Report

MT 313 IC ENGINES LECTURE NO: 04 (24 Feb, 2014) Khurram [email protected] 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 • 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 Air Standard Cycle • Thermal efficiency ηℎ = − = − = − = • Thermal efficiency is also called air standard efficiency ηa Important Formulas • Swept Volume 2 = 4 • Clearance Volume = + • Compression Ratio + = • Clearance ratio = Ideal cycles are simplified Otto Cycle, ideal for spark ignition engines OTTO CYCLE • Process No 1-2 – Isentropic Expansion 1 pVγ = c P1 2 P2 γ= V1 V2 p– V diagram • OTTO CYCLE Process No 1-2 – Reversible Adiabatic or Isentropic Expansion 1 T1 T2 2 T – S diagram No Heat is added or rejected Q 1-2 = 0 S1, S2 OTTO CYCLE • Process No 2-3 – Constant volume cooling process 2 P2 P3 3 V2 p– V diagram Heat is rejected by air getting cooled from T2 to T3 • OTTO CYCLE Process No 2-3 – Constant volume cooling process T2 2 T3 S3 , S4 T – S diagram S1, S2 Heat is rejected by air getting cooled from T2 to T3 OTTO CYCLE • Process No 3-4 – Isentropic Compression 4 pVγ = c P4 P3 γ= 3 V4 p– V diagram No heat is added or rejected V3 OTTO CYCLE • Process No 3-4 – Reversible Adiabatic or Isentropic Expansion T4 4 T3 3 S3 , S4 T – S diagram No Heat is added or rejected Q 3-4 = 0 OTTO CYCLE • Process No 4-1 – Constant volume heating process 1 P1 P4 4 V2 p– V diagram Heat is absorbed by air getting heated from T4 to T1 • OTTO CYCLE Process No 4-1 – Constant volume heating process 1 T1 T4 4 S3 , S4 S1, S2 T – S diagram Heat is absorbed by air getting heated from T4 to T1 OTTO Cycle • Process 1-2 No heat is added or rejected 1−2 = 0 • Process 2-3 Heat is rejected by air getting cooled from temperature T2 to T3 1−2 = 2 − 3 • Process 3-4 No heat is added or rejected 3−4 = 0 • Process 4-1 Heat is absorbed by air getting heated from temperature T4 to T1 4−1 = 1 − 4 OTTO Cycle • Work Done = Heat absorbed – Heat rejected • Work Done = 4−1 − 1−2 • Work Done = 1 − 4 - 2 − 3 •η= •η= 1 −4 − 2 −3 1 −4 OTTO Cycle •η= 1 −4 − 2 −3 1 −4 • η=1− • η=1− • η=1− 2 −3 1 −4 2 −3 1 −4 2 3 3 1 4 4 −1 −1 OTTO Cycle • For reversible adiabatic expansion process 1-2 • 2 1 = 1 γ−1 [ ] 2 • where expansion ratio = 2 1 OTTO Cycle • For reversible adiabatic expansion process 3-4 • 3 4 = 4 γ−1 [ ] 3 • where expansion ratio = 3 4 • 3 4 = OTTO Cycle 2 1 • η=1− • η=1 − • η=1− 3 4 2 1 1 γ−1 OTTO CYCLE • Process No 1-2 – Isentropic Expansion 1 pVγ = c P1 2 P2 γ= V1 V2 p– V diagram OTTO Cycle 2 1 • 1−2 = • γ = • 1−2 = • 1−2 = • 1−2 = 2 1 γ 1 1 −2 2 γ−1 (1 −2 ) γ−1 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 Solution • Swept Volume • = • = 4 4 2 13.72 ∗ 13 • = 1916 cm • Clearance Volume • = • 4 ∗ = 297.7 cm3 Solution • Compression ratio • = + • = 7.85 • Air Standard efficiency • η=1− • 1 γ−1 = 56.2% 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 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 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. 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 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