- 1. CHAPTER 5 MEC 451 Thermodynamics Air Standard Cycle Lecture Notes: MOHD HAFIZ MOHD NOH HAZRAN HUSAIN & MOHD SUHAIRIL Faculty of Mechanical Engineering Universiti Teknologi MARA, 40450 Shah Alam, Selangor For students EM 220 and EM 221 only
- 2. ηth net in W Q = ηth Carnot L H T T , = − 1 Upon derivation the performance of the real cycle is often measured in terms of its thermal efficiency The Carnot cycle was introduced as the most efficient heat engine that operate between two fixed temperatures TH and TL. The thermal efficiency of Carnot cycle is given by Review – Carnot Cycle 2
- 3. The ideal gas equation is defined as mRT PV or RT Pv = = where P = pressure in kPa v = specific volume in m3 /kg (or V = volume in m3 ) R = ideal gas constant in kJ/kg.K m = mass in kg T = temperature in K Review – Ideal Gas Law 3 3
- 4. The Δu and Δh of ideal gases can be expressed as ) ( 1 2 1 2 T T C u u u v − = − = ∆ ) ( 1 2 1 2 T T C h h h P − = − = ∆ Δu - constant volume process Δh - constant pressure process 4
- 5. Process Description Result of IGL isochoric constant volume (V1 = V2 ) isobaric constant pressure (P1 = P2 ) isothermal constant temperature (T1 = T2 ) polytropic -none- isentropic constant entropy (S1 = S2 ) According to a law of constant = n V P 2 2 1 1 T P T P = 2 2 1 1 T V T V = 2 2 1 1 V P V P = 1 2 1 1 2 2 1 − = = n n n T T V V P P Review – Thermodynamics Processes 5
- 6. R = 0.2871 kJ/kg.K Cp = 1.005 kJ/kg.K Cv = 0.718 kJ/kg.K k = 1.4 where R = ideal gas constant Cp = specific heat at constant pressure Cv = specific heat at constant volume k = specific heat ratio Review – Properties of Air 6
- 7. IC Engine – combustion of fuel takes place inside an engine’s cylinder. Introduction 7
- 8. Air continuously circulates in a closed loop. Always behaves as an ideal gas. All the processes that make up the cycle are internally reversible. The combustion process is replaced by a heat-addition process from an external source. Air-Standard Assumptions 8
- 9. A heat rejection process that restores the working fluid to its initial state replaces the exhaust process. The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25o C or 77o F). No chemical reaction takes place in the engine. Air-Standard Assumptions 9
- 10. Top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure Terminology for Reciprocating Devices 10
- 11. The compression ratio r of an engine is defined as r V V V V BDC TDC = = max min The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. MEP W V V w v v net net = − = − max min max min11
- 12. Otto Cycle The Ideal Cycle for Spark-Ignition Engines 12
- 13. The processes in the Otto cycle are as per following: Process Description 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection 13
- 14. Related formula based on basic thermodynamics: 1 2 1 1 2 2 1 − = = n n n T T V V P P 1 2 1 1 2 2 1 − = = n n n T T V V P P ( ) 3 2 in v Q mC T T = − ( ) 4 1 out v Q mC T T = − 14
- 15. Thermal efficiency of the Otto cycle: ηth net in net in in out in out in W Q Q Q Q Q Q Q Q = = = − = − 1 Apply first law closed system to process 2-3, V = constant. Thus, for constant specific heats Q U Q Q mC T T net net in v , , ( ) 23 23 23 3 2 = = = − ∆ ,23 ,23 23 3 ,23 ,23 ,23 2 0 0 net net net other b Q W U W W W PdV − = ∆ = + = + = ∫ 15
- 16. Apply first law closed system to process 4-1, V = constant. Thus, for constant specific heats, Q U Q Q mC T T Q mC T T mC T T net net out v out v v , , ( ) ( ) ( ) 41 41 41 1 4 1 4 4 1 = = − = − = − − = − ∆ The thermal efficiency becomes ηth Otto out in v v Q Q mC T T mC T T , ( ) ( ) = − = − − − 1 1 4 1 3 2 ,41 ,41 41 1 ,41 ,41 ,41 4 0 0 net net net other b Q W U W W W PdV − = ∆ = + = + = ∫ 16
- 17. ηth Otto T T T T T T T T T T , ( ) ( ) ( / ) ( / ) = − − − = − − − 1 1 1 1 4 1 3 2 1 4 1 2 3 2 Recall processes 1-2 and 3-4 are isentropic, so Since V3 = V2 and V4 = V1, 3 3 2 4 1 4 1 2 T T T T or T T T T = = 1 1 3 2 1 4 1 2 4 3 k k T T V V and T V T V − − = = ÷ ÷ 17
