Chapter 9
M1A2 Abrams, with current and proposed 1500 hp turbine engine. Photos courtesy of United States Military Academy Gas Power Systems ENGINEERING CONTEXT The vapor power systems studied in Chap. 8 use working fluids that are alternately vaporized and condensed. The objective of the present chapter is to study power systems utilizing working fluids that are always a gas. Included in this group are gas turbines and internal combustion engines of the spark- ignition and compression-ignition types. In the first part of the chapter, internal combustion engines are considered. Gas turbine power plants are discussed in the second part of the chapter. The chapter concludes with a brief study of compressible flow in nozzles and diffusers, which are components in gas turbines for aircraft propulsion and other devices of practical importance. Piston-Cylinder Engines
4 Stroke Engine Process • *Intake Stroke • Compression Stroke • Power Stroke (Expansion) • Exhaust Stroke
* For Spark Ignition engines, intake is of an air/fuel mixture. For Diesel engines, intake is air only. Compression Ratio, r r = [Volume (BDC) / Volume (TDC)] > 1
Mean Effective Pressure, mep
Mep = Net work for one cycle/ Displacement Volume
Simplify Actual “Cycle” Otto Cycle • Process 1–2 is an isentropic compression of the air as the piston moves from bottom dead center to top dead center. • Process 2–3 is a constant-volume heat transfer to the air from an external source while the piston is at top dead center. This process is intended to represent the ignition of the fuel–air mixture and the subsequent rapid burning. • Process 3–4 is an isentropic expansion (power stroke). • Process 4–1 completes the cycle by a constant- volume process in which heat is rejected from the air while the piston is at bottom dead center. Air Standard Analysis The following assumptions are made: • Air, an Ideal Gas, is the working fluid • Combustion is replaced with Heat Addition (see Chap 13 for details) • No exhaust and intake strokes – constant volume heat rejection • All processes are internally reversible (Q ~ T-S area & W ~ P-V area)
For Cold-Air Standard, Specific Heats are also assumed constant Cycle Analysis PG Region: Far from Dome • Process 1–2 is an isentropic compression of the air as the piston moves from bottom dead center to top dead center.
1 Isentropic => Adiabatic, Reversible Q12 In = Stored + Out 0
QW12+ 12=∆ Emuu =() 2 − 1 W12 2 W Q 12 =−()uu 12 m 21
W12 Note: Work convention non-standard • Process 2–3 is a constant-volume heat transfer to the air from an external source while the piston is at top dead center. This process is intended to represent the ignition of the fuel–air mixture and the subsequent rapid burning.
2 3 Constant Volume: dV = dv = 0
In = Stored + Out Q 23 0 QW23+ 23=∆ Emuu =() 3 − 2
W23 Q 23 =−()uu m 32 • Process 3–4 is an isentropic expansion (power stroke).
Isentropic => Adiabatic, Reversible Q 3 34 In = Stored + Out
W34 0
QEWmuuW34= ∆+ 34 =() 4 − 3 + 34 4 W34 =−()uu34 Q34 m
Note: Work convention is standard
W34 • Process 4–1 completes the cycle by a constant-volume process in which heat is rejected from the air while the piston is at bottom dead center.
Constant Volume: dV = dv = 0 4 1 In = Stored + Out
Q41 0 0()= ∆+EQ41 + W 41 = muu 1 − 4 + Q 41 Q 41 =−()uu m 41 W41 PowerPoint frozen? Click here and try again Otto Cycle
*Cycle Analysis: W 12 = uu− m 12 W 34 = uu− m 34 Q 23 = uu− m 32 4 Internally Reversible Processes: Q • Isentropic Compression 41 = uu14− • Constant Volume Heat Addition m • Isentropic Expansion * Sign Conventions (Work in • Constant Volume Heat Rejection negative, etc.) are sometimes changed for cycle applications Analysis & Performance Parameters
W W W cycle =−=−−−34 12 ()()uu uu mmm 34 21 W Q Q cycle =23 −41 =−−−()()uu uu mmm 32 41
WWW− (uu−−−) ( uu) uu− η ==net 34 12 =34 21 =−1 41 QQin in () uu32− uu 32−
uu− η =−1 41 uu32− How would these equations be used?
