Answer:
a) T_3 = 1539 K , P_3 = 3898 KPa
b) w_net,out = 392.4 KJ/kg
c) n_th = 0.523 - 52.3 %
d) mep = 495 KPa
Explanation:
Given:
- Sate 1: P_1 = 95 KPa, T_1 = 27 C
- q_in = 750 KJ/kg (constant volume process)
- R_air = 0.287 KJ/kgK
- r = 8 (compression ratio)
Find:
a) Pressure and Temperature (P_3 & T_3) at the end of Heat addition process.
Analysis:
Path 1 to 2 (isentropic compression):
T_1 = 300 K ----------> u_1 = 214.07 KJ/kg , v_r1 = 621.2
v_r2 = v_r1 * (V_2 / V_1) = 621.2 *(1 / 8) = 77.65
v_r2 = 77.65 -----------> u_2 = 491.2 KJ/kg , T_2 = 673.1 K
Use Ideal Gas Law for states 1 and 2:
P_2 * v_2 / T_2 = P_1 * v_1 / T_1
P_2 = P_1 * (v_1 /v_2) * ( T_2 / T_1 )
P_2 = 95 KPa * (8)*(673.1/300) = 1705 KPa
Path 2 to 3 (constant volume Heat - Addition):
q_in = (u_3 - u_2 )
u_3 = q_in + u_2
u_3 = 750 +491.2 = 1241.2 KJ/kg
u_3 = 1241.2 KJ/kg ----------> T_3 = 1539 K , v_r3 = 6.588
Use Ideal Gas Law for states 2 and 3:
P_3 * v_3 / T_3 = P_2 * v_2 / T_2
P_3 = P_2 * ( T_3 / T_2 )
P_3 = 1705 KPa*(1539/673.1) = 3898 KPa
Answer: T_3 = 1539 K and P_3 = 3898 KPa
Path 3 to 4 (isentropic expansion):
v_r4 = v_r3 * (V_1 / V_2) = 6.588 *(8) = 52.7
v_r4 = 52.70 -----------> u_4 = 571.69 KJ/kg , T_4 = 774.5 K
Path 4 to 1 (constant volume Heat - Rejection):
q_out = (u_4 - u_1 )
q_out = 571.69 - 214.07
q_out = 357.62 KJ/kg
b) The net work output w_net,out
Analysis:
w_net, out = q_in - q_out
w_net, out = 750 - 357.62
w_net, out = 392.4 KJ/kg
Answer: The net work output w_net,out = 392.4KJ/kg
c) Thermal Efficiency n_th
Analysis:
n_th = w_net, out / q_in
n_th = 392.4 / 750
n_th = 0.523 ~ 52.3%
Answer: = Thermal Efficiency n_th = 52.3%
d) mean effective pressure for the cycle.
Use ideal gas Law at state 1:
v_1 = R*T_1 / P_1 = 0.287 * 300 / 95
v_1 = v_max= 0.906 m^3 /kg
v_min = v_max / r
mep = w_net,out / (v_1-v_2) = w_net,out / v_1*(1 - 1 / r)
mep = 392.4 / 0.906 * (1 - 1/8)
mep = 495.0 KPa
Answer: mean effective pressure for the cycle mep = 495.0 KPa
