APPLIED THERMODYNAMICS 18ME42 Module 01 question no 1a & 1b
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1.0 Air standard cycles: Definitions 1.1 Carnot, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.2 Otto, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.3 Diesel, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.4 Dual and Stirling cycles, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.5 Comparison of Otto and Diesel cycles. 1.6 Solved Previous Year Question Papers
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APPLIED THERMODYNAMICS 18ME42 Module 01 question no 1a & 1b
- 1. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 1 APPLIED THERMODYNAMICS 18ME42 Course Coordinator Mr. THANMAY J. S Assistant Professor Department of Mechanical Engineering VVIET Mysore Module 01: Question Number 1a & 1b: Air standard cycles Course Learning Objectives To understand the applications of the first and second laws of Thermodynamics to various gas processes and cycles. Course Outcomes At the end of the course the student will be able to: CO1: Apply thermodynamic concepts to analyze the performance of Air Standard cycles.
- 2. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 2 Contents Appendix 01: Basics of Thermodynamic Process 1.0 Air standard cycles: Definitions 1.1 Carnot, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.2 Otto, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.3 Diesel, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.4 Dual and Stirling cycles, description, p-v and T -s diagrams, efficiencies, mean effective pressures. 1.5 Comparison of Otto and Diesel cycles. 1.6 Solved Previous Year Question Papers
- 3. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 3 Appendix 01: Basics of Thermodynamic Process a Real gas Constant 𝑹 = 𝑪𝒑 − 𝑪𝒗 Cp= Specific Heat at Constant Pressure Cv= Specific Heat at Constant Volume b Ratio of Specific Heat 𝜸 = 𝑪𝒑 𝑪𝒗 Charecteristic of Gas Equation PV = mRT c Constant Volume (Isochoric) Process P1 T1 = P2 T2 = 𝑃 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐐 = 𝐦 𝐂𝐯 [𝐓𝟐 − 𝐓𝟏] d Constant Pressure (Isobaric) Process V1 T1 = V2 T2 = 𝑉 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐐 = 𝐦 𝐂𝐩 [𝐓𝟐 − 𝐓𝟏] e Constant Temperature (Isothermal) Process 𝑃1𝑉1 = 𝑃2𝑉2 = 𝑃𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑾 = 𝑸 = 𝒎 𝑹 𝑻 𝒍𝒏 [ 𝑽𝟐 𝑽𝟏 ] = 𝒎 𝑹 𝑻 𝒍𝒏 [ 𝑷𝟏 𝑷𝟐 ] f Isentropic or Adiabatic Process Heat Supplied Q = 0 P1𝑉1𝛾 = P2𝑉2𝛾 = Constant T1 𝑉1𝛾−1 = T2 𝑉2𝛾−1 = Constant 𝑃1𝛾−1 𝑇1𝛾 = 𝑃2𝛾−1 𝑇2𝛾 = Constant Adiabatic Expansion or Adiabatic Compression 𝑻𝟏 𝑻𝟐 = [ 𝑽𝟐 𝑽𝟏 ] 𝜸−𝟏 = [ 𝑷𝟐 𝑷𝟏 ] 𝜸−𝟏 𝜸 or 𝑽𝟏 𝑽𝟐 = [ 𝑻𝟐 𝑻𝟏 ] 𝟏 𝜸−𝟏 = [ 𝑷𝟐 𝑷𝟏 ] 𝟏 𝜸 or 𝑷𝟏 𝑷𝟐 = [ 𝑽𝟐 𝑽𝟏 ] 𝜸 = [ 𝑻𝟏 𝑻𝟐 ] 𝜸 𝜸−𝟏 g
- 4. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 4 1.0 Air standard cycles: Definitions Air standard cycles are theoretical cycles used to study the operations and performances of Internal Combustion Engines, they are represented with theoretical engines operating on thermodynamic cycles and these theoretical engines are referred to as air standard engines. They are called air standard engines because their working fluid is taken to be majorly air. In these engines, heat is added from an external source as opposed to burning fuel and a heat sink is provided as opposed to exhaust, thus returning the air back to its original state. The following assumptions are made for the air standard cycle: Air continuously circulates in a closed loop. Always behaves as an ideal gas. All the processes that make up the cycle are internally reversible. The combustion process is replaced by a heat-addition process from an external source. (Heating TDC by External Heat Source). A heat rejection process that restores the working fluid to its initial state replaces the exhaust process. (Heat Sink or Cooling of Engine TDC) The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25oC or 77oF). No chemical reaction takes place in the engine. Figure 01: Example for Air Standard Cycle Analysis
