Question

The wall of a heat exchanger separates hot water at Tin 90 C with h." 80 w/m2.K from cold water at Tout 10 C with h,ut 60 W/m2.K. The thermal conductivity of the heat exchanger is 180 W/m.K To extend the heat transfer area, two-dimensional ridges are machined on the cold side of the wall. This geometry causes non-uniform thermal stresses, which may become critical for crack initiation along the lines between two ridges. To predict thermal stresses, the temperature field inside the wall must be determined. Convection coefficients are high enough so that the surface temperature is equal to that of the water on each side of the wa!!. (a) Identify the smallest section of the wall that can be analyzed in order to find the temperature field in the whole wall, (b) For the domain found in part (a), obtain the temperatures Col d ater hast 10 nunn S 2 10 san 2.

Answer 1

Solufion'- Giiven dato The pchile goum in the ue cage of tho dimensiona! heat 'onsken. since he geo asodient n those uo kinds ot plares, Hence no heat ffoue can consid them os insulates suaces and psoced out aral^sis AsSumptíons: stete cordition I-steads 2. Tumensonal corditl on (o,2 C0j0)

flow anabys's oooss the left swfoce. Hence it can be considered ag solectead nodles, we aroa NOtü, Arno knouo the ten e teimpouhiores 24 1203 So, cue Sust have o find out Bihsee tem pernkoies here; Tu, To and Toz Awalusis A Node C,): 4 3 2. Ho,109,十92+934% :0 ( stady skale. Ti': constant.)

Toit 110-4Tuno Tol-uTu_ -110-10 b) Node (o,i)'- go- To AY Co,i) o2-Toi AY (ax/2) Cpt Toi) c) Node Co, 2):- AY BX Ay

2 so, we have D,②'C CYuagong and Voor ables Values of Tol -46.92ΐ