1 / 30

MT 313 IC ENGINES

MT 313 IC ENGINES. LECTURE NO: 04 (24 Feb, 2014) Khurram engr014@yahoo.com Yahoo Group Address: ICE14. Air Standard Cycle. The air as the working fluid follows the perfect gas law The working fluid is homogeneous throughout and no chemical reaction takes place

nieve
Download Presentation

MT 313 IC ENGINES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MT 313 IC ENGINES LECTURE NO: 04 (24 Feb, 2014) Khurram engr014@yahoo.com Yahoo Group Address: ICE14

  2. Air Standard Cycle • The air as the working fluid follows the perfect gas law • The working fluid is homogeneous throughout and no chemical reaction takes place • Specific heats of air do not vary with temperature • The mass of air in the cycle remains fixed • The exhaust process is replaced by an equivalent heat rejection process • The combustion process is replaced by an equivalent heat addition process • All processes are internally reversible

  3. Air Standard Cycle • Thermal efficiency • Thermal efficiency is also called air standard efficiency ηa

  4. Important Formulas • Swept Volume • Clearance Volume • Compression Ratio • Clearance ratio

  5. Ideal cycles are simplified

  6. Otto Cycle, ideal for spark ignition engines

  7. OTTO CYCLE • Process No 1-2 – Isentropic Expansion p– V diagram 1 pVγ = c P1 2 P2 V1 V2

  8. OTTO CYCLE • Process No 1-2 – Reversible Adiabatic or Isentropic Expansion T – S diagram No Heat is added or rejected Q 1-2 = 0 1 T1 T2 2 S1, S2

  9. OTTO CYCLE • Process No 2-3 – Constant volume cooling process p– V diagram Heat is rejected by air getting cooled from T2to T3 2 P2 P3 3 V2

  10. OTTO CYCLE • Process No 2-3 – Constant volume cooling process T – S diagram Heat is rejected by air getting cooled from T2 to T3 T2 2 T3 S3 , S4 S1, S2

  11. OTTO CYCLE • Process No 3-4 – Isentropic Compression p– V diagram No heat is added or rejected 4 pVγ = c P4 P3 3 V4 V3

  12. OTTO CYCLE • Process No 3-4 – Reversible Adiabatic or Isentropic Expansion T – S diagram No Heat is added or rejected Q3-4 = 0 T4 4 T3 3 S3 , S4

  13. OTTO CYCLE • Process No 4-1 – Constant volume heating process p– V diagram Heat is absorbed by air getting heated from T4to T1 1 P1 P4 4 V2

  14. OTTO CYCLE • Process No 4-1 – Constant volume heating process T – S diagram Heat is absorbed by air getting heated from T4 to T1 1 T1 T4 4 S3 , S4 S1, S2

  15. OTTO Cycle • Process 1-2 No heat is added or rejected • Process 2-3 Heat is rejected by air getting cooled from temperature T2to T3 • Process 3-4No heat is added or rejected • Process 4-1 Heat is absorbed by air getting heated from temperature T4to T1

  16. OTTO Cycle • Work Done = Heat absorbed – Heat rejected • Work Done = • Work Done = -

  17. OTTO Cycle

  18. OTTO Cycle • For reversible adiabatic expansion process 1-2 • where expansion ratio =

  19. OTTO Cycle • For reversible adiabatic expansion process 3-4 • where expansion ratio =

  20. OTTO Cycle

  21. OTTO CYCLE • Process No 1-2 – Isentropic Expansion p– V diagram 1 pVγ = c P1 2 P2 V1 V2

  22. OTTO Cycle

  23. Problem 1 • Calculate the air standard efficiency of a four stock Otto cycle engine with the following data Piston diameter (bore) = 13.7 cm Length of stock = 13.0 cm Clearance volume = 14.6 % • Diagram

  24. Solution • Swept Volume • = 1916 cm • Clearance Volume • = 297.7 cm3

  25. Solution • Compression ratio • = 7.85 • Air Standard efficiency • = 56.2%

  26. Problem 2 • In an Otto cycle the compression ratio is 6 . The initial pressure and temperature of the air are 1 bar and 100˚C. the maximum pressure in the cycle is 35 bar. Calculate the parameter at the salient points of the cycle. What is the ratio of heat supplied to heat rejected • How does air standard efficiency of the cycle compares with that of a Carnot cycle working within the same extreme temperature limits? Explain the difference between the two values

  27. Problem 2 • If the engine has a relative efficiency of 50 % determine the fuel consumption per kWh. Assume the fuel used has a calorific value of 42,000 kJ/kg

  28. Problem 3 • An Otto cycle working on air has a compression ratio of 6 and starting condition are 40˚C and 1 bar. The peak pressure is 50 bar. Draw the cycle on p-v and T-S coordinates if compression and expansion follow the law pV1.25 = C. Calculate mean effective pressure and heat added per kg of air.

  29. Problem 4 • An Otto cycle has compression ratio of 8 and initial conditions are 1 bar and 15˚C. Heat added during constant volume process is 1045 kJ/kg. Find : • Maximum cycle temperature • Air standard efficiency • Work done per kg of air • Heat rejected • Take cv = 0.7175 kJ/kg-K and γ = 1.4

  30. Problem 5 • Find out the compression ratio in an Otto for maximum work output • An Otto cycle engine has the following data. Calculate compression ratio, air standard efficiency and specific fuel consumption. Piston diameter = 13.7 Length of stock = 13 cm Clearance volume = 280 cm3 Relative efficiency = 60 % Lower calorific volume of petrol = 41900kJ/kg

More Related