Working fluid is Air- assumed to behave as an Ideal Gas
Ideal Gas Review given in Table 9.1
Works, Heat Transfers & Thermal Efficiency: based on u values
Internal energy (U, u) is only a function of T for ideal gas
Make use of isentropic relations for Ideal Gas for two processes Apply Air Standard Otto Cycle Analysis
Here the temperature dependence of the specific heats is accounted for by using Table A22 or A22E Typical Problem-Solving Approach Know the inlet pressure and temperature: get u1 and vr1
Know compression ratio, r = V2/V1 = vr2/vr1
Process 1 -> 2 isentropic: Table A22- get vr2, T2 and u2
Use PV = mRT to get P2: P2 = P1(T2/T1)(V1/V2)
V3 = V2, so use PV= mRT to get P3: P3 = P2(T3/T2)
Know maximum temperature, T3: Table A22- get u3 and vr3
Process 3 -> 4 isentropic: get vr4, T4 and u4
All W, Q and thermal efficiency in terms of known u’s
m = PV/RT
Mep = Wcycle/(V1 – V2)
Note: Temperature dependence of Cp & Cv accounted for When specific heats can be assumed constant we apply the so-called Cold Air-Standard Analysis
In this case simple isentropic relations for PG are used instead of Table A22 or A22E Typical Problem-Solving Approach
Know the inlet pressure and temperature: get u1
Know compression ratio, r = V2/V1 = V4/V3
Process 1 -> 2 isentropic: T2/T1= (V1/V2)^k-1 = r^(k-1) gets T2 and u2
Use PV = mRT to get P2: P2 = P1(T2/T1)(V1/V2)
V3 = V2, so use PV= mRT to get P3: P3 = P2(T3/T2)
Know maximum temperature, T3: Table A22- get u3
Process 3 -> 4 isentropic: T4/T3= (V3/V4)^k-1 = 1/(r^(k-1)) gets T4 and u4
All W, Q and thermal efficiency in terms of known u’s
m = PV/RT Mep = Wcycle/(V1 – V2) Effect of Compression Ratio, r on Performance 3’
2’ 3
2
Increasing the Compression Ratio Increases Average Temperature of Heat Addition with Same Heat Rejection Temperature Will Increase the Thermal Efficiency Define: Thus: (uu41− ) cTv ( 41− T) η =−1 η =−1 () ()uu32− cTv 32− T
Algebra to: Where s= const: And:
T4 k −1 k −1 T1 −1 TV 1 11 T4 V3 1 T1 == == η =−1 TV rk −1 k −1 22 TV34 r T3 T2 −1 T c 2 k = p c Thus: v T 1 T4 T3 1 = η =1− η =−1 k −1 TT12 T2 r Efficiency of Cold-Standard Otto Cycle (for Air with k =1.4)
c k ==p 1.4 cv Analysis & Performance Parameters Air Standard Analysis Cold Air Standard Analysis
T vr1 2 = rk−1 v = T r2 r 1
T 1 4 = vrv= k−1 rr43 Tr3
uu41− η =−1 1 uu− η =−1 32 rk−1 Diesel Cycles
*Cycle Analysis: W 12 = uu− m 12 W 34 = uu− m 34 W 23 =−Pv() v m 23 2 4 Internally Reversible Processes: Q23 • Isentropic Compression = hh32− • Constant Pressure Heat Addition m • Isentropic Expansion Q • Constant Volume Heat Rejection 41 = uu− m 14 Diesel Cycles
Differs from Otto (Spark Ignition Cycle) by model for heat addition
For Diesel it is constant Pressure rather than constant Volume
Diesel Otto Leads to Difference in T-S Diagram
Diesel Otto Cycle Analysis W 3 23 ==−Pdv P v v ∫ 23() 2 mu( 3−=− u 2) Q 23 W 23 m 2 Q 23 =−+()()()uu Pvv −=−+ uu PvPv − m 32 232 32 3322
Q23 Q41 =−()hh =−()uu41 m 32 m