- 5. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 5 1.1 Carnot, description, p-v and T -s diagrams, efficiencies, mean effective pressures. Process 1-2: isothermal heat transfer or Heat sinking (condenser or heat rejection): heat is rejected at constant temperature TL 𝐈𝐬𝐨𝐭𝐡𝐞𝐫𝐦𝐚𝐥 𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝑾 = 𝑸𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 = 𝒎 𝑹 𝑻𝟏 𝒍𝒏 [ 𝑽𝟏 𝑽𝟐 ] Process 2-3: isentropic compression (work in): the air compresses isentropically to the high pressure and temperature. 𝐈𝐬𝐞𝐧𝐭𝐫𝐨𝐩𝐢𝐜 𝐨𝐫 𝐀𝐝𝐢𝐚𝐛𝐚𝐭𝐢𝐜 𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝑻𝟐 𝑻𝟑 = [ 𝑽𝟑 𝑽𝟐 ] 𝜸−𝟏 𝑸 = 𝟎 Process 3-4: isothermal heat transfer or Isothermal expansion (heat supply): heat is supplied at constant temperature TH 𝐈𝐬𝐨𝐭𝐡𝐞𝐫𝐦𝐚𝐥 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝑾 = 𝑸𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 = 𝒎 𝑹 𝑻𝟑 𝒍𝒏 [ 𝑽𝟒 𝑽𝟑 ] Process 4-1: isentropic expansion (work output): air expands isentropically from the high pressure and temperature to the low pressure and temperature 𝐈𝐬𝐞𝐧𝐭𝐫𝐨𝐩𝐢𝐜 𝐨𝐫 𝐀𝐝𝐢𝐚𝐛𝐚𝐭𝐢𝐜 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝑻𝟒 𝑻𝟏 = [ 𝑽𝟏 𝑽𝟒 ] 𝜸−𝟏 𝑸 = 𝟎
- 6. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 6 Air Standard Efficiency of Carnot cycle: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜂 (𝑐𝑎𝑟𝑛𝑜𝑡) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑠 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 − 𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑦𝑝𝑝𝑙𝑦 = 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 = 1 − 𝑄𝑅 𝑄𝑆 = 1 − 𝒎 𝑹 𝑻𝟏 𝒍𝒏 [ 𝑽𝟏 𝑽𝟐] 𝒎 𝑹 𝑻𝟑 𝒍𝒏 [ 𝑽𝟒 𝑽𝟑] = 1 − 𝑇1 𝑇3 𝐵𝑒𝑐𝑎𝑢𝑠𝑒 𝑉1 𝑉2 = 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑎𝑛𝑑 𝑉4 𝑉3 = 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑠𝑎𝑚𝑒 𝑇𝐿 = 𝑇1 = 𝑇2 = 𝐿𝑜𝑤𝑒𝑠𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑇𝐻 = 𝑇3 = 𝑇4 = 𝐻𝑖𝑔ℎ𝑒𝑠𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 ∴ 𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 = 𝜼 (𝒄𝒂𝒓𝒏𝒐𝒕) = 𝟏 − 𝑻𝑳 𝑻𝑯 Mean Effective Pressure for Carnot Cycle: (MEP = Pm) Mean effective pressure 𝐏𝐦 = Net Work done / Cycle Swept Volume of the Cylinder Ratio of isentropic (or adiabatic) compression = 𝑉2 𝑉3 Ratio of isothermal expansion = 𝑉4 𝑉3 Stroke Volume 𝑉𝑠 = 𝑉1 − 𝑉3 𝑽𝟏 𝑽𝟐 = 𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐫𝐚𝐭𝐢𝐨 = 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐫𝐚𝐭𝐢𝐨 = 𝑽𝟒 𝑽𝟑 In general, for solving Problems direct substitutions are used 𝐏𝐦 = Qsupplied − Q rejected Stroke Volume
- 7. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 7 1.2 Otto, description, p-v and T -s diagrams, efficiencies, mean effective pressures. Process 1-2: represents the adiabatic compression of air due to which p1, V1 and T1 change to p2, V2 and T2 respectively. 𝐈𝐬𝐞𝐧𝐭𝐫𝐨𝐩𝐢𝐜 𝐨𝐫 𝐀𝐝𝐢𝐚𝐛𝐚𝐭𝐢𝐜 𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝑻𝟐 𝑻𝟏 = [ 𝑽𝟏 𝑽𝟐 ] 𝜸−𝟏 𝑸 = 𝟎 Process 2-3: shows the supply of heat to the air at constant volume so that p2 and T2 change to p3 and T3 (V3 being the same as V2). 𝐇𝐞𝐚𝐭 𝐒𝐮𝐩𝐩𝐥𝐢𝐞𝐝 𝐐𝐒 = 𝐦 𝐂𝐯 [𝐓𝟑 − 𝐓𝟐] Process 3-4 represents the adiabatic expansion of the air. During expansion p3, V3 and T3 change to a final value of p4, V4 or V1 and T4, respectively. 𝐈𝐬𝐞𝐧𝐭𝐫𝐨𝐩𝐢𝐜 𝐨𝐫 𝐀𝐝𝐢𝐚𝐛𝐚𝐭𝐢𝐜 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝑻𝟑 𝑻𝟒 = [ 𝑽𝟒 𝑽𝟑 ] 𝜸−𝟏 𝑸 = 𝟎 Process 4-1: shows the rejection of heat by air at constant volume until original state (point 1) reaches. 𝐇𝐞𝐚𝐭 𝐑𝐞𝐣𝐞𝐜𝐭𝐞𝐝 𝐐𝐑 = 𝐦 𝐂𝐯 [𝐓𝟒 − 𝐓𝟏]
- 8. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 8 Air Standard Efficiency of Otto cycle: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜼 (𝑶𝒕𝒕𝒐) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑠 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 − 𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑦𝑝𝑝𝑙𝑦 = 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 = 1 − 𝑄𝑅 𝑄𝑆 = 1 − 𝐦 𝐂𝐯 [𝐓𝟒 − 𝐓𝟏] 𝐦 𝐂𝐯 [𝐓𝟑 − 𝐓𝟐] = 1 − [𝐓𝟒 − 𝐓𝟏] [𝐓𝟑 − 𝐓𝟐] 𝜼 (𝑶𝒕𝒕𝒐) = 1 − [T4 − T1] [T3 − T2] = 1 − [ 𝑇4 𝑇1 − 1] 𝑇1 [ 𝑇3 𝑇2 − 1] 𝑇2 we know that 𝑇2 𝑇1 = [ 𝑉1 𝑉2 ] 𝛾−1 but V2 = V3 and V1 = V4 ∴ 𝑇2 𝑇1 = [ 𝑉4 𝑉3 ] 𝛾−1 but from Process 3 − 4 𝑇3 𝑇4 = [ 𝑉4 𝑉3 ] 𝛾−1 ∴ we have 𝑻𝟑 𝑻𝟒 = 𝑻𝟐 𝑻𝟏 𝒐𝒓 𝑻𝟒 𝑻𝟏 = 𝑻𝟑 𝑻𝟐 Substituting T4 T1 = T3 T2 in 𝜼 (𝑶𝒕𝒕𝒐) = 1 − [ 𝑇4 𝑇1 − 1] 𝑇1 [ 𝑇3 𝑇2 − 1] 𝑇2 𝑤𝑒 𝑔𝑒𝑡 𝜼 (𝑶𝒕𝒕𝒐) = 𝟏 − 𝑻𝟏 𝑻𝟐 𝐛𝐮𝐭 𝐟𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟏 − 𝟐 𝐰𝐞 𝐠𝐞𝐭 T2 T1 = [ V1 V2 ] γ−1 𝐰𝐡𝐞𝐫𝐞 𝐕𝟏 𝐕𝟐 = 𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐑𝐚𝐭𝐢𝐨 (𝐑𝐂) that is 𝑻𝟐 𝑻𝟏 = [𝑹𝒄]𝜸−𝟏 𝒐𝒓 𝑻𝟏 𝑻𝟐 = [ 𝟏 𝑹𝒄 ] 𝜸−𝟏 = [ 𝟏 (𝑹𝒄)𝜸−𝟏 ] ∴ 𝜼 (𝑶𝒕𝒕𝒐) = 𝟏 − 𝟏 (𝑹𝒄)𝜸−𝟏
