Internal Combustion Engines

V2

V1 V1 Compression Ratio, rc = V2

Bottom Top Dead Dead Center Center BDC TDC

1 Lenoir Cycle 1860

Intake Combustion Power TDC Exhaust TDC 0-1 1-2 2-3 3-0 p=constant v=constant Q=0 p=constant one revolution 2 Q Air Standard p v=c Q=0 Lenoir Cycle 0 3 1 Q p=c 2 v Air Standard Lenoir Cycle

2 Closed Thermodynamic System − quantity of mass Q q = Δu + w Q=0 p v=c 2 − 3 Combustion Process 0 v = constant ⇒ w = 0

3 q = u 2 − u1 1 Q p=c 2 − 3 Expansion Process, Power Stroke q = 0,adiabticprocess k k −1 k−1 T3  p3   v2  =   =   T2  p2   v1 

w = −(u 2 − u3 )

3−1 Exhaust Process p = c q = Δu + w

q = (u1 − u3 ) + p3 (v1 − v3 ) = h1 − h3

w = q31 − (u3 − u1)

3 OTTO CYCLE , 2 Revolutions/cycle

Compression Combustion Power Exhaust Exhaust Intake 1-2 2-3 3-4 4-1 1-0 0-1 TDC BDC TDC BDC revolution 1 revolution 2 3 p OTTO CYCLE spark ignition, SI fuel injection + SI 2 4 0 1 v 4 4 CYCLE, 2 REVOLUTIONS

Revolution 1 Revolution 2 5 Air Standard Otto Cycle Closed System- a quantity of mass p p adiabatic and isentropic

W Q v ignition exhaust opens W atmospheric pressure intake opens intake closes exhaust closes W v 6 AIR STANDARD OTTO CYCLE, 2 Revolutions/cycle

Q in Qout V=c

V=c

W Win out Compression Combustion Expansion Exhaust 1-2 2-3 Power 4-1 3-4 p p Q adiabatic in W v=const isentropic 4 adiabatic 4 2 2 isentropic v=const Qout W 1 1 7 v v “COLD” Compression 1 → 2, Δs = 0, pvk = c, q = 0, w = Δu k k Air Standard Otto Cycle p1v1 = p2v2 k−1  v   1  3 T2 = T1   v2  Δs = 0 w 2−1 = Δu = u 2−u1 = cv (T2 − T1 ) T qin pvk = c Combustion 2 → 3, v = c, w = pdv = 0, q =ΔU Q = 0, ∫

q2−3 = Δu = u3 − u 2 = cv (T3 − T2 ) 2 k 4 Expansion 3 → 4, Δs = 0, pv = c, q = 0, w = ΔU k−1 q  v  out T = T  3  4 3 v  1  4  v w3−4 = Δu = u3 − u 4 = cv (T3 − T4 ) CLOSED SYSTEM Exhaust 4 →1, v = c, w = ∫ pdv = 0, q =ΔU

