L/O/G/O EKNIK KONVERSI & KONSERVASI ENERGI F 091324 / 4 SKS / Smt. 6 Dosen : Syamsul Arifin syamsul@ep.its.ac.id K5 hermodynamics Fundamental for KE http://share.its.ac.id
eknik Konversi Energi Fasilitator Syamsul Arifin ( syamp3ai@its.ac.id ) 2011 http://share.its.ac.id
Syamsul Arifin Energy Conversion
Syamsul Arifin Energy Source and Conversion Processes
Syamsul Arifin Energy Conversion echnologies
Syamsul Arifin Laws of hermodynamics
Syamsul Arifin Entropy
Syamsul Arifin Carnot Heat Engine
Syamsul Arifin he Carnot Cycle
Q1 Q2
Syamsul Arifin he Carnot Cycle.1.2.3
Syamsul Arifin he Carnot Cycle
Syamsul Arifin Stirling Engine
Stirling Engine
Stirling Engine
Syamsul Arifin hermodynamic Processes in β Configuration
Syamsul Arifin hermodynamic Processes in β Configuration
Syamsul Arifin hermodynamic Processes in a Ideal Stirling Cycle Q H W out V min H V max L
Syamsul Arifin Efficiency of an Ideal Stirling Cycle
Motor Bakar Dalam
Phase 1: Virtually all the gas is in the compression space at ambient temperature and the displacer is in the tip of the cold finger. In this phase the pistons are driven inwards, compressing the gas. his proces is nearly isothermal, the heat output Qc being dissipated via heat sinks around the compressor and the base of the cold finger. Phase 2: he pistons have reached the end of the compression stroke, the gas in the compression space is at ambient temperature and the displacer has not yet moved. his is the situation at the start of Phase II. hroughout this phase the pistons remain stationary and hence the total volume of gas remains constant. he displacer moves downwards as its spring compresses and gas flows through the regenerator, giving up heat Qr in the proces. his heat is stored in the renegerator until later in the cycle. Phase 3: he pistons are driven outwards and the gas expands. his expansion process, too, is nearly isothermal, the heat input Qe being drawn from the surroundings of the expansion space. As a result refrigeration occurs at the tip of the cold finger. Phase 4: hroughout this phase the pistons remain stationary. he displacer, however, moves upwards because of the lower gas pressure in the expansion space. Gas from the expansion space therefor flows back through the regenerator, taking up the stored heat Qr in the process and re-entering the compression space at ambient temperature.
Figure 1 - he Sunpower EG-1000 free-piston Stirling engine/generator he linear electrical generator is comprised of powerful rare-earth magnets in the piston cutting a magnetic circuit and coils in the cylinder. his produces 240 Volts at 50 Herz - designed for operation in Europe, and is capable of producing more than one kilowatt of electrical power output at around 90% efficiency. he hot water is provided by operating the cooling water at a temperature of 50 C.
In this photograph we see the Sunpower EG-1000 being demonstrated using sawdust pellets as the fuel, and generating more than 1000W of electricity to a light panel. his was done at the Sustainability Fair in the Fairgrounds of Athens Ohio, 2001. A closeup photograph of the basic system is shown. Notice the closed cycle radiator and vibration pump used in the water cooling system.
Syamsul Arifin Stirling Engine Analysis
Syamsul Arifin Stirling Engine Analysis..1
Syamsul Arifin Stirling Engine Analysis..2..3..4..5
Syamsul Arifin Stirling Engine Analysis..6..7..8..9..10..11..12
Syamsul Arifin Stirling Engine Analysis..13..14..15..16..17..18
Syamsul Arifin Stirling Engine Analysis..19..20..21..22
Syamsul Arifin Stirling Engine Analysis..23..24..25
Stirling Engine Analysis..26..27..28 Syamsul Arifin
Syamsul Arifin Stirling Engine Analysis
Syamsul Arifin Solar Dish Stirling System
Syamsul Arifin Solar Dish Stirling System Efficiency
Rankin Cycle Engine Syamsul Arifin
Syamsul Arifin Rankine Cycle Efficiency
Syamsul Arifin Ideal Reheat Rankine Cycle
