EGR 334 Thermodynamics Chapter 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

Two types of internal combustion engine 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.

Intake stroke : fill cylinder Sec 9.1 : Introducing Engine Terminology Four Stroke Cycle : Two revolutions : Combusts mix 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

Two Stroke Cycle : Two revolutions Power/Exhaust: 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. http://www.drivegreen.com/Auto_Basics_Engines.htm Two Stroke Animation: http://library.thinkquest.org/C006011/english/sites/2_taktmotor.php3?v=2

Mean Effective Pressure (mep) 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 Review State Equation: Energy Relationships: Chap 3: Quality Polytropic Process Ideal Gas Model Relations: Ideal Gas Model Review State Equation: 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 ( assuming constant specific heats) ( vr and pr for use with Table A-22)

Assumption: At top dead center, heat addition occurs instantaneously 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

Otto Cycle analysis Processes 1–2: ∆s = 0 and Q = 0 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

Otto Cycle Thermal Efficiency: 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. State 1 2 3 4 T (K) 305 960 p (kPa) 85 u (kJ/kg) v T3= 960 K p1=85 kPa T1=305 K

State 1: Using Ideal Gas Law: and Table A-22: Example (9.11): State 1 2 3 4 T (K) 305 960 p (kPa) 85 u (kJ/kg) v 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: 

State 3: using v3 = v2, ideal gas law, and Table A22: Example (9.11): State 1 2 3 4 T (K) 305 960 p (kPa) 85 u (kJ/kg) 217.67 v 1.0298 0.1373 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:

then for isentropic process 1-2 using k = 1.361 and Table A-22: Example (9.11): State 1 2 3 4 T (K) 305 960 p (kPa) 85 2006.7 u (kJ/kg) 217.67 725.02 v 1.0298 0.1373 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:

Heat added during process 2-3: Example (9.11): State 1 2 3 4 T (K) 305 631 960 464 p (kPa) 85 1319 2006.7 129.3 u (kJ/kg) 217.67 458.55 725.02 328.5 v 1.0298 0.1373 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:

Net Work over the cycle: Example (9.11): State 1 2 3 4 T (K) 305 631 960 464 p (kPa) 85 1319 2006.7 129.3 u (kJ/kg) 217.67 458.55 725.02 328.5 v 1.0298 0.1373 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

Mean Effective Pressure: Example (9.11): State 1 2 3 4 T (K) 305 631 960 464 p (kPa) 85 1319 2006.7 129.3 u (kJ/kg) 217.67 458.55 725.02 328.5 v 1.0298 0.1373 The heat rejection, in kJ. The net work, in kJ. The thermal efficiency. The mean effective pressure, in kPa. Mean Effective Pressure: where

End of Slides for Lecture 33