- 9. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 9 Mean Effective Pressure for Otto Cycle: (MEP = Pm) Mean effective pressure 𝐏𝐦 = Net Work done / Cycle Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 Swept Volume of the Cylinder = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 − 𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 Swept Volume of the Cylinder = 𝑄𝑆 − 𝑄𝑅 𝑉1 − 𝑉2 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 = 𝑄𝑆 − 𝑄𝑅 = 𝐦 𝐂𝐯 [𝐓𝟑 − 𝐓𝟐] − 𝐦 𝐂𝐯 [𝐓𝟒 − 𝐓𝟏] ∴ 𝑃𝑚 = 𝐦 𝐂𝐯 [𝐓𝟑 − 𝐓𝟐] − 𝐦 𝐂𝐯 [𝐓𝟒 − 𝐓𝟏] 𝑉1 − 𝑉2 … … … … … … (1) Process 1-2: represents the adiabatic compression Isentropic or Adiabatic Compression 𝑇2 𝑇1 = [ 𝑉1 𝑉2 ] 𝛾−1 ≫ 𝑉1 𝑉2 = 𝑅𝑐 ∴ 𝑻𝟐 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏 Process 2-3: supply of heat to the air at constant volume P1 T1 = P2 T2 = 𝑃 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ≫ 𝑃3 𝑃2 = 𝑇3 𝑇2 ≫ 𝑻𝟑 = 𝑻𝟐 ( 𝑷𝟑 𝑷𝟐 ) 𝑷𝟑 𝑷𝟐 = 𝜶 𝑬𝒙𝒑𝒍𝒐𝒔𝒊𝒐𝒏 𝒓𝒂𝒕𝒊𝒐 𝒐𝒓 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒓𝒂𝒕𝒊𝒐 ∴ 𝑇3 = 𝑇2 ( 𝑃3 𝑃2 ) ≫ 𝑻𝟑 = 𝑻𝟐 (𝜶) 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑇2 𝑣𝑎𝑙𝑢𝑒 𝑤𝑒 𝑔𝑒𝑡 ≫ 𝑻𝟑 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏(𝜶) Process 3-4 represents the adiabatic expansion of the air. Isentropic or Adiabatic Expansion 𝑻𝟑 𝑻𝟒 = [ 𝑽𝟒 𝑽𝟑 ] 𝜸−𝟏 𝑏𝑢𝑡 𝑉4 = 𝑉1 𝑎𝑛𝑑 𝑉3 = 𝑉2 ∴ 𝑇3 𝑇4 = [ 𝑉4 𝑉3 ] 𝛾−1 ≫ 𝑻𝟑 𝑻𝟒 = [ 𝑽𝟏 𝑽𝟐 ] 𝜸−𝟏 ≫ 𝑻𝟑 𝑻𝟒 = [𝑹𝒄]𝜸−𝟏 ≫ 𝑻𝟒 = 𝑻𝟑 [𝑹𝒄]𝜸−𝟏 Substituting T3 Value for 𝑇4 𝑤𝑒 𝑔𝑒𝑡 𝑻𝟒 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏(𝜶) [𝑹𝒄]𝜸−𝟏
- 10. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 10 We know that 𝑷𝟏𝑽𝟏 = 𝒎𝑹𝑻𝟏 ∴ 𝑽𝟏 = 𝒎𝑹𝑻𝟏 𝑷𝟏 We also know that 𝑽𝟏 𝑽𝟐 = 𝑹𝒄 𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒓𝒂𝒕𝒊𝒐 ∴ 𝑉2 = 𝑉1 𝑅𝑐 𝑆𝑢𝑏𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑉1 𝑣𝑎𝑙𝑢𝑒 𝑤𝑒 𝑔𝑒𝑡 𝑽𝟐 = 𝒎𝑹𝑻𝟏 𝑷𝟏𝑹𝒄 Substituting T2, T3, T4, V1 and V2 values in MEP (Pm) we get ∴ 𝑃𝑚 = m Cv [T3 − T2] − m Cv [T4 − T1] 𝑉1 − 𝑉2 𝑎𝑠 ∴ 𝑃𝑚 = 𝐏𝟏 𝐂𝐯 [𝛂 − 𝟏] [𝐑𝐜𝛄−𝟏 − 𝟏] 𝑹 [ 𝑹𝒄 − 𝟏 𝑹𝒄 ] We know that 𝑹 = 𝑪𝒑 − 𝑪𝒗 ≫ ∴ 𝑹 𝑪𝒗 = 𝑪𝒑 𝑪𝒗 − 𝟏 ≫ 𝑹 𝑪𝒗 = 𝜸 − 𝟏 𝑛𝑜𝑤 𝑠𝑢𝑏𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑅 𝐶𝑣 𝑖𝑛 𝑀𝐸𝑃 (𝑃𝑚)𝑤𝑒 𝑔𝑒𝑡 ∴ 𝑃𝑚 = 1 [𝛾 − 1] P1 Rc [α − 1] [Rcγ−1 − 1] [𝑅𝑐 − 1] ∴ 𝑷𝒎 = 𝐏𝟏 𝐑𝐜 [𝛂 − 𝟏] [𝐑𝐜𝛄−𝟏 − 𝟏] [𝜸 − 𝟏][𝑹𝒄 − 𝟏]
- 11. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 11 1.3 Diesel, description, p-v and T -s diagrams, efficiencies, mean effective pressures. Process 1-2: Isentropic Compression, in this process the piston moves from BDC to TDC and compression of air takes place isentropically. Here the air is compressed so the pressure increases from P1 to P2, volume decreases from V1 to V2, Temperature increases from T1 to T2 and entropy remains constant ( i.e. S1 = S2). Process 2-3: Constant Volume Heat Addition, in this process the, the hot body is kept in contact with the cylinder and heat addition to the air takes place at constant pressure. During this process, the piston rest for a moment at TDC. The pressure remains constant (i.e. P2 = P3), volume increases from V2 to V3, temperature increases from T2 to T3, entropy increases from S2 to S3. Process 3-4: Isentropic Expansion, in this process, after heat addition, the expansion of air takes place isentropically and work is obtained from the system. The piston moves downward during this process and reaches to BDC. The pressure falls from P3 to P4, Volume increases from V3 to V4, the temperature falls from T3 to T4 and entropy remains constant (i.e. S3=S4). Process 4-1: Constant Volume Heat Rejection, in this process, the piston rest at BDC for a moment and the cold body is brought in contact with the cylinder and the heat rejection takes place at constant volume. The pressure decreases from P4 to P1, temperature decreases from T4 to T1, entropy decreases from S4 to S1 and volume remains constant (i.e.V4 = V1).