q = Δu + w q4−1 = Δu = u 4 − u1 = cv (T4 − T1 ) "COLD"⇒ Cycle

constant c c at T . qout q4-1 cv (T4 − T1 ) p, v 1 ηcycle =1− = 1− = 1− 8 qin q3-2 cv (T3 − T2 ) 3 “COLD”Air Standard Otto Cycle CLOSED SYSTEM, Q = ΔU + W "COLD"⇒ T Compression 1 → 2, Δs = 0, Q = 0, W = ΔU constant cp, cv at T1 Combustion 2 → 3, v = c, W = ∫ pdv = 0, Q =ΔU Expansion 3 → 4, Δs = 0, Q = 0, W = ΔU 2 4 Exhause 2 → 3, v = c, W = ∫ pdv = 0, Q =ΔU 1.0 ∆s2→3 = Δs4→1 1.0 1  T   v   T   v   3   3   4   4  cvln  − R ln  = cvln  − R ln   T2   v2   T1   v1  V2 V1 T T V v 3 = 4 Compression Ratio = r = 1 = 1 T T c V v 2 1 2 2 k−1 T  v  Q − Q Q 1 2 η =1− in out =1− out − =   T2  v1  Qin Qin 1 1 1 u 4 − u1 cv (T4 − T1 ) η =1− k−1 =1− k−1 =1− k−1 η =1− =1−  v  compression (r) u − u c T − T 1 3 2 v ( 3 2 )      v2   ratio   T   4  w = p × V − V  −1 net mean ( 2 1 ) T4 − T1 T1  T1  η =1− =1− wnet T − T T   Mean Effective Pressure (mep) = 3 2 2 T3 (V − V ) 9  −1 1 2  T2  What is the thermal efficiency k−1  v  of a COLD air standard Otto cycle T = T  1  = 559.7×6.4 =1146o R operating between 2800 F and 2 1   v2  100 F with a compression ratio of 6? w 2−1 = u 2−u1 = cv (T2 − T1 ) w = .171×(1146− 559.7) =100.3BTU/lb Closed System 2−1 q = u − u Q = ΔU + W 2−3 3 1 Compression 1→ 2, Δs = 0, q = 0, w = Δu q2−3 = .171×(3259.7 −1146) = 361.4BTU/lb k−1 .4 Combustion 2 → 3,v = c, w = pdv = 0, q =Δu  v   1  ∫  3  o T4 = T3  = 3259.7  =1591.9 R Expansion 3 → 4, Δs = 0, q = 0, w = Δu  v4   6  Exhaust 2 → 3, v = c, w = pdv = 0, q =ΔU ∫ w3−4 = u3 − u 4 w = .171(3259.7 −1591.9) = 285.2BTU/lb 3 3−4 2800 F q4−3 = u 4 − u3

q4−3 = .171(1591.9− 559.7) =176.5BTU/lb p W Q  w   285.2−100.3  net  ηcycle =   =   = 51.1%  qin   361.4  2 4 Q 1 1 W η =1− =1− = 51.1% cycle k−1 61.4−1 1  v1  100F   10  v2  v AIR TABLE - Tables A-22, A-22E specific heats are integrated as variables of T- page 113, 230

1) Table base 0o F, 0o C 5) Entropy - ideal gas  T   v   2   2  s2 − s1 = cvln   + R ln  2) Isentropic Process,  T1   v1  pvk = constant  T   p  s − s = c ln  2  − R ln 2  p (p ) 2 1 p     1 = r 1  T1   p1  p (p ) 2 r 2 Entropy - Using Table A − 22 and Table A 22E, v (v ) 1 = r 1 dT v (v ) s = c (T) + R ln (volume ratio) 2 r 2 ∫ v T dT s = c (T) − R ln (pressure ratio) 3) Enthalpy ∫ p T

