LESSON - 45 AIR CYCLES- DUAL CYCLE, AND THE COMPARISON OF OTTO, DIESEL AND DUAL CYCLES

45.1. DUAL CYCLE OR LIMITED PRESSURE AIR CYCLE

The combustion process in a spark ignition engine does not occur exactly at constant volume, nor does the combustion process in an actual compression ignition engine occur exactly at constant pressure, therefore another idealized cycle known as Dual cycle has been developed that more closely approximate the actual spark-ignition and compression-ignition engines. The p-v and T-s diagrams of Dual cycle are shown in Fig. 45.1. In this cycle, part of heat addition occurs at constant volume while the rest is at constant pressure. The dual cycle is also called mixed or limited pressure cycle.

Fig. 45.1. P-v and T-s diagram for Dual cycle

The process description of Dual cycle is as below:

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

Consider 1 kg of air

Heat addition at constant volume in process 2-3

= Cv (T3 − T2)

Heat addition at constant pressure in process 3-4

= Cp (T4 − T3)

Total heat added in process 2-3  and 3-4,

qin = [Cp (T3 −T2) − Cv (T4 – T1)]

Heat rejection at constant volume in process 5-1,

qout  = Cv (T5 – T1)

Net work done during cycle,

wnet  = qin  − qout

  = [Cv (T3 − T2) + Cp (T4 − T3)] − Cv (T5 − T1)

Air Standard Efficiency:

 ηth  =

                   

                                    …………….(45.1)

This thermal efficiency can be converted in terms of measurable quantities as it was done for Otto and Diesel cycle.

The Compression ratio (rv) is given by

                                                                                       ……………………(45.2)

Pressure ratio (rp) is given by

      and

Cutoff ratio (rc) is given by


For isentropic compression (1-2)

For constant volume process (2-3)

For constant pressure process (3-4)

For isentropic expansion process (4-5)

Put T2, T3, T4, and T5in equation (45.1)

Thermal efficiency of dual cycle is reduces to

Mean effective pressure, pm:

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

                          bar    

Now, express mean effective pressure in easily measurable quantities as given below:

                                             .......................................... (45.3)   

But from equation (45.2), we have  

Put T2, T3, T4, T5 and (v1-v2) in equation (45.3)

Therefore,   pm

  

  where,  rv = compression ratio = v1/v2

rc = cut-off ratio  = v4/v3

rp = explosion ratio =  p3/p2

Important points

1. Increase in compression ratio (CR) increases Efficiency and Mean effective pressure.

As efficiency and mean effective pressure increase in both Otto and Diesel cycles with increase in CR, they have to increase in dual cycle as well. This can be proved on the same lines as it was done for Otto and Diesel cycles.

2. For dual cycle, efficiency depends not only on compression ratio but also on how much heat is added at constant volume and constant pressure. For same amount of heat added, efficiency will increase if its share in constant volume heat addition is increased and vice versa.

Proof:

To see the above effect, compare cycle 1-2-3-4-5-1, Cycle 1-2-3’-4’-5’-1 and Cycle 1-2-3”-4”-5”-1 of  Fig. 44.2.

So we can say that if portion of heat added at constant volume is increased (total heat added remaining same), the efficiency increases & vice-versa.

For diesel cycle, we noticed that with increase in cut-off ratio the efficiency decreases. But we need some time for solid injection of diesel oil into cylinder. In dual cycle, injection of fuel starts little before piston reaches TDC. So cut-off ratio of dual cycle is less than that of diesel cycle so its efficiency will be higher than diesel cycle for same compression ratio.  For rp =1, efficiency becomes equal to that of diesel cycle. For rc =1, efficiency becomes equal to that of Otto cycle.

3. Increase in heat input either due to increase in rp or rc increases mean effective pressure ‘pm’.

This can be seen by drawing comparative p-v diagrams of dual cycle with different heat inputs.