- 18. The Otto cycle efficiency becomes ηth Otto T T , = − 1 1 2 Since process 1-2 is isentropic, where the compression ratio is r = V1/V2 and ηth Otto k r , = − − 1 1 1 1 2 1 1 2 1 1 1 2 2 1 1 k k k T V T V T V T V r − − − = ÷ = = ÷ ÷ 18
- 19. An Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P1 = 95 kPa, T1 = 17°C, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, ηth, the back work ratio and the mean effective pressure. Example 5.1 Solution: Data given: 1 1 2 23 1 1 290 9 7.5 95 3.8 T K V V Q kJ P kPa V Litres = = = = = 19
- 20. Example 5.1 ( ) ( ) ( ) ( ) 1 0.4 2 1 2 1 2 1 1.4 2 1 2 1 2 290 9 698.4 95 9 Pr 1 2 Pr 2 3 . 2059 1 : k k st net net T V T K T V P V P kPa P V law Q W ocess isentropiccompression ocess Const volumeheat addition − − = ⇒ = = ÷ − − = ⇒ = = ÷ − ( ) ( ) 3 0 23 3 2 1 1 1 1 23 1 23 23 1 0.2871 290 : 0.875 95 1727 v m kg kJ kg U Q mC T T IGL Pv RT v Q v q Q m V = ∆ = − = ⇒ = = = = = 20
- 21. Example 5.1 ( ) ( ) ( ) ( ) ( ) 3 2 3 2 23 3 2 3 2 3 3 3 1 0.4 3 4 4 3 3 4 1.4 3 4 4 3 3 4 : 0.718 698.4 9.15 3103.7 1/ 9 1288.8 1/ 9 422 Pr 3 4 exp v k k ocess isentr Back to IGL ButV V P P q C T T T T T P MPa T K V T T T opic ansi K T V V P P P kPa P on V − = = − = = − = = = ⇒ = = ÷ = ⇒ = = ÷ − 21
- 22. Example 5.1 ( ) ( ) ( ) ( ) 41 4 1 41 4 1 0.718 1288.8 290 717. Pr 4 1 . 1 v v kJ kg Q mC T ocess Const volumeheat rejec T q C T tion T = − = − − = − = Then: ( ) 23 41 , 1009.6 0.585 58.5% net in out kJ kg net th Otto in W q q q q W q η = − = − = = = 22
- 23. Example 5.1 What else? ( ) ( ) ( ) max min max min 1 2 1 2 1 1 1 12 exp 34 1 / 1009.6 1298 1 0.875 1 1/ 9 net net net net net r v compr bw ans W w MEP V V v v w w v v v v v w kPa v C w u r w u = = − − = = − − = = = − − ∆ = = = −∆ ( ) 2 1 v T T C − ( ) ( ) 3 4 0.225 22.5% T T − = 23
- 24. Supplementary Problems 5.1 1. An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during the constant-volume heat- addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heat addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle. [(a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent,(d ) 495 kPa] 2. The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100 kPa, 35°C, and 600 cm3 . The temperature at the end of the isentropic expansion process is 800 K. Using specific heat values at room temperature, determine (a) the highest temperature and pressure in the cycle; (b) the amount of heat transferred in, in kJ; (c) the thermal efficiency; and (d) the mean effective pressure. [(a) 1969 K, 6072 kPa,(b) 0.59 kJ, (c) 59.4 percent, (d) 652 kPa] 24
- 25. The processes in the Diesel cycle are as per following: Diesel Cycle 25
- 27. Related formula based on basic thermodynamics: 1 2 1 1 2 2 1 − = = n n n T T V V P P 1 2 1 1 2 2 1 − = = n n n T T V V P P ( ) 3 2 in P Q mC T T = − ( ) 4 1 out v Q mC T T = − 27
- 28. Thermal efficiency of the Diesel cycle ηth Diesel net in out in W Q Q Q , = = − 1 Apply the first law closed system to process 2-3, P = constant. Thus, for constant specific heats Q U P V V Q Q mC T T mR T T Q mC T T net net in v in p , , ( ) ( ) ( ) ( ) 23 23 2 3 2 23 3 2 3 2 3 2 = + − = = − + − = − ∆ ( ) ,23 ,23 23 3 ,23 ,23 ,23 2 2 3 2 0 0 net net net other b Q W U W W W PdV P V V − = ∆ = + = + = = − ∫ 28
- 29. Apply the first law closed system to process 4-1, V = constant Q U Q Q mC T T Q mC T T mC T T net net out v out v v , , ( ) ( ) ( ) 41 41 41 1 4 1 4 4 1 = = − = − = − − = − ∆ Thus, for constant specific heats The thermal efficiency becomes ηth Diesel out in v p Q Q mC T T mC T T , ( ) ( ) = − = − − − 1 1 4 1 3 2 ,41 ,41 41 1 ,41 ,41 ,41 4 0 0 net net net other b Q W U W W W PdV − = ∆ = + = + = ∫ 29
- 30. PV T PV T V V T T P P 4 4 4 1 1 1 4 1 4 1 4 1 = = = where Recall processes 1-2 and 3-4 are isentropic, so PV PV PV PV k k k k 1 1 2 2 4 4 3 3 = = and Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain Therefore, 3 4 4 2 k k c V P r T V = = ÷ ( ) , 1 1 1 1 1 k c th Diesel k c r r k r η − − = − − 30
- 31. An air-standard Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2.5. The state at the beginning of compression is fixed by P = 0.9 bar ant T = 300K. Calculate: i. the thermal efficiency of the cycle, ii. the maximum pressure, Pmax, and iii. The mean effective pressure. Example 5.2 Solution: Data given: 1 2 3 2 18 2.5 V V V V = = 31