- 12. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 12 Air Standard Efficiency of Diesel cycle: Heat addition at constant pressure in process 2-3, QS = Cp (T3 – T2) Heat rejection at constant volume in process 4-1, QR =Cv (T4 – T1) Net work done during cycle, Wnet = QS – QR = Cp (T3-T2) – Cv (T4 – T1) 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜂 (𝐷𝑖𝑒𝑠𝑒𝑙) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑠 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑−𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑦𝑝𝑝𝑙𝑦 = 𝑄𝑆−𝑄𝑅 𝑄𝑆 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 = 1 − 𝑄𝑅 𝑄𝑆 = 1 − m Cv [T4 − T1] m Cp [T3 − T2] = 1 − 1 𝛾 [T4 − T1] [T3 − T2] 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝜸 [𝐓𝟒 − 𝐓𝟏] [𝐓𝟑 − 𝐓𝟐] 𝐅𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟏 − 𝟐 𝑇2 𝑇1 = [ 𝑉1 𝑉2 ] 𝛾−1 𝒐𝒓 𝑇2 𝑇1 = [𝑅𝑐]𝛾−1 𝑻𝟐 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏 𝐅𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟑 − 𝟒 𝑇3 𝑇4 = [ 𝑉4 𝑉3 ] 𝛾−1 but V4 = V1 so 𝑇3 𝑇4 = [ 𝑉1 𝑉3 ] 𝛾−1 𝐫𝐞 − 𝐰𝐫𝐢𝐭𝐢𝐧𝐠 𝑇3 𝑇4 = [ 𝑉1 𝑉2 × 𝑉2 𝑉3 ] 𝛾−1 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑇3 𝑇4 = [𝑅𝑐 × 1 𝜌 ] 𝛾−1 ≫ 𝑇3 𝑇4 = 𝑅𝑐𝛾−1 × 1 𝜌𝛾−1 𝒘𝒉𝒆𝒓𝒆 𝑽𝟑 𝑽𝟐 = (𝝆) = 𝑪𝒖𝒕 − 𝒐𝒇𝒇 𝒓𝒂𝒕𝒊𝒐 ∴ 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑻𝟒 = 𝑻𝟑 ( 𝝆𝜸−𝟏 𝑹𝒄𝜸−𝟏) 𝐅𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟐 − 𝟑 𝑉 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉2 𝑇2 = 𝑉3 𝑇3 ≫ 𝑇3 𝑇2 = 𝑉3 𝑉2 ≫ 𝑻𝟑 𝑻𝟐 = 𝝆 ∴ we have T3 = T2 (ρ) but 𝑇2 = 𝑇1[𝑅𝑐]𝛾−1 ∴ 𝐓𝟑 = 𝐓𝟏 (𝛒) [𝑹𝒄]𝜸−𝟏 Substituting 𝐓𝟑 and 𝐓𝟒 Values we get 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝜸 [𝐓𝟒 − 𝐓𝟏] [𝐓𝟑 − 𝐓𝟐] = 𝟏 − 𝟏 𝛄(𝐑𝐜𝛄−𝟏) [ 𝛒𝛄 − 𝟏 (𝛒 − 𝟏) ]
- 13. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 13 Mean Effective Pressure for Diesel cycle: (MEP = Pm) Mean effective pressure 𝐏𝐦 = Net Work done / Cycle Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 − 𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = 𝑄𝑆 − 𝑄𝑅 𝑉1 − 𝑉2 (𝑀𝐸𝑃)𝑃𝑚 = 𝐦𝐂𝐩 (𝐓𝟑 – 𝐓𝟐) − 𝐦𝐂𝐯 (𝐓𝟒 – 𝐓𝟏) 𝑽𝟏 − 𝑽𝟐 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑹𝑪 = 𝑽𝟏 𝑽𝟐 and cut − off ratio 𝝆 = 𝑽𝟑 𝑽𝟐 𝑽𝟏 − 𝑽𝟐 = 𝑚 𝑅 𝑇1 𝑃1 − 𝑚 𝑅 𝑇1 𝑃1𝑅𝑐 ≫ 𝒎 𝑹 𝑻𝟏 𝑷𝟏 [𝟏 − 𝟏 𝑹𝒄 ] 𝑻𝟐 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏 , 𝐓𝟑 = 𝐓𝟏 (𝛒) [𝑹𝒄]𝜸−𝟏 𝒂𝒏𝒅 𝑻𝟒 = 𝑻𝟑 ( 𝝆𝜸−𝟏 𝑹𝒄𝜸−𝟏 ) 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (𝑉1 − 𝑉2), 𝑇2, 𝑇3 𝑎𝑛𝑑 𝑇4 𝑖𝑛 (𝑴𝑬𝑷)𝑷𝒎 𝑚𝑎𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑒 𝑔𝑒𝑡 𝑴𝑬𝑷 (𝑷𝒎) = 𝑷𝟏 𝑹𝒄 {[𝜸 𝑹𝒄𝜸−𝟏(𝝆𝜸 − 𝟏) − (𝝆𝜸 − 𝟏)]} (𝜸 − 𝟏)(𝑹𝒄 − 𝟏)
- 14. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 14 1.4 Dual cycles, description, p-v and T -s diagrams, efficiencies, mean effective pressures. The process description of Dual cycle is as below: Process 1-2: Reversible adiabatic compression. Process 2-3: Constant volume heat addition. Process 3-4: Constant pressure heat addition. Process 4-5: Reversible adiabatic expansion. Process 5-1: Constant volume heat reject Air Standard Efficiency of Dual cycle: Heat addition at constant pressure in process 3-4 = Cp (T4 − T3) Total heat added in process 2-3 and 3-4, QS = [Cp (T3 −T2) − Cv (T4 – T1)] Heat rejection at constant volume in process 5-1, QR = Cv (T5 – T1) Net work done during cycle, Wnet = QS − QR = [Cv (T3 − T2) + Cp (T4 − T3)] − Cv (T5 − T1) 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜼 (𝑫𝒖𝒂𝒍) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑠 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑−𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑦𝑝𝑝𝑙𝑦 = 𝑄𝑆−𝑄𝑅 𝑄𝑆 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 = 1 − 𝑄𝑅 𝑄𝑆 = 1 − Cv (T5 – T1) Cv (T3 – T2) + Cp (T4 – T3) 𝜼 (𝑫𝒖𝒂𝒍) = 𝟏 − (𝐓𝟓 – 𝐓𝟏) (𝐓𝟑 − 𝐓𝟐) + 𝜸 (𝐓𝟒 − 𝐓𝟑)