Wcycle Q 41 ()uu− η ==−=−mm1141 QQ() 23 23 hh32− mm
(uu− ) η =−1 41 ()uu32− Thermal Efficiency Increases with Compression Ratio
Need properties at all states to calculate efficiency This means we need temperatures at all states V Assume we know T1 (vr1) and compression ratio, r r = 1 V2 V3 And Cut-Off Ratio, defined as: rc = V2 V2 1 To get T2 (u2, h2): vTrrr22()== v 1 v 1 Vr1 To get T3 (u3, h3): PV= mRT
PP32= PV33 TVmR V 33= = 3 TTrT322==c TV22PV22 V 2 mR Vr vT()==4 v vT() For s = constant process, 3 -> 4: rrr44 3 33 Vr3 c V To get T4 (u4, h4): vT()= 4 vT() rr44V 33 3 VVV Which can be expressed conveniently in terms of 442= Compression Ratio and Cut-Off Ratio: VVV323 V Where: V1 3 VV= r = rc = Cut-Off Ratio 41 V V2 2 VVVVVr 44212= == VVVVVr32323c
Thus, r vTrr44()= vT 33() rc
Note: Variable specific heats accounted for If Constant Specific Heats Is Assumed: Cold Air-Standard Analysis V Assume we know T1 and compression ratio, r r = 1 V2 V3 And Cut-Off Ratio, defined as: rc = V2 k −1 TV k −1 21== r T2, h2, u2 known TV12
V 3 T3, h3, u3 known From PG: TTrT322==c V2
k −1 k −1 T4 Vr3 c == T4, h4, u4 known TV34 r Effect of Compression Ratio on Performance
Use of Simple Cold Air-Standard Analysis Defines Basic Performance
Diesel: k 1 rc −1 η =−1 k −1 rkr()c −1
k is constant
Otto: 1 η =−1 r k −1
Thus, for same compression ratio, the efficiency of Otto cycle superior to that of Diesel cycle Analysis & Performance Parameters Air Standard Analysis Cold Air Standard Analysis v r1 T2 k−1 vr2 = = r r T1
TrT32= c k −1 Tr r 4 = c vv= rr43 Tr3 rc
k uu41− 11 r − η =−1 η =−1 c hh− k−1 32 rkr ()c −1 Real Process Idealized Processes
Otto
Diesel
Pressure-Volume diagrams of real internal combustion engines not described well by Otto or Diesel Introduce the Air-Standard Dual Cycle
The Dual Cycle is a combination in sense that Heat Addition is modeled as occurring in two steps, Constant-Volume, then Constant-Pressure Dual Cycle *Cycle Analysis: W 12 = uu− m 12 Q 23 = uu− m 32 W 34 =−Pv() v m 34 3 5 Internally Reversible Processes: • Isentropic Compression Q34 = hh43− • Constant Volume Heat Addition m • Constant Pressure Heat Addition • Isentropic Expansion W 45 = uu− • Constant Volume Heat Rejection m 45 uu− η =−1 51 Q ()()uu−+− hh 51 = uu− 32 43 m 15 21 MW
Turbojet-powered Raptor fighter aircraft Gas Turbines
Open System Closed System
Assumptions for the basic gas turbine cycle: • Air, as an ideal gas, is the working fluid throughout • Combustion is replaced with heat transfer from an external source PowerPoint frozen? Click here and try again The Air-Standard Brayton Cycle
Cycle Analysis: • W 12 • = hh12− m
•
Q23 • = hh32− m
• W 34 • = hh34− m
•
Q41 • = hh14− m Compressor & Turbine Assumed Adiabatic W W T − C mm()()hh34−−− hh 21 η == () Qhhin 32− m WC m ()hh− bwr == 21 () WT hh34− m Typical bwr of gas turbines is 40% to 80% Compared to 1 or 2% for vapor power cycles (Rankine)
