本研究是以Angstrom method理論為基礎，開發熱擴散率量測儀器，實驗方面以定義不同之厚寬比τ，以快速並且準確地測量金屬材料、石墨片、均溫板以及熱管的熱擴散率α。校正方面以銅、錫、鋁合金、銀、黃銅、氮化鋁機板等六種金屬進行機台校準測試，以一維不同的修正模型為主，比較實驗值與標準值誤差。對於一維修正模型，由厚寬比τ=T/W做區分，當τ小於0.02，使用一維分析2D模型；當τ介於0.02與0.075間則用一維分析1D模型；當τ大於0.075則用一維分析0.5D模型。
另外測量廠商提供之石墨片(T68AP)，石墨片是由三層構造所組成的，測得αT68AP=4.12。進而推算其k值，整體石墨片ktot,T68AP=590W/m．K。本文以均質模型為基礎，經由並聯熱阻的關係，算出石墨層之熱傳導係數kc,T68AP=1060.9 W/m．K。
在熱管方面，熱管量測熱擴散值結果為: αA=115cm2/s、αB=167cm2/s。進而算得kHP,A,TDMI=18843 W/m‧K 、kHP,B,TDMI=29696W/m‧K。以熱管性能測試儀器(HPPT)量測同一熱管之最大熱傳量，數據顯示熱管A最大熱傳量Qmax,A為20W，熱阻Rth,A為1.17℃/W，算得kHP,A,HPPTS=4536W/m‧℃；熱管B最大熱傳量Qmax,B為40W，熱阻Rth,B為0.8℃/W，算得kHP,B,HPPTS=6634 W/m‧℃。理論上在求得熱管最大熱傳量Qmax後，可以求得熱管之熱阻值。如果利用此熱阻值再利用傅立葉公式得到熱管之熱傳導係數會遠小於利用TDMI量測之熱傳導係數，也較文獻所記載熱管之熱傳導係數14000 ~ 20000 W/m‧K小的很多。根據實驗結果，總結熱擴散量測儀器TDMI量測散熱材料或者是均溫板、熱管都是比較可靠且精準的量測儀器。

This paper is a development of Thermal Diffusivity Measurement Instrument (TDMI) based on Angstrom method . In terms of the experiment , for measuring the thermal diffusivity α we delimit different aspect of ratio τ , make us be able to measure materials、graphite sheets、heat pipes、vapor chamber quickly and accurately. In calibration aspect, choosing copper、tin、aluminum、silver、brass、aluminum nitride to take test and make sure the accuracy of TDMI , including one-dimension different factor thermal diffusivity theory and compare the error between the experience and the standard . In regards to one-dimension revise factor, as distinguish it by ratio of thick and width(τ ) : when τ is lower than 0.02, use one-dimension 2D mode ; when τ is between 0.02 and 0.075 use one-dimension 1D mode；when τ larger 0.075 use one-dimension 0.5D mode.
In addition, after measuring graphite sheet (T68AP) which is composed of three structures supplied by the T-global technology, obtain data as below: αG=4.12cm2/s . According to this data, calculated the total thermal conductivity ktot,E=590W/m．K. Based on parallel thermal resistance, reach calculated that the thermal conductivity of graphite sheet kc,G=1060.9 W/m．K.The TDMI measuring for the graphite sheet is valid.
In the aspect of heat pipes, after measuring the thermal diffusion shows data : αA=115cm2/s、αB=167cm2/s ,then is able to calculate out the thermal conductivity as kHP,A,TDMI=18843W/m．K and kHP,B,TDMI=29696W/m‧K. By Heat-Pipes Performance Testing System (HPPTS), the data shows that the maximum heat transfer of heap pipe A Qmax,A=20W, the thermal resistance Rth,A=1.17℃/W, and the thermal conductivity kHP,A,HPPTS=4536 W/m‧℃ and the maximum heat transfer of heat pipe B Qmax,B=40W, the thermal resistance Rth,B=0.8℃/W, the thermal conductivity kHP,A,HPPTS=6634W/m‧℃. Theoretically, after obtaining the maximum heat transfer (Qmax) of the heat pipe, the thermal resistance can be obtained. Some people can use this thermal resistance data and Fourier formula to obtain the heat transfer coefficient .The problem is that the heat transfer coefficient obtained by the Fourier formula the heat conduct material must be in a solid state, and this is different from the known heat pipe that used to be utilizing liquid evaporation and convection to transfer the heat. Therefore, the obtained solid heat transfer coefficient must be much smaller then the heat transfer coefficient of the actual liquid convection heat transfer mode.

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