h = cp (T) dT  v  ∫ 0 0  2  s2 − s1 = s (T2 ) − s (T1) + R ln   v1  4) Internal Energy  p  s − s = s0 (T ) − s0 (T ) − R ln 2  u = c (T) dT 2 1 2 1   11 ∫ v  p1  @ 300o K, v = 621.2, u = 214.07 r1 1 9.1 RT .286×300 3 v1 = = = .858 m3/kg p p 100 1400 r=8.5 v v  v  kJ/kg r2 = 2 , v = v  2  = 621.2/8.5 = 73.08 v v r2 r1 v  4 r1 1  1  2 T vr u 75.5 496.62 1 73.08 503.06 kJ/g = u 3 100 kPa 72.56 504.45 v 300 K 73.08 − 72.56 .52 interpolation ratio = = = .1769 75.5 − 72,56 2.94 a) W = Qin − Qout =1400 − 674.25 W = 725.75kJ/kg Qin = u3 − u 2 = 1400 kJ/kg u3 =1400 − 503.6 =1903.06 kJ/kg  Qout  725.75 b) η =1−   =1− = 51.8 % InterpolateTableA − 22 near u3 =1903.6 kJ/kg  Qin  1400 vr3 =1.9176 W W c) mep = =  v  V − V  v   4  1 2  2  vr4 = vr3  =1.9176×8.5 =16.3 v11−   v3   v1  Interpolate TableA − 22 near v =16.3 725.75 r4 mep == = 958.7 psi u 4= 888.32 kJ/kg  1  .858×1−  12 Qout = (u3−u 4 ) = (888.32 − 214.07) = 674.25kJ/kg  8.5  The initial temperature and pressure of Process 1-2 air inan ideal Cold Otto cycle having a compression ratio of10, is 60 F and 14.7 psia isentropic, adiabatic with pvk = constant respectively. Heat isadded in the constant k−1  V  volume process at a rateof 800 BTU/lbm.  1  .4 o T2 = T1 ×  = 520×10 =1305 Consider air to be an idea gas.  V2  a) Calculate: a) The change in internal energy per pound of air ΔU =cv (T2 −T1 ) = .17×(1305 −520) =134.2BTU/lb during the compression process. b) b) The maximum temperature and pressure of the cycle Combustion Heat,Q = cv (T3 −T2 ) k =1.4 800 = .17×(T3 −1305) cv =.171BTU/lbm−R T = 5893o R cp =.24BTU/lbm−R 3 Since V = V , from the ideal gas law Pv = mRT, 3 2 3 Expansion  T   3  p3 = p2   T2  p Since 1− 2 is isentropic and adiabatic, Combustion k 2 Q=800 BTU/lb 4  V  p = p  1  =14.7×101.4 = 367psia Compression Exhaust 2 1   V2  1 14.7 psia 520 R  5983  13 .1×V1 V1 p3 = 367×  =1685psia V 1305  a) A 265 cubic in V8 gasoline engine (1500 −1200) BTU/power stroke/cylinder ×8 cylinders × 4600 rpm runs at 4600 rpm. Compression work HP = is 1200 ft-lb, expansion work is 550 ft − lb/HP × 2 rev/ power stroke × 60 sec/min 1500 ft-lb and heat input is 1.27 BTU HP = 167.3 HP per piston per cycle. The atmosphere b) w (1500 −1200) is 14.7 psia and 70 F. The air fuel η = net = = 30.4% ratio is 20:1. The heating value of q in 1.277 × 778 gasoline is 18,900 BTU/lb. Find c) a) indicated HP b) thermal efficiency 1.27 BTU/power stroke/cylinder ×8 cylinders × 4600 × 60 Fuel = c) gasoline consumption per hr 18,900 × 2 rev/power stroke d) specific fuel consumption. Fuel = 74.18lb/hr d) 74.18 lb/hr Specific Fuel Consumptio n = 167.3 HP 265cu in Specific Fuel Consumptio n = .44lb fuel/HP hr

14.7 psia Intake 70 F

Compression Q=1.27 We=1500 ---1st Revolution p Power

Wc=1200 Exhaust v 14 ---2nd Revolution 2 CYCLE, 1 Revolution/cycle

Compression Ignition Power Exhaus Intake 1-2 2-3 3-4 4-1 1 3 p

2 4

1 15 v Diesel Cycle Compression - isentropic, Q=0 Combustion -constant pressure Expansion - isentropic, Q=0 Exhaust - constant volume 2 3 p W1→2 = ΔU Closed system Q2→3 = ΔH 4 Q4→1 = ΔU 1 V Compression ratio, r = 1 v V2

V3 Cut off ratio, rc = V2 1  rk −1  η =1−  c  cycle k−1   16 (r)  k(rc −1) What is the thermal efficiency RT1 53.35 × 535 3 of a COLD air standard diesel v1 = = = 13.94 ft /lb cycle operating on 14.7 psia air p1 14.2 ×144 at 75 F? The temperature of the 1 2.5 air before and after heat addition  T  k−1  535   1  3 are 750 F and 2900 F respectively. v 2 = v1  = 13.94  = 1.812 ft /lb  T2  1210   T   3360   2  3 v3 = v 2   = 1.812  = 5.032 ft /lb  T1   1210  k−1 .4 1210 R 3360 R  v   5.032  2 3  3  o T4 = T3   = 3360  = 2235 R v 4  13.94  Qin  

qin = cp (T3 − T2 ) = .241(3360 −1210) p 4 qin = 518.15 BTU/lb

Qout q out = c v (T4 − T1 ) = .171(2235 − 535) = 1 75 F q out = 290.7 BTU/lb 14.7 psia qin − qout 518.15 − 290.7 v η = = = 43.9% cycle 17 qin 518.15 7-21 AIR STANDARD DUAL CYCLE 4 V 3 Cut off ratio, r = 4 c V p adiabatic and 3 isentropic V Compression ratio, r = 2 2 V1

