An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 105 kPa and 27°C. Assume constant specific heats. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

a) T_3 = 1724.8 K

b) n_th = 56.3 %

c) MEP = 746.8 KPa      

Explanation:

Given:

- Pressure at state 1 P_1 = 105 KPa

- Temperature at state 1 T_1 = 300 K

- Compression ratio r = 16

- Cutoff ratio n = 2

-cp = 1.005 kJ/kg·K,

-cv = 0.718 kJ/kg·K,

-R = 0.287 kJ/kg·K,

-k = 1.4.

Find:

- The temperature at the start of Heat addition process

- Thermal Efficiency n_th

- Mean Effective Pressure

Solution:

Process 1 - 2 : Isentropic compression

- Using the Temperature at state 1, and air property tables we have:

 T_1 = 300 K      > u_1 = 214.07 KJ/kg , v_r1 = 621.2

- Using isentropic compression ratio:

                                v_1*v_r2 = v_2*v_r1

                                v_r2 = ( v_2 / v_1 ) * v_r1

                                v_r2 = (1 / r) * v_r1 = (1 / 16) * 621.2

                                v_r2 = 38.825

- Evaluating state 2 using air property tables:

v_r2 = 38.825   > T_2 = 862.4 K , h_2 = 890.9 KJ/kg

Process 2 - 3 : Constant pressure Heat Addition ( P_2 = P_3 )

- Using ideal gas relation as follows:

                                  P_3*v_3 / T_3 = P_2*v_2 / T_2

                                   T_3 = ( v_3 / v_2 ) * T_2 = n * T_2

                                   T_3 = 2*T_2 = 2*862.4

                                    T_3 = 1724.8 K

- Evaluating state 3 using air property tables:

T_3 = 1724.8 K   > v_r3 = 4.546 K , h_3 = 1910.6 KJ/kg

- Evaluate the amount of heat added q_in:

                                     q_in = h_3 - h_2

                                     q_in = 1910.6 - 890.9

                                      q_in = 1019.7 KJ/kg

Process 3 - 4 : Isentropic expansion

- Using isentropic compression ratio:

                                v_3*v_r4 = v_4*v_r3

                                v_r4 = ( v_4 / v_3 ) * v_r3

                                v_r4 = (r/2) * v_r3 = (16 / 2) * 4.546

                                v_r4 = 36.37

- Evaluating state 4 using air property tables:

v_r4 = 36.37   >  u_4 = 659.7 KJ/kg

Process 4 - 1 : Constant Volume Heat Rejection ( v_4 = v_1 )

- Evaluate the amount of heat rejected q_out:

                                   q_out = u_4 - u_1

                                   q_out = 659.7 - 214.07

                                   q_out = 445.63 KJ/kg

- Evaluate the thermal efficiency n_th:

                                   n_th = 1 - q_out / q_in

                                   n_th = 1 - 445.63/1019.7

                                   n_th = 56.3 %

- Apply energy balance for entire system or cycle:

                              w_net, out = q_in - q_out

                              w_net, out = 1019.7 - 445.63

                              w_net, out = 574.07 KJ/kg

- Compute the maximum volume @ state 1 or 4, v_1 or v_4. Use ideal gas law:

                               v_1 = R*T_1 / P_1

                               v_1 = 0.287*300 / 105

                               v_1 = 0.82 m^3 / kg = v_max

- Compute v_min @state 2 using compression ratio:

                               v_min = v_2 = v_max / r

                               v_min = 0.82 / 16

                               v_min = 0.05125 m^3 / kg

- The mean effective pressure can now be calculated from definition:

                               MEP = w_net,out / ( v_1 - v_2)

                               MEP = 574.07 / ( 0.82 - 0.05125)      

                               MEP = 746.8 KPa                

answer: 500 wph.

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