LESSON - 43 AIR CYCLE - OTTO CYCLE

 

43.1. OTTO CYCLE OR CONSTANT VOLUME AIR CYCLE: THE IDEAL CYCLE FOR SPARK-IGNITION IC ENGINE

The Otto Cycle is the Ideal Cycle that closely approximates a spark-ignition IC engine. The close approximation of Otto cycle to the actual spark-ignition four stroke IC engine can be better understood by details of each stroke for both Otto cycle and the actual spark-ignition IC engine given in the following table:

Actual spark-ignition IC engine	Otto cycle
Fig 43.1. The execution of actual cycle in an actual four-stroke spark-ignition IC engine together with a P-V diagram.	Fig 43.2. The execution of Otto cycle in a piston-cylinder arrangement of petrol engine along with p-v and T-s diagrams.
Initially, both the intake and the exhaust valves are closed, and the piston is at its lowest position (BDC).	Initially,the piston is at its lowest position (BDC) and the cylinder is full of air (represent as point 1 on p-v and T-s diagram).
Compression: During the compression stroke, the piston moves upward from BDC position to TDC position, compressing the air–fuel mixture which has entered during its previous intake stroke.	Isentropic Compression: The piston moves from BDC position to TDC position,compression of the air occurs isentropically ending at point 2 and thus raising the pressure and temperature of the air (represent as process 1-2 on p-v and T-s diagram). This process is taken as adiabatic because the time taken during compression is quite small due to high speed of engines.
Combustion: Shortly before the piston reaches its highest position (TDC) and heat is added by igniting the fuel air mixture by firing the spark plug thus increasing the pressure and temperature of the combustion product. As the combustion occurs very rapidly and movement of piston during combustion is very small therefore the volume will remain almost constant during heat addition due to combustion.	Constant-volume heat addition: The piston is momentarily at rest at TDC position and heat is added instantaneously in the air of cylinder from external reservoir for a short period so that both pressure and temperature rise to highest at a constant volume (represent as process 2-3 on p-v and T-s diagram).
Expansion (power stroke): The product of combustion at high pressure forces the piston down from TDC to BDC which in turn forces the crankshaft to rotate, producing a useful work output. It is also called power stroke.At the end of this stroke, the piston is at its lowest position,and the cylinder is filled with combustion products.	Isentropic Expansion (power stroke): The air at high pressure and temperature expands isentropically causing the piston to move from TDC position to BDC position (represent as process 3-4 on p-v and T-s diagram).It is also called power stroke because work is done by the system. Like compression this process is also adiabatic as the time taken during expansion is quite small due to high speed of engines.
Initial exhaust: The exhaust valve opens shortly before the end of the expansion stroke (i.e. shortly before piston reaches the BDC position) and consequently most of the exhaust gases leave the cylinder, thus rejecting heat to the atmosphere at constant volume.	Constant-volume heat rejection:The piston is momentarily at rest at BDC position and heat is rejected instantaneously from the air of cylinder to external reservoir for a short period so that both pressure and temperature decreases at a constant volume to its initial lowest point 1 (represent as process 4-1 on p-v and T-s diagram).
Remaining exhaust: The piston moves upward from BDC position to TDC position, removing the remaining exhaust gases through the exhaust valve. It is called the exhaust stroke. Intake: The piston finally moves from TDC position to BDC position, drawing in fresh air–fuel mixture through the intake valve (the intake stroke). Notice that the pressure in the cylinder is slightly above the atmospheric value during the exhaust stroke and slightly below during the intake stroke.	There are no exhaust and intake processes in the Otto cycle as the engine is imagined to operate in a closed cycle so that working fluid is restored to its initial state at the end of each cycle and the mass of air in the cycle remains fixedthroughout the entire cycle

From the above details it is clear that the Otto cycleconsists of four reversible processes as given below:

1-2     Isentropic (adiabatic) compression

2-3    Constant volume heating.

3-4    Isentropic (adiabatic) expansion

4-1    Constant volume cooling

Air Standard Efficiency and Mean effective pressure of Otto cycle can be calculated as follows:

Consider mass of air = 1 kg

Heat addition at constant volume in process 2-3,

q_in= C_v (T₃ –T₂)

Heat rejection at constant volume in process 4-1,

q_out=C_v (T₄ –T₁)

Net work done during cycle,

w_net  = q_in  - q_out

         = C_v (T₃ –T₂) - C_v (T₄ –T₁) 

       

              

       

Air Standard Efficiency:

η_th  = 

                                                      ………….   (43.1)

Now convert above equation into easily measurable quantities.

