1. Engine Cycle Analysis

2. Air Standard Otto Cycle

4. ► Starting with the piston at bottom dead center (BDC), compression proceeds isentropically from state 1 to state 2. ► Heat is added at constant volume from state 2 to state 3. ► Expansion occurs isentropically from state 3 to state 4. ► Heat is rejected at constant volume from state 4 to state 1

5. Air Standard Otto Cycle

6. ► Thermal Efficiency (η th ) = W Net /Q in ► Heat Added at constant volume from state 2 to state 3 Q 2-3 = U 3 – U 2 = mc v (T 3 -T 2 ) ► Heat Rejected at constant volume from state 4 to state 1 Q 4-1 = U 1 – U 4 = -mc v (T4 – T 1 ) ► Specific Heat (C v ) = 0.1714 Btu/lbm-°R ► Mean Effective Pressure (mep) = W Net / Displacement Volume = W Net / V 1 – V 2

7. Air Standard Otto Cycle ► Compression ratio (r) = Vol at BDC/Vol at TDC ► r = V 1 /V 2 = V 4 /V 3 ► η th = 1 – (1/(r) k-1 ) ► Specific Heat Ratio (k) = C p /C v (for air k = 1.4) ► Percentage of clearance (c) = Clearance Volume/Displacement volume = V 2 /(V 1 – V 2 )

8. Air Standard Diesel Cycle ► Starting with the piston at bottom dead center (BDC), compression occurs isentropically from state 1 to state 2 ► Heat is added at constant pressure from state 2 to state 3 ► Expansion occurs isentropically from state 3 to state 4 ► Heat rejection occurs at constant volume from state 4 to state 1

9. Air Standard Diesel Cycle

11. ► Heat supplied at constant pressure for a closed system, Q 2-3 = H 3 – H 2 = mc p (T 3 – T 2 ) ► Heat rejected at constant volume, Q 4-1 = U 1 – U 4 = mc v (T 4 – T 1 ) ► Specific Heat (C p ) = 0.24 Btu/lbm-°R ► η th = 1-1/k((T 4 – T 1 )/(T 3 – T 2 ))

12. Air Standard Diesel Cycle ► Cutoff ratio (r c ) = Vol at the end of heat addition/Vol at the start of heat addition = V 3 /V 2 ► Cutoff percentage (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► η th = 1 – (1/(r) k-1 )[(r c k – 1)/(k(r c – 1))]

13. Polytropic Process for a Closed System ► Pv = RT ► Specific Gas Constant (R) = for air = 53.34 ft-lb f /lbm-°R ► p 1 = RT 1 /v 1 ► p 2 = RT 2 /v 2 ► P 1 /P 2 = (V 2 /V 1 ) k ► P 2 /P 1 = (V 1 /V 2 ) k ► T 1 /T 2 = (V 2 /V 1 ) k-1 ► T 2 /T 1 = (V 1 /V 2 ) k-1

14. Polytropic Process for a Closed System ► V 2 /V 1 = (T 1 /T 2 ) 1/k-1 ► V 1 /V 2 = (P 2 /P 1 ) 1/k ► T 1 /T 2 = (P 2 /P 1 ) (1-k)/k = (P 1 /P 2 ) (k-1)/k ► P 1 /P 2 = (T 1 /T 2 ) k/(k-1)

15. Air Standard Otto Cycle ► A air standard Otto cycle has an initial temperature of 100 °F, a pressure of 14.7 psia, and compression pressure P 2 = 356 psia. The pressure at the end of heat addition is 1100 psia. Determine: ► (A) The compression ratio (r) ► (B) The thermal efficiency η thermal ► (C) The percentage of clearance (c) ► (D) The maximum temperature T max and the remaining Pressures, Temperature and Volumes

16. Air Standard Otto Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm) 156014.7 2356 31100 4

17. Air Standard Otto Cycle ► r = V 1 /V 2 ► V 1 /V 2 = (P 2 /P 1 ) 1/k ► V 1 /V 2 = (356/14.7) 1/1.4 ► r = 9.74 ► η th = 1 – (1/(r) k-1 ) ► η th = 1 – (1/(9.74) 1.4-1 ) ► η th = 0.598 or 59.8 %

18. Air Standard Otto Cycle ► P 1 = RT 1 /V 1 ► V 1 = RT 1 /P 1 ► V 1 = ((53.34 ft-lb f /lbm-°R)(560 °R)) / ((14.7 lb f /in 2 )(144 in 2 /ft 2 )) ► V 1 = 14.11 ft 3 /lbm ► r = V 1 /V 2 ► V 2 = V 1 /r ► V 2 = 14.11/9.74 ► V 2 = 1.449 ft 3 /lbm

19. Air Standard Otto Cycle ► Percentage of clearance (c) = V 2 /(V 1 – V 2 ) ► Percentage of clearance (c) = 1.449/(14.11 – 1.449) ► Percentage of clearance (c) = 0.114 or 11.4 % 11.4 % ► P 2 = RT 2 /V 2 ► T 2 = P 2 V 2 /R

