Question

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Use the pre-lab data to calculate the thermal conductivity of the Aluminum sample measured in a composite wall experiment. Mid-section material: Aluminum Ambient temperature: 27°C T.(°C) Tz(°C) Power(W) 96.1 84.5 43.0 T3(°C) T(C) Ts(°C) T.(°C) T (°C) 72.9 62. 6 52.2 41.6 30.6 Distance 0.0 0.02 0.04 0.06 0.08 0.10 0.12 from T(m) PAGE: Given Data o Thermal conductivity of the brass (type CZ121) is 123 W/m.K o Thermal convection of air is 20 W/ m K O Area of all composite walls is 7.07x10-m'. O Thickness of the middle section is 20 mm. O L=60mm, L2=20mm, and L3-60mm. O Sections 1 and 3 are made of brass. Formulas to use R-7-1. R = R +R,+R. Rumah yang

Answer 1

In given question we are asked to find out the thermal conductivity of aluminum from composite wall section of two brass and one Al at mid section of 20 mm .

In table temperature at different location of section is given

Lab data Tq Mid Section-Al 1927' Tres Tz | T3 | Tu To fowe wlas 96.184.5 12.962,6 52.2 43.0 Distance from Tiemy 0.0 0.02 0.04 0.06 0.08 To *41.6 | 3006 0.10 0.12 A kg='123 19= Kg K3 7123 she ! MK arrangement composite way diogram is shown in 0.06m 0.06 mm m 70.02% 0.04m! Ligure , we have to find thermal Conductivity of Aluminum 43.0W) (Brass) (AU (Brassi The temperature vallees are given in table only up to Tz (m) 2 Tol Tu To To which only at oodym thickness of thind brces well so we have to talhe hecat flow upto 0-12m finom (Section 3) Section & temp. Te I L = 6mm Now total resistance from 0.0 to 0.12 m T. L₂ = 20 mm . . LI's Rtotal = 1 R2 A2 R3A3 - A=A1 - A2=A3 = 7,07x10 m² 10.J2m 12 1 1'3 - Yomm L3= 60 mm 0.04 Rent forum et om een + 09na 1 Ti-T7 KF K3= 123 W/onk 1992.waw Onde +0.02 Out + 0 0 =65.3xA 43" =96.16 4 = 306 °C 123 K2 ik, 1 kg = 75.77 Wimp 75.77 Wim 0.06 0.02 0.06 Til Bril Brit eraler But if we consider it as and temp. at distance 0.14m from Ti of then outlet + og , 07x7°=) 1K2 = 197. 37 Wmke

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