Chapter 12.6, Problem 75P

a)

To determine

The exit velocity of the nozzle.

Answer to Problem 75P

The exit velocity of the nozzle is $1738\text{ft}/\text{s}$.

Explanation of Solution

Write the 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 change in internal energy of system is $\Delta {\dot{E}}_{system}$.

Substitute $h_1+\left(\frac{V_1^2}{2}\right)$ for $E_{in}$, $h_2+\left(\frac{V_2^2}{2}\right)$ for $E_{out}$, 0 for $V_1^{}$ and $0$ for $\Delta E_{system}$ in Equation (VI).

$\begin{array}{l}{h}_1+\left(\frac{V_1^2}{2}\right)-h_2-\left(\frac{V_2^2}{2}\right)=0\\ V_2=\sqrt{2\left(h_1-h_2\right)}\end{array}$                                                                                  (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 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 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 $M$.

Conclusion:

Refer table A-19E, “Ideal properties of oxygen”, obtain the enthalpy of inlet and entropy at initial temperature of $1060\text{R}$ as $7543.1\text{Btu}/\text{lbmol}$ and $53.921\text{Btu}/\text{lbmol}\cdot \text{R}$.

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

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).

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

Refer table A-19E, “Ideal properties of oxygen”, obtain the enthalpy of inlet and final temperature at final entropy of $\text{51}\text{.836 Btu}/\text{lbmol}\cdot \text{R}$ as $5614.1\text{Btu}/\text{lbmol}$ and $802\text{R}$.

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

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).

$\begin{array}{c}{V}_2=\sqrt{2\times 60.30\text{Btu}/\text{lbm}}\\ =\sqrt{2\times 60.30\text{Btu}/\text{lbm}\times \left(\frac{25037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ =1738\text{ft}/\text{s}\end{array}$

Thus, the exit velocity of the nozzle is $1738\text{ft}/\text{s}$.

b)

To determine

The exit velocity of the nozzle.

Answer to Problem 75P

The exit velocity of the nozzle is $1740\text{ft}/\text{s}$.

Explanation of Solution

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 $P_1$.

Calculate the change in enthalpy $\left(h_2-h_1\right)$ using generalized 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 $R$.

Conclusion:

Refer table A-1E, “Molar mass, gas constant and critical properties table”, obtain the molar mass, critical temperature, critical pressure, and gas constant of oxygen as $\text{31}\text{.999 }\text{lbm}/\text{lbmol}$, $278.6\text{ R}$, $736\text{psia}$, and $0.06206\text{Btu}/\text{lbmol}\cdot \text{R}$.

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”, select the compressibility factor $Z_{h1}$ at initial state of reduced pressure and temperature of $3.805$ and $0.272$ as $0.000759$.

$Z_{h1}=0.000759$

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”, select the compressibility factor $Z_{h2}$ at final state of reduced pressure and temperature of 2.879 and $0.0951$ as 0 and 0.00894.

$Z_{h2}=0.00894$

Substitute $c_p\left(T_2-T_1\right)$ for ${\left(h_2-h_1\right)}_{ideal}$.

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

Here, specific heat at constant pressure is $c_p$, final temperature is $T_2$, and initial temperature is $T_1$.

Substitute $0.00894$ for $Z_{h2}$, $0.000759$ for $Z_{h1}$, $0.06206\text{Btu}/\text{lbmol}\cdot \text{R}$ for $R$, $60\text{Btu}/\text{lbm}$ ${\left(h_2-h_1\right)}_{Ideal}$, and 278.6 R for $T_{cr}$ in Equation (X).

$\begin{array}{c}\left(h_2-h_1\right)=\left\{-0.06206\text{Btu}/\text{lbmol}\cdot \text{R}\left(278.6\text{R}\right)\left(0.000759-0.00894\right)-60\text{Btu}/\text{lbm}\right\}\\ =-60.44\text{Btu}/\text{lbm}\end{array}$

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

$\begin{array}{c}{V}_2=\sqrt{2\times 60.44\text{Btu}/\text{lbm}}\\ =\sqrt{2\times 60.44\text{Btu}/\text{lbm}\times \left(\frac{25037{\text{ft}}^2/{\text{s}}^2}{1\text{Btu}/\text{lbm}}\right)}\\ =1740\text{ft}/\text{s}\end{array}$

Thus, the exit velocity of the nozzle is $1740\text{ft}/\text{s}$.