformly, (3) Negligible convection effects, (4) A5A5A A5B ANALYSIS: To determine the net radiation heat transfer from each surface, find the surface radiosities using Eq. 13.20 To determine the value of Ji, energy balances must be written for each of the five surfaces. For surfaces 1, 2, and 3, the form is given by Eq. 13.21. Ehi-J i= 1,2, and 3 1 AF For the insulated or adiabatic surfaces, Eq. 13.22 is appropriate with q 0; that is 4 and 5 In order to write the energy balances by Eq. (2) and (3), we will need to know view factors. Using Fig. 13.4 (parallel rectangles) or Fig. 13.5 (perpendicular rectangles) find: F12 F21 0.39 F13 F14 0.19 F34 F43 0.19 F24 F13 0.19 ZX-4/10 0.4, x/L = 10/6 = 1.66, Z/X -4/10 0.4, Y/L = 4/6 = 0.67 YX 6/10 0.6 Note the use of symmetry in the above relations. Using reciprocity, find, F2 = A2F23 =A2F3 =40×0.19-: 0.285; F1 = ALPİ,-48 × 0.23:0 Fis 1-2-3-4-1-0.39-0.19-0.19 Fi=A2F3 = 40×0.19 = 0.285 3 Fs =--× 0.25 = 0.208 48 From the summation view factor relation, 0.23 F35 1-1 F32 F341-0.285-0.285-0.19 0.24 Continued...

Using Eq. (2), now write the energy balances for surfaces 1, 2, and 3. (Note Eb oT) 544.2-J l-0.8 / 0.8×60 1/60×0.39 1/60×0. 19 1/60×0.19 1/60×0.23 -1.2500, + 0.0975J2+ 0.0475, + 0.57ОЈ5--544.2 617.2-J 1-0.9/0.9×60 2-J3 1/60×0. 19 2-J5 1/60×0.23 234 1/60×0.19 1/60×0.39 0.0433J1 1.111J20.02111J30.02111J4 0.02556J5617.2 390.1-J 1-0.7 / 0.7×40 1/40×0.285 1/40×0.285 1/40×0.19 1/40×0.24 0.1221J0.1221J2- 1.4284J3 +0.08143J4 0.1028J5390.1 Using Eq. (3), now write the energy balances for surfaces 4 and 5 noting q4 q50 1/40×0.285 1/40×0.285 1/40×0.19 1.40×0.24 0.285J1 + 0.285J2 +0.19J3 - 1.0J40.24J5 0 J5-J2 1/48×0.288 5-J3 1/48×0.208 1/48×0.288 1/48×0.208 0.288J1 + 0.288J2 +0.208J3 +0.208J4-0.992J5-0 Note that Eqs. (4) - (8) represent a set of simultaneous equations which can be written in matrix notation following treatment of Section 13.3.2. That is, [A] [JC] with 544.2 -617.2 -1.250 0.0975 0.0475 0.0475 0.057 0.0433-.11 0.02111 0.02111 0.02556 545.1 607.9 A=| 0.1221 0.1221-1.4284 0.08 143 0.1028 C390.J 441.5 W/m 0.285 0.285 0.190 -1.000 0.240 0.288 0.288 0.208 0.208 0.992 542.3 5410 where the J, were found using a computer routine. The net radiation heat transfer from each of the surfaces can now be evaluated using Eq. (1) qi 60 m [0.39(545.1 - 607.9) +0.19(545.1-441.5) 0.19(545.1-542.3) +0.23(545.1-541.0)] W/m200 W < q2 60 m [0.39(607.9 - 545.1) +0.19607.9-441.5) +0.19607.9-542.3) +0.23(607.9-541.0) W/m 5037 W < q3 = 40 m-[0.285(44 1.5-545.1) + 0.285(441.5-607.9) +0.19(441.5-542.3) 0.24(441.5 541.0)] W/m4,799 W Since A4 and A5 are insulated (adiabatic), q4 95 0

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