- 15. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 15 cut − off ratio: 𝝆 = 𝑽𝟒 𝑽𝟑 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑜𝑟 𝐶𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜: 𝑹𝒄 = 𝑪𝒊𝒍𝒊𝒏𝒅𝒆𝒓 𝑽𝒐𝒍𝒖𝒎𝒆 𝑪𝒍𝒆𝒂𝒓𝒆𝒏𝒄𝒆 𝑽𝒐𝒍𝒖𝒎𝒆 = 𝑽𝟏 𝑽𝟐 Explosion ratio or Pressure ratio: 𝜶 = 𝑷𝟑 𝑷𝟐 From Process 1-2: Reversible adiabatic compression. 𝑇2 𝑇1 = ( 𝑉1 𝑉2 ) 𝛾−1 ≫ 𝑇2 = 𝑇1 ( 𝑉1 𝑉2 ) 𝛾−1 ≫ 𝑻𝟐 = 𝑻𝟏 (𝑹𝒄𝜸−𝟏 ) From Process 2-3: Constant volume heat addition. P1 T1 = P2 T2 = P3 T3 = 𝑃 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴ P2 T2 = 𝑃3 𝑇3 ≫ T3 T2 = 𝑃3 𝑃2 = α Pressure ratio ∴ 𝐓𝟑 = 𝐓𝟐 (𝛂) ∴ 𝑏𝑦 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑻𝟐 𝑉𝑎𝑙𝑢𝑒 𝑤𝑒 𝑔𝑒𝑡 𝐓𝟑 = 𝑻𝟏 (𝑹𝒄𝜸−𝟏 )(𝛂) Process 3-4: Constant pressure heat addition. V1 T1 = V2 T2 = 𝑉 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴ V3 T3 = V4 T4 ≫ 𝐓𝟒 𝐓𝟑 = 𝐕𝟒 𝐕𝟑 = (𝛒) Cut − off ratio ∴ 𝐓𝟒 = 𝐓𝟑 (𝛒) ∴ 𝑏𝑦 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑻𝟑 𝑉𝑎𝑙𝑢𝑒 𝑤𝑒 𝑔𝑒𝑡 𝐓𝟒 = 𝑻𝟏(𝑹𝒄𝜸−𝟏 )(𝛂)(𝛒) Process 4-5: Reversible adiabatic expansion. 𝑇4 𝑇5 = ( 𝑉5 𝑉4 ) 𝛾−1 𝑏𝑢𝑡 𝑉5 = 𝑉1 ∴ 𝑇5 𝑇4 = ( 𝑉1 𝑉4 ) 𝛾−1 𝑟𝑒 − 𝑤𝑟𝑖𝑡𝑖𝑛𝑔 𝑇4 𝑇5 = ( 𝑉1 𝑉2 × 𝑉2 𝑉4 ) 𝛾−1 𝑎𝑛𝑑 𝑉2 = 𝑉3 𝑎𝑙𝑠𝑜 𝑽𝟒 𝑽𝟑 = 𝝆 𝑐𝑢𝑡 − 𝑜𝑓𝑓 𝑟𝑎𝑡𝑖𝑜, 𝑠𝑜 𝑽𝟑 𝑽𝟒 = 𝟏 𝝆 𝑇4 𝑇5 = (𝑅𝑐 × 1 𝜌 ) 𝛾−1 ≫ 𝑇5 𝑇4 = (𝑅𝑐𝛾−1) ( 1 𝜌 ) 𝛾−1 ≫ 𝑻𝟓 𝑻𝟒 = (𝑹𝒄𝜸−𝟏)(𝝆)𝟏−𝜸 𝑻𝟓 = 𝑻𝟒 (𝑹𝒄𝜸−𝟏)(𝝆)𝟏−𝜸 ∴ 𝑏𝑦 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑻𝟒 𝑉𝑎𝑙𝑢𝑒 𝑤𝑒 𝑔𝑒𝑡 T5 = 𝑇1 (𝑅𝑐𝛾−1)(α)(ρ)(𝑅𝑐𝛾−1)(𝜌)1−𝛾 = 𝑇1 𝛼 𝜌𝛾 𝑏𝑦 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡 𝐓𝟓 = 𝑻𝟏 𝜶 𝝆𝜸
- 16. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 16 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑇2, 𝑇3, 𝑇4 𝑎𝑛𝑑 𝑇5 𝑣𝑎𝑙𝑢𝑒𝑠 𝑡𝑜 𝜂 (𝐷𝑢𝑎𝑙) = 1 − (T5 – T1) (T3 − T2) + 𝛾 (T4 − T3) 𝑤𝑒 𝑔𝑒𝑡 𝜼 (𝑫𝒖𝒂𝒍) = 𝟏 − 𝟏 𝑹𝒄𝜸−𝟏 [ (𝜶)(𝝆𝜸) − 𝟏 (𝜶 − 𝟏) + 𝜶 𝜸 (𝝆 − 𝟏) ] 𝑖𝑓 𝝆 = 𝟏 𝑡ℎ𝑒𝑛 𝜂 (𝐷𝑢𝑎𝑙) = 1 − 1 (𝑅𝑐)𝛾−1 = 𝜂 (𝑂𝑡𝑡𝑜) 𝑖𝑓 𝜶 = 𝟏 𝑡ℎ𝑒𝑛 𝜂 (𝐷𝑢𝑎𝑙) = 1 − 1 γ(Rcγ−1) [ ργ − 1 (ρ − 1) ] = 𝜂 (𝐷𝑖𝑒𝑠𝑒𝑙) Mean effective pressure of Dual cycle can be calculated as follows: Mean effective pressure 𝐏𝐦 = Net Work done / Cycle Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 Swept Volume of the Cylinder = (𝑀𝐸𝑃)𝑃𝑚 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑−𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 Swept Volume of the Cylinder (𝑀𝐸𝑃)𝑃𝑚 = [𝑄𝑆2−3 + 𝑄𝑆3−4] − 𝑄𝑅 𝑉1 − 𝑉2 (𝑀𝐸𝑃)𝑃𝑚 = 𝐦𝐂𝐯 (𝐓𝟑 – 𝐓𝟐) + 𝐦𝐂𝐩 (𝐓𝟒 – 𝐓𝟑) − 𝐦𝐂𝐯 (𝐓𝟓 – 𝐓𝟏) 𝑽𝟏 − 𝑽𝟐 𝑻𝟐 = 𝑻𝟏 (𝑹𝒄𝜸−𝟏 ); 𝐓𝟑 = 𝑻𝟏 (𝑹𝒄𝜸−𝟏 )(𝛂); 𝐓𝟒 = 𝑻𝟏(𝑹𝒄𝜸−𝟏 )(𝛂)(𝛒); 𝐓𝟓 = 𝑻𝟏 𝜶 𝝆𝜸 and 𝑽𝟏 − 𝑽𝟐 = 𝒎 𝑹 𝑻𝟏 𝑷𝟏 [𝟏 − 𝟏 𝑹𝒄 ] By substituting these values we get (𝑴𝑬𝑷)𝑷𝒎 = 𝐏𝟏 𝐑𝐜 (𝑹𝒄 − 𝟏)(𝜸 − 𝟏) [𝑹𝒄𝜸−𝟏(𝜶 − 𝟏) + 𝜶 𝜸 𝑹𝒄𝜸−𝟏(𝝆 − 𝟏) − (𝜶 𝝆𝜸 − 𝟏)]
- 17. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 17 1.5 Stirling cycles, description, p-v and T -s diagrams, efficiencies, mean effective pressures. Figure: Experimental Modal of Stirling Cycle Stirling Cycle P-V Graph Stirling Cycle T-S Graph Stirling Cycle Processes: 1. The air is compressed isothermally from state 1 to 2 heat is rejected from the working medium to external sink (TL to TH). 2. The air at state-2 is passed into the regenerator from the top at a temperature T1. The air passing through the regenerator matrix gets heated from TL to TH. 3. The air at state-3 expands isothermally in the cylinder until it reaches state-4. 