Question 6.41: Nitrogen gas at 60 kPa and 7°C enters an adiabatic diffuser ......
Nitrogen gas at 60 \mathrm{kPa} and 7{ }^{\circ} \mathrm{C} enters an adiabatic diffuser steadily with a velocity of 200 \mathrm{~m} / \mathrm{s} and leaves at 85 \mathrm{kPa} and 22^{\circ} \mathrm{C}. Determine (a) the exit velocity of the nitrogen and (b) the ratio of the inlet to exit area A_{1} / A_{2}.
Nitrogen is decelerated in a diffuser from 200 \mathrm{~m} / \mathrm{s} to a lower velocity. The exit velocity of nitrogen and the ratio of the inlet-to-exit area are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Nitrogen is an ideal gas with variable specific heats. \mathrm{3} Potential energy changes are negligible. \mathrm{4} The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.
Properties The molar mass of nitrogen is M=28 \mathrm{~kg} / \mathrm{kmol} (Table A-1). The enthalpies are (from nitrogen ideal gas tables-not available in this text)
\begin{aligned} & T_{1}=7^{\circ} \mathrm{C}=280 \mathrm{~K} \rightarrow \bar{h}_{1}=8141 \mathrm{~kJ} / \mathrm{kmol} \\ & T_{2}=22^{\circ} \mathrm{C}=295 \mathrm{~K} \rightarrow \bar{h}_{2}=8580 \mathrm{~kJ} / \mathrm{kmol} \end{aligned}
Analysis (a) There is only one inlet and one exit, and thus \dot{m}_{1}=\dot{m}_{2}=\dot{m}. 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} \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 {sytem }} {}^\nearrow {}^{0 \text { (steady) }}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0 \\ \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(\text { since } \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}=\frac{\bar{h}_{2}-\bar{h}_{1}}{M}+\frac{V_{2}^{2}-V_{1}^{2}}{2}, \end{aligned}
Substituting,
0=\frac{(8580-8141) \mathrm{kJ} / \mathrm{kmol}}{28 \mathrm{~kg} / \mathrm{kmol}}+\frac{V_{2}^{2}-(200 \mathrm{~m} / \mathrm{s})^{2}}{2}\left\lgroup\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^{2} / \mathrm{s}^{2}}\right\rgroup
It yields
V_{2}=93.0 \mathrm{~m} / \mathrm{s}
(b) The ratio of the inlet to exit area is determined from the conservation of mass relation,
\frac{1}{{v}_{2}} A_{2} V_{2}=\frac{1}{{v}_{1}} A_{1} V_{1} \longrightarrow \frac{A_{1}}{A_{2}}=\frac{{v}_{1}}{{v}_{2}} \frac{V_{2}}{V_{1}}=\left\lgroup\frac{R T_{1} / P_{1}}{R T_{2} / P_{2}}\right\rgroup \frac{V_{2}}{V_{1}}
or,
\frac{A_{1}}{A_{2}}=\left\lgroup\frac{T_{1} / P_{1}}{T_{2} / P_{2}}\right\rgroup \frac{V_{2}}{V_{1}}=\frac{(280 \mathrm{~K} / 60 \mathrm{kPa})(93.0 \mathrm{~m} / \mathrm{s})}{(295 \mathrm{~K} / 85 \mathrm{kPa})(200 \mathrm{~m} / \mathrm{s})}=\mathrm{0 . 6 2 5}
Alternative Solution Using constant specific heats for \mathrm{N}_{2} at room temperature (Table A-2a), from the energy balance we obtain
\begin{aligned} & 0=c_{p}\left(T_{2}-T_{1}\right)+\frac{V_{2}^{2}-V_{1}^{2}}{2} \\ & 0=(1.039 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(22-7)^{\circ} \mathrm{C}+\frac{V_{2}^{2}-(200 \mathrm{~m} / \mathrm{s})^{2}}{2}\left\lgroup\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^{2} / \mathrm{s}^{2}}\right\rgroup \\ & V_{2}=94.0 \mathrm{~m} / \mathrm{s} \end{aligned}
The result is practically identical to the result obtained earlier.