Extract
[...] A cycle corresponds to a rotation of the electric motor, therefore to a duration = 1 / where f is the rotation frequency of the motor. Therefore, for a cycle, we have : P = Wf / f I 3 During a cycle completed by the engine, the total mechanical energy corresponds to that transferred during the relaxation phases and compression car W = 0 during each isochoric phase : Wtot = WC + WD The total thermal energy of the system can be written as : Qtot = Q1 + Qf + 3 + C The thermal exchanges at the level of the regenerator are normally compensated so : 1 + Q3 = 0 So we have : Qtot = Qf + QC Qf is the heat received by the system, unlike QC which is ceded by the gas I 3 According to the 1st principle of thermodynamics, the variation of the internal energy of the system (gas) is equal to : = Wtot + Qtot In a closed system, like the case for this engine, = 0 : So Wtot + Qtot = 0 so it Wtot = - Qtot During an isothermal transformation (case of the relaxation and compression phase), we also have = 0 During a transformation at constant volume, we will have W = 0 so = Q According to the second principle of thermodynamics, the coefficient of performance (COP) of this heat engine will therefore be: - Let the ratio of the heat recovered (thus given by the system) to the amount of mechanical energy consumed by the system, in the case of the heat pump: In this case, the heat recovered is that given by the gas QC : . [...]
[...] / = 9,2.10-4 kg.s-1 Caue : specific heat capacity of water = 4.19 kJ. kg?1. K?1 Wf is required in Joules so we convert Cwater = J. kg?1. K?1 (le is a temperature difference so the result is the same in °C or in K. II 4 The work of forces of frottements augmente regularly with the speed of rotation, over this frequency range, but the variation does not seem linear. [...]
[...] Their minimum values are obtained for a frequency of the order of around 3.5 Hz, which corresponds to the maximum value of the COPcold , which would correspond to optimal operation in a refrigeration machine, for this range of tested frequencies. III 3 The temperature measurements were not made beyond 2 min 47 s. It is observed that the temperature curve bends over time, after a faster drop from 0 to 1 minute. It would have been necessary to continue the measurements over a longer period of time until the temperature stabilizes to well determine the minimum temperature: it is therefore below 8°C. [...]
[...] On measured the average volumetric flow rate: 3.31.10-6 m3.s- On converted it to the same mass flow rate 3.31.10-3 kg.s-1 m / = 3,31.10-3 so m = 3,31.10- m is the mass of cooling water displaced on a stirling cycle of duration = 1/f so m = 3,31.10-3 / f where QC = 3,31.10-3 .Caue.?T/f The total work W is W = QC - Qf (according to II 3 COPcold = 1 + Qf /QC (according to II 3 Qc and W follow the same non-linear variations with the COPfrost evolves in the opposite way. The maximum values of Qc and W are obtained for a frequency of the order of 2.4 Hz, which would correspond to optimal operation in a heat pump, for this tested frequency range. (COPcold minimal in this case). [...]
[...] the mass of the hot source ? the specific heat capacity of the hot source : II 1 Volume Temps Débit Incertitude =volume/temps = measured flow rate - average flow rate / average flow rate Measure 1 10mL = 10-5 m3 10, 35 9,7.10-7 0,05 Measure 2 10-5 9,58 10,4.10-7 0,13 Measure 3 10-5 13,55 7,4.10-7 0,19 Debit mean : 9,2.10-7 m3.s-1 II 2 T = 28°C II 3 Frequency Variation de la temperature ?Tmax.(°C) Work of forces of frottement The initial temperature is that of the cooling water upstream (measured in III = 28°C for the 4 calculations The final temperature is that measured upstream after stabilization See the calculations below 1 28 - 28 = 0 0,00 2 28,4 - 28 = 0,4 0,08 3 28,6 - 28 = 0,6 0,08 4 29,1 - 28 = 1,1 0,11 P = Wf / f P = / ?t).Ceaù.?T with m / corresponding to the mass flow rate of water (in kg / s-1). [...]
