Problema 3.ppt Jorc.pptx

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3. Un acero inoxidable (alambre de 3.2 mm de diΓ‘metro) y 30 cm de largo, esta sometido a voltje de 10 volts. La temperatura de superficie exterior se mantiene a 93 ℃ . Calcular la temperatura en el centro del alambre, la resistencia del alambre es de 70πœ‡Ξ© Ξ€π‘π‘š y K=22.5π‘€Ξ€π‘šβ„ƒ (7π‘₯10βˆ’7 π΄π‘šπ‘)(0.3π‘š) R= 8.042π‘₯10βˆ’6π‘š2 = 0.026Amp/m=0.026 10π‘£π‘œπ‘™π‘‘π‘‘π‘  i= 0.026Ξ© = 384.615π΄π‘šπ‘π‘ 

D = 3.2 mm L=30 cm

V=10 volts

Ξ©

𝑃 = 𝑖 2 𝑅 = (384.615π΄π‘šπ‘π‘ )2 (0.026Ξ©)=3846.146w

Tw=93℃ T=Centro = X Resistividad ( 𝑒) 70πœ‡Ξ© Ξ€π‘π‘š

K=22.5π‘€Ξ€π‘šβ„ƒ

70πœ‡Ξ© Ξ€π‘π‘š

10βˆ’6 Ξ©

1π‘š βˆ’7 amp =7x10 1πœ‡Ξ© 100π‘π‘š

A=πœ‹π‘Ÿ 2 =πœ‹ = (1.6π‘₯10βˆ’3 π‘š)2 = 8.042x10βˆ’6 π‘š2

𝑝 3846.146𝑀 βˆ’9 𝑀 °𝑔 = 2 = = 1.594π‘₯10 ΰ΅—π‘š3 πœ‹π‘Ÿ 𝐿 πœ‹ 2.56π‘₯10βˆ’6 0.3π‘š (1.594π‘₯10βˆ’9 π‘€Ξ€π‘š3 )(1.6π‘₯10βˆ’3 π‘š)2 𝑇𝑐 = + 93℃ 4(22.5 π‘€ΰ΅—π‘šβ„ƒ)

𝑇𝑐 = 138.340℃

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