Answer
$\dot{m}=0.2976\text{ kg/s}$
$\dot{S}_{\mathrm{gen}}=0.0854\text{ kW/K}$
Work Step by Step
(a) There is only one inlet and one exit, and thus $\dot{m}_1=m_2=\dot{m}$. We take nozzle 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
$
\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\text {in }}=\underbrace{\Delta \dot{E}_{\text {stem }} \lambda_0}_{\text {system (steady) }}=0
$
Rate of net energy trasfer Rate of changein intemal, kinetic, byheat, work, mind mess potentialetce. energies
$
\begin{aligned}
\dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\
\dot{m}\left(h_1+V_1^2 / 2\right) & =\dot{m}\left(h_2+V_2^2 / 2\right) \quad(\sin c \dot{Q} \cong \dot{\mathrm{W}} \cong \Delta \mathrm{pe} \cong 0) \\
0 & =h_2-h_1+\frac{V_2^2-V_1^2}{2}
\end{aligned}
$
Substituting,
$
h_2=3137.7 \mathrm{~kJ} / \mathrm{kg}-\frac{(390 \mathrm{~m} / \mathrm{s})^2-(55 \mathrm{~m} / \mathrm{s})^2}{2}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=3063.1 \mathrm{~kJ} / \mathrm{kg}
$
Thus,
The mass flow rate of steam is
$
\dot{m}=\frac{1}{v_1} A_1 V_1=\frac{1}{0.13860 \mathrm{~m}^3 / \mathrm{kg}}\left(7.5 \times 10^{-4} \mathrm{~m}^2\right)(55 \mathrm{~m} / \mathrm{s})=0.2976 \mathrm{~kg} / \mathrm{s}
$
(b) Again we take the nozzle to be the system. Noting that no heat crosses the boundaries of this combined system, the entropy balance for it can be expressed as
$
\begin{aligned}
& \underbrace{\dot{S}_{\text {in }}-\dot{S}_{\text {out }}}_{\begin{array}{c}
\text { Rateof net entropy tensfer } \\
\text { by heat and mass }
\end{array}}+\underbrace{\dot{S}_{\text {gen }}}_{\begin{array}{c}
\text { Rate of entropy } \\
\text { generation }
\end{array}}=\underbrace{\Delta \dot{S}_{\text {systema }}}_{\begin{array}{c}
\text { Rate of change } \\
\text { of entropy }
\end{array}}=0 \\
& \dot{m} s_1-\dot{m} s_2+\dot{S}_{\mathrm{gen}}=0 \\
& \dot{S}_{\mathrm{gen}}=\dot{m}\left(s_2-s_1\right) \\
&
\end{aligned}
$
Substituting, the rate of entropy generation during this process is determined to be
$
\dot{S}_{\mathrm{gen}}=\dot{m}\left(s_2-s_1\right)=(0.2976 \mathrm{~kg} / \mathrm{s})(7.2454-6.9583) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}=\mathbf{0 . 0 8 5 4} \mathbf{~k W} / \mathbf{K}
$
