Thermodynamic Cycles • Look at Different Cycles That Approximate Real Processes • You Can Categorize These Processes Several Different Ways • Power Cycles Vs

Thermodynamic Cycles • Look at different cycles that approximate real processes • You can categorize these processes several different ways • Power Cycles vs. Refrigeration • Gas vs. Vapor • Closed vs. open • Internal Combustion vs. External Combustion Power Cycles • Otto Cycle • Spark Ignition • Diesel Cycle • Brayton Cycle • Gas Turbine •RankineCycle These are all heat engines. They convert heat to work, so the efficiency is: Wnet ηth = Qin Ideal Cycles • We’ll be using ideal cycles to analyze real systems, so lets start with the only ideal cycle we’ve studied so far Carnot Cycle W Q Q-W=0 Æ Q=W In addition, we know that the efficiency for a Carnot Cycle is: TL ηth, Carnot =−1 TH Carnot Cycle is not a good model for most real processes • For example • Internal combustion engine •Gas turbine • We need to develop a new model, that is still ideal Air-Standard Assumptions • Air continuously circulates in a closed loop and behaves as an ideal gas • All the processes are internally reversible • Combustion is replaced by a heat- addition process from the outside • Heat rejection replaces the exhaust process • Also assume a constant value for Cp, evaluated at room temperature Terminology for Reciprocating Devices V max V Compression Ratio r ==BDC V min VTDC Mean Effective Pressure 2 W = PdV ∫1 W = P∆V v v 1-2 Isentropic Compression 2-3 Constant Volume Heat Addition 3-4 Isentropic Expansion 4-1 Constant Volume Heat Rejection Thermal Efficiency of the Otto Cycle Wnet Qnet QQin− out Qout ηth === =−1 Qin Qin Qin Qin Apply First Law Closed System to Process 2-3, V = Constant Qnet,23 −Wnet,23 = ∆U 23 3 Wnet,23 = Wother ,23 +Wb,23 = 0 + PdV = 0 ∫2 QUnet , 23= ∆ 23 QQmCTTnet, 23== in v () 3 − 2 Apply First Law Closed System to Process 4-1,V = Constant Qnet,41 −Wnet,41 = ∆U 41 1 Wnet,41 = Wother ,41 +W b,41= 0 + PdV = 0 ∫4 QUnet , 41= ∆ 41 QQmCTTnet, 41=− out = v () 1 − 4 Qout=− mCTT v()14 − = mCTT v () 41 − Qout ()TT41− ηth, Otto =−1 ηth, Otto =−1 Qin ()TT32− Æ mC() T− T TT14(/ T 1−1 ) =−1 v 41 =−1 TT(/ T−1 ) mCv () T32− T 23 2 Recall processes 1-2 and 3-4 are isentropic, so k −1 k −1 T ⎛ v ⎞ T ⎛ v ⎞ T2 T3 2 = ⎜ 1 ⎟ and 3 = ⎜ 4 ⎟ = ⎜ ⎟ ⎜ ⎟ T1 T4 T1 ⎝ v2 ⎠ T4 ⎝ v3 ⎠ Æ or T4 T3 v3 = v2 and v4 = v1 = T1 T2 ()TT41− ηth, Otto =−1 ()TT32− TT(/ T−1 ) =−1 14 1 TT23(/ T 2−1 ) 1 T1 ηth, Otto =−1 T2 Is this the same as the Carnot efficiency? Efficiency of the Otto Cycle vs. Carnot Cycle • There are only two temperatures in the Carnot cycle • Heat is added at TH • Heat is rejected at TL • There are four temperatures in the Otto cycle!! • Heat is added over a range of temperatures • Heat is rejected over a range of temperatures Since process 1-2 is isentropic, k −1 T ⎛V ⎞ 1 1 = ⎜ 2 ⎟ = ⎜ ⎟ k −1 1 T2 ⎝ V1 ⎠ r ηth, Otto =−1 Æ r k −1 T1 ηth, Otto =−1 T2 Typical Compression Ratios for Gasoline Engines Increasing Compression Ratio Increases the Efficiency Why not use higher compression Ratios? • Premature Ignition • Causes “Knock” • Reduces the Efficiency • Mechanically need a better design Diesel Engines •No spark plug • Fuel is sprayed into hot compressed air State Diagrams for the Diesel Cycle Diesel Cycle Otto Cycle The only difference is in process 2-3 Consider Process 2-3 • This is the step where heat is transferred into the system • We model it as constant pressure instead of constant volume qin,23 − wb,out = ∆u = u3 − u2 qin,23 = ∆u + P∆v = ∆h = C p (T3 −T2 ) Consider Process 4-1 • This is where heat is rejected • We model this as a constant v process • That means there is no boundary work q41 − w41 = ∆u q41 = −qout = ∆u = Cv (T1 −T4 ) qout = Cv (T4 −T1 ) As for any heat engine… wnet qout ηth = =1− qin qin qout = Cv ()T4 −T1 and qin = C p (T3 −T2 ) Cv (T4 −T1) (T4 −T1 ) nth,diesel =1− =1− C p ()T3 −T2 k()T3 −T2 Rearrange T1(T4 T1 −1) nth,diesel =1− kT2 ()T3 T2 −1 PV33 PV22 == where PP32 T3 T2 T3 V3 ==rc T2 V2 rc is called the cutoff ratio – it’s the ratio of the cylinder volume before and after the combustion process T1(T4 T1 −1) nth,diesel =1− kT2 ()rc −1 PV44 PV11 == where VV41 T4 T1 T P 4 = 4 T1 P1 T1(TP44 TP11 −1) nth,diesel =1− kT2 ()rc −1 Since Process 1-2 and Process 3-4 are both isentropic k k P1V1 = P2V2 and k k P4V4 = P3V3 k P V k P V k ⎛V ⎞ 4 4 = 3 3 = ⎜ 3 ⎟ = r k k k ⎜ ⎟ c P1 V1 P2 V2 ⎝V2 ⎠ 11 k T1(rc −1) nth,diesel =1− kT2 ()rc −1 Finally, Since process 1-2 is isentropic k −1 The volume ratio from 1 to 2 is the T1 ⎛ v2 ⎞ = ⎜ ⎟ compression ratio, r T2 ⎝ v1 ⎠ k −1 T ⎛ 1 ⎞ 1 = ⎜ ⎟ T2 ⎝ r ⎠ k 1 ⎡ rc −1 ⎤ ηth,diesel =1− k −1 ⎢ ⎥ r ⎣k()rc −1 ⎦ k=1.4 The efficiency of the Otto cycle is always higher than the Diesel cycle Why use the Diesel cycle? Because you can use higher compression ratios.

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