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There rods of material X and there rods of material Y are connected as show in Fig. 7(e).21. All the rods are of identical length and cross-sectional area. If end A is maintained at 60^(@)C and junction E at 10^(@)C, Calculate the temperature of the junctions B,C and D. Given thermal conductivity of X is 0.92 cal cm^(-1) s^(-1) .^(@)C^(-1) and of Y is 0.46 cal cm^(-1) s^(-1) .^(@)C^(-1). |
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Answer» Let l be the LENGTH each rod and A be the area of cross-section of each rod. At STEADY state, heat entering a junction per second is equal to heat leaving that junction per second. For junction B, Heat entering the junction B per second `=(K_(y)A(T_(A)-T_(B)))/(l)`.....(i) Heat leaving the junction B per second `(K_(x)A(T_(B)-T_(C)))/(l)+(K_(Y)A(T_(B)-T_(D)))/(l)`....(ii) Equating (i) and (ii), we have `(K_(Y)A(T_(A)-T_(B)))/(l)=(K_(X)A(T_(B)-T_(C)))/(l)+(K_(Y)A(T_(B)-T_(D)))/(l)` or `K_(Y)(T_(A)-T_(B))=K_(X)(T_(B)-T_(C))+K_(Y)(T_(B)-T_(D))` or `0.46(60-T_(B))=0.92(T_(B)-T_(C))+0.46(T_(B)-T_(D))` or `(60-T_(B))=2 (T_(B)-T_(C))+(T_(B)-T_(D))` or `4 t_(B)-2 T_(C)+T_(D)=-10`...(III) For the junction C, Proseeding as above, we get `K_(X)(T_(B)-T_(C))=K_(X)(T_(C)-T_(D))+K_(X)(T_(C)-T_(E))` or `T_(B)-3 T_(C)+T_(D)= -T_(E)` or `T_(B)-3 T_(C)+T_(D)= -10`.....(iv) For the junction D, Prosseding as above, we obtain `K_(Y) (T_(B)-T_(D))=K_(X)(T_(D)-T(C))+K_(Y)(T_(D)-T_(E))` or `0.46 (T_(B)-T_(D))=0.92(T_(D)-T_(C))+0.46(T_(D)-T_(E))` or `T_(B)-T_(D)=2(T_(D)-T_(C))+(T_(D)-T_(E)` or `T_(B)+2 T_(C) -4T_(D)= -10`.....(V) On solving the equation (iii), (iv) and (v), we get `T_(B)=30^(@)C, T_(C)=20^(@)C=T_(D)` |
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