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##### EGR 334 Thermodynamics Chapter 9: Sections 1-2

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**EGR 334 ThermodynamicsChapter 9: Sections 1-2**Lecture 33: Gas Power Systems: The Otto Cycle Quiz Today?**Today’s main concepts:**• Understand common terminology of gas power cycles. • Be able to explain the processes of the Otto Cycle • Be able to perform a 1st Law analysis of the Otto Cycle and determine its thermal efficiency. • Be able to discuss limitations of the Otto cycle compared to real spark ignition power systems. • Be able to state the assumptions of standard air analysis. Reading Assignment: Read Chapter 9, Sections 3-4 Homework Assignment: Problems from Chap 9: 1, 4, 11, 14**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.**Sec 9.1 : Introducing Engine Terminology**Four Stroke Cycle : Two revolutions : Combustsmix of hydrocarbons + O2 • Intake stroke : fill cylinder • Spark cycle : fill with fuel and air mixture • Compression cycle : fill with air • Compression stroke : p , T , V , Win • Spark cycle : spark near end of stroke • Compression cycle : inject fuel • Power stroke : gas expands • Exhaust stroke : spent gas is exhausted U--Tube video of 4 stroke: http://www.youtube.com/watch?v=2Yx32F1cncg**Sec 9.1 : Introducing Engine Terminology**Two Stroke Cycle : Two revolutions • Power/Exhaust: • The piston is forced down • @ exhaust port, spent gas leaves • Piston continues down and compresses air/fuel in crank case • Compressed charge enters cylinder • Intake/Compression • Piston moves up compressing charge • Draws vacuum in crank case. Two Stroke Animation: http://library.thinkquest.org/C006011/english/sites/2_taktmotor.php3?v=2**Sec 9.1 : Introducing Engine Terminology**Mean Effective Pressure (mep) Air –standard Analysis (A simplification used to allow for thermodynamic analysis) Assumptions: -- Fixed amount of air modeled as closed system-- Air is treated as Ideal Gas -- Constant cp (cold air-standard) -- Combustion is modeled as a heat transfer to system, Exhaust as heat flow out of system -- All processes internally reversible**Ideal Gas Model Relations:**Chap 3: Quality Polytropic Process State Equation: Ideal Gas Model Review Energy Relationships: where or look up values for k, cv, and cp on Table A-20 Entropy Relationships: where so values are found on Table A-22 Special case: isentropic process where s1 = s2 then ( vr and pr for use with Table A-22) ( assuming constant specific heats)**Sec 9.2 : Air-Standard Otto Cycle**Assumption: At top dead center, heat addition occurs instantaneously Otto Cycle: comprised of 4 internally reversible processes Process 1 – 2 : Isentropic compression of air (compression stroke). Process 2 – 3 : Constant volume heat transfer to the air from an external source while piston is at top dead center (ignition) Process 3 – 4 : Isentropic expansion (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 ignition power ignition compression power compression exhaust exhaust**Sec 9.2 : Air-Standard Otto Cycle**Otto Cycle analysis Closed system energy balance : • Processes 1–2: ∆s = 0 and Q = 0 Process 3–4: ∆s = 0 and Q = 0 Processes 2–3 : ∆V = 0 and W = 0 Processes 4-1 : ∆V = 0 and W = 0**Sec 9.2 : Air-Standard Otto Cycle**Otto Cycle Thermal Efficiency: Thermal Efficiency can also be related to the compression ratio: clearance displacement As the compression ration, r, , the efficiency, η, **Example (9.11): An air-standard Otto cycle has a compression**ratio of 7.5. At the beginning of compression, p1 = 85 kPa and T1 = 32°C. The mass of air is 2 g, and the maximum temperature in the cycle is 960 K. Determine • The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • T3= 960 K • p1=85 kPa • T1=305 K**Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • State 1: Using Ideal Gas Law: • and Table A-22: • State 2: Using compression ratio, ideal gas law, and Table A-22: • **Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • State 3: using v3 = v2, ideal gas law, and Table A22: • State 4: Using v4=v1:**Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure • , in kPa. • also knowing for isentropic processes • then for isentropic process 1-2 using k = 1.361 and Table A-22: • and for isentropic process 3-4:**Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • Heat added during process 2-3: • Heat rejected during process 4-1:**Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • Net Work over the cycle: • Cycle Efficiency: • Compare to**Example (9.11):**• The heat rejection, in kJ. • The net work, in kJ. • The thermal efficiency. • The mean effective pressure, in kPa. • Mean Effective Pressure: • where