1. EGR 334 Thermodynamics Chapter 9: Sections 3-4 Lecture 32: Gas Power Systems: The Diesel Cycle Quiz Today?

2. Today’s main concepts: Understand common terminology of a Diesel engine Be able to explain the processes of the Diesel Cycle Be able to perform a 1 st Law analysis of the Diesel Cycle and determine its thermal efficiency. Be able to discuss limitations of the Diesel cycle compared to real Diesel engines. Reading Assignment: Homework Assignment: Read Chapter 9, Sections 5-6 Problems from Chap 9: 17, 20, 24, 38

3. 3 Two types of internal combustion engine Spark Ignition (lower power & lighter) Compression Ignition (spontaneous combustion) Terminology Stroke : The distance the piston moves in one direction Top Dead Center : The piston has minimum volume at the top of the stroke. Bottom Dead Center : The piston has maximum volume at the bottom of the stroke. Clearance Volume : Min vol Displacement Volume : Max vol. – Min vol. Compression Ratio: Max vol. / min vol. Sec 9.1 : Introducing Engine Terminology

4. 4 Sec 9.3 : Air-Standard Diesel Cycle The Diesel cycle can operate using either the 4 or 2 stroke engine. Has a lower efficiency than the Otto cycle for the same compression ratio. Compression Ignition More difficult to start when cold Can have higher compression ratio 16 <r Diesel <25 vs r≈11 for Otto cycle to avoid pre-ignition Produces less CO, because excess air, but produces soot. Diesel Engine Animation at http://www.animatedengines.com/diesel.html

5. 5 Sec 9.3 : Air-Standard Diesel Cycle For the Otto cycle, the Q in occurs at constant volume (compressed cylinder) For the diesel cycle, the fuel is injected while the cylinder is expanding, so the Q in occurs at constant pressure Process 1 – 2 : Isentropic compression of air (compression stroke). Process 2 – 3 : Constant pressure heat transfer to the air from an external source while moves down (ignition and part of power stroke) Process 3 – 4 : Isentropic expansion (remainder of power stroke) Process 4 – 1 : Completes cycle by a constant volume process in which heat is rejected from the air while piston is at bottom dead center

6. 6 Diesel Cycle analysis: Use W 23 with energy balance to find Q 23. Sec 9.3 : Air-Standard Diesel Cycle Process 2 – 3 : Constant pressure heat transfer to the air from an external source while moves down (ignition and part of power stroke) Remaining processes analysis is the same as for Otto cycle.

7. 7 Find State 1 Properties Sec 9.3 : Air-Standard Diesel Cycle Given T 1 and r  use table to find u 1 & h 1. Find state 2: for isentropic process. were r c is the cutoff ratio: Find state 3: Use ideal gas law with p 3 = p 2. Find state 4. For Cold-Air Standard analysis: (isentropic processes 1-2 and 3-4) For state 2.For state 4. or

8. 8 A couple words about how and when to use relative pressure and relative spec. volume. only for use between isentropic processes.

9. 9 Example (9.26): An air-standard Diesel cycle has a compression ratio of 16 and a cut-off ratio of 2. At the beginning of the compression, p 1 = 14.2 psi, V 1 = 0.5 ft 3, T 1 = 520°R. Calculate (a)The heat added, in Btu. (b)The max T. (c)The thermal efficiency. (d)The mean effective pressure, State1234 T (R)520 p (psi)14.2 u (Btu/lb m ) h (Btu/lb m ) vrvr prpr Given info: Diesel Cycle State 1: p 1 =14.2 psi, T 1 =520 R V 1 = 0.5 ft 3 State 2: s 2 = s 1 State 3: p 3 = p 2 State 4: s 4 =s 3 and v 4 =v 1 Compression ratio: r = V 1 /V 2 =16 Cutoff ratio: r c = V 3 /V 2 = 2

