Calorimetric measurements on losses in superconducting materials

Abstract

AbrtractWe are constructing a cryogenic dual compensated calorimeter for loss measurements in superconducting materials. We plan to measure the losses due to incomplete Meissner effect in slowly varying small magnetic fields appropriate for levitation. These measurements bear importance for the realization of the kilogram using the "superconducting magnetic levitation method".

Arguments for making use of a differential gas thermometer in cryogenic microcalorimetry

“…That set-up involved a simple differential gas thermometer to measure the temperature drop along the sample, provided the heat flux through the sample is known. Such a thermometer consists of two reservoirs filled with 4 He gas and connected by capillary tubes to either input of a differential pressure meter. With this arrangement any difference in gas temperatures of the reservoirs causes a pressure difference to be measured by the meter.…”

“…Neglecting the small amount of gas inside the capillary, the gas column of the arm is divided into two volumes: the working volume V t having a gas temperature T l and the dead volume V d of the pressure sensor. We assume the 4 He gas to be ideal 3 . The equation of state of the gas inside the column can then be written as:…”

“…Also, to make c as close to unity as possible one should choose low mass samples and calorimeter construction materials of low specific heat capacities. This is feasible for a variety of calorimetric problems, particularly by using single crystal sapphire for the sample holder and walls of the thermometer reservoir 4 . Where this is the case, and τ is chosen to be, say, 60 s the estimate for δP from (3) is about 0.1 nW.…”

“…This limit could probably be achieved in practice with a thorough calorimeter design. Note that for the cryogenic calorimeter having a dual-compensated structure with resistive thermometers of 470 K −1 sensitivity, τ = 200 s and R h = 2×10 4 K W −1 , the optimal estimate gives a 1 nW measurement uncertainty [4].…”

“…The working volume of the gas thermometer is V t = V d T l /T r = 0.1 ml. Assume for simplicity that the system is filled with 4 He gas at room temperature and atmospheric pressure p a . Then, the resulting pressure in the gas thermometer reservoir cooled to 4.2 K is determined by [2]…”