Answer
$
r_c=1+\frac{q_{\text {in }}}{c_p r^{k-1} T_1}
$
Work Step by Step
Employing the isentropic process equations, $$
T_2=T_1 r^{k-1}
$$ while the ideal gas law gives $$
T_3=T_2 r_c=r_c r^{k-1} T_1
$$ When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is $$
q_{\text {in }}=c_p\left(T_3-T_2\right)=c_p\left(r_c r^{k-1} T_1-r^{k-1} T_1\right)
$$ When this is solved for cutoff ratio, the result is $$
r_c=1+\frac{q_{\text {in }}}{c_p r^{k-1} T_1}
$$