4. The air coming out of the engine at temperature TH (condition 4) enters into regenerator from the bottom and gets cooled while passing through the regenerator matrix at constant volume and it comes out at a temperature TL, at condition 1 and the cycle is repeated. It can be shown that the heat absorbed by the air from the regenerator matrix during the process 2- 3 is equal to the heat given by the air to the regenerator matrix during the process 4-1, and then the exchange of heat with external source will be only during the isothermal processes. Note: A regenerator is a component in a Stirling engine that stores heat from one cycle so it can be used in the next cycle. Regenerators are often made of sheets of foil, steel wool, or a metallic sponge. The hot working gas flows over the regenerator (storing some of its heat there) on it’s way to the cold zone. When the cold gas returns, it flows back over the regenerator and is pre-heated before it goes to the hot zone.
- 18. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 18 Air Standard Efficiency of Stirling cycle: Process 1-2: Isothermal Compression. 𝑄𝑅 = 𝑃1 𝑉1 𝑙𝑛 [ 𝑉1 𝑉2 ] = 𝑚 𝑅 𝑇1 ln(𝑅𝑐) But T1 = TL (Low temperature) = 𝒎 𝑹 𝑻𝑳 𝐥𝐧(𝑹𝒄) Process 2-3: Constant volume heat addition. 𝑸𝟐−𝟑 = 𝒎 𝑪𝒗 [𝑻𝟑 − 𝑻𝟐] Process 3-4: Constant pressure heat addition. 𝑄𝑆 = 𝑃3 𝑉3 𝑙𝑛 [ 𝑉4 𝑉3 ] = 𝑏𝑢𝑡 𝑉4 = 𝑉1 & 𝑉3 = 𝑉2 𝑃3 𝑉3 𝑙𝑛 [ 𝑉4 𝑉3 ] = 𝑃3 𝑉3 𝑙𝑛 [ 𝑉1 𝑉2 ] = 𝑚 𝑅 𝑇3 ln(𝑅𝑐) But T3 = TH (High temperature) = 𝒎 𝑹 𝑻𝑯 𝐥𝐧(𝑹𝒄) Process 4-1: Constant volume heat reject 𝑸𝟒−𝟏 = 𝒎 𝑪𝒗 [𝑻𝟒 − 𝑻𝟏] 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜼 (𝑺𝒕𝒊𝒓𝒍𝒊𝒏𝒈) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑠 = 𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑−𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝐻𝑒𝑎𝑡 𝑆𝑦𝑝𝑝𝑙𝑦 = 𝑄𝑆−𝑄𝑅 𝑄𝑆 𝑄𝑆 − 𝑄𝑅 𝑄𝑆 = 1 − 𝑄𝑅 𝑄𝑆 = 1 − 𝒎 𝑹 𝑻𝑳 𝐥𝐧(𝑹𝒄) 𝒎 𝑹 𝑻𝑯 𝐥𝐧(𝑹𝒄) 𝜼 (𝑺𝒕𝒓𝒊𝒍𝒊𝒏𝒈) = 𝟏 − 𝑻𝑳 𝑻𝑯
- 19. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 19 1.6 Comparison of Otto and Diesel cycles. Sl.no OTTO CYCLE DIESEL CYCLE 1. Heat addition takes place at constant volume. Heat addition takes place at constant pressure. 2. Petrol engines work on this cycle. Diesel engines work on this cycle. 3. At constant volume, heat rejection takes place. In diesel cycle also the heat rejection takes place at constant volume. 4. Compression ratio is less. It is 7:1 to 10:1. Compression ratio is more. It is 11:1 to 22:1. 5. Efficiency is less. Efficiency is more. 6. Adiabatic expansion takes place during the complete backward stroke of the piston. After the heat addition is cut-off in the backward stroke, the adiabatic expansion takes place during the remaining portion of stroke. Differences between otto and diesel cycle: 1: For a given compression ratio V1/V2 , the efficiency of otto cycle is higher than diesel cycle. 2: The combustion process occurs at constant volume in otto cycle and occurs at constant pressure in diesel cycle. 3: Combustion in otto cycle occurs due to generation of spark in the cylinder when the piston reaches TDC and is instantaneous. Whereas in case of diesel cycle combustion is time consuming and occurs due to injection of fuel in cylinder which is filled with hot compressed air (at point 2 in figure) due to which auto ignition of fuel takes place. 4: Due to the fact that we compress air in diesel cycle rather than fuel(in otto cycle) , the air can be compressed to very high pressures and then the fuel can be injected accordingly , thus incresing compression ratio and higher efficiency of diesel cycle. 5: So for same max pressure of cycle, diesel cycle is more efficient that otto cycle. 6: The compression ratio in case of otto cycle is however limited because in otto cycle , air fuel mixture is being compressed which can lead to very high temperature of air fuel mixture and can cause auto ignition of fuel before the power stroke. 7: The compression ratio in diesel cycle is also limited due to the metallurgical reasons , air is compressed to such temperatures which will not harm the engine parts.
