Engine Cycle Analysis. Air Standard Otto Cycle.

Slides:



Advertisements
Similar presentations
ME 200 L36 Ground Transportation (Continued) (Air Standard Otto Cycle) 9.1 and 9.2 Kim See’s Office ME Gatewood Wing Room 2172 Examination and Quiz grades.
Advertisements

Lecture 13 Use of the Air Tables.
Ideal Intake and Exhaust Strokes
ISENTROPIC EFFICIENCY CALCULATIONS POSITIVE-DISPLACEMENT
EGR 334 Thermodynamics Chapter 9: Sections 3-4 Lecture 32: Gas Power Systems: The Diesel Cycle Quiz Today?
THERMAL ENGINEERING (ME 2301 )
This Week > POWER CYCLES
Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 28 Internal Combustion Engine Models The Otto Cycle The Diesel.
Four Stroke Cycle Engine
İsmail ALTIN, PhD Assistant Professor Karadeniz Technical University Faculty of Marine Sciences Department of Naval Architecture and Marine Engineering.
Internal Combustion Engine Theory
Gas Power Cycles Cengel & Boles, Chapter 8 ME 152.
GAS POWER CYCLES Chapter 9. Introduction Two important areas of application for thermodynamics are power generation and refrigeration. Two important areas.
EGR 334 Thermodynamics Chapter 9: Sections 1-2
Diesel Engine Classification
Thermodynamics II Chapter 3 Compressors
Ship Propulsion Systems STEAMDIESELOTHERConventionalNuclear Slow Speed Medium Speed Gas Turbine Combined Cycle DIESEL PROPULSION.
Internal Combustion Engines. Ideal Diesel Cycle Ideal Diesel Cycle.
Shaft Power Generation Devices - 1
Diesel / Brayton Cycles
Specific Heat Thermodynamics Professor Lee Carkner Lecture 8.
CHAPTER 9 Gas Power Cycles.
Thermodynamic Analysis of Internal Combustion Engines P M V SUBBARAO Professor Mechanical Engineering Department IIT Delhi Work on A Blue Print Before.
For next time: Read: § 8-6 to 8-7 HW11 due Wednesday, November 12, 2003 Outline: Isentropic efficiency Air standard cycle Otto cycle Important points:
Gas Power Cycle - Internal Combustion Engine
Engines, Motors, Turbines and Power Plants: an Overview Presentation for EGN 1002 Engineering Orientation.
INTERNAL COMBUSTION ENGINES (reciprocating). Geometry.
Applied Thermodynamics
Gas Turbines By: Katie Steddenbenz.
Thermodynamic Cycles Air-standard analysis is a simplification of the real cycle that includes the following assumptions: 1) Working fluid consists of.
ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma.
EGR 334 Thermodynamics Chapter 9: Sections 5-6
Thermodynamic Cycles for CI engines In early CI engines the fuel was injected when the piston reached TC and thus combustion lasted well into the expansion.
8 CHAPTER Gas Power Cycles.
Gas Power Cycles.
CHAPTER 8 Gas Power Cycles. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8-1 FIGURE 8-1 Modeling is a.
Chapter 9 Gas Power Systems.
Thermodynamics The First Law of Thermodynamics Thermal Processes that Utilize an Ideal Gas The Second Law of Thermodynamics Heat Engines Carnot’s Principle.
Thermodynamic Cycles for CI engines
ME 200 L35 Ground Transportation (Air Standard Otto Cycle) 9.1 and 9.2 Material not picked up this week may be recycled! ME 200 L35 Ground Transportation.
TEKNIK PERMESINAN KAPAL II (Minggu – 3) LS 1329 ( 3 SKS) Jurusan Teknik Sistem Perkapalan ITS Surabaya.
Gas Power Cycles Thermodynamics Professor Lee Carkner Lecture 17.
ENGINE DESIGN AND OPERATION. ENGINE CLASSIFICATIONS n VALVE ARRANGEMENT n CAMSHAFT LOCATION n IGNITION TYPE n CYLINDER ARRANGEMENT n NUMBER OF CYLINDERS.
Chapter 9 Gas Power Cycles Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition by Yunus A. Çengel and Michael.
MT 313 IC ENGINES LECTURE NO: 04 (24 Feb, 2014) Khurram Yahoo Group Address: ICE14.
AR Thermodynamics I Fall 2004 Course # 59:009 Chapter 9, Section 2 Professor Ratner.
Diesel Cycle and the Brayton Cycle Chapter 9b. Rudolph Diesel  German inventor who is famous for the development of the diesel engine  The diesel engine.
Presentation on HEAT ENGINE PREPARED BY: CHAUHAN SATISH(EN. NO: ) GAUTAM ASHISH(EN. NO: ) KETUL PATEL(EN. NO: ) SUB:
Internal combustion engines
APPLIED THERMODYNAMICS UNIT- 2 Gas power cycle 1 Department of Mechanical Engineering,A.I.E.T.,Mijar 3)Air Standard Diesel Cycle/ Constant Pressure cycle:
ET375 Applied Thermodynamics 09 Thermodynamic Cycles Introduction to Gas Cycles 12/1/131rm.
CHAPTER 9 Gas Power Cycles.
8. GAS POWER CYCLES. Objectives Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle. Develop.
Difference Between otto and diesel cycle
Prepared by, Brijrajsinh Sarvaiya(13ME548) Jaypalsinh Jadeja(13ME517) Pradipsinh Jadeja(13ME518) Virendrasinh Parmar(13ME539) Gas power cycle.
Analysis of Diesel Cycle and Scope for Modification P M V Subbarao Professor Mechanical Engineering Department Creation of Rational Models for Engines…
LECTURE 1.
THERMODYNAMIC ANALYSIS OF IC ENGINE Prepared by- Sudeesh kumar patel.
Basic Engine Terms. Dead Center –The term identifies the relation between the positions of the crankshaft and the piston. When the piston is at its outward.
Gas Power Cycles.
A. Diesel cycle : The ideal cycle for CI engines
ES 211:Thermodynamics Tutorial 10
Gas Power Cycle - Internal Combustion Engine
Engineering Thermodynamics ME-103
SI Engine Cycle Actual Cycle Intake Stroke Compression Power Exhaust
THE OTTO CYCLE (engine)
Thermo-Economic Analysis of Otto Cycle
Ideal Diesel and Dual Cycles for I.C. Engines
Ideal Diesel and Dual Cycles for I.C. Engines
Thermodynamic Analysis of Internal Combustion Engines
Presentation transcript:

