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 − 1 1 − − − − (/ ) (/ ) 41 32 14 1 23 2 TT TT TT T () () TT T 3 3 4 2 T T T T 1 1 = = =− =− 1 1 2 4 T T T T or , th Otto

η Æ 1 − k Æ ⎞ ⎟ ⎟ ⎠ 4 3 v v ⎛ ⎜ ⎜ ⎝ = 4 3 T T 3-4 are isentropic, so − − 41 32

1 () () v v = v and in out 4

Q mC T T 1 Q mC T T − k ⎞ ⎟ ⎟ ⎠ and v 1 1 2 1 v v 2 =− =− ⎛ ⎜ ⎜ ⎝ = = v 3 2 1 , Recall processes 1-2 and v T T th Otto

η 1 2 T T 1 1 1 1 =− − − , − − (/ ) (/ ) 41 32 th Otto 14 1 23 2 TT TT TT T () TT T () η 1 1 =− =− Is this the same as Carnot efficiency? , th Otto

η 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