5 w = u − u q = 0 1→2 2 1 1→2 w = u − u q = u − u 1 2→3 2 3 3→4 3 2 v w = p(v − v ) q = h − h Closed System 3→4 4 3 3→4 3 4 − quantity of mass w 4→5 = u 4 − u5 q4→5 = 0 w = 0 q = u − u q = Δu + w 5→1 5→1 5 1

w p=constant = p(Δv) 18 qp=constant = Δh Stirling and Erickson Cycles

19 1 2 Stirling Cycle, Closed System T Process 1 → 2

qin at conatant TH , v increases 4 3  v  w = RT ln 2  v out 1    v1  Process 2 → 3 v = constant

q regenerator = Δu Process 3 → 4 q at conatant T , v decreases W out H out  v   4  win = R T3 ln   v3  Process 4 →1 v = constant

q regenerator = Δu Win 20 Simple Brayton, Gas Turbine, Cycle

Qin 3 T 2 combustor p=c 3 2

turbine 4 1 W Net compressor Work 1 compressor Output 4 s Compressor, Combustion,and Turbine are Δs = 0, isentropic Open Thermodyanic Systems Q = 0, adiabatic Steady Flow Energy Equation Form of First Law pvk = constant reversible 21 Q = Δ(h + KE)+ Wnet Propulsion Gas Turbine

22 Qin

W W = W net output compressor turbine

Gas Power Generator Turbine

Aero-derivative Gas Turbine

23 Figure 9.4 Air Standard Brayton Cycle

24 Brayton Cycle with Real Compression and Expansion

Turbine Efficiency turbine actual work combustor η = t isentropic work

h3 − h 4 ηt = h3 − h 4s Compressor Efficiency compressor isentropic work η = c actual work

h 2s − h1 ηc = h 2 − h1

25 Brayton Air Standard Cycle

Air Standard Actual Process Process Steady Flow, Open System - region in space Steady Flow Energy Equation V2 Q = m× ∆(u + pv + + ρgh) + W 2 shaft

Compression Process, Q = 0, W = m(h 2 − h1 )

Combustion Process, W = 0, Q = m(h3 − h 2 )

Expansion Process, Q = 0, W = m(h3 − h 4 ) 26 Exhause Process, W = 0, Q = m(h 4 − h1 ) In an air COLD standard gas k−1  p  k turbine, 60 F and 14.7 psia air is T = T  2  = (460+60)×10.2857 =1003.9o R 2i 1 p  compressed through a pressure  1  c T −T ratio of 10. Air enters at 1540 F Wideal h2is −h1 p ( 2is 1 ) ηcompressor = = = =.83 and expands to14.7 psia. If the Wactual h2 −h1 cp (T2 −T1) isentropic efficiency of the T −T 1003.9−520 T = T + 2is 1 = 520+ =1103.01o R compressor and turbine are 83% 2 1 .83 .83

and 93% respectively.What is the q2−3 = h3 −h2 = cp (T3 −T2 ) thermal efficiency of the cycle? q2−3 =.241(2000−1103.01) = 215.26BTU/lb

