Answer:
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
