Respuesta :
Answer:
a) 23.19
b) 2.19
c) 0.597 ( 59.7%)
d) 975 KPa
Explanation:
Solution:-
- We will determine the properties of the working fluid (Air) at each state value.
- We will seek help from air standard property tables ( A-22 ) and write down the following:
State 1:
P1 = 95 KPa , T1 = 300 K
u1 = 214.07 KJ/kg , vr1 = 621.2 , Pr1 = 1.386
- Using relations for isentropic compression we determine the reduced pressure value at state 2:
State 2:
P2 = P3 = 7200 KPa
[tex]P_r_2 = P_r_1*(\frac{P_2}{P_1} )\\\\P_r_2 = 1.386*(\frac{7200}{95} ) = 105.04\\[/tex]
T2 = 979.6 K , vr2 = 26.793 , h2 = 1022.82 KJ/kg
State 3:
P3 = 7200 KPa , T3 = 2150 K
h3 = 2440.3 KJ/kg , vr3 = 2.175 , Pr1 = 1.386
- Using relations for isentropic expansion we determine the reduced volume value at state 4:
State 4:
[tex]\frac{v_4}{v_3} = \frac{v_1}{v_2}*\frac{v_2}{v_3} = \frac{v_r_1}{v_r_2}*\frac{T_2}{T_3} \\\\\frac{v_4}{v_3} = \frac{621.2}{26.793}*\frac{979.6}{2150} = 10.56[/tex]
[tex]v_r_4 = \frac{v_4}{v_3}*v_r_3 = 10.56*2.175\\\\v_r_4 = 22.98[/tex]
T4 = 1031 K , u4 = 785.75 KJ/kg
a) the compression ratio ( r )
- It is the ratio of volume decreased in the compression stroke ( State 1 -> state 2 ):
[tex]r = \frac{v_1}{v_2} = \frac{v_r_1}{v_r_2} =\frac{621.2}{26.793}\\\\r = 23.19[/tex]
b) the cutoff ratio ( rc )
- It is the ratio of volume increased in the heat-addition process ( State 2 -> state 3 ):
[tex]r_c = \frac{v_3}{v_2} = \frac{T_3}{T_2}\\\\r_c = \frac{2150}{979.6} = 2.19[/tex]
c) the thermal efficiency of the cycle ( nth )
- It is the ratio of total work-output from the cycle and the heat addition (Q23). The net work of the cycle is comprised of both heat addition(St2 - St3) and exhaust (St4 - St1 ). Hence, the work done by the cycle is:
[tex]w_n_e_t = w_e - w_c\\\\w_n_e_t = (h_3 - h_2) - (u_4 - u_1 )\\\\w_n_e_t = (2440.3 - 1022.82) - (785.75 - 214.07 )\\\\w_n_e_t = 845.8 KJ/kg[/tex]
- The thermal efficiency of the cycle would be:
[tex]n_t_h = \frac{w_n_e_t}{h_3 - h_2} = \frac{845.8}{2440.3-1022.82} \\\\n_t_h = \frac{845.8}{1417.48}\\\\n_t_h = 0.597[/tex]
d) the mean effective pressure, mep
- It is the ratio of net work done by the cycle and the displaced volume of the compression stroke ( V1 - V2 ).
[tex]mep = \frac{w_n_e_t}{V_1 - V_2} = \frac{w_n_e_t}{v_1( 1 - \frac{v_2}{v_1} )} = \frac{w_n_e_t}{v_1( 1 - \frac{1}{r} )} \\[/tex]
- We need to determine the specific volume of air at state 1. We will use ideal gas equation to determine.
[tex]v_1 = \frac{RT_1}{P_1} = \frac{0.287*300}{95} \\\\v_1 = 0.9063 \frac{m^3}{kg}[/tex]
- Now we can use the calculated specific volume ( v1 ), compression ratio ( r ) and the net cycle work ( wnet ) to determine the ( mep ):
[tex]mep = \frac{845.8}{0.9063*(1-\frac{1}{29.13}) } \\\\mep= 975 KPa[/tex]