- 32. Example 5.2 ( ) ( ) ( ) ( ) ( ) 1 0.4 2 1 2 1 2 3 3 2 2 3 3 2 2 3 2 4 1 2 3 2 3 4 Pr 1 2 Pr 2 3 300 18 953.3 2383. . Pr 3 4 exp 3 . 18 1/ 2.5 7.2 k ocess isentropiccompression ocess Const pressureheat addition ocess isentropic ansio T V T K T V V V V P P T T K T T V V V V V V V T n − = ⇒ = = ÷ = ⇒ = ⇒ = = ÷ = = = − − − ( ) 1 0.4 3 4 3 4 2383.3 1/ 7.2 1082 k V T K T V − = ⇒ = = ÷ 32
- 33. Example 5.2 ( ) ( ) ( ) ( ) 23 3 2 3 2 41 4 1 4 1 1437.15 561.48 875.67 kJ in P in p kg kJ out P out p kg kJ net in out kg Q Q mC T T q C T T Q Q mC T T q C T T w q q = = − ⇒ = − = = = − ⇒ = − = = − = What we need? ( ) ( ) ( ) ( ) ( ) ( ) ( ) , max 2 3 1 2 2 2 max 1 1 1 0.6093 60.93% 5148 875.67 969.1 1 1/ 0.9566 1 1/18 net th diesel in k k net w i q ii P P P P T P kPa P P T w iii MEP kPa V r η − = = = = = ⇒ = ÷ ÷ = = = − − 33
- 34. Supplementary Problems 5.2 1. An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and 20°C. If the maximum temperature in the cycle is not to exceed 2200 K, determine (a) the thermal efficiency and (b) the mean effective pressure. Assume constant specific heats for air at room temperature. [ (a) 63.5 percent, (b) 933 kPa] 2. An ideal diesel cycle has a compression ratio of 16 to 1. The maximum cycle temperature is 1700°C and the minimum cycle temperature is 15°C. Calculate: i. the specific heat transfer to the cycle ii. the specific work of the cycle iii. the thermal efficiency of the cycle 34
- 35. Dual cycle gives a better approximation to a real engine. The heat addition process is partly done at a constant volume and partly at constant pressure. From the P-v diagram, it looks like the heat addition process is a combination of both Otto and Diesel cycles. Dual Cycle 35
- 36. The same procedure as to Otto and Diesel cycles can be applied to Dual cycle. Upon substitutions, the thermal efficiency of Dual cycle becomes ( ) ( ) [ ] 1 1 1 1 1 − − + − − − = k v c p p k c p th r r c k r r r η Dual Cycle 36
- 37. At the beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is 300 K and the pressure is 1 bar. The pressure ratio for the constant volume part of the heating process is 1.5 to 1. The volume ratio for the constant pressure part of the heating process is 1.2 to 1. Determine (a) the thermal efficiency and (b) the mean effective pressure. (WRONG SOLUTION!!) Example 5.3 Solution: 1 1 2 2 4 1 3 1 18 1.5 300 1.2 1 V P V P V T K V P bar = = = = = Data given: 37
- 38. ( ) ( ) ( ) ( ) ( ) 1 0.4 2 1 2 1 2 3 3 2 2 3 3 2 2 3 2 4 1 2 3 2 3 4 Pr 1 2 Pr 2 3 300 18 953.3 2383. . Pr 3 4 exp 3 . 18 1/ 2.5 7.2 k ocess isentropiccompression ocess Const pressureheat addition ocess isentropic ansio T V T K T V V V V P P T T K T T V V V V V V V T n − = ⇒ = = ÷ = ⇒ = ⇒ = = ÷ = = = − − − ( ) 1 0.4 3 4 3 4 2383.3 1/ 7.2 1082 k V T K T V − = ⇒ = = ÷ 38
- 39. ( ) ( ) ( ) 1 1 1 5 3 4 4 4 5 4 4 4 5 5 3 5 0.4 1 18 1715.94 1.2 5 Pr 4 5 84.85 exp k k k ocess isentropic ansio T V V V V T T T V n T V V V K − − − = ⇒ = = ÷ ÷ ÷ ÷ = = − Information needed? ( ) ( ) ( ) 51 5 1 23 34 3 2 4 3 204.52 629.65 kJ out v kg in v p kJ kg Q Q C T T Q Q Q C T T C T T m m m = = − = = + = − + − = 39
- 40. Answer the questions ? ( ) ( ) ( ) ( ) ( ) 1 1 1 18 204.52 1 1 0.675 67.5% 629.65 1 425.13 0.8613 1 522.63 net in out out th in in in net r W Q Q Q a Q Q Q W b MEP v kPa η − = = = − = − = = − = − = 40
- 41. Indicated power (IP) Brake power (bp) Friction power (fp) and mechanical efficiency, ηm Brake mean effective pressure (bmep), thermal efficiency and fuel consumption Volumetric efficiency, ηv Criteria of Performance 41
- 42. Defined as the rate of work done by the gas on the piston as evaluated from an indicator diagram obtained from the engine using the electronic engine indicator. 2 LANn p IP i = For four-stroke engine, And for two-stroke engine, LANn p IP i = Indicated Power 42 ip = work done per cycle × cycle per minute n is the number of cylinders.
