Maquinas sincronicas

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Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions 9-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot be approximated using the hardware of actual power producing devices. 9-2C It is less than the thermal efficiency of a Carnot cycle. 9-3C It represents the net work on both diagrams. 9-4C Thecold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature. 9-5C Under the air standard assumptions, the combustion process is modeled as a heat addition process, and the exhaust process as a heat rejection process. 9-6C The air standard assumptions are: (1) the working fluid is air which behaves as anideal gas, (2) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state. 9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as thepiston moves between the top dead center and the bottom dead center. 9-8C It is the ratio of the maximum to minimum volumes in the cylinder. 9-9C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. 9-10C Yes. 9-11C Assuming no accumulation of carbon deposits on the pistonface, the compression ratio will remain the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car gets older as a result of wear and tear. 9-12C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines. 9-13CStroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the minimum volume formed in the cylinder.


9-14 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and thethermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) The properties of air at various states are
T1 = 300K  → Pr 2 = h1 = 300.19 kJ/kg Pr1 = 1.386

2 1

u =389.22 kJ/kg P2 800 kPa (1.386) = 11.088  Pr1 = → 2 T2 = 539.8 K 100 kPa P1 u 3 = 1487.2 kJ/kg Pr3 = 1310



T3 = 1800 K  →


P3v 3 P2v 2 T 1800 K (800 kPa ) = 2668 kPa =  → P3 = 3 P2 = 539.8 K T3 T2 T2 P 100 kPa (1310) = 49.10  h4 = 828.1 kJ/kg → Pr 4 = 4 Pr3 = 2668 kPa P3


2 4 1

From energy balances,
q in = u 3 − u 2 = 1487.2 − 389.2 = 1098.0 kJ/kg qout = h4 − h1 = 828.1 − 300.19 = 527.9 kJ/kg wnet,out = q in − q out = 1098.0 − 527.9 = 570.1 kJ/kg


(c) Then the thermal efficiency becomes

η th =

wnet,out q in


570.1kJ/kg = 51.9% 1098.0kJ/kg


9-15 EES Problem 9-14 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-sand P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] P[2] = 800 [kPa] T[3]=1800 [K] P[4] = 100 [kPa] "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of...