Otto Cycle, ideal for spark ignition engines
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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 ππ = ππ π • 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 ππππ π·πππ ηπ‘β = π»πππ‘ ππ’ππππππ π»πππ‘ ππ’ππππππ − π»πππ‘ π πππππ‘ππ = π»πππ‘ ππ’ππππππ ππ − ππ = ππ ππ − ππ = ππ • Thermal efficiency is also called air standard efficiency ηa Important Formulas • Swept Volume π 2 ππ = π πΏ 4 • Clearance Volume π = ππ + ππ • Compression Ratio ππ + ππ π= ππ • Clearance ratio ππ π= ππ 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= ππ ππ£ 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= ππ ππ£ 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 π1−2 = 0 • Process 2-3 Heat is rejected by air getting cooled from temperature T2 to T3 π1−2 = πππ£ π2 − π3 • Process 3-4 No heat is added or rejected π3−4 = 0 • Process 4-1 Heat is absorbed by air getting heated from temperature T4 to T1 π4−1 = πππ£ π1 − π4 OTTO Cycle • Work Done = Heat absorbed – Heat rejected • Work Done = π4−1 − π1−2 • Work Done = πππ£ π1 − π4 - πππ£ π2 − π3 •η= •η= ππππ π·πππ π»πππ‘ π΄ππ πππππ πππ£ π1 −π4 − πππ£ π2 −π3 πππ£ π1 −π4 OTTO Cycle •η= πππ£ π1 −π4 − πππ£ π2 −π3 πππ£ π1 −π4 • η=1− • η=1− • η=1− πππ£ πππ£ π2 −π3 π1 −π4 π2 −π3 π1 −π4 π2 π3 π3 π1 π4 π4 −1 −1 OTTO Cycle • For reversible adiabatic expansion process 1-2 • π2 π1 = π£1 γ−1 [ ] π£2 • where expansion ratio = π£2 π£1 OTTO Cycle • For reversible adiabatic expansion process 3-4 • π3 π4 = π£4 γ−1 [ ] π£3 • where expansion ratio = π£3 π£4 • π£3 π£4 = OTTO Cycle π£2 π£1 • η=1− • η=1 − • η=1− π3 π4 π2 π1 1 π γ−1 OTTO CYCLE • Process No 1-2 – Isentropic Expansion 1 pVγ = c P1 2 P2 γ= ππ ππ£ V1 V2 p– V diagram OTTO Cycle 2 πππ 1 • π1−2 = • ππ γ = πΆπππ π‘πππ‘ • π1−2 = πΆ • π1−2 = • π1−2 = 2 ππ 1 πγ π1 π1 −π2 π2 γ−1 π (π1 −π2 ) 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 • ππ = • = π 4 π 4 π·2πΏ 13.72 ∗ 13 • = 1916 cm • Clearance Volume • ππ = • π 4 ∗ ππ = 297.7 cm3 Solution • Compression ratio • ππ = ππ +ππ ππ • = 7.85 • Air Standard efficiency • η=1− • 1 π 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 100ΛC. 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 40ΛC 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 15ΛC. 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