1540F w1−2 = h2 −h1

3 w1−2 = cp (T2 −T1 ) =.241(1103.01−520) =139.92BTU/lb k−1 2 .2857 2i  p  k  1  T = T  4  = 2000× =1035.9o R η = 93% 4i 3    t  p3  10 T w h −h c (T −T ) 1959.7−T η = actual = 3 4 = p 3 4 = 4 =.93 turbine w h −h c T −T 1959.7−1029.5 ηc = 83% ideal 3 4i p ( 3 4i ) T = T −.92×(T −T ) = 2000−.93×(2000−1035.9) =1113.03o R 1 4 4 3 3 4is 4i w = h −h =.241(2000−1110.03) = 213.59BTU/lb 60 F 3−4 3 4 14.7 psia q4−1 = h4 −h1 =.241(T4 −T1) =.241(1113.03-520.) =142.33BTU/lb h −h 1113.03−520 s η =1− 4 1 =1− = 33.9% cycle h −h 2000−1103.01 Check : ∑ w = ∑q 3 2 w 215.26−139.92 213.59 -139.92 = 215.26 −142.33 η = net = = 33.9% cycle q 215.26 27 73.67 = 72.93 in 7-23 o In an air standard gas at 520 R, pr1 =1.2147, h1 =124.27 BTU/lb

turbine, 60 F and 14.7 psia air is T pr h  p2is  compressed through a pressure p = p   =1.2147×10 =12.147 980 10.61 236.02 r2is r1 p  ratio of 10. Air enters at 1540 F  1  998 12.147 240.48 and expands to 14.7 psia. If the h 2is = 240.48 BTU/lb isentropic efficiency of the 1000 12.30 240.98 W h − h compressor and turbine are 83% ideal 2is 1 .17 ηcompressor = = = .83 ratio = .1006 and 93% respectively.What is the Wactual h 2 − h1 1.69 thermal efficiency of the cycle? h − h 240.48−124.27 h = h + 2is 1 =124.27 + 2 1 .83 .83

h 2 = 264.28 BTU/lb o 1540F At 2000 R, pr3 =174., h 3 = 504.71 3  p  T p = p  4is  =174.×1/10 =17.4 2 r4is r3 p  2i  1  by interpolation at p =1.74, h = 265.99 BTU/lb ηt =93% r4is 4is Wactual h3 − h 4 ηturbine = = = .93 Wideal h3 − h 4i ηc =83% h 4 = h3 −.92×(h3 − h 4is ) = 282.7 BTU/lb 4 q h h 504.7 264.28 240.42 BTU/lb 1 4i in = 3 − 2 = − =

60 F qout = h 4 − h1 = 282.7 −124.27 =155.43 BTU/lb 14.7 psia Qin 155.43 η =1− =1− = 35.3% 28 s Qout 240.42 Regenerated Brayton Cycle

Actual HeatTransfer Regenerator Effectiveness = Ideal Heatr Transfer h − h Regenerator Effectiveness = 5 2 h 4 − h 2 h − h Regenerator Effectiveness = 4 6 h 4 − h 2

29 6

8 7 9 T 5 4 2 10 Brayton Cycle with: Intercooling 1 Regeneration 3 Reheat s 30 A gas turbine cycle operates with 2 stages of inter-cooled compression, a regenerator, and two stages of expansion with reheat between them. Air enters the compressor at 100 kpa, 300 K at a volume flow of 5 cubic meters/ sec. The turbine and compressor stages have an equal pressure ratio, isentropic efficiencies of 80 % and the overall cycle pressure ratio is 8. The regenerator effectiveness is 80%. Turbine inlet temperature is 1400 K. Determine: a) the cycle efficiency.

Pt T p (p ) h r Q 1 300 100 1.386 300.19 in Qin 2s 403.29 283 3.920 404.31 8 2 283 430.34 3 300 100 1.386 300.19 7 3s 403.29 283 3.920 404.31 4 283 430.34 5 9 5 800 1061.2 T 10 6 1400 800 450.50 1515.42 4 2 7s 1105.9 283 159.3 1144. 73 7 283 1218.9 Q 8 1400 800 450.50 1515.42 out 1 8s 1105.9 283 159.3 1144. 73 3 Q 9 283 1218.9 out 10 100 588.05 s 31 Point 1 v = RT/p = .287 × 300/100 = .861 kg/sec 3 m = 5m /sec/v = 5.81 kg/sec Q Table A − 22 @ T = 300, h = 300.19 kJ/kg, (p ) = 1.386 in r1 Qin Point 2s 8 stage pressure ratio = pr × pr = 8, pr = 2.83 (p ) = (p )× p = 3.92 r2s r1s r 7 @ (pr2s ) = 3.92, h2s = 404.31 Point 3 5 9 h = h + (h − h )/.8 = 430.34 T 2 1 2s 1 2 10 Point 6 4

