找到了一份Icepak TEC Macro的相關資料
給大家做參考
簡單的說
在Icepak中,要先使用TEC Macro建立出TEC幾何,與原本的系統結合
然後使用RUN TEC Macro,設定TEC的Seeback coefficient等參數(這是TEC的重要參數,必填)
然後,最重要的是,要開始求解必須要點選RUN TEC Macro的"accept"
而非平常習慣的"run solution"
tec-modeling.rar (461.41 KB)
Thermoelectric Cooler Modeling
What is thermoelectric cooler?
A thermoelectric cooler (TEC) is a solid state heat pump that utilizes the Peltier effect
The Peltier effect: when an electric current passes through the junction of two dissimilar electrical
conductors, it causes heat to be absorbed or released depending on the direction of current flow.
Practical application requires the development of materials that are good conductors of electricity but
poor conductors of heat
Bismuth telluride is primarily used as the semiconductor material
Why use thermoelectric cooler?
Keep temperature below ambient
Conventional fan + heat sink cooling technology can only keep device temperature above ambient
Precise temperature control
TEC has a feedback control circuit that adjusts its cooling capability based on the temperature monitored.
TEC performance formula
Terminology
TH Hot Side Temperature (Kelvin)
TC Cold Side Temperature (Kelvin)
DT TH - TC (Kelvin)
Tave 1/2 (TH + TC) (Kelvin)
G Area / Length of T.E. Element (cm)
N Number of Thermocouples
I Current (amps)
COP Coefficient of Performance (QC / IV)
a Seebeck Coefficient (volts / Kelvin)
r Resistivity (ohm cm)
k Thermal Conductivity (watt / (cm Kelvin))
Z Figure of Merit (a2 / (p k)) (Kelvin-1)
S Device Seebeck Voltage (2 a N) (volts / Kelvin)
R Device Electrical Resistance (2 p N / G) (ohms)
K Device Thermal Conductance (2 k N G) (Watt / Kelvin)
Material property
Typical material parameters (@ T = 296 K)
a = 2.0 x 10-4 (volts / Kelvin)
r = 1.0 x 10-3 (ohm cm)
k = 1.5 x 10-2 (watt / (cm Kelvin))
Z = 2.67 x 10-3 (Kelvin-1)
Material Property Coefficients
a = (a0 + a1 x Tave + a2 x Tave^2) x 10-9 (volts / Kelvin)
a0 = 22224.0
a1 = 930.6
a2 = -0.9905
r = (r0 + r1 x Tave + r2 x Tave^2) x 10-8 (ohm cm)
r0 = 5112.0
r1 = 163.4
r2 = 0.6279
k = (k0 + k1 x Tave + k2 x Tave^2) x 10-6 (watt / (cm Kelvin))
k0 = 62605.0
k1 = -277.7
k2 = 0.4131
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