The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are95 kPa and 300 K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and thetemperature is 2150 K. Allowing for variable specific heats, Determine: (a) the compression ratio, r. (b) the cutoff ratio, rc. (c) the thermal efficiency of the cycle,? th (%). (d) the mean effective pressure, mep (kPa).

r = 22.01

r_c = 2.081

mep = 853 KPa

Explanation:

Given:

- State 1: P_1 = 95 KPa, T_1 = 300 K

- State 2 and 3: P_2 = P_3 = 7200 KPa, T_3 = 2150 K

- k = 1.4

- c_p = 1.005 KJ/kgK, c_v = 0.718 KJ/kgK

Find:

a) compression ratio, r

b) cutoff ratio, rc

c) mean effective pressure, mep

Solution:

- Compute T_2:

T_2 = T_1 * (P_2 / P_1) ^ (k - 1) / k

T_2 = 300 * (7200/95) ^0.4/1.4

T_2 = 1033.1 K

- Compute (V_4 / V_3):

(V_4 / V_3) = (V_1 / V_2) * (V_2 / V_3)

(V_4 / V_3) = (T_2 / T_1) ^ (1/k-1) * (T_2 / T_3)

(V_4 / V_3) = (1033.1 / 300) ^ (2.5) * (1033.1 / 2150)

(V_4 / V_3) = 10.57

- Compute T_4:

T_4 = T_3 * (V_3 / V_4) ^ (k - 1)

T_4 = 2150 * (1/10.57) ^0.4

T_4 = 837 K

- Compression ratio:

r = (V_1 / V_2) = (T_2 / T_1) ^ (1/k-1)

r = (1033.1/300) ^2.5 = 22.01

- Cut - Off ratio:

r_c = (V_3 / V_2) = (T_3 / T_2)

r_c = (2150/1033.1) = 2.081

- mep

mep = w_net / (V_1 - V_2)

mep = (c_p * (T_3 - T_2) - c_v * (T_4 - T_1)) / (V_1 - V_2)

V_1 = R*T_1 / P_1 = 0.2866 * (300/95) = 0.90505 m^3 / kg

V_2 = 0.90505 / 22.01 = 0.04112 m^3 / kg

mep = (1.005 * (2150 - 1033.1) - 0.718 * (837-300)) / (0.905-0.04112)

mep = 853 KPa