Engine Cycle Analysis

Air Standard Otto Cycle

► Starting with the piston at bottom dead center (BDC), compression proceeds isentropically from state 1 to state 2. ► Heat is added at constant volume from state 2 to state 3. ► Expansion occurs isentropically from state 3 to state 4. ► Heat is rejected at constant volume from state 4 to state 1

Air Standard Otto Cycle

► Thermal Efficiency (η th ) = W Net /Q in ► Heat Added at constant volume from state 2 to state 3 Q 2-3 = U 3 – U 2 = mc v (T 3 -T 2 ) ► Heat Rejected at constant volume from state 4 to state 1 Q 4-1 = U 1 – U 4 = -mc v (T4 – T 1 ) ► Specific Heat (C v ) = Btu/lbm-°R ► Mean Effective Pressure (mep) = W Net / Displacement Volume = W Net / V 1 – V 2

Air Standard Otto Cycle ► Compression ratio (r) = Vol at BDC/Vol at TDC ► r = V 1 /V 2 = V 4 /V 3 ► η th = 1 – (1/(r) k-1 ) ► Specific Heat Ratio (k) = C p /C v (for air k = 1.4) ► Percentage of clearance (c) = Clearance Volume/Displacement volume = V 2 /(V 1 – V 2 )

Air Standard Diesel Cycle ► Starting with the piston at bottom dead center (BDC), compression occurs isentropically from state 1 to state 2 ► Heat is added at constant pressure from state 2 to state 3 ► Expansion occurs isentropically from state 3 to state 4 ► Heat rejection occurs at constant volume from state 4 to state 1

