A method for reducing axial heat conduction of regenerator regenerator material of regenerative cryogenic refrigerator
A low-temperature refrigerator and cold storage material technology, applied in refrigerators, superheaters, refrigeration components, etc., can solve the problems of large heat conduction loss, large thermal conductivity, and low thermal conductivity of regenerators, and achieve sufficient heat exchange and flow. Low cost, simple and reliable method
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Embodiment 1
[0027] Step 1. The inner diameter of the regenerator of the regenerative cryogenic refrigerator is Φ5mm and the length is 50mm. According to the total thickness of the thermal barrier is 3% of the total length of the regenerator cylinder, the total thickness of the thermal barrier is 1.5mm , a total of 3 thermal barriers with a thickness of 0.5 mm need to be added, and the cold storage material of the regenerator is divided into 3 parts evenly. The outer diameter is closely matched with the inner hole of the regenerator, and the holes are evenly drilled on the thermal barrier. %,
[0028] Calculation formula (I) for the ratio of the hollow area to the cross-sectional area of the thermal barrier:
[0029] η = n × d 2 2 d 1 2 × 100 % - ...
Embodiment 2
[0045] Step 1. The inner diameter of the regenerator of the regenerative cryogenic refrigerator is Φ7mm and the length is 55mm. According to the total thickness of the thermal barrier is 3% of the total length of the regenerator cylinder, the total thickness of the thermal barrier is 1.65mm , it is necessary to add 3 pieces of 0.5mm thick thermal barriers, divide the heat storage material of the regenerator into 3 uniform parts, and determine the outer diameter of the thermal barrier as Φ7mm according to the inner diameter of the regenerator as Φ7mm, so that the thermal barrier The outer diameter and the inner hole of the regenerator are closely matched, and the holes are evenly drilled on the thermal barrier. %,
[0046] Calculation formula (I) for the ratio of the hollow area to the cross-sectional area of the thermal barrier:
[0047] η = n × d 2 2 ...
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