20. Air Standard Otto Cycle ► T 2 = (((356 lb f /in 2 )(144 in 2 /ft 2 )(1.449 ft 3 /lbm)) / 53.34 ft-lb f /lbm-°R) ► T 2 = 1393 °R

21. Air Standard Otto Cycle

22. ► Since V 2 = V 3 ► V 2 = RT 2 /P 2 ► V 3 = RT 3 /P 3 ► RT 2 /P 2 = RT 3 /P 3 ► T 2 /P 2 = T 3 /P 3 ► T 3 = T max = T 2 (P 3 /P 2 ) ► T 3 = 1393 °R (1100/356) ► T 3 = 4304.2 °R

23. Air Standard Otto Cycle ► Find P 4 and T 4 ► (P 3 /P 4 ) = (V 4 /V 3 ) k ► (V 4 /V 3 ) = r = 9.74 ► (9.74) 1.4 = (P 3 /P 4 ) ► (9.74) 1.4 = (1100/P 4 ) ► P 4 = 45.4 psia

24. Air Standard Otto Cycle ► T 3 /T 4 = (V 4 /V 3 ) k-1 ► T 3 /T 4 = (r) k-1 ► 4302/T 4 = (9.74) 1.4-1 ► T 4 = 4302/(9.74) 1.4-1 ► T 4 = 1731.2 °R

25. Air Standard Otto Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm) 156014.714.11 213933561.449 3430211001.449 41731.245.414.11

26. Air Standard Diesel Cycle ► An air standard Diesel Cycle operates on 1 ft 3 of air at 14.5 psia and 140 °F. The compression ratio (r) is 14, and the cutoff is 6.2 % of the stroke. Determine: ► (A) The temperatures, pressures and volumes around the cycle ► (B) The net work ► (C) The heat added ► (D) The efficiency

27. Air Standard Diesel Cycle

28. State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm) 160014.51 2 3 4

29. Air Standard Diesel Cycle ► P 1 = mRT 1 /V 1 ► m = P 1 V 1 /RT 1 ► m = (((14.5 lb f /in 2 )(144 in 2 /ft 2 )(1 ft 3 )) / ((53.34 ft-lb f /lbm-°R)(600 °R))) ► m = 0.0652 lbm ► T 2 /T 1 = (V 1 /V 2 ) k-1 ► T 2 = T 1 (V 1 /V 2 ) k-1 ► r = (V 1 /V 2 )

30. Air Standard Diesel Cycle ► T 2 = (600 °R)(14) 0.4 ► T 2 = 1724.2 °R ► P 2 /P 1 = (V 1 /V 2 ) k ► P 2 = P 1 (V 1 /V 2 ) k ► P 2 = (14.5 psia)(14) 1.4 ► P 2 = 583.4 psia ► r = (V 1 /V 2 )

31. Air Standard Diesel Cycle ► V 2 = (V 1 /r) ► V 2 = (1/14) ► V 2 = 0.0714 ft 3

32. Air Standard Diesel Cycle

33. ► Since the process between P 2 to P 3 is constant pressure P 3 = 583.4 psia. ► Cutoff percentage (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► 0.062 = ((V 3 – 0.0714)/(1 – 0.0714)) ► V 3 = 0.129 ft 3

34. Air Standard Diesel Cycle ► P 2 = RT 2 /V 2 ► P 3 = RT 3 /V 3 ► Since P 2 = P 3 ► RT 2 /V 2 = RT 3 /V 3 ► T 3 = T 2 (V 3 /V 2 ) ► T 3 = (1724.2 °R)(0.129/0.0714) ► T 3 = 3115 °R

35. Air Standard Diesel Cycle ► T 4 /T 3 = (V 3 /V 4 ) k-1 ► T 4 = T 3 (V 3 /V 4 ) k-1 ► T 4 = (3115 °R)(0.129/1.00) 0.4 ► T 4 = 1373 °R ► P 4 /P 3 = (V 3 /V 4 ) k ► P 4 = P 3 (V 3 /V 4 ) k ► P 4 = 583.4(0.129/1.0) 1.4 ► P 4 = 33.2 psia

36. Air Standard Diesel Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm) 160014.51 21724.2583.40.0714 33115583.40.129 4137333.21

37. Air Standard Diesel Cycle

38. ► The heat added is: ► Q 2-3 = H 3 – H 2 = mC p (T 3 – T 2 ) ► Q 2-3 = (0.0652 lbm)(0.24 Btu/lbm-°R)(3115 – 1742.2 °R) ► Q 2-3 = 21.76 Btu

39. Air Standard Diesel Cycle ► The heat out is: ► Q Out = Q 4-1 = mC v (T 1 – T 4 ) ► Q 4-1 = (0.0652)(0.1714)(600 – 1373) ► Q 4-1 = - 8.64 Btu ► W net = ∑Q = Q in + Q out ► W net = 21.76 Btu + - 8.64 ► W net = 13.12 Btu

40. Air Standard Diesel Cycle ► Thermal Efficiency (η th ) = W net /Q in ► η th = 13.12/21.76 ► η th = 0.603 = 60.3 %