Answer
$1.4842$ Btw/lbmâ‹…R
Work Step by Step
Because the separator divides the inlet stream into the liquid and vapor portions,$$
\begin{aligned}
& \dot{m}_2=x \dot{m}_1=0.9 \dot{m}_1 \\
& \dot{m}_3=(1-x) \dot{m}_1=0.1 \dot{m}_1
\end{aligned}
$$ According to the water property tables at $100 \mathrm{psia}$ (Table A-5E),
$$
s_1=s_f+x s_{f s}=0.47427+0.9 \times 1.12888=1.4903\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}
$$ When the increase in entropy principle is adapted to this system, it becomes
$$
\begin{aligned}
\dot{m}_2 s_2+\dot{m}_3 s_3 & \geq \dot{m}_1 s_1 \\
x \dot{m}_1 s_2+(1-x) \dot{m}_1 s_3 & \geq \dot{m}_1 s_1 \\
0.9 s_2+0.1 s_3 & \geq s_1 \\
& \geq 1.4903\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}
\end{aligned}
$$ To test this hypothesis, let's assume the outlet pressures are $110 \mathrm{psia}$. Then, $$
\begin{aligned}
& s_2=s_g=1.5954\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& s_3=s_f=0.48341\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathbf{R}
\end{aligned}
$$ The left-hand side of the above equation is $$
0.9 s_2+0.1 s_3=0.9 \times 1.5954+0.1 \times 0.48341=1.4842\ \mathrm{Btw} / \mathrm{lbm} \cdot \mathbf{R}
$$ It is less than the minimum possible specific entropy. Hence, the outlet pressure cannot be 110 psia. Inspection of the water table in light of above equation proves that the pressure at the separator outlet cannot be greater than that at the inlet.
