suposse that the thermal conductivity of copper is twice that of the aluminium y four times that of brass. Three metal rods, made of copper, aluminium and brass respectively, have 6 inches of lenght and 1inch of diameter.The rods are matched side by side of their ends, with the aluminium rod in the middle. The free ends of the copper and brass rods are at 100?C and 0?C respectively. Calculate the temperatures of equilibrium of the union between copper and aluminium rods, and of the aluminium and brass ones.
The heat conduction equation is: heat flux q - k * (dT/dx) The three rods (equal cross-section (R 0.5 inch), equal length (L 6 inches): - Copper: kc 4a - Aluminium: ka 2a - Brass: kb a - The heat transferred along each rod is the cross-sectional area (pi*(0.5)^2) multiplied by the heat flux. - Since heat is provided/taken out only at the end points (T_h 100 ?C, T_c 0 ?C), the heat transfer must be the same at each junction: nothing goes missing. - Since the cross-sectional areas are all the same, the heat-flux values must also be the same. - The heat flux depends on the temperature gradient over each bar: but since the length of each bar is the same, the temperature gradient depends only on the temperature drop over each bar: q -k * dT/dx -k * ΔT/L So if T(0) T_h at the left end of the copper rod T(L) temperature at the copper/aluminium junction T(2L) temperature at the aluminium/brass junction T(3L) T_c at the right end of the brass rod we conclude that if all the q's are equal, (eqn 1): kc * (T(L) - T(0)) ka * (T(2L) - T(L)) 4a * (T(L) - T(0)) 2a * (T(2L) - T(L)) 2*T(L) - 2*T(0) T(2L) - T(L) (eqn a): T(2L) 3*T(L) - 2*T(0) and (eqn 2): ka * (T(2L) - T(L)) kb * (T(3L) - T(2L) 2a * (T(2L) - T(L)) a * (T(3L) - T(2L) 2*T(2L) - 2*T(L) T(3L) - T(2L) (eqn b): 3*T(2L) T(3L) + 2*T(L) (eqn a) and (eqn b) imply: T(3L) + 2*T(L) 3*T(2L) 3*( 3*T(L) - 2*T(0)) 9*T(L) - 6*T(0) T(3L) + 6*T(0) 7*T(L) T(L) (T(3L) + 6*T(0))/7 and T(2L) 3*T(L) - 2*T(0) (3/7)*(T(3L) + 6*T(0)) - 2*T(0) (3/7)*T(3L) + ((18-14)/7)*T(0) (9*T(3L) + 4*T(0))/7 Therefore: T(0) T_h 100?C T(L) (T_c + 6*T_h)/7 (6/7)*100 85.7?C T(2L) (9*T_c +4*T_h)/7 (4/7)*100 57.1?C T(3L) T_c 0?C T(L) is the temperature of the copper/aluminium junction, T(2L) is the temperature of the aluminium/brass junction.
