We have already seen that the thermodynamic cycle of the Hummingbird consists of two adiabatic tran

The adiabatic transformation The adiabatic expansion of the vapor in the Mollier diagram The isobaric transformation The vapor pressure of water The isothermal transformation The transformation isochoric The engine ultravent Manson free piston double-acting The density of the free piston engine Manson - Hummingbird monoeffetto biellato Episode 07 - Episode 4
We have already seen that the thermodynamic cycle of the Hummingbird consists of two adiabatic transformations and two isochoric transformations. In this post are analyzed the performance ultravent in the case of steam supply, ultravent even if this engine could exploit the expansion of the air, or in general of any gas. ENGINE DATA Hummingbird ultravent free piston monoeffetto Volume PMI PMI = V = 100 cc = 0.10 dm 3 volume at TDC TDC = V = 20cc = 0.02 dm 3 Compression ratio = 5: 1 OPERATING CONDITIONS Working fluid: Saturated steam at 10 bar (vaporization temperature: 179.9 C) Inlet pressure = P = 10 bar injection pressure at the exhaust Exhaust = P = 1 bar (atmospheric operation) ANALYSIS enthalpy of saturated liquid at 1 bar = H Liquid Saturated @ 1bar = 417.5 kJ / kg enthalpy of saturated steam at 10 bar H = Saturated Steam @ 10bar = 2777.1 kJ / kg Density of saturated steam at 10 bar = ρ Saturated Steam @ 10bar = 5.15 kg / m 3 below the flowchart in the PV plane.
Pressure at the end of adiabatic expansion: P Fine Adiabatic Expansion = 1.6 bar fraction of steam discharged PMS V = {- [(P * Exhaust SME V range) / P injection] (1 / gamma)} / V = PMS = {0.02 dm 3 - [(1 bar * (0.10 dm 3) 1.138) ultravent / 10 bar] (1 / 1,138)} / 0.02 = 0.339 dm 3 (33.9%) of 17 CORRECTION / 11/2012 (see note at end of post) Mass of steam consumed per cycle = m vap = Fraction of gas discharged PMS * V * ρ Saturated Steam @ 10bar = = 0.339 * 0.02 * 5.15 dm 3 kg / m 3 = 0.03492 ultravent g Heat supplied = m * vap (H Saturated Steam @ 10 bar - Liquid Saturated H @ 1 bar) = g * = 0.03492 (2777.1 kJ / kg - 417.5 kJ / kg) = 82, 4 J Work adiabatic expansion L = AB = 28.8 J Jobs in adiabatic compression CD = L = - 18.0 J Labor useful engine = AB + L L J CD = 28.8 - 18.0 J = 10, 8 J Efficiency ultravent = Work useful engine / Heat supplied = 10.8 J / 82.4 J = 13.1% efficiency Rankine (theoretical maximum ultravent with steam) = 16.6% of Carnot efficiency (theoretical maximum for any heat engine) = 17.7% COMMENTS With an operating temperature of 180 C hot that allows for saturated steam at a pressure of 10 bar and in the presence of atmospheric release, the theoretical yield of thermo-mechanical conversion of the Hummingbird ultravent is 13.1%. As the Carnot efficiency ultravent under the same conditions of temperature (T = 179.9 C hot and cold T = 99.6 C) that is 17.7% means that the engine pulls 74.0% of the theoretical maximum. Also by way of comparison, the performance of the Rankine cycle under the same conditions is equal to 16.6%. In this case the Hummingbird it is able to extract 78.9%. The useful work per cycle is 10,8J, a low value for a steam engine of 100cc displacement that operates with power to 10bar. Note however that the power developed, ie the work done in the unit of time, depends linearly on the operating ultravent frequency. Considering the constructional features of this engine is reasonable to assume that they are easily ultravent accessible operating frequencies of at least 50Hz (3000rpm) and in this case, the power would be amplified by 50 times (10,8J * 540W = 50Hz). CONCLUSIONS Hummingbird unite respectable performance in a device of extreme simplicity. This combination, absent in most of the external combustion engines, ultravent makes it particularly attractive to a potential practical use. CORRIGENDUM 17/11/2012 the formula previously given to evaluate the fraction of escaping vapor was wrong. Are shown below for possible comparisons. Fraction of steam discharged = 1 - (P Exhaust / P Fine Adiabatic Expansion) = 1 - (1 bar / 1.6 bar) = 0.375 (37.5%) All calculations employees have been revised and updated.
Hello Yuz, The isochores are always irreversible transformations in theory, that deadweight losses. In the Stirling cycle is possible during the isochoric keep and transfer the heat of isotherms through the regenerator. In the Brayton cycle to adiabatic interrupted and in the Ericsson cycle (two isobars and two isochoric), recovery is possible against-heat in the exhaust gases. In the case of the Hummingbird, the isochoric BC is still minimal ultravent loss but it is absolutely essential because it is responsible for the escape of steam expanded through the valve end lights. In practice, we save the building and the movement of the valve, the heart and soul of Uniflow. ultravent Very important however is the loss that occurs ultravent thermo-dynamic nell'isocora DA, one that allows the filling