LESSON - 44 AIR CYCLE- DIESEL CYCLE

 44.1. DIESEL CYCLE OR CONSTANT VOLUME AIR CYCLE: THE IDEAL CYCLE FOR COMPRESSION-IGNITION IC ENGINE

The Diesel Cycle is the ideal cycle that closely approximates a compression-ignition IC engine.The Diesel Cyclelike Otto cycle consists of four reversible processes.Since the compression-ignition engineis very similar to the spark-Ignition engine and differs mainly in the combustion process therefore  out of the four processes, only one process (i.e. heat addition process)of Otto and the Diesel cycles is different and the remaining three processes (i.e. process 1-2 is isentropic compression, 3-4 is isentropic expansion, and 4-1 is constant-volume heat rejection) are same. The similarity between the two cycles is also apparent from the P-v and T-s diagrams of the Diesel cycle, shown in Fig. 44.1.

Fig. 44.1. P-v and T-s diagram for Ideal Diesel cycle

As in actual diesel engines,the fuel injection and burning process starts when the piston approaches TDC and continues during the first part of the power stroke. During injection and burning of fuel, the volume of mixture changes itself and pressure remains constant.The heat addition in ideal diesel cycle from an external source is not taken as constant volume and it is approximated as a constant-pressure heat-addition process.

From the above discussion it is clear that the Diesel cycle consists of four reversible processes as shown in Fig 44.1 and given below:

Process Description

1-2       Isentropic compression with compression ratio rv=  v₁/v₂

2-3       Heat addition is at constant pressure. It starts at point 2 and cuts off at point 3.

3-4       Isentropic expansion till piston reaches B.D.C.

4-1       Heat rejection of at constant volume.

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

Consider 1 Kg of air

Heat addition at constant pressure in process 2-3,

q_in = C_p (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_p (T₃-T₂) – C_v (T₄ – T₁)

Air Standard Efficiency :

 

                                                             ..……..(44.1)

Now convert above equation into easily measurable quantities.

Based on heat addition start at point 2 and heat addition cut-off at 3, we now define a new quantity, the cut-off ratio r_c, as the ratio of the cylinder volumes after and before the combustion process: 

Compression ratio (r_v) is given by

                                                                                 …………..(44.2)

 Using above definition and the isentropic ideal-gas relations for processes 1-2 and 3-4, we now express thermal efficiency in easily measurable quantities.

For adiabatic compression 1-2

                                                                        ……………………………………..(44.3)

For constant pressure heating   (2-3)

or            

or             T₃ = r_cT₂                                                                                    …………………………………     (44.4)

For adiabatic expansion  (3-4)

 

or                  

or                 

or            

Using equation (44.4),       T₄ = r_c T₂                                                                   ………………………..(44.5)

Substituting (44.3), (44.4) and (44.5) in (44.1), we obtain

                       ………………………………..(44.6)

Mean effective pressure, p_m:

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

                        bar             

                                                                                               …………..(44.7)

But from equation (44.2), we have

Put values of (v₁-v₂) in equation (44.7), we have

Important Points

1. Efficiency of diesel cycle will always be less than that of Otto cycle for a given compression ratio

Proof:

                              ……………………………..(44.8)         

                                                   ……………………………..(44.9)

As in Diesel engine v₃ > v₂  so  r_c> 1,

Therefore, value of  is always greater than 1.

Consequently, from equations (44.8) and 44.9) η_th(Diesel) will always less than that of η_th(Auto) for given compression ratio.

However, in practice the diesel engine have always higher compression ratio (r_v) than petrol engine based on Otto cycle because in diesel engine only air and not air-fuel mixture is compressed.

2. Mean Effective Pressure increases and efficiency decreases with increase in heat input (r_c) with compression ratio being same.

Proof:

To see the effect of ‘r_c’ on mean effective pressure and efficiency, compare cycle 1-2-3-4-1 and cycle 1-2-3’-4’-1 of  Fig. 44.2.

a)  Effect of ‘r_c’ on mean effective pressure:

       Thus the mean effective pressure increases with increase in heat input, r_c.This is also shown in Fig. 44.4.

b)  Effect of ‘r_c’ on efficiency:

     With the increase in heat input (r_c), the increase in area (3-3’-4-4’) corresponding to incremental work is lesser than the increase in area (C-3-3’-D) corresponding to incremental heat input.

