Answer
$
\frac{\dot{m}_A}{\dot{m}_B}=\frac{h_3-h_4}{h_2-h_1}
$
Work Step by Step
Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the
heat exchanger at state 1 and leaves at state 2. The working fluid of the bottoming cycle (cycle B) enters at state 3 and
leaves at state 4. Neglecting any changes in kinetic and potential energies, and assuming the heat exchanger is well
insulated, the steady-flow energy balance relation yields $$
\begin{aligned}
\dot{E}_{\text {in }}-\dot{E}_{\text {out }} & =\Delta \dot{E}_{\text {system }} \dot{y}^{(\text {steady })}=0 \\
\dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\
\sum \dot{m}_e h_e & =\sum \dot{m}_i h_i \\
\dot{m}_A h_2+\dot{m}_B h_4 & =\dot{m}_A h_1+\dot{m}_B h_3 \text { or } \dot{m}_A\left(h_2-h_1\right)=\dot{m}_B\left(h_3-h_4\right)
\end{aligned}
$$ Thus, $$
\frac{\dot{m}_A}{\dot{m}_B}=\frac{h_3-h_4}{h_2-h_1}
$$
