Thermodynamics HW Solutions 29

Thermodynamics HW Solutions 29 - Chapter 1 Basics of Heat...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 1 Basics of Heat Transfer 1-69 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well- insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis For each sample we have & / (. . . Q A T == =−= ° 28 2 14 01 001 82 74 8 W m)( m) m C 2 Δ Then the thermal conductivity of the material becomes & & ( ( . )(8 ) Qk A T L k QL AT =⎯ ⎯= = ° Δ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

Ask a homework question - tutors are online