TableA − 22 @ T = 1400, h = 11414.42 kJ/kg, (pr1 ) = 450.5 (p ) = (p )/p = 159.3 r7s r6s r Qout @ (p ) = 159.3, h = 1144.73, T = 1105.9 r7s 1 h7 = h6 − (h6 − h7s )/.8 = 1218.9 kJ/kg 3 Qout Recuperator Energy Balance s Q h − h h − h ε = actual = 5 4 = 9 10 Qmax h9 − h4 h9 − h4 Point 5

h5 = h4 + ε × (h9 − h4 ) = 1061.2 kJ/kg Point 10

h10 = h9 − ε × (h9 − h4 ) = 588.05 kJ/kg 32 Q Q (h − h )+ (h − h ) in η = 1− out = 1− 10 1 2 3 Q cycle Q (h − h )+ (h − h ) in in 6 5 2 3 8 combustor + reheat ηcycle = 1− exhaust + intercooling T 7 288.05+130.15 ηcycle = 1− = 44.3 % 454.22 + 296.52 5 9 2 10 W = Turbine Work − Compressor Work 4

W = m ×[(h6 − h7 ) + (h8 − h9 ) − (h2 − h1) − (h4 − h3)] W = 5.81× 2 ×[(1515.42 −1218.9) − (430.34 − −300.19)] Qout W = 5.81× 2 ×[(296.52) − (130.15)] 1 W = 1921.6 kJ/sec = 1921.6 KW 3 Qout s

Back Work Ratio = Wcompressor /Wturbine Back Work Ratio = 130.15/296.52 = .438

33 2single 2issingle stage stage Intercooled Compression

T w1→2 = h 2 − h1 discharge pressure q2→3 = h 2 − h3

w 3→4 = h 4 − h3 inlet pressure k−1  p  k s T = T  2  2is 1 p  discharge  1  pressure T 4 4is 2is 2 W W

Q inlet 1 pressure 3 34 s k−1  p  k Compare compression over  2  .2857 o T2is = T1  = 530(2) = 646 R a pressure ratio of 4 from  p1  2single 14.7 psia, 530 R with T2is − T1 646 − 530 o stage T2 = T1 + = 530 + = 675 R compression in two .8 .8

stages of equal pressure w12 = cp (T2 − T3 ) = .24×(675 − 530) = 34.80BTU / lb

ratio intercooled q2→3 = cp (T2 − T3 ) = .24×(646 − 530) = 32.64 BTU/lb to 510 R. The k−1  p  k efficieny of 2issingle  4  .2857 o T4is = T1  = 510(2) = 621.69 R all stages stage  p3  T − T 621.69 − 510 is 80% 2 T = T + 4is 3 = 510 + = 649.61o R 4 3 .8 .8 w = c (T − T ) = .24×(649.61− 510) = 33.51 BTU/lb 4 3→4 p 4 3 2 w = 34.8 + 33.51 = 68.31 BTU/lb 2is inter 4is cooled T k−1 k   .2857 o T2is single = T1p2 single / p1  = 530×(4) = 787.56 R W Q W stage  stage  T2is single − T1 stage o Tsingle = T1 + = 851.95 R 3 1 stage .8 w single = cp (T4 − T3 ) = .24×(851.95 − 530) = 77.68 BTU/lb s stage 35 11.4% higher INTEGRAL GEAR CENTRIFUGAL COMPRESSOR

36 Supercharger Turbocharger 1 4

2 3

Intercooled Supercharger

Q

37 Air Standard Otto Cycle with a Turbocharger Turbocharger 1 4 engine 2 3 combustion power v=constant stroke T 3 P atm engine 4 compression W stroke 2

1

38 ProCharger Supercharger

39 TurboCharger Air Inlet

Compressed Air to Engine

Engine Exhaust

40