Question

1. Steady-state temperatures at selected nodal points of the symmetrical section of a flow channel are known to be, T, = 95.47°C,T, = 117.3°C, T, = 79.79°C,T, = 77.29°C,T, = 87.28°C.T. = 77.65°C.The wall experiences uniform volumetric heat generation of q'=106 W/m and has a thermal conductivity of k = 10"-. The inner and outer surfaces of the channel experience convection with fluid m, K temperatures of T = 50°C, T = 25°C and convection coefficients of h = 500 a) Determine the temperatures at nodes 1, 4 (10 points). b) Calculate the heat rate per unit length from the inner fluid to surface B (10 points). c) Calculate the heat rate per unit length (W/m) from the outer surface A to the adjacent fluid (15 points) Insulation Surface B 5 -6.9 6 -Symmetry plane Ar=Ay - 25 mm 101 Surface A

answer a, b, and c

Answer 1

Date: Date: | Page no: Page no: Soll Dlow the schematic Diagram of a flow channel: TAY=A2- 25 mm Too,i = 50°C hi=500 W/m²K 1918 surface of Insulation - 8 = 100 W/m² K = 10 W/m-k <symmetry plane / T8 l Tg Tio surface a Todo = 25°c ho=25 W/m²K Given : Steady-state temperatures at selected nodal points are, T = 95.47°6, T = 117.3°6 T5 = 79. 19°C 1. To = 77-29°C, T8 = 87028°C, Tio= 77 65°C Assumptions: a steady state, two dimensional Conduction 2 Constant Properties.

Date:___ _ Page no: 2 Analysis : (a) The nodal binite - difference equations are obtained from energy Balances on control volumes about the nodes shown in schematic below. Т2 Node 1 : qa't qo't vot vd + F g = 0 10tk (en) "The + K ( 22 ) Toy Tu 40 44(14.09) | T, = :( Tz + 3) + 4 6x² By substitutiny values ; Ti = 95.47 +117.3)°C Hloo w/m3) (25X25 )x10 m2 4x10 W/m ok 1 T = 122.010 ot 122°C Node 4 : votan't wc + vat qe tvit Eg=0 K(27) 744 thi (%)(Tosi Te + 8 (AX) (Tan-Ta) + KDV 3 To + KCAK) T-T2 + k (DY) T3-TF + 9 (30%A4) = 0 AX. Ay ΔΧ

Date: Page no: (3) 74 = 12 +263 + T5 +218 +2 Chi DX/K2 Toni +(39 02°/2k)] [6+2 (hi DX/K) T4 = 94 50°C Ans The heat rate per unit length from the inner fluid. to the surface B VB is the sum of the convection theat rates from the inner surfaces of nodes 2,4,5 46 1 2 2 98 = hi[(04) (Too,i-T) + ( 14 + ) (Tooi - T4)+ ax Tool - Is) +10x) (Too, i To)7 198 =500m ) (50-95.47) + (25 + 25 )(50-94-50) +25(50-79.79) +(25)<50-7329)]x109 m.k qB = - 1383 W/m Ans. © The heat zate Pex unit length from the outer surface to the adjacent bluid, un, is the sum of the convection heat rates from the outer surfaces of nodes 7, 8, 9410; 194 = he (AX) (T7 - Too,0) + 0%(T8-T2,0) +-% (T9 - T0,0) + (19/2) (J10-T00,027 2 mak 2 VA = 25 W (25)(95.80-25) +25 (87.28 -25)+25 (19-67-25) +(25/2) (77.65-25)710 Y = 111.7 W/m กด