Chapter 12.6, Problem 77P

To determine

The velocity of the oxygen at nozzle exit by treating the oxygen as an ideal gas and using enthalpy departure charts.

Answer to Problem 77P

The exit velocity of the nozzle by treating the oxygen as an ideal gas is $1738\text{ft}/\text{s}$.

The exit velocity of the nozzle by using enthalpy departure charts is $1740\text{ft}/\text{s}$.

Explanation of Solution

Write the general formula energy balance equation for closed system.

${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$                                                                                           (I)

Here, rate of energy transfer into the system is ${\dot{E}}_{in}$, rate of energy transfer from the system is ${\dot{E}}_{out}$, and rate of change in net energy of system is $\Delta {\dot{E}}_{system}$.

At the steady state, the rate of change of net energy of the system is zero.

$\Delta {\dot{E}}_{system}=0$

Since the inlet velocity is negligible and the Equation (I) is rewritten as follows for the nozzle.

$\begin{array}{l}\left[h_1+\left(\frac{V_1^2}{2}\right)\right]-\left[h_2+\left(\frac{V_2^2}{2}\right)\right]=0\\ h_1+\left(\frac{V_1^2}{2}\right)-h_2-\left(\frac{V_2^2}{2}\right)\\ V_2=\sqrt{2\left(h_1-h_2\right)}\end{array}$                                                                            (II)

Here, inlet velocity is $V_1^{}$, exit velocity is $V_2^{}$, enthalpy at initial state is $h_1$ and enthalpy at final state is $h_2$.

Write the formula for change in entropy equation $\left(s_2^{°}-s_1^{°}\right)$.

$s_2^{°}-s_1^{°}=R_u\ln \left(\frac{P_2}{P_1}\right)$ (III)

Here, universal gas constant is $R_u$, final pressure is $P_2$, and initial pressure is $P_1$.

Write the change in enthalpy equation per mole basis.

${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}={\overline{h}}_{2,Ideal}-{\overline{h}}_{1,Ideal}$ (IV)

Here, Ideal enthalpy at final state is ${\overline{h}}_{2,Ideal}$ and Ideal enthalpy at initial state is ${\overline{h}}_{1,Ideal}$.

Write the formula for change in enthalpy equation in mass basis.

${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}=\frac{{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}}{M}$ (V)

Here, molar mass is $M$.

Calculate the reduced temperature $\left(T_{R1}\right)$ at initial state.

$T_{R1}=\frac{T_1}{T_{cr}}$ (VI)

Here, critical temperature is $T_{cr}$ and initial temperature is $T_1$.

Calculate the reduced pressure $\left(P_{R1}\right)$ at initial state.

$P_{R1}=\frac{P_1}{P_{cr}}$ (VII)

Here, critical pressure is $P_{cr}$ and initial pressure is $P_1$.

Calculate the reduced temperature $\left(T_{R2}\right)$ at final state.

$T_{R2}=\frac{T_2}{T_{cr}}$ (VIII)

Here, critical temperature is $T_{cr}$ and final temperature is $T_2$.

Calculate the reduced pressure $\left(P_{R2}\right)$ at final state.

$P_{R2}=\frac{P_2}{P_{cr}}$ (IX)

Here, critical pressure is $P_{cr}$ and final pressure is $P_1$.

Write the formula for change in enthalpy $\left(h_2-h_1\right)$ using generalized enthalpy departure chart relation.

$h_2-h_1=R{T}_{cr}\left(Z_{h1}-Z_{h2}\right)+{\left(h_2-h_1\right)}_{ideal}$ (X)

Here, change in enthalpy of ideal gas is ${\left(h_2-h_1\right)}_{ideal}$ and gas constant is $R$.

Refer Table A-19E, “Ideal properties of oxygen”.

The inlet enthalpy $\left(h_1\right)$ and entropy $\left(s_1\right)$ corresponding to the temperature of $1060\text{R}$ is $7543.1\text{Btu}/\text{lbmol}$ and $53.921\text{Btu}/\text{lbmol}\cdot \text{R}$ respectively.

$\begin{array}{l}{\overline{h}}_{1,Ideal}=7543.6\text{Btu}/\text{lbmol}\\ s_1^{°}=53.921\text{Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Refer table A-1E, “Molar mass, gas constant and critical properties table”.

The molar mass of oxygen is $\text{31}\text{.999 }\text{lbm}/\text{lbmol}$.

The critical temperature of oxygen is $278.6\text{ R}$.

The critical pressure of oxygen is $736\text{psia}$.

The gas constant of oxygen is $0.06206\text{Btu}/\text{lbmol}\cdot \text{R}$.

Conclusion:

Substitute $53.921\text{Btu}/\text{lbmol}\cdot \text{R}$ for $s_1^{°}$, $\text{1}\text{.9858 Btu}/\text{lbmol}\cdot \text{R}$ for $R_u$, $70\text{}\text{psia}$ for $P_2$, and $200\text{}\text{psia}$ for $P_1$ in Equation (III).

$\begin{array}{l}{s}_2^{°}-53.921\text{Btu}/\text{lbmol}\cdot \text{R}=\text{1}\text{.9858 Btu}/\text{lbmol}\cdot \text{R}\times \ln \left(\frac{70\text{}\text{psia}}{200\text{}\text{psia}}\right)\\ s_2^{°}=53.921\text{Btu}/\text{lbmol}\cdot \text{R}-\text{2}\text{.085 Btu}/\text{lbmol}\cdot \text{R}\\ s_2^{°}=\text{51}\text{.836 Btu}/\text{lbmol}\cdot \text{R}\end{array}$

Refer Table A-19E, “Ideal properties of oxygen”.