Air Standard Diesel Cycle

► Heat supplied at constant pressure for a closed system, Q 2-3 = H 3 – H 2 = mc p (T 3 – T 2 ) ► Heat rejected at constant volume, Q 4-1 = U 1 – U 4 = mc v (T 4 – T 1 ) ► Specific Heat (C p ) = 0.24 Btu/lbm-°R ► η th = 1-1/k((T 4 – T 1 )/(T 3 – T 2 ))

Air Standard Diesel Cycle ► Cutoff ratio (r c ) = Vol at the end of heat addition/Vol at the start of heat addition = V 3 /V 2 ► Cutoff percentage (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► η th = 1 – (1/(r) k-1 )[(r c k – 1)/(k(r c – 1))]

Polytropic Process for a Closed System ► Pv = RT ► Specific Gas Constant (R) = for air = ft-lb f /lbm-°R ► p 1 = RT 1 /v 1 ► p 2 = RT 2 /v 2 ► P 1 /P 2 = (V 2 /V 1 ) k ► P 2 /P 1 = (V 1 /V 2 ) k ► T 1 /T 2 = (V 2 /V 1 ) k-1 ► T 2 /T 1 = (V 1 /V 2 ) k-1

Polytropic Process for a Closed System ► V 2 /V 1 = (T 1 /T 2 ) 1/k-1 ► V 1 /V 2 = (P 2 /P 1 ) 1/k ► T 1 /T 2 = (P 2 /P 1 ) (1-k)/k = (P 1 /P 2 ) (k-1)/k ► P 1 /P 2 = (T 1 /T 2 ) k/(k-1)

Air Standard Otto Cycle ► A air standard Otto cycle has an initial temperature of 100 °F, a pressure of 14.7 psia, and compression pressure P 2 = 356 psia. The pressure at the end of heat addition is 1100 psia. Determine: ► (A) The compression ratio (r) ► (B) The thermal efficiency η thermal ► (C) The percentage of clearance (c) ► (D) The maximum temperature T max and the remaining Pressures, Temperature and Volumes

Air Standard Otto Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm)

Air Standard Otto Cycle ► r = V 1 /V 2 ► V 1 /V 2 = (P 2 /P 1 ) 1/k ► V 1 /V 2 = (356/14.7) 1/1.4 ► r = 9.74 ► η th = 1 – (1/(r) k-1 ) ► η th = 1 – (1/(9.74) ) ► η th = or 59.8 %

Air Standard Otto Cycle ► P 1 = RT 1 /V 1 ► V 1 = RT 1 /P 1 ► V 1 = ((53.34 ft-lb f /lbm-°R)(560 °R)) / ((14.7 lb f /in 2 )(144 in 2 /ft 2 )) ► V 1 = ft 3 /lbm ► r = V 1 /V 2 ► V 2 = V 1 /r ► V 2 = 14.11/9.74 ► V 2 = ft 3 /lbm

Air Standard Otto Cycle ► Percentage of clearance (c) = V 2 /(V 1 – V 2 ) ► Percentage of clearance (c) = 1.449/(14.11 – 1.449) ► Percentage of clearance (c) = or 11.4 % 11.4 % ► P 2 = RT 2 /V 2 ► T 2 = P 2 V 2 /R

Air Standard Otto Cycle ► T 2 = (((356 lb f /in 2 )(144 in 2 /ft 2 )(1.449 ft 3 /lbm)) / ft-lb f /lbm-°R) ► T 2 = 1393 °R

Air Standard Otto Cycle

► Since V 2 = V 3 ► V 2 = RT 2 /P 2 ► V 3 = RT 3 /P 3 ► RT 2 /P 2 = RT 3 /P 3 ► T 2 /P 2 = T 3 /P 3 ► T 3 = T max = T 2 (P 3 /P 2 ) ► T 3 = 1393 °R (1100/356) ► T 3 = °R

