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

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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 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 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 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/11/3335677/slides/slide_4.jpg", "name": "4 Sec 9.3 : Air-Standard Diesel Cycle The Diesel cycle can operate using either the 4 or 2 stroke engine.", "description": "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

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

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 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 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 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 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 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 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:

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