- 20. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 20 Previous Year Questions for 1a & 1b Modal Question paper 1 1 b) D= 20 cm = 0.2 m L = 30 cm = 0.3 m VC = 4.2 x 10-4 m3 VSa =5% VS stroke Cut off is delayed from 5% Air standard Efficiency =? We know that 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑹𝒄 = 𝑽𝟏 𝑽𝟐 = ( 𝑽𝑪+𝑽𝑺 𝑽𝑪 ) 𝑽𝑺 = 𝝅 𝟒 𝑫𝟐 × 𝑳 cut − off ratio 𝝆 = 𝑽𝟑 𝑽𝟐 = (𝑽𝑪 + 𝟓% 𝑽𝑺𝒂 𝑽𝑪 𝑨𝒔𝒔𝒖𝒎𝒆 𝜸 = 𝟏. 𝟒 𝒂𝒏𝒅 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒆 𝑹𝒄 𝒂𝒏𝒅 𝝆 𝑽𝒂𝒍𝒖𝒆𝒔 𝒇𝒊𝒏𝒅 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝛄(𝐑𝐜𝛄−𝟏) [ 𝛒𝛄 − 𝟏 (𝛒 − 𝟏) ] … … … … … … … … . (𝟏) cut − off ratio 𝝆 = 𝑽𝟑 𝑽𝟐 = (𝑽𝑪 + 𝟖% 𝑽𝑺𝒂 𝑽𝑪 𝑨𝒔𝒔𝒖𝒎𝒆 𝜸 = 𝟏. 𝟒 𝒂𝒏𝒅 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒆 𝑹𝒄 𝒂𝒏𝒅 𝝆 𝑽𝒂𝒍𝒖𝒆𝒔 𝒇𝒊𝒏𝒅 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝛄(𝐑𝐜𝛄−𝟏) [ 𝛒𝛄 − 𝟏 (𝛒 − 𝟏) ] … … … … … … … … . (𝟐) 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒊𝒏 𝒕𝒉𝒆 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝒕𝒉𝒂𝒕 𝒊𝒔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝟏) ≈ (𝟐)
- 21. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 21 Modal Question paper 2 1b Formulas 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑹𝒄 = 𝑽𝟏 𝑽𝟐 = 16 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑻𝟏 = 𝟏𝟓 °𝑪 𝑎𝑛𝑑 𝑻𝟑 = 𝟏𝟒𝟖𝟎°𝑪 𝐟𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟏 − 𝟐 𝑇2 𝑇1 = [ 𝑉1 𝑉2 ] 𝛾−1 𝒐𝒓 𝑇2 𝑇1 = [𝑅𝑐]𝛾−1 𝑻𝟐 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏 𝐅𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟐 − 𝟑 𝑉 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉2 𝑇2 = 𝑉3 𝑇3 ≫ 𝑇3 𝑇2 = 𝑉3 𝑉2 ≫ 𝑻𝟑 𝑻𝟐 = 𝝆 𝒄𝒖𝒕 𝒐𝒇𝒇 𝒓𝒂𝒕𝒊𝒐 ∴ 𝐓𝟑 = 𝐓𝟐 × (𝛒) = 𝐓𝟏 (𝛒) [𝑹𝒄]𝜸−𝟏 Heat added QS = Cp (T3 – T2) 𝐜𝐮𝐭 − 𝐨𝐟𝐟 𝐫𝐚𝐭𝐢𝐨 𝝆 = 𝑽𝟑 𝑽𝟐 = 𝑻𝟑 𝑻𝟐 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝛄(𝐑𝐜𝛄−𝟏) [ 𝛒𝛄 − 𝟏 (𝛒 − 𝟏) ] 𝑷𝒎 = 𝑷𝟏 𝑹𝒄 {[𝜸 𝑹𝒄𝜸−𝟏(𝝆𝜸 − 𝟏) − (𝝆𝜸 − 𝟏)]} (𝜸 − 𝟏)(𝑹𝒄 − 𝟏)
- 22. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 22 Question paper Jan 2020 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑹𝒄 = 𝑽𝟏 𝑽𝟐 = 14 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑻𝟏 = 𝟏𝟎𝟎 °𝑪 𝐟𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟏 − 𝟐 𝑇2 𝑇1 = [ 𝑉1 𝑉2 ] 𝛾−1 𝒐𝒓 𝑇2 𝑇1 = [𝑅𝑐]𝛾−1 𝑻𝟐 = 𝑻𝟏[𝑹𝒄]𝜸−𝟏 𝐅𝐫𝐨𝐦 𝐏𝐫𝐨𝐜𝐞𝐬𝐬 𝟐 − 𝟑 𝑉 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉2 𝑇2 = 𝑉3 𝑇3 ≫ 𝑇3 𝑇2 = 𝑉3 𝑉2 ≫ 𝑻𝟑 𝑻𝟐 = 𝝆 𝒄𝒖𝒕 𝒐𝒇𝒇 𝒓𝒂𝒕𝒊𝒐 = 𝟐. 𝟐 ∴ 𝐓𝟑 = 𝐓𝟐 × (𝛒) = 𝐓𝟏 (𝛒) [𝑹𝒄]𝜸−𝟏 ∴ 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑻𝟒 = 𝑻𝟑 ( 𝝆𝜸−𝟏 𝑹𝒄𝜸−𝟏 ) Process 1-2 𝑷𝟏 𝑷𝟐 = [ 𝑽𝟐 𝑽𝟏 ] 𝜸 = [ 𝑻𝟏 𝑻𝟐 ] 𝜸 𝜸−𝟏 find P2 P2 = P3 Pressure constant Process 342 𝑷𝟑 𝑷𝟒 = [ 𝑽𝟒 𝑽𝟑 ] 𝜸 = [ 𝑻𝟑 𝑻𝟒 ] 𝜸 𝜸−𝟏 find P4