- 43. Indicated Power 43 constant diagram of length diagram of area net × = i p Indicated mean effective pressure, pi given by:, For one engine cylinder Work done per cycle = pi × A × L Where A = area of piston L = length of stroke time unit per cycle AL P ip i × = Power output = (work done per cycle) x (cycle per minute) For four-stoke engines, the number of cycles per unit time is N/2 and for two-stroke engines the number of cycles per unit time is N, where N is the engine speed. volume nt displaceme cycle per done work = i p
- 44. Brake power is a way to measure the engine power output. The engine is connected to a brake (or dynamometer) which can be loaded so that the torque exerted by the engine can be measured. The torque is obtained by reading off a net load, w at known radius, r. Wr = τ Brake Power 44
- 45. Therefore τ πN bp 2 = 2 2 LANn P LANn p IP bp b i m m = = = η η Brake power is also can be expressed as Then the brake mean effective pressure (Pb) is i m b P P η = 45
- 46. Friction Power 46 The difference between the Ip and bp is the friction power (fp). It is the power that overcome the frictional resistance of the engine parts. bp IP fp − =
- 47. Power input to the shaft is usually bigger than the indicated power due to frictional losses or the mechanical efficiency. power indicated power brake mech = η Mechanical Efficiency 47
- 48. Brake Mean Effective Pressure 48 From the definition of Brake power IP BP m η = Since 2 LANn p IP i = for 4 stroke engine and 2 2 LANn P LANn p bp b i m = = η Since and Pi are difficult to obtain, they may be combined and replaced by a brake mean effective pressure, Pb m η m η Equating this equation to another definition of bp: NT LANn Pb η 2 2 = T LAn Pb η 4 = So: Its observed that bmep is proportional to torque.
- 49. Brake Thermal Efficiency 49 The power output of the engine is obtained from the chemical energy of the fuel supplied. The overall engine efficiency is given by the brake thermal efficiency, m η v net f p fe p bp Q m b P b , power equivalent fuel power brake given power power brake = = = = η mf = mass flow fuel , Qnet,v = net calarofic value of the fuel.
- 50. sfc is the mass flow rate of fuel consumed per unit power output and is a criterion of economical power production. bp m sfc f = Specific Fuel Consumption 50
- 51. Volumetric efficiency is only used with four-stroke cycle engine, which have a distinct induction process. The parameter used to measure the effectiveness of an engine’s induction process is the volumetric efficiency. s V V V = η Volumetric Efficiency 51
- 52. An engine operating at 2400 rpm consumes 12 ml of fuel (s.g. 0.85) in 60 second. The engine indicates a load of 30 N on the pony brake system and the brake’s torque arm is 20 cm. Determine (a) the brake power, (b) the mass flow rate of fuel, and (c) the specific fuel consumption. Example 5.4 Solution: 52
- 53. A four-cylinder petrol engine has a bore of 57 mm and a stroke of 90 mm. Its rated speed is 2800 rpm and it is tested at this speed against a brake which has a torque arm of 0.356 m. The net brake load is 155 N and the fuel consumption is 6.741 l/h. The specific gravity of the petrol used is 0.735 and it has a net calorific value of 44,200 kJ/kg. The engine is tested in an atmospheric condition at 101.325 kPa and 15 oC at air-fuel ratio of 14.5/1. Calculate for this speed, the engine torque, the bmep, the brake thermal efficiency, the specific fuel consumption and the volumetric efficiency of the engine. Example 5.4 Solution: 53
- 54. 2 LANn p IP i = τ πN bp 2 = Real Case 54