For isentropic compression (1-2) and isentropic expansion (3-4)

     and                                                                        ………….   (43.2)

But compression ratio (r_v) is given by

r_v  =                                                                            …………..(43.3)

From equations (43.2) and (43.3), we have

                                                                                                                           …………..(43.4)

or           T₂ = T₁ r_v^γ-1            and          T₃ = T₄r_v^γ-1    , put in equation (43.1)

               η_th =     =  

Mean effective pressure, p_m:

Generally, it is defined as the ratio of the net workdone to the displacement/swept volume of the piston.

p_m  =                             bar    

                                                                                                     …………..(43.5)

But from equation (43.3), we have    

Put values of T₂, T₃ and (v₁-v₂) in equation (43.5), we have     

                                                                                   …………..(43.6)

For constant volume process 2-3,                                                  …………..(43.7)

But pressure ratio (r_p) is given by

                                                                                                                               …………..(43.8)

From equations (43.7) and (43.8)

   

From equation (43.4), we have      

From above two relations, we have  

Put T₄ in equation (43.6), we have

p_m  =         bar           

Important points

1. Ideal thermal efficiency is function of compression ratio (r_v) alone. The efficiency goes on increasing with increase in r_v as shown in Fig. 43.3.

 

Fig. 43.3. η_th vs r_v(Compression ratio) of Ideal Otto cycle

Proof:  

The Ideal thermal efficiency goes on increasing with increase in compression ratio (CR), r_v:

To see the effect of ‘CR’  on thermal efficiency, keep heat addition ‘Q₁’ constant and increase ‘CR’. Compare cycle 1-2-3-4-1 and cycle 1-2’-3’-4’-1 in Fig. 43.4.

 Therefore, cycle with higher CR gives higher efficiency.

2. Ideal thermal efficiency is not function of q_in,  T₃,  or  r_p= p₃/p₂ (called pressure ratio). 

Proof:

   The Ideal thermal efficiency does not change with increase in q_in,  T₃   or   r_p= p₃/p₂ (called pressure ratio):

   To see the effect of heat addition ‘q_in’on thermal efficiency_,keep CR constant, compare cycle 1-2-3-4-1 and cycle 1-2-3”-4”-1 in T-s diagram of Fig. 43.4.

 

   Therefore efficiency remains same irrespective of change in heat addition as far as CR is constant.

3.  Mean effective pressure (p_m) or Specific work output (Net work output / mass flow of working fluid) increases with increase in heat input q_in,  r_p= p₃/p₂ (called pressure ratio), or T₃.  

Proof:

To see the effect of heat addition on mean effective pressure,keep ‘CR’constant and change heat input, compare cycle 1-2-3-4-1 and cycle 1-2-3”-4”-1 of Fig. 43.4.

         Therefore for same CR, increase in heat addition results in increase in p_m. 

• Otto cycle 1-2-3-4-1 ;

• Otto cycle with more heat input & same CR (1-2-3”-4”-1);

• Otto cycle with same heat input & higher CR (1-2’-3’-4’-1)

Fig. 43.4. Effect of  CR and q_in  on p_m & η_th

4.  Mean effective pressure increases with increase in compression ratio ‘CR’:

Proof:

To see the effect of ‘CR’ on mean effective pressure, keep same heat  input and change ‘CR’, compare cycle 1-2-3-4-1 and cycle 1-2-3’-4’-1 of  Fig. 43.4.

        Thus the mean effective pressure increases with increase in compression ratio.

5. The ideal thermal efficiency of Otto cycle is always less than that of Carnot cycle working between same maximum and minimum temperatures.

    Proof: Say Otto cycle and Carnot cycle work between T₃ maximum and T₁ minimum temperature as shown in Fig. 43.5.

Fig 43.5. T-s diagram of the Otto and Carnot cycle

For Otto cycle from eqn. (43.1)

Also from eqn. (43.2)

Therefore ,          

or                         

Now carnot efficiency based on maximum and  minimum temperature is  

But efficiency of Otto cycle  is 

Now compare thermal efficiencies of Otto and Carnot cycle, Since T₄> T₁, the Otto cycle efficiency is less than Carnot efficiency. 

The reason for this is that for Otto cycle heat addition & rejection is not at constant temperature.

Problem 43.1:  Calculate the thermal efficiency of an engine working on the otto cycle. The bore and stroke of the cylinder are 17 & 30 cm. Clearance volume is 0.002025 m³. γ  = 1.4

Solution:

Given: Bore of the cylinder = 17 cm = 0.17 m;

            Stroke of the cylinder, L = 30 cm = 0.3 m;

            Clearance volume = 0.002025 m³

            γ  = 1.4

Determine the thermal efficiency of an engine working on the otto cycle, η_th:

Formula:   

Finding unknown, r_v ;

                      The compression ratio, r_v = 

Finding unknown, cylinder volume;

                        Swept volume, V_s =  (Cylinder volume  - Clearance volume)

                                                        =   = 0.006810 m³

                         Cylinder volume = (V_s + Clearance volume) = 0.002025 + 0.006810

                                                                                                    = 0.008835 m³

                            r_v =   = 4.35

Answer:       η_th =  =   0.4445