Air Standard Otto Cycle ► Find P 4 and T 4 ► (P 3 /P 4 ) = (V 4 /V 3 ) k ► (V 4 /V 3 ) = r = 9.74 ► (9.74) 1.4 = (P 3 /P 4 ) ► (9.74) 1.4 = (1100/P 4 ) ► P 4 = 45.4 psia

Air Standard Otto Cycle ► T 3 /T 4 = (V 4 /V 3 ) k-1 ► T 3 /T 4 = (r) k-1 ► 4302/T 4 = (9.74) ► T 4 = 4302/(9.74) ► T 4 = °R

Air Standard Otto Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm)

Air Standard Diesel Cycle ► An air standard Diesel Cycle operates on 1 ft 3 of air at 14.5 psia and 140 °F. The compression ratio (r) is 14, and the cutoff is 6.2 % of the stroke. Determine: ► (A) The temperatures, pressures and volumes around the cycle ► (B) The net work ► (C) The heat added ► (D) The efficiency

Air Standard Diesel Cycle

State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm)

Air Standard Diesel Cycle ► P 1 = mRT 1 /V 1 ► m = P 1 V 1 /RT 1 ► m = (((14.5 lb f /in 2 )(144 in 2 /ft 2 )(1 ft 3 )) / ((53.34 ft-lb f /lbm-°R)(600 °R))) ► m = lbm ► T 2 /T 1 = (V 1 /V 2 ) k-1 ► T 2 = T 1 (V 1 /V 2 ) k-1 ► r = (V 1 /V 2 )

Air Standard Diesel Cycle ► T 2 = (600 °R)(14) 0.4 ► T 2 = °R ► P 2 /P 1 = (V 1 /V 2 ) k ► P 2 = P 1 (V 1 /V 2 ) k ► P 2 = (14.5 psia)(14) 1.4 ► P 2 = psia ► r = (V 1 /V 2 )

Air Standard Diesel Cycle ► V 2 = (V 1 /r) ► V 2 = (1/14) ► V 2 = ft 3

Air Standard Diesel Cycle

► Since the process between P 2 to P 3 is constant pressure P 3 = psia. ► Cutoff percentage (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► (R c ) = ((V 3 – V 2 )/(V 1 – V 2 )) x 100 ► = ((V 3 – )/(1 – )) ► V 3 = ft 3

Air Standard Diesel Cycle ► P 2 = RT 2 /V 2 ► P 3 = RT 3 /V 3 ► Since P 2 = P 3 ► RT 2 /V 2 = RT 3 /V 3 ► T 3 = T 2 (V 3 /V 2 ) ► T 3 = ( °R)(0.129/0.0714) ► T 3 = 3115 °R

Air Standard Diesel Cycle ► T 4 /T 3 = (V 3 /V 4 ) k-1 ► T 4 = T 3 (V 3 /V 4 ) k-1 ► T 4 = (3115 °R)(0.129/1.00) 0.4 ► T 4 = 1373 °R ► P 4 /P 3 = (V 3 /V 4 ) k ► P 4 = P 3 (V 3 /V 4 ) k ► P 4 = 583.4(0.129/1.0) 1.4 ► P 4 = 33.2 psia

Air Standard Diesel Cycle State Point Temperature (°R) Pressure (psia) Volume (ft 3 /lbm)

Air Standard Diesel Cycle

► The heat added is: ► Q 2-3 = H 3 – H 2 = mC p (T 3 – T 2 ) ► Q 2-3 = ( lbm)(0.24 Btu/lbm-°R)(3115 – °R) ► Q 2-3 = Btu

Air Standard Diesel Cycle ► The heat out is: ► Q Out = Q 4-1 = mC v (T 1 – T 4 ) ► Q 4-1 = (0.0652)(0.1714)(600 – 1373) ► Q 4-1 = Btu ► W net = ∑Q = Q in + Q out ► W net = Btu ► W net = Btu

Air Standard Diesel Cycle ► Thermal Efficiency (η th ) = W net /Q in ► η th = 13.12/21.76 ► η th = = 60.3 %