If temperatures are known at all states, then A22 (or A22E) can be used to get specific enthalpies OR Cold Standard-Air analysis with constant specific heats used Ideal Brayton Cycle Assume that Irreversibilities are Not Present for Ideal Engine
1 to 2: Rev, Adiabatic (Isentropic) Compression 2 to 3: Reversible, Isobaric Heating 3 to 4: Rev, Adiabatic (Isentropic) Expansion 4 to 1: Reversible, Isobaric Cooling Recall Work Expression and Representation for mass passing through a control volume (Open System)
Internally Reversible Process
W 2 CV =− vdP ∫ m IntRe v 1
Remember the difference between this and Closed System
2 WPdV= 12 ∫ 1 Ideal Brayton Cycle Assume that Irreversibilities are Not Present for Ideal Engine
1 to 2: Rev, Adiabatic (Isentropic) Compression Work is area: 1-2-a-b-1 3 to 4: Rev, Adiabatic (Isentropic) Expansion Work is area: 3-4-b-a-3 Ideal Brayton Cycle Assume that Irreversibilities are Not Present for Ideal Engine
1 to 2: Rev, Adiabatic (Isentropic) Compression 2 to 3: Reversible, Isobaric Heating 3 to 4: Rev, Adiabatic (Isentropic) Expansion 4 to 1: Reversible, Isobaric Cooling Area Representation of Heat Transfer Ideal Brayton Cycle Assume that Irreversibilities are Not Present for Ideal Engine
2 to 3: Reversible, Isobaric Heating Heat Transfer Added is Area 2-3-a-b-2 4 to 1: Reversible, Isobaric Cooling Heat Transfer Removed is Area 1-4-a-b-1 Cycle Analysis Ideal Brayton Cycle Assume that Irreversibilities are Not Present for Ideal Engine
Typical case, knowns are: T1, P1, Compressor pressure Ratio and turbine inlet temperature, T3 We know T1, Compressor ratio (P2/P1) and that Process 1 -> 2 is isentropic: Defines u1, h1
P2 pTrr22()= pT 11 () Defines T2, u2, h2 P1 We know T3, and that Process 3 -> 4 is isentropic: Defines u3, h3 P pT= pT 4 rr44() 33 ()But, PP23= PP14= P3
Therefore, PP41 pTrrr44()== p 3 pT 33() PP32 Defines T4, u4, h4
Accounts for temperature Dependence of specific heats When Specific Heats Can Be Assumed Constant: Cold Air-Standard Basis Analysis
T1 and T3 are known, so all properties are known at states 1 and 3
k −1 k P2 Where pressure ratio and k are known TT21= So all properties are known for state 2 P1
k −1 k −1 k k PP41 TT43== T 3 PP32
Where T3, pressure ratio and k are known So all properties are known for state 4 Effect of Compressor Ratio on Ideal Brayton Cycle
Higher Compression to 2’ and 3’ Opens up area and Increases Thermal Efficiency Gas Turbine Irreversibilities and Losses
Friction produces entropy in Heat Exchangers (not large)
Friction produces entropy In Compressor & Turbine (Important)
Isentropic efficiencies of 80% to 90% possible in Compressors & Turbines
Entropy generation in the Combustion process is most significant (Chapter 13) W Wc T m ()hh− m ()hh34− s 21s η == ηc == T () () W hh− Wc hh21− T 34s m m s In actual case, pressure drops and friction occur in both heat exchangers, the compressor and the turbine
In model case, assume no pressure drops in heat exchangers. Include entropy generation in the compressor and turbine. Use isentropic efficiencies END