10. 10 Example (9.26): An air-standard Diesel cycle has a compression ratio of 16 and a cut-off ratio of 2. At the beginning of the compression, p 1 = 14.2 psi, V 1 = 0.5 ft 3, T 1 = 520°R. Calculate (a)The heat added, in Btu. (b)The max T. (c)The thermal efficiency. (d)The mean effective pressure, State1234 T (R)520 p (psi)14.2 u (Btu/lb m )88.62 h (Btu/lb m ) vrvr 158.58 prpr 1.2147 Determine State Properties: For State 1: Using Table A-22E: for T 1 Read u 1 = 88.62 Btu/lb m h 1 = 124.27 Btu/lb m p r1 = 1.2147 and v r1 = 158.58 State1234 T (R)520 p (psi)14.2 u (Btu/lb m ) h (Btu/lb m ) vrvr prpr Also using Ideal Gas Equation:

11. 11 Example (9.26): For State 2: (process 1-2 is isentropic) Using compression ratio, r Next use v r2 and Table A-22E to find other state properties: T 2 = 1502.5°R u 2 = 266.84 Btu/lbm p r2 = 56.27 h 2 = 369.85 Btu/lb m State1234 T (R)5201502.5 p (psi)14.2657.8 u (Btu/lb m )88.62266.84 h (Btu/lb m )124.27369.85 vrvr 158.589.911 prpr 1.214756.27 Then: State1234 T (R)520 p (psi)14.2 u (Btu/lb m )88.62 h (Btu/lb m )124.27 vrvr 158.58 prpr 1.2147

12. 12 Example (9.26): For State 3: Then using Table A-22E, use T 3 to find properties u 3 = 586.16 Btu/lb m h 3 = 792.03 Btu/lb m p 3r = 948.36 v 3r = 1.174 State1234 T (R)5201502.53005 p (psi)14.2657.8 u (Btu/lb m )88.62266.84586.16 h (Btu/lb m )124.27369.85792.03 vrvr 158.589.9111.174 prpr 1.214756.27948.36 Using ideal Gas process 2-3 constant pressure State1234 T (R)5201502.5 p (psi)14.2657.8 u (Btu/lb m )88.62266.84 h (Btu/lb m )124.27369.85 vrvr 158.589.911 prpr 1.214756.27

13. 13 Example (9.26): For State 4: therefore: State1234 T (R)5201502.530051530.8 p (psi)14.2657.8 41.9 u (Btu/lb m )88.62266.84586.16272.38 h (Btu/lb m )124.27369.85792.03377.47 vrvr 158.589.9111.1749.392 prpr 1.214756.27948.3660.46 then (process 3-4 is isentropic and V 4 = V 1 ) Then using Table A-22E, use v r4 to find properties u 4 = 272.38Btu/lb m h 4 = 377.47 Btu/lb m p r4 = 60.46 T 4 = 1530.8 R State1234 T (R)5201502.53005 p (psi)14.2657.865.65 u (Btu/lb m )88.62266.84586.16 h (Btu/lb m )124.27369.85792.03 vrvr 158.589.9111.174 prpr 1.214756.27948.36

14. 14 Example (9.26): (a)The heat added, in BTU. (b)The max T. (c)The thermal efficiency. (d)The mean effective pressure, Heat is added during process 2-3 State1234 T (R)5201502.530051530.8 p (psi)14.2657.8 41.9 u (Btu/lb m )88.62266.84586.16272.38 h (Btu/lb m )124.27369.85792.03377.47 vrvr 158.589.9111.1749.392 prpr 1.214756.27948.3660.46 Applying the 1 st Law Maximum Temperature is at Sate 3: T max = 3005 o R

15. 15 Example (9.26): (a)The heat added, in BTU. (b)The max T. (c)The thermal efficiency. (d)The mean effective pressure, State1234 T (R)5201502.530051530.8 p (psi)14.2657.8 41.9 u (Btu/lb m )88.62266.84586.16272.38 h (Btu/lb m )124.27369.85792.03377.47 vrvr 158.589.9111.1749.392 prpr 1.214756.27948.3660.46 Thermal Efficiency: Mean Effective Pressure: where:

16. End of Slides for Lecture 34 16