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EGR 334 Thermodynamics Chapter 9: Sections 3-4

EGR 334 Thermodynamics Chapter 9: Sections 3-4. Lecture 32: Gas Power Systems: The Diesel Cycle. Quiz Today?. Today’s main concepts:. Understand common terminology of a Diesel engine Be able to explain the processes of the Diesel Cycle

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EGR 334 Thermodynamics Chapter 9: Sections 3-4

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  1. EGR 334 ThermodynamicsChapter 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 1st 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: Read Chapter 9, Sections 5-6 Homework Assignment: Problems from Chap 9: 17, 20, 24, 38

  3. Sec 9.1 : Introducing Engine Terminology 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.

  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 <rDiesel <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. Sec 9.3 : Air-Standard Diesel Cycle 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 • For the Otto cycle, the Qin occurs at constant volume (compressed cylinder) • For the diesel cycle, the fuel is injected while the cylinder is expanding, so the Qin occurs at constant pressure

  6. Sec 9.3 : Air-Standard Diesel Cycle Diesel Cycle analysis: Process 2 – 3 : Constant pressure heat transfer to the air from an external source while moves down (ignition and part of power stroke) • Use W23 with energy balance to find Q23. Remaining processes analysis is the same as for Otto cycle.

  7. Sec 9.3 : Air-Standard Diesel Cycle Find State 1 Properties Given T1 and r  use table to find u1 & h1. Find state 2: for isentropic process. or Find state 3: Use ideal gas law with p3 = p2. • were rc is the cutoff ratio: Find state 4. For Cold-Air Standard analysis: (isentropic processes 1-2 and 3-4) For state 2. For state 4.

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

  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, p1= 14.2 psi, V1= 0.5 ft3, T1= 520°R. Calculate • The heat added, in Btu. • The max T. • The thermal efficiency. • The mean effective pressure, • Given info: Diesel Cycle • State 1: p1=14.2 psi, T1=520 R • V1 = 0.5 ft3 • State 2: s2 = s1 • State 3: p3 = p2 • State 4: s4 =s 3 and v4=v1 • Compression ratio: r = V1/V2 =16 • Cutoff ratio: rc = V3/V2 = 2

  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, p1= 14.2 psi, V1= 0.5 ft3, T1= 520°R. Calculate • The heat added, in Btu. • The max T. • The thermal efficiency. • The mean effective pressure, Determine State Properties: For State 1: Using Table A-22E: for T1 Read u 1 = 88.62 Btu/lbm h1 = 124.27 Btu/lbm pr1 = 1.2147 and vr1 = 158.58 Also using Ideal Gas Equation:

  11. Example (9.26): • For State 2: • (process 1-2 is isentropic) • Using compression ratio, r • Next use vr2 and Table A-22E to find other state properties: • T2 = 1502.5°R u2= 266.84 Btu/lbm • pr2= 56.27h2 = 369.85 Btu/lbm • Then:

  12. Example (9.26): • For State 3: • Using ideal Gas • process 2-3 constant pressure • Then using Table A-22E, use T3 to find properties • u3 = 586.16 Btu/lbm h3 = 792.03 Btu/lbm • p3r = 948.36 v3r = 1.174

  13. Example (9.26): • For State 4: • (process 3-4 is isentropic • and V4 = V1) • therefore: • Then using Table A-22E, use vr4 to find properties • u4 = 272.38Btu/lbm h4 = 377.47 Btu/lbm • pr4 = 60.46 T4 = 1530.8 R • then

  14. Example (9.26): • The heat added, in BTU. • The max T. • The thermal efficiency. • The mean effective pressure, • Heat is added during process 2-3 • Applying the 1st Law • Maximum Temperature is at Sate 3: Tmax = 3005 oR

  15. Example (9.26): • The heat added, in BTU. • The max T. • The thermal efficiency. • The mean effective pressure, • Thermal Efficiency: • Mean Effective Pressure: • where:

  16. End of Slides for Lecture 34

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