- 23. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 23 Question paper July 2019 Process 1-2 𝑷𝟏 𝑷𝟐 = [ 𝑽𝟐 𝑽𝟏 ] 𝜸 = [ 𝑻𝟏 𝑻𝟐 ] 𝜸 𝜸−𝟏 Process 3-4 𝑷𝟑 𝑷𝟒 = [ 𝑽𝟒 𝑽𝟑 ] 𝜸 = [ 𝑻𝟑 𝑻𝟒 ] 𝜸 𝜸−𝟏
- 24. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 24 Question paper Sept 2020 Process 1-2 𝑷𝟏 𝑷𝟐 = [ 𝑽𝟐 𝑽𝟏 ] 𝜸 = [ 𝑻𝟏 𝑻𝟐 ] 𝜸 𝜸−𝟏 ∴ P2 T2 = 𝑃3 𝑇3 ≫ T3 T2 = 𝑃3 𝑃2 = α Pressure ratio ∴ 𝐓𝟑 = 𝐓𝟐 (𝛂) Process 3-4 𝑷𝟑 𝑷𝟒 = [ 𝑽𝟒 𝑽𝟑 ] 𝜸 = [ 𝑻𝟑 𝑻𝟒 ] 𝜸 𝜸−𝟏
- 25. Asst Proff Mr THANMAY J S, Department of Mechanical Engineering VVIET Mysore Page 25 Important Formulas to remember Cycle Air Standard Efficiency General Air Standard Efficiency Mean Effective Pressure Carnot cycle 𝜼 (𝒄𝒂𝒓𝒏𝒐𝒕) = 𝟏 − 𝑻𝟏 𝑻𝟑 𝜼 (𝒄𝒂𝒓𝒏𝒐𝒕) = 𝟏 − 𝑻𝑳 𝑻𝑯 𝐏𝐦 = 𝐐𝐬𝐮𝐩𝐩𝐥𝐢𝐞𝐝 − 𝐐 𝐫𝐞𝐣𝐞𝐜𝐭𝐞𝐝 𝐒𝐭𝐫𝐨𝐤𝐞 𝐕𝐨𝐥𝐮𝐦𝐞 Otto cycle 𝜼 (𝑶𝒕𝒕𝒐) = 𝟏 − [𝐓𝟒 − 𝐓𝟏] [𝐓𝟑 − 𝐓𝟐] 𝜼 (𝑶𝒕𝒕𝒐) = 𝟏 − 𝟏 (𝑹𝒄)𝜸−𝟏 𝑷𝒎 = 𝐏𝟏 𝐑𝐜 [𝛂 − 𝟏] [𝐑𝐜𝛄−𝟏 − 𝟏] [𝜸 − 𝟏][𝑹𝒄 − 𝟏] Diesel Cycle 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝜸 [𝐓𝟒 − 𝐓𝟏] [𝐓𝟑 − 𝐓𝟐] 𝜼 (𝑫𝒊𝒆𝒔𝒆𝒍) = 𝟏 − 𝟏 𝛄(𝐑𝐜𝛄−𝟏) [ 𝛒𝛄 − 𝟏 (𝛒 − 𝟏) ] 𝑷𝒎 = 𝑷𝟏 𝑹𝒄 {[𝜸 𝑹𝒄𝜸−𝟏(𝝆𝜸 − 𝟏) − (𝝆𝜸 − 𝟏)]} (𝜸 − 𝟏)(𝑹𝒄 − 𝟏) Dual Cycle 𝜼 (𝑫𝒖𝒂𝒍) = 𝟏 − (𝐓𝟓 – 𝐓𝟏) (𝐓𝟑 − 𝐓𝟐) + 𝜸 (𝐓𝟒 − 𝐓𝟑) 𝜼 (𝑫𝒖𝒂𝒍) = 𝟏 − 𝟏 𝑹𝒄𝜸−𝟏 [ (𝜶)(𝝆𝜸) − 𝟏 (𝜶 − 𝟏) + 𝜶 𝜸 (𝝆 − 𝟏) ] (𝑴𝑬𝑷)𝑷𝒎 = 𝐏𝟏 𝐑𝐜 (𝑹𝒄 − 𝟏)(𝜸 − 𝟏) [𝑹𝒄𝜸−𝟏(𝜶 − 𝟏) + 𝜶 𝜸 𝑹𝒄𝜸−𝟏(𝝆 − 𝟏) − (𝜶 𝝆𝜸 − 𝟏)] Striling Cycle 𝜼 (𝑺𝒕𝒓𝒊𝒍𝒊𝒏𝒈) = 𝟏 − 𝒎 𝑹 𝑻𝑳 𝐥𝐧(𝑹𝒄) 𝒎 𝑹 𝑻𝑯 𝐥𝐧(𝑹𝒄) 𝜼 (𝑺𝒕𝒓𝒊𝒍𝒊𝒏𝒈) = 𝟏 − 𝑻𝑳 𝑻𝑯 General Formula 𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒓𝒂𝒕𝒊𝒐 𝑹𝒄 = 𝑽𝟏 𝑽𝟐 𝐜𝐮𝐭 − 𝐨𝐟𝐟 𝐫𝐚𝐭𝐢𝐨 𝝆 = 𝑽𝟑 𝑽𝟐 𝐄𝐱𝐩𝐥𝐨𝐬𝐢𝐨𝐧 𝐫𝐚𝐭𝐢𝐨 𝐨𝐫 𝐏𝐫𝐞𝐬𝐬𝐮𝐫𝐞 𝐫𝐚𝐭𝐢𝐨: 𝜶 = 𝑷𝟑 𝑷𝟐 Mean effective pressure 𝐏𝐦 = 𝐍𝐞𝐭 𝐖𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 / 𝐂𝐲𝐜𝐥𝐞 𝐒𝐰𝐞𝐩𝐭 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐂𝐲𝐥𝐢𝐧𝐝𝐞𝐫 Assume CP = 1.00 kJ/kg.K, Cv = 0.718 kJ/kg. K, and 𝜸 = 1.4. 𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 = 𝑾𝒐𝒓𝒌 𝑫𝒐𝒏𝒆 𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒔 = 𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 − 𝑯𝒆𝒂𝒕 𝑹𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝑯𝒆𝒂𝒕 𝑺𝒚𝒑𝒑𝒍𝒚 = 𝑸𝑺 − 𝑸𝑹 𝑸𝑺