Question 3.18: 10 mol/s of gas enters a steady state, adiabatic nozzle at T...

Set up an energy balance around the nozzle.

\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\left\{N\left(\underline{\rm U}+\frac{\rm v^2}{2}+\rm gh\right) \right\} \\& =\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\text {out }}\left(\underline{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+\mathrm{gh}_{\text {out }}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}

Cancelling terms as in (Example 3.4)

0=\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\mathrm{out}}\left(\underline{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right)

This balance is valid regardless of the specific gas travelling through the nozzle.

A) Steam at 300^{\circ} \mathrm{C} and 5 bar \rightarrow\left(2675.8 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)

Steam at 100^{\circ} \mathrm{C} and 1 \mathrm{bar} \rightarrow 3064.6 \frac{\mathrm{kJ}}{\mathrm{kg}}

(Even though the given flow rate is in mol/sec, this number cancels out, so it isn’t necessary to convert specific enthalpy into molar enthalpy.)

\begin{aligned}& 0=10 \frac{\mathrm{mol}}{\mathrm{sec}}\left(2675.8 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-10 \frac{\mathrm{mol}}{\mathrm{sec}}\left(3064.6 \frac{\mathrm{kJ}}{\mathrm{kg}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}\right)\\& -\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right)=\left(2675.8 \frac{\mathrm{kJ}}{\mathrm{kg}}-3064.6 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)\\& \sqrt{\rm v_{\text {out }}^{2}}=\sqrt{779.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\left(\frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}}\right)\left(\frac{\mathrm{kg} \,\mathrm{\textrm {m } ^ { 2 }}}{\mathrm{J} \,\mathrm{\textrm {sec } ^ { 2 }}}\right)}=\bf 882.7 \frac{\mathbf{m}}{\mathbf{s e c}}\end{aligned}

B)

\begin{aligned}& 0=\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\mathrm{out}}\left(\underline{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}\right)\\& \frac{\mathrm{v}_{\text {out }}^{2}}{2}=\underline{\mathrm{H}}_{\mathrm{in}}-\underline{\mathrm{H}}_{\mathrm{out}}\end{aligned}

Use Appendix D to solve for \underline{H}_{\text {in }}-\underline{\mathrm{H}}_{\text {out }}

\begin{aligned}& \mathrm{d} \underline{\mathrm{H}}=\mathrm{C}_{\mathrm{P}} \mathrm{dT}\\& \frac{C_{P}^{*}}{R}=A+B T+C T^{2}+D T^{3}+E T^{4}\\& d \underline{\mathrm{H}}=\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+\mathrm{RET}^{4}\right) \mathrm{dT}\\& \int_{\text {out }}^{\text {in }} \mathrm{d} \underline{\mathrm{H}}=\int_{\text {out }}^{\text {in }}\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+\mathrm{RET}^{4}\right) \mathrm{dT}\\&\left(\underline{\mathrm{H}}_{\text {out }}-\underline{\mathrm{H}}_{\text {in }}\right)= \mathrm{RA}\left(\mathrm{T}_{\text {out }}-\mathrm{T}_{\text {in }}\right)+\frac{\mathrm{RB}}{2}\left(\mathrm{~T}_{\text {out }}{ }^2-\mathrm{T}_{\text {in }}{ }^2\right)+\frac{\mathrm{RC}}{3}\left(\mathrm{~T}_{\text {out }}{ }^3-\mathrm{T}_{\text {in }}{ }^3\right) \\ & \qquad\qquad\qquad +\frac{\mathrm{RD}}{4}\left(\mathrm{~T}_{\text {out }}{ }^4-\mathrm{T}_{\text {in }}{ }^4\right)+\frac{\mathrm{RE}}{5}\left(\mathrm{~T}_{\text {out }}{ }^5-\mathrm{T}_{\text {in }}{ }^5\right)\\& \mathrm{T}_{\text {out }}=373 \mathrm{~K} \\& \mathrm{~T}_{\mathrm{in}}=573 \mathrm{~K}\end{aligned}

\begin{aligned}& \left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)=\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol}~ \mathrm{K}}\right)\{(3.539)(373 \mathrm{~K}-573 \mathrm{~K}) \\& \qquad\qquad\qquad +\frac{1}{2}\left(-2.61 \times 10^{-4}\right)\left(373 \mathrm{~K}^{2}-573 \mathrm{~K}^{2}\right) \\& \qquad\qquad\qquad +\frac{1}{3}\left(7.00 \times 10^{-8}\right)\left(373 \mathrm{~K}^{3}-573 \mathrm{~K}^{3}\right) \\& \qquad\qquad\qquad +\frac{1}{4}\left(1.57 \times 10^{-9}\right)\left(373 \mathrm{~K}^{4}-573 \mathrm{~K}^{4}\right) \\& \qquad\qquad\qquad \left.+\frac{1}{5}\left(-9.9 \times 10^{-13}\right)\left(373 K^{5}-573 K^{5}\right)\right\} \\& \left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)=-5904.6 \frac{\mathrm{kJ}}{\mathrm{kmol}}\left(\frac{\mathrm{kmol}}{28.02 \mathrm{~kg}}\right)=-210.7 \frac{\mathrm{kJ}}{\mathrm{kg}} \\& \frac{\mathrm{v}_{\text {out }}^{2}}{2}=\underline{H}_{\mathrm{in}}-\underline{\mathrm{H}}_{\mathrm{out}} \\& \frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}=210.7 \frac{\mathrm{kJ}}{\mathrm{kg}}\left(\frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}}\right)\left(\frac{\mathrm{kg} \mathrm{m}^{2}}{\mathrm{J~sec}^{2}}\right)=210700 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}} \\& \mathrm{v}_{\mathrm{out}}=\bf 649 \frac{\mathbf{m}}{\mathbf{sec}}\end{aligned}

C) 

\begin{aligned} & 0=\dot{\mathrm{n}}_{\mathrm{in}}\left(\underline{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{n}}_{\text {out }}\left(\underline{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right) \\& \frac{\mathrm{v}_{\text {out }}^{2}}{2}=\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\end{aligned}

Since this is an ideal gas, we can use:

\begin{aligned} & \mathrm{dH}=\mathrm{C}_{\mathrm{P}}^{*} \mathrm{dT}\\& \underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}=\left(\frac{7}{2}\right)\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol} \,\mathrm{K}}\right)(300 \mathrm{~K}-100 \mathrm{~K})=5819.8 \frac{\mathrm{kJ}}{\mathrm{kmol}}\\& \frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}=\left(5819.8 \frac{\mathrm{kJ}}{\mathrm{kmol}}\right)\left(\frac{\mathrm{kmol}}{28 \mathrm{~kg}}\right)\left(\frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}}\right)\left(\frac{\mathrm{kg} \,\mathrm{m}^{2}}{\mathrm{~J} \,\mathrm{sec}^{2}}\right)\\& \mathrm{v}_{\mathrm{out}}=\bf 644.7 \frac{\mathrm{m}}{\mathrm{sec}}\end{aligned}

C_P^* = (7/2)R is sometimes used as a temperature independent approximation for the heat capacity of nitrogen. In this case comparing b and c illustrates that it produces a velocity that is ~3% different from that obtained from the more rigorous expression.