  • 1-2-3-4-5-1 Dual cycle,

  • 1-2-3’-4’-5’-1 Dual cycle with same heat input but higher rpand lower rc ,

  • 1-2-3”-4”-5”-1 Dual cycle with same heat input but lower rpand higher rc

Fig. 45.2. Effect of ‘CR’, ‘rp’& ‘rc’ on mean effective pressure & thermal efficiency.

45.2. COMPARISON OF OTTO, DIESEL & DUAL CYCLES ALL HAVING SAME COMPRESSION RATIO (CR) AND HEAT INPUT

       The Otto cycle is most efficient cycle followed by dual & diesel cycles,all having same CR and heat input.

Proof:

To proof the above statement, comparethe Otto cycle 1-2-3-4, Dual cycle 1-2-2’-3’-4’ and Diesel cycle1-2-3”-4”in Fig. 45.3.

 

 Therefore, Otto cycle is most efficient cycle followed by dual & diesel cycles.

            • 1-2-3-4 Otto cycle ,

            • 1-2-3”-4” Diesel cycle,

            • 1-2-2’-3’-4’   Dual cycle

Fig. 45.3.  Otto, Diesel & Dual cycles all having same CR and heat input

Problem 45.1: An engine working on dual cycle has bore of 20 cm & stroke of 40 cm. Compression ratio is 14.5 & pressure ratio at constant volume heat addition process is 1.5. The constant volume heat addition cut-off takes place at 4.9 % of the stroke. Assume  = 1.4, find air standard efficiency.

Solution:

Given:   Bore of the cylinder, d = 20 cm = 0.2 m;

              Stroke of the cylinder, L = 40 cm = 0.4 m;

Stroke volume, Vs = (V1-V2) =   x (d)2 x L 

                                =    x (20)2 x 40 = 12571.4 cm3

              Compression ratio, rv =  = 14.5;

              Pressure ratio, rp =  =  1.5

              (V4-V3) = 4.9% of the stroke volume

                              = 0.049 Vs                                       

               γ  = 1.4

Determine the air standard efficiency,  ;

METHOD-I

       Formula:   

Finding unknown rc ;

               

Finding unknown V3  and V4 ;

= 14.5

= 14.5

1 = 14.5

= 14.5- 1 = 13.5

 V3 = = = 930 cm3

                     (V4-V3) =  0.049 Vs                                       

            or       V4 = V3 + 0.049 Vs

                            = V3 + 0.049 x 12571.4 = 1545 cm3 ;

                      rc = =  = 1.66

Answer:   = 1 −

                          = 1 −
     

                          = 0.627   or 62.7%

  METHOD-II

Formula:  Air standard efficiency, 

                         

           or                          

            [Hint: Find all temperatures ‘T1, T2, T3, T4 and T5’ as a function of ‘T’ then substitute them in the above equation and T will get cancelled].

Finding unknown, T1, T2, T3, T4 and T5  as a function of T;

                              Assume T3 = T ;   

                             For constant pressure process (3-4), we have

                            or      

                            or       

Finding unknown V3  and V4 ;

                                       = 14.5

                                         = 14.5

                                        1 + = 14.5

                                      = 14.5- 1 = 13.5

                                        V3 = 

                                              = = 930 cm3

                                         Also  (V4-V3) =  0.049 Vs                                       

                               or       V4 = V3 + 0.049 Vs

                                               = V3 + 0.049 x 12571.4 = 1545 cm3

                                                      

                 For isentropic expansion (4-5), we have

Finding unknown, V5 ;  We have, 

V5 = V1 = V3 + Vs = 930 + 12571.4 = 13501.4 cm3

 

    = 1.66 × T × (0.1144)0.41 = 0.68244 T 

For constant volume process (2-3), we have

                                 

        or   

                              

For isentropic compression (1-2), we have

 

     

T1 = 

     =    = 0.23 T

                 Answer:   

              = 1−

              = 1−

                                                    = 0.639    or 63.9%

Last modified: Wednesday, 11 September 2013, 6:42 AM