Respuesta :
Answer:
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