     Thus the efficiency decreases with increase in heat input (r_c).This is also shown in Fig. 44.3 and Fig 44.4.

3. Mean Effective Pressure & Efficiency increase with increase in compression ratio (CR) with Heat Input being same.

 Proof:

To see the effect of ‘CR’ on mean effective pressure and efficiency, compare cycle 1-2-3-4-1 and cycle 1-2’-3”-4”-1 of  Fig. 44.2.

         So higher CR results in higher Mean Effective Pressure and higher efficiency.

         Variation of thermal efficiency and mean effective pressure with variation of compression ratiois shown in Fig. 44.4.

 

• 1-2-3-4-1 Diesel cycle.

• 1-2’-3”-4”-1 Diesel cycle with higher CR,

• 1-2-3’-4’-1 Diesel cycle with more heat input

Fig. 44.2. Effect of  compression ratio and heat input  on mean effective pressure&Thermal efficiency

Fig. 44.3. Thermal efficiency vs compression ratio of Ideal Diesel cycle

 

 Fig. 44.4. Thermal efficiency / p_m vs compression ratio/ heat addition

Problem 44.1: A Diesel air standard cycle has a compression ratio of 15. The lowest and highest temperature of the cycle are 27^°C and 1627 °C respectively. The pressure at the beginning of compression is 1 bar. Calculate (a) the pressures and temperatures at the salient points of the cycle, (b) the heat supplied; (c) the heat rejected; (d) net work output, (e) efficiency; (f) clearance, (g) cut-off and (h) mep.

Solution: 

Given:   Compression ratio, r_v = = 15;

Lowest temperature of the cycle, t₁ = 27^°C            or     T₁ = 27+273 = 300 K;

Highest temperature of the cycle, t₃ = 1627 °C     or     T₃= 1627+273 = 1900 K;

               Pressure at the beginning of compression, p₁ = 1 bar.

 

(a) Determine the pressures and temperatures at the salient points of the cycle:

       For isentropic process 1-2, we have

T₂ = T₁ r_v^γ-1  = 300 x (15) ^0.4 = 886 K or 613°C

                                

p₂ = P₁ r^γ= 1 x 15^1.4 = 44.3 bar

For constant pressure process 2-3, we have

         p_3 = p₂ = 44.3 bar

 For isentropic process 3-4, we have

         

or      

or       

                  T₄ = T₃  r_v^γ-1  = 1900  x  (0.143)^0.4 = 873 K   or   600 ^oC

                               

Finding unknown,  ;

          = 0.143

Therefore,  

p₄ = 44.3 x 0.143^1.4 = 2.91 bar

(b)   Determine the heat supplied q_in ;

 Formula:  Heat supplied, q_in = Cp (T₃ - T₂)

       Answer:   Heat supplied, q_in = Cp (T₃ - T₂) = 1.005 (1900 - 886) = 1019.07 kJ/kg/cycle

(c)    Determine the heat rejected, q_out ;

Formula: Heat rejected = q_out = Cv (T₄ - T₁)

       Answer:   Heat rejected = q_out = Cv (T₄ - T₁) = 0.718 (873-300) = 411.41 kJ/kg/cycle

(d)   Determine the net work output, w ;

Formula: Net work output, w = q_in – q_out

       Answer:   Net work output, w = q_in – q_out = 1019.07 - 411.41 = 607.66 kJ/kg/cycle

(e)    Determine thermal efficiency,  ;

Formula: Efficiency,  

       Answer:   Efficiency,  = 59.63%

(f)    Determine the clearance,

Formula:    Clearance =  

Finding unknown, v₁ and v₂ :

v₁ =  =  = 0.861 m³/kg

r_v = = 15   

or        v₂ =  =  = 0.0574 m³/kg

      Answer:   Clearance =    = 0.07143 of stroke

(g)   Determine the cut-off

Formula: cut-off =  

 Finding unknown, v₃ :

         For constant pressure process 2-3, we have

or     = 2.145

                     v₃ = 2.145 v₂ = 2.145 × 0.0574= 0.123 m³/kg

    Answer:   cut-off =  = 0.0818 of stroke

(h) Determine mean effective pressure, m.e.p. or p_m ;

     Formula:     m.e.p.,  p_m =  

     Answer:      m.e.p.,   p_m =    = 756 kN/m²