Answer
$\dot{m}=6.129\text{ lbm/s}$
$\dot{S}_{\text {gen }}=0.0395\text{ Btu/sā
R}$
Work Step by Step
(a) The mass flow rate of the steam can be determined from its definition to be
$
\dot{m}=\frac{1}{v_2} A_2 V_2=\frac{1}{16.316 \mathrm{ft}^3 / \mathrm{lbm}}\left(1 \mathrm{ft}^2\right)(100 \mathrm{f} / \mathrm{s})=6.129\ \mathrm{lbm} / \mathrm{s}
$
(b) We take diffuser 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}
& \dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\
& \dot{m}\left(h_1+V_1^2 / 2\right)-\dot{Q}_{\text {out }}=\dot{m}\left(h_2+V_2^2 / 2\right) \quad(\text { since } \mathrm{W} \cong \Delta \mathrm{pe} \cong 0) \\
& \dot{Q}_{\text {out }}=-\dot{m}\left(h_2-h_1+\frac{V_2^2-V_1^2}{2}\right) \\
&
\end{aligned}
$
Substituting, the rate of heat loss from the diffuser is determined to be
$
\dot{Q}_{\text {oat }}=-(6.129 \mathrm{lbm} / \mathrm{s})\left(1160.5-1162.3+\frac{(100 \mathrm{fi} / \mathrm{s})^2-(900 \mathrm{fi} / \mathrm{s})^2}{2}\left(\frac{1 \mathrm{Btu} / \mathrm{lbm}}{25,037 \mathrm{fi}^2 / \mathrm{s}^2}\right)\right)=108.42\mathrm{~
Btu} / \mathrm{s}
$
The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the diffuser and its immediate surroundings so that the boundary temperature of the extended system is $77^{\circ} \mathrm{F}$ at all times. It gives
$
\begin{aligned}
& \underbrace{\dot{S}_{\text {in }}-\dot{S}_{\text {out }}}_{\begin{array}{c}
\text { Rateof net entropy tunsfer } \\
\text { byheat and mass }
\end{array}}+\underbrace{\dot{S}_{\text {gen }}}_{\begin{array}{c}
\text { Rateof entropy } \\
\text { genention }
\end{array}}=\underbrace{\Delta \dot{S}_{\text {system }}}_{\begin{array}{c}
\text { Rateof change } \\
\text { of entropy }
\end{array}}=0 \\
& \dot{m} s_1-\dot{m} s_2-\frac{\dot{Q}_{\text {out }}}{T_{\text {bsum }}}+\dot{S}_{\text {gen }}=0 \\
&
\end{aligned}
$
Substituting, the total rate of entropy generation during this process becomes
$
\dot{S}_{\text {gen }}=\dot{m}\left(s_2-s_1\right)+\frac{\dot{Q}_{\text {out }}}{T_{\mathrm{b} \text { sur }}}=(6.129 \mathrm{lbm} / \mathrm{s})(1.7141-1.7406) \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}+\frac{108.42 \mathrm{Btu} / \mathrm{s}}{537 \mathrm{R}}=\mathbf{0 . 0 3 9 5} \mathbf{~B t u} / \mathbf{s} \cdot \mathbf{R}
$
