MT 313 IC ENGINES

LECTURE NO: 04(24 Feb, 2014)

Khurramengr014@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

• 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

• Thermal efficiency is also called air standard efficiency ηa

Important Formulas• Swept Volume

• 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

p– V diagram

1

2

V2V1

P2

P1 pVγ = c

OTTO CYCLE• Process No 1-2 – Reversible Adiabatic or Isentropic

Expansion

T – S diagramNo Heat is added or rejected Q 1-2 = 0

1

2

S1, S2

T2

T1

OTTO CYCLE• Process No 2-3 – Constant volume cooling process

p– V diagramHeat is rejected by air getting cooled from T2 to T3

3

2

V2

P2

P3

OTTO CYCLE• Process No 2-3 – Constant volume cooling process

T – S diagram

Heat is rejected by air getting cooled from T2 to T3

2

S1, S2

T2

S3 , S4

T3

OTTO CYCLE• Process No 3-4 – Isentropic Compression

p– V diagramNo heat is added or rejected

4

3

V3V4

P3

P4

pVγ = c

OTTO CYCLE• Process No 3-4 – Reversible Adiabatic or Isentropic Expansion

T – S diagram

No Heat is added or rejected Q 3-4 = 0

S3 , S4

T3

T4

3

4

OTTO CYCLE• Process No 4-1 – Constant volume heating process

p– V diagramHeat is absorbed by air getting heated from T4 to T1

4

1

V2

P1

P4

OTTO CYCLE• Process No 4-1 – Constant volume heating process

T – S diagramHeat is absorbed by air getting heated from T4 to T1

1

4

S1, S2

T4

T1

S3 , S4

OTTO Cycle• Process 1-2 No heat is added or rejected

• Process 2-3 Heat is rejected by air getting cooled from temperature T2 to T3

• Process 3-4 No heat is added or rejected

• Process 4-1 Heat is absorbed by air getting heated from temperature T4 to T1

OTTO Cycle• Work Done = Heat absorbed – Heat rejected

• Work Done =

• Work Done = -

OTTO Cycle

OTTO Cycle• For reversible adiabatic expansion process

1-2

•

• where expansion ratio =

OTTO Cycle• For reversible adiabatic expansion process

3-4

•

• where expansion ratio =

OTTO Cycle

OTTO CYCLE• Process No 1-2 – Isentropic Expansion

p– V diagram

1

2

V2V1

P2

P1 pVγ = c

OTTO Cycle

Problem 1• Calculate the air standard efficiency of a four

stock Otto cycle engine with the following data

Piston diameter (bore)= 13.7 cmLength of stock = 13.0 cmClearance volume = 14.6 %

• Diagram

Solution

• Swept Volume• = 1916 cm• Clearance Volume

• = 297.7 cm3

Solution• Compression ratio• = 7.85• Air Standard efficiency• = 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 = 13.7Length of stock = 13 cmClearance volume = 280 cm3

Relative efficiency = 60 %Lower calorific volume of petrol = 41900kJ/kg