Answer:
the maximum permissible compression ratio (pressure ratio across the compressor) r = 11.525 and the cycle efficiency (η) = 0.5
Explanation:
[tex]T_1 =300K\\ P_1=100kPa\\T_3=1400K\\q=800kJ/kq[/tex]
adiabatic exponent k = 1.4 and air specific heat [tex]C_p= 1.004kJ/kgK[/tex]
The specific heat transfer q is given by the equation:
[tex]q=C_p(T_3-T_2)\\T_2=T_3-\frac{q}{C_p}[/tex]
Substituting values:
[tex]T_2=T_3-\frac{q}{C_p}= 1400-\frac{800}{1.004}= 603.18K[/tex]
the maximum permissible compression ratio (pressure ratio across the compressor) r is given by:
[tex]r = (\frac{T_2}{T_1} )^\frac{k}{k-1}[/tex]
Substituting values:
[tex]r = (\frac{T_2}{T_1} )^\frac{k}{k-1} = (\frac{603.18}{300} )^\frac{1.4}{1.4-1}= 11.525[/tex]
r = 11.525
Exhaust temperature ([tex]T_4[/tex]) = [tex]T_3(r^\frac{1-k}{k})[/tex]
[tex]T_4=T_3(r^\frac{1-k}{k})=1400(11.525^\frac{1-1.4}{1.4}) =696.3K[/tex]
the cycle efficiency (η) = [tex]\frac{C_p(T_3-T_4)}{q} - \frac{C_p(T_2-T_1)}{q}[/tex]
η = [tex]\frac{C_p(T_3-T_4)}{q} - \frac{C_p(T_2-T_1)}{q}= \frac{1.004(1400-696.3)}{800} - \frac{10004(603.18-300)}{800}=0.5[/tex]