An energy balance of a Stirling engine cycle can be do through an indicator diagram or trend pressure inside a Stirling engine which is calculated. Results of this energy balance are an input heat, a rejection heat, a regenerated heat, an internal work of the cycle and an internal thermal efficiency.
This article describes analytical calculation through equations, which are derived from the equations of the first law of thermodynamic. Through these derived equations can be calculated energy flows and theirs connection with others quantities.
The equations of this energy balance are true under certain simplifying assumptions. Calculated results through these equation can vary from reality with increasing of differences between simplified and real processes:
(1) Working gas is ideal gas. (2) Stirling engine is perfect sealed. (3) There is no pressure loss, pressure of working gas is same in whole working volume. (4) Regenerator has perfect thermal isolation. (5) Pressure as function of angle of rotation (p(φ)), working volume on hot side as function of angle of rotation (VT(φ)) and working volume on cold side as function of angle of rotation (VS(φ)) are known. (6) Stirling engine cycle is steady (the same cycle repeats).1.459 The simplifying assumptions of the solving.
The internal work of the engine is function the pressure and the working volume:
AT [J] 1382,2113 AS [J] -506,0131 A [J] 876,1982Problem 1: summary of entries and results.
Inside the Stirling engine is transformed heat to work. For this transformation are true the rules of heat cycle, it is means, only a piece of the heat input to the engine is transformed to work and others the heat is necessary rejected from the engine. For input and output of the heat to/from the engine between the working gas and a heat transfer surface must be a temperature gradient. This temperature gradient is vary during one cycle, because working gas temperature is changed also (see chapter 34. Temperature change of working gas inside Stirling engine). Therefore in all internal sections of the engine the heat is inputted and outputted to/from working gas during one cycle. It is evident (from the principle of the Stirling engine) the heat balance during one cycle of the hot side is positive (the heat is inputted to the working gas) and the heat balance during one cycle of the cold side is negative (the heat is outputted from the working gas), the heat balance of the regenerator during one cycle must be neutral:
The heat ΔI inputs through the cycle on the hot side and it increases of internal energy of the working fluid and other part of the cycle outputts on the cold side from the engine (through the temperature gradient between the hot and cold side). This heat decreasing a dimension capacity-rating of the heat transfer surface and a thermal efficiency of the cycle. On the other hand the heat ΔI develops the temperature gradient between the hot and the cold side even in for case of the regenerator with small capacity or for case it is not installation inside the engine. For cases isothermal processes on the hot and the cold side (e.g. Schmidt theory) must be the change of the enthalpy ΔI equal zero.
The internal work of the engine can be measured indirectly, e.g. by indication of the pressure of the working gas or by measuring work shaft (in this case is necessary to know mechanical losses of the engine). The internal work of the engine can be approximately calculated of the pressure trend by the procedure which is shown in article 35. Stirling engine cycle (calculation without losses) or in article 36. Losses in Stirling engines (calculation with losses).
Values of heats QT and QS can to get from a measuring of heat flow to/from the engine or it can be approximately calculated through an estimate the internal thermal efficiency:
The internal thermal efficiency of the Stirling engine is ratio between its the internal work A and heat, which inputted to the working gas from a surroundings during one cycle QD* respectively this definition is the same as definition of efficiency of heat cycle, because inside engine is performed complet cycle. For the conditions, which are shown on List 1 can be descripted equality of heat flows:
The internal thermal efficiency can be estimated through similarities of the Stirling engines similar construction:
C [-] 0,65...0,75 ΔI [J] 746,03...462,42 QS [J] -1252,05...-968,43 ηt [-] 0,41...0,48 QT [J] 2128,24...1844,63Problem 2: summary of entries and results.
The regenerated heat inside the regenerator can be calculated from a function which describes an amount of a heat transfered inside the regenerator of the working gas from a start of the cycle to any point of the cycle. The regenerated heat is equal of a difference between the maximum and the minimum of this function:
The enthalpy of the working gas inside the hot or the cold side in case isothermal processes is not changed respectively in Equation 5 is true IT,x=0; IT,x=0.
