Question

Problem V(15); The temperature at the following nodal points of the symmetrical section (right side of sketch) of a flow channel ( left side of sketch)under steady state are shown on sketch (right side bottom). The inner surface of channel is at a uniform temperature of 600 K and its outer surface is exposed to a fluid at 300 K with h=50 W/m^2/K i) Determine the temperature at nodes 2,4, and 7 ii) Determine the heat rate per unit length from the upper surface of the section T- 300 K A= 50 Wim?. K Az Ay = 0.01 m T-600 K k-1 Wim-K T-T,- 600 K T- 492 K T.- 430 K T- 394 K

Answer 1

ANSWER! Given that the temperature at the following nodal points of the Symmetrical Section A uniform temperature = 6ook. the inner Surface of Channel 13 Gt a uniforro temperature 600k Outer Surface is Exposed to a fluid = 300k, - ha 50 w162lk. e Symmetty adiabak k=1w/mk Axsay=00m Node TIK) 430 394 492 600 ASSUMPHONS! ( Steady state conditions (a) TOO dimensional Conduct 100.

(3) NO internal volumetric generazion. (6) {on stant properties, Analysis: (a) Define control volunes about the nodes 214 and 7, taking, advantage Of Symmelty where appropriate and performing Energy balances Ein Eout =0 with An-Ay. - Noder-2 zarab+26+ id=0 Tzyb T ela • I I .13 eid L. 216 lac har (te=12) +K(44b) CE-TO) + KA2 TGT2 +* (09/) (TT TR= [0.5 T, +0-5 T3 +Tst ( hox/k) T] [+ (hau/K)] To = 0.5(430) +015 (394) +499 + (50 10.08) 1300 2+50 (0:01) Ti 215+ 197 +992 +150

T = 1054 25 T2 = 4216 429K, Node!-4 12h น Suomekry plane qatalb + 2'c=0 h(Axi2) (99-04) +0+* (ayla) zo -To [Tat (hax/k) 7o) [1+ har 74 394+0:5X300 1+0.5 TU 394+150 1.5 - T4= 544 T4 = 362-66 - 363k. . "T4 = 363K frtro the first schemat C ie Cogn 2) that the diagonal 19 ou an olayloy no a Symmetry adiabat, we can treat hode hence T7= 0.95 (T3773 +16 +76)

gubsu tute tee value T7 = 0.95(394+394. +4924492) "T+ 20.95( 177a) T1 = 443 K. (b) the heat transfer loss from the uppey surface can be expressed as the sum of the convection rates from Each node cy i Mustaated in the first schematic s'cv= 81 +82 +3'9 +24 acv = hlAxle) (72-72) at h(41) (12-Ta) tha1 (43-12)+ h (axb) (793). plev= 50x0.01 [ 430-300 + (422-300) + (394.-300) + 363 POJK Pevo 0.5 ( 130 + 122 +94 + 627 O'CU = 0:5( 65+192 +94 +315] El CV=0.5 312.5] 2'cv = 156.95 wlm.