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Ejercicio pared compuesta

Pared plana sin generación interna de calor. Ejercicio pared compuesta. Calcúlese el flujo de calor a través del muro de la figura. A. C. D. λ A = 75 W / m K λ B = 58 W / m K λ C = 60 W / m K λ D = 20 W / m K. A = 2 m 2. a. θ 1 = 500 ºC. El circuito eléctrico equivalente será:.

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Ejercicio pared compuesta

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  1. Pared plana sin generación interna de calor Ejercicio pared compuesta Calcúlese el flujo de calor a través del muro de la figura A C D λA = 75 W / m K λB = 58 W / m K λC = 60 W / m K λD = 20 W / m K A = 2 m2 a θ1 = 500 ºC El circuito eléctrico equivalente será: B QC θ4 = 100 ºC QD QA a RC = LC / AcλC QB RA = LA / A λA RD = LD / A λD 20 40 25 RB = LB / ABλB cm

  2. Pared plana sin generación interna de calor Ejercicio pared compuesta Resolviendo el circuito: Q RC A C D R = RA + [ RB·RC / ( RB + RC ) ] + RD RD RA RC·RB / (RC + RB ) a θ1 = 500 ºC RB RA = LA / (A λA)= 0’2 / (A·75)= 0’00267/A ºC / W RB = LB / (ABλB )= 0’25 / [(A/2) 58] = 0’00862/A ºC / W B RC = LC / (AcλC )= 0’25 / [(A/2)·60] = 0’00834/A ºC / W θ4 = 100 ºC RD = LD / (A λD )= 0’4 / (A·20) = 0’02/A ºC / W a R = RA + [ RB·RC / ( RB + RC ) ] + RD = (1/A)·[0’00267+ [ 0’00862·0’00834 / (0’00862 + 0’00834) ] + 0’02 = 0’0269/A ºC / W Q = ( θ1 - θ4 ) / RD = ( 500 – 100 ) / (0’0269/A) = 29.730 W 20 40 25 cm

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