The amount of the heat QReg can be calculated approximately if the amount of the regenerated heat QReg is much more than the amount of the heat ΔI:
QReg [J] 6421,6506 QReg/ΔI [-] 8,6077 φ [°] V [cm3] QR,x [J] φ [°] V [cm3] QR,x [J] -------------------------- -------------------------- 0 377,1489 0 190 508,8472 2477,7114 10 365,4078 359,5118 200 519,8974 2005,3991 20 355,8134 818,4803 210 528,337 1577,1375 30 348,5937 1377,206 220 533,859 1193,4843 40 343,9179 2026,5137 232,5 536,3626 773,8312 52,5 341,8098 2931,351 240 535,4585 551,9241 60 342,5702 3495,834 250 531,4778 288,1827 70 345,9296 4228,6122 259,647 524,7618 66,2469 80 351,8947 4883,2127 270 514,6757 -138,2063 87,0244 357,5704 5264,0468 280 502,4646 -303,3848 100 371,0266 5733,5592 285 495,5851 -373,7416 105 377,1489 5822,4453 300 472,5932 -532,2328 120 398,1411 5777,4868 310 455,9721 -588,8656 130 413,8632 5516,86 320 438,9569 -599,2053 140 430,4707 5117,7967 340 405,8632 -447,5454 150 447,4797 4626,2816 350 390,7495 -265,9646 160 464,3663 4085,5746 360 377,1489 0 180 495,5851 2989,632Problem 3: summary of entries and results.
During the solving of previous problem was neglected an influence of the change of enthalpy of the working gas on the hot and cold side respectively the heat ΔI. For actually case the influence of the change enthalpy is increased with decrease of the temperature ration τ respectively the ration between the heat QReg and the heat QD significantly decreasing and this procedure of the calculate used in Problem 3 increases its inaccuracy. This fact must be taken into account when designing the size of the regenerator:
|7.197 The ratio between the heat QReg and the heat ΔI as function the temperature ratio τ.|
This curve is true for the cycle from the Problem 2 and Problem 3.
An equation of a change of specific entropy of the working gas can be derived from First law of thermodynamics for closed system:
Through last Equation can be constructed a T-s diagram of the cycle. From T-s diagram of cycle can be identified losses of the cycle. In praxis can be measured perfectly only pressure of the working gas as function of angle of rotation. Exact amount of the working gas inside of the engine and its exact mean temperature of the working gas is impossible to measure. Therefore is constructed Θ-s diagram*, where Θ is ratio between the mean temperature of working gas inside of the engine and maximum temperature of the working gas inside of the engine. The Θ-s diagram can be constructed only from the measured pressure.
Θ-s diagram, which is designed from a measurement, can show losses and weaknesses of the engine:
cv [J·kg-1·K-1] 3116,168 r [J·kg-1·K-1] 2077,22 sx-s0 sx-s0 sx-s0 [J·kg-1·K-1] Θ [-] [J·kg-1·K-1] Θ [-] [J·kg-1·K-1] Θ [-] 0 0,6853 1269,3741 0,9934 712,0368 0,6852 29,309 0,7066 1296,7358 0,9767 634,0198 0,674 88,6894 0,733 1300,7945 0,9526 546,6878 0,6639 177,8893 0,7647 1285,939 0,9239 459,8105 0,6561 294,1768 0,801 1256,5622 0,8929 415,9258 0,6529 469,1112 0,8507 1168,7142 0,8313 285,7881 0,6463 583,793 0,8813 1115,3026 0,8029 203,8984 0,6447 739,4541 0,9204 1057,5191 0,7769 130,0204 0,6458 888,758 0,9546 995,9703 0,7536 23,1234 0,6575 984,6345 0,974 930,7996 0,7329 -0,9958 0,6691 1132,4227 0,9965 844,0397 0,7105 0 0,6853 1177,3692 1 789,0023 0,6989Problem 4: summary of entries and results.
|Problem 4: summary of entries and results.|
The fact, minimal entropy corresponding with the state p0, V0 is only a random.
The amount of the working gas inside of the Stirling engine can be computed by the equation of state [2, p. 67 (cz)] for each working volumes. The amount of the working gas is computed for a known state of the working gas. If real mean temperature of the working gas inside individual volumes is not known, then can be use of computed of temperature change of the working gas inside Stirling engine:
m [kg] 5,92337E-3
|Problem 5: summary of entries and results.|
This document is English version of the original in Czech language: ŠKORPÍK, Jiří. Energetická bilance oběhu Stirlingova motoru, Transformační technologie, 2009-07, [last updated 2012-04]. Brno: Jiří Škorpík, [on-line] pokračující zdroj, ISSN 1804-8293. Dostupné z http://www.transformacni-technologie.cz/35.html. English version: Energy balance of Stirling engine cycle. Web: http://www.transformacni-technologie.cz/en_35.html.