The exit enthalpy $\left(h_2\right)$ and exit temperature $\left(T_2\right)$ at final entropy of $\text{51}\text{.836 Btu}/\text{lbmol}\cdot \text{R}$ is $5614.1\text{Btu}/\text{lbmol}$ and $802\text{R}$ respectively.

$\begin{array}{l}{\overline{h}}_{2,Ideal}=5614.1\text{Btu}/\text{lbmol}\\ T_2=802\text{R}\end{array}$

Substitute $5614.1\text{Btu}/\text{lbmol}$  ${\overline{h}}_{2,Ideal}$ and $7543.6\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{1,Ideal}$ in Equation (IV).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}=5614.1\text{Btu}/\text{lbmol}-7543.6\text{Btu}/\text{lbmol}\\ =-1929.5\text{Btu}/\text{lbmol}\end{array}$

Substitute $31.999\text{lbm}/\text{lbmol}$ for $M$ and $-1929.5\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}$ in

Equation (V).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{Ideal}=\frac{-1929.5\text{Btu}/\text{lbmol}}{31.999\text{lbm}/\text{lbmol}}\\ =-60.30\text{Btu}/\text{lbm}\end{array}$

Substitute $60.30\text{Btu}/\text{lbm}$ for $\left(h_2-h_1\right)$ in Equation (II).

$\begin{array}{c}{V}_2=\sqrt{2\times 60.30\text{Btu}/\text{lbm}}\\ =\sqrt{2\times 60.30\text{Btu}/\text{lbm}\times \left(\frac{25037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ =\sqrt{3019462.2{\text{ft}}^2/{\text{s}}^2}\\ =1737.6899\text{ft}/\text{s}\end{array}$

$\simeq 1738\text{ft}/\text{s}$

Thus, exit velocity of the nozzle by treating the oxygen as an ideal gas is $1738\text{ft}/\text{s}$.

Substitute 1060 R for $T_1$ and 278.6 R for $T_{cr}$ in Equation (VI).

$\begin{array}{c}{T}_{R1}=\frac{1060\text{R}}{278.6\text{R}}\\ =3.805\end{array}$

Substitute $200\text{psia}$ for $P_1$ and $736\text{psia}$ for $P_{cr}$ in Equation (VII).

$\begin{array}{c}{P}_{R1}=\frac{200\text{psia}}{736\text{psia}}\\ =0.272\end{array}$

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”.

Obtain the enthalpy departure factor $\left(Z_{h1}\right)$ at initial state corresponding to the reduced pressure $\left(P_{R1}\right)$ and temperature $\left(T_{R1}\right)$ is $0.000759$.

$Z_{h1}=0.000759$

Substitute 802 R for $T_2$ and 278.6 R for $T_{cr}$ in Equation (VIII).

$\begin{array}{c}{T}_{R2}=\frac{\text{802 R}}{278.6\text{ R}}\\ =2.879\end{array}$

Substitute $70\text{psia}$ for $P_2$ and $736\text{psia}$ for $P_{cr}$ in Equation (IX).

$\begin{array}{c}{P}_{R2}=\frac{70\text{psia}}{736\text{psia}}\\ =0.0951\end{array}$

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”.

Obtain the enthalpy departure factor $\left(Z_{h2}\right)$ at initial state corresponding to the reduced pressure $\left(P_{R2}\right)$ and temperature $\left(T_{R2}\right)$ is $0.00894$.

$Z_{h2}=0.00894$

Substitute $0.00894$ for $Z_{h2}$, $0.000759$ for $Z_{h1}$, $0.06206\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$, $60.30\text{Btu}/\text{lbm}$ ${\left(h_2-h_1\right)}_{Ideal}$, and 278.6 R for $T_{cr}$ in Equation (X).

$h_2-h_1=R{T}_{cr}\left(Z_{h1}-Z_{h2}\right)+{\left(h_2-h_1\right)}_{ideal}$

$\begin{array}{c}\left(h_2-h_1\right)=\left\{-60.30\text{Btu}/\text{lbm}+\left[\begin{array}{l}\left(0.06206\text{Btu}/\text{lbmol}\cdot \text{R}\right)\left(278.6\text{R}\right)\\ \left(0.00894-0.000759\right)\end{array}\right]\right\}\\ =-60.30\text{Btu}/\text{lbm}+0.1414\text{Btu}/\text{lbm}\\ =60.44\text{Btu}/\text{lbm}\end{array}$

Substitute $60.44\text{Btu}/\text{lbm}$ for $\left(h_2-h_1\right)$ in Equation (II).

$\begin{array}{c}{V}_2=\sqrt{2\times 60.44\text{Btu}/\text{lbm}}\\ =\sqrt{2\times 60.44\text{Btu}/\text{lbm}\times \left(\frac{25037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ =\sqrt{3026472.56{\text{ft}}^2/{\text{s}}^2}\\ =1739.6759\text{ft}/\text{s}\end{array}$

$\simeq 1740\text{ft}/\text{s}$

Thus, the exit velocity of the nozzle by using enthalpy departure charts is $1740\text{ft}/\text{s}$.