Syamsul Arifin Organic Rankine Cycle
Syamsul Arifin emperature Dependence of Rankine Cycle Devices
Syamsul Arifin Cycle Efficiencies with Heat Engines
Syamsul Arifin Organic Rankine Cycle
Syamsul Arifin Externally Heated Systems
Syamsul Arifin Cogeneration
Spark ignition Internal Combustion Engines Air Standard Cycles
Number of Strokes Four stroke wo stroke
Cylinder Design Small engines usually have one or two cylinders, but may have as many as four. hree Common Cylinder Orientations For Single Cylinder Engines Vertical Horizontal Slanted
Cylinder Design-cont. hree common cylinder configuration in multiple cylinder engines: V Horizontally opposed In-line
Cylinder Design-cont. Small gas engines use three crankshaft orientations: Hor izontal Multi-position Vertical
Otto Cycle P-V & -s Diagrams Pressure-Volume emperature-entropy
Otto Cycle Derivation hermal Efficiency: η th = Q H Q - Q H L = 1- Q Q L H For a constant volume heat addition (and rejection) process; Q in = m C Assuming constant specific heat: v Q Rej = m C v η th = 1 - m C m C v v ( ( 4 3 - - 1 2 ) = 1- ) 1 2 4 1 3 2-1 -1
For an isentropic compression (and expansion) process: where: γ = C p /C v hen, by transposing, = V V = V V = 4 3 3 4-1 2 1-1 1 2 γ γ = 1 4 2 3 Otto Cycle Derivation = 1-2 1 η th Leading to
Otto Cycle Derivation he compression ratio (r v ) is a volume ratio and is equal to the expansion ratio in an otto cycle engine. Compression Ratio r v = V V 1 2 = V V 4 3 where Compression ratio is defined as r v = otal volume Clearance volume = v s + v v cc cc r v = v v s cc +1
hen by substitution, 1 1-γ 2 =( rv 1 ) γ 2 1 Otto Cycle Derivation V = V he air standard thermal efficiency of the Otto cycle then becomes: η th 1-γ 1 = 1-( rv ) = 1- γ -1 ( rv )
Otto Cycle Derivation Summarizing ηth = Q H Q - Q H L = 1- Q Q L H where Q= m C v η th = 1-1 2 4 1 3 2-1 -1 and 3 2 = 4 1 then η th = 1 1 2 Isentropic behavior 1 1-γ 2 =( rv 1 ) γ 2 1 V = V 1-γ 1 ηth= 1 - ( rv ) = 1 - γ -1 ( rv )
Otto Cycle Derivation Heat addition (Q) is accomplished through fuel combustion Q = Lower Heat Value (LHV) BU/lb, kj/kg also Q in cycle = m a F A Q fuel Q in = m C v
Otto Cycle Analysis
Otto Cycle P & Prediction Determine the temperatures and pressures at each point in the Otto Cycle. Compression Ratio = 9.5:1 1 temperature = 25 o C = 298 o K P 1 pressure = 100 kpa
Diesel Cycle Derivation
Diesel Cycle P-V & -s Diagrams
hermal Efficiency (Diesel): Diesel Cycle Derivation η th = Q H Q - Q H L = 1- Q Q L H For a constant pressure heat addition process; For a constant volume heat rejection process; Q= m C p Q= m C v Assuming constant specific heat: η = 1- th m C m C v p ( ( 4 3 - - 1 2 4 1 ) 1 = 1 - ) γ 2-1 -1 3 2 where: γ = C p /C v
For an isentropic compression (and expansion) process: However, in a Diesel he compression ratio (r v ) is a volume ratio and, in a diesel, is equal to the product of the constant pressure expansion and the expansion from cut-off. = V V V V = 4 3 3 4-1 2 1-1 1 2 γ γ V V V V =V V 3 4 2 1 4 1 Diesel Cycle Derivation
Compression Ratio Diesel Cycle Derivation V 1 V 4 rv c = hen by substitution, V 2 V 3 r v c = r cp re= V V 2 3 V 3 v4 1 1-γ 2 =( rv 1 ) γ 2 1 V = V η th = 1- ( r v 1 γ -1 ) ( ) rcp γ ( r γ cp - 1-1)
Diesel Cycle Analysis
Diesel Cycle P & Prediction Determine the temperatures and pressures at each point in the Diesel Cycle. Compression Ratio = 20:1 1 temperature = 25 o C = 298 o K P 1 pressure = 100 kpa
Otto-Diesel Cycle Comparison
Dual Cycle P-V Diagrams:
Dual Cycle Efficiency Dual Cycle hermal Efficiency = m C ( - )+ m C ( - Qin v 2.5 2 p 3 2.5 ( ) Q = m Cv Rej 4 1 ) η = 1-1 CR ( γ -1) γ α β - 1 ( α -1)+ γα( β -1) α = P P 3 2 β = V V 3 2.5 where: γ = C p /C v
Diesel Cycle Derivation Critical Relationships in the process include 1 1-γ 2 =( rv 1 ) γ 2 1 V = V P P 2 1 γ γ = ( rv) 1 2 V = V Q F Q= m C p = ma Q fuel Q= m Cv cycle A 1 η = 1 - th ( rv ) γ -1 ( ) rcp γ ( r γ cp - 1-1)