Thermodynamics HW Solutions 28

Thermodynamics HW Solutions 28 - Chapter 1 Basics of Heat...

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Chapter 1 Basics of Heat Transfer 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 The electrical power consumed by the heater and converted to heat is & () ( . ) WV I e == = 110 0 6 66 V A W Q 3 cm 3 cm The rate of heat flow through each sample is & & Q W e = 2 66 33 W 2 W Then the thermal conductivity of the sample becomes A D Qk A T L k QL AT = ⎯→ ⎯= = ° ππ 22 4 004 4 0 001257 33 0 001257 10 (. ) . & & ( ) ( ) m m = W)(0.03 m) mC 2 2 Δ Δ 78.8 W / m. C 1-68 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.
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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.

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