Answer
$w_{\mathrm{P}, \text { in }} =1.319\text{ kJ/kg}$
Work Step by Step
We take the entire unit as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as
$$
\begin{aligned}
\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c}
\text { Rate of net energy transfer } \\
\text { by heat, work, and mass }
\end{array}} & =\underbrace{\Delta \dot{E}_{\text {system }} 80 \text { (steady) }}_{\begin{array}{c}
\text { Rate of change in intemal, kinetic, } \\
\text { potential etc. energies }
\end{array}}=0 \\
\dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\
\dot{m}_1 h_1+\dot{m}_3 h_3+\dot{m}_3 w_{\mathrm{P}, \text { in }} & =\dot{m}_6 h_6 \\
\dot{m}_1 h_1+\dot{m}_3 h_3+\dot{m}_3 w_{\mathrm{P}, \text { in }} & =\left(\dot{m}_1+\dot{m}_3\right) h_6
\end{aligned}
$$ Solving this for $\dot{m}_3$, $$
\dot{m}_3=\dot{m}_1 \frac{h_6-h_1}{\left(h_3-h_6\right)+w_{\mathrm{P}, \text { in }}}=(1 \mathrm{~kg} / \mathrm{s}) \frac{1267.5-1134.8}{2969.5-1267.5+1.319}=\mathbf{0 . 0 7 7 9} \mathbf{k g} / \mathbf{s}
$$ where $$
\begin{aligned}
w_{\mathrm{P}, \text { in }} & =v_4\left(P_5-P_4\right)=v_{f @ 6000 \mathrm{kPa}}\left(P_5-P_4\right) \\
& =\left(0.001319 \mathrm{~m}^3 / \mathrm{kg}\right)(7000-6000) \mathrm{kPa}\left(\frac{1 \mathrm{kPa}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& =1.319 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$
