Answer:
Maximum temperature of the cycle is 2231.3 K
Explanation:
See table (values there do not assume constant specific heat) and figure attached.
Assuming ideal gas behaviour, p1*v1 = p2*v2, rearranging p2/p1 = v1/v2
Data
[tex]p_1 = 1 bar [/tex]
[tex]T_1 = 300 K [/tex]
[tex] \frac{v_1}{v_2} = 8.5 [/tex] (compression ratio)
[tex] \frac{Q_{23}}{m} = 1400 kJ/kg [/tex] (heat addition)
We can use the following relationship for air
[tex] \frac{v_1}{v_2} = \frac{v_{r1}}{v_{r2}} [/tex]
[tex] v_{r1} [/tex] is only function of temperature and can be taken from table. In this case:
[tex] v_{r1} = 621.2 [/tex]
Rearranging previous equation
[tex] v_{r2} = v_{r1} \times \frac{v_2}{v_1} [/tex]
[tex] v_{r2} = 621.2 \times \frac{1}{8}[/tex]
[tex] v_{r2} = 73.082 [/tex]
Interpolating from table
[tex] u_2 = 503.06 kJ/kg [/tex]
Energy balance in the process 2-3 gives
[tex] \frac{Q_{23}}{m} = u_3 - u_2 [/tex]
[tex] u_3 = \frac{Q_{23}}{m} + u_2 [/tex]
[tex] u_3 = 1400 kJ/kg + 503.06 kJ/kg [/tex]
[tex] u_3 = 1903.06 kJ/kg [/tex]
Interpolating from table
[tex] T_3 = 2231.3 K [/tex]
