Chapter 5.5, Problem 32P

Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa and 500°C with a mass flow rate of 6000 kg/h and leaves at 100 kPa and 450 m/s. The inlet area of the nozzle is 40 cm². Determine (a) the inlet velocity and (b) the exit temperature.

To determine

The inlet velocity.

Answer to Problem 32P

The inlet velocity is $60.8416\text{m/s}$.

Explanation of Solution

The carbon dioxide flows through the nozzle at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for specific volume $\left(v\right)$ at the inlet.

$v_1=\frac{R{T}_1}{P_1}$ (I)

Here, the gas constant of carbon dioxide is $R$, the temperature is $T$ and the pressure is $P$; the suffix 1 indicates the inlet condition.

Write the formula for mass flow rate.

$\dot{m}=\frac{A_1{V}_1}{v_1}$ (II)

Here, the cross-sectional area is $A$ and the velocity is $V$; the suffix 1 indicates the inlet state.

Rearrange the Equation (II) to obtain the inlet velocity $\left(V_1\right)$.

$V_1=\frac{\dot{m}{v}_1}{A_1}$ (III)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of carbon dioxide is, $R=0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

Conclusion:

Substitute $0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $500°\text{C}$ for $T_1$ and $1\text{MPa}$ for $P_1$ in

Equation (I).

$\begin{array}{c}{v}_1=\frac{\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(500°\text{C}\right)}{1\text{MPa}}\\ =\frac{\left(0.1889\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(500+273\right)\text{K}}{1\text{MPa}\times \frac{1000\text{kPa}}{1\text{MPa}}}\\ =\frac{146.0197\text{kPa}\cdot {\text{m}}^3\text{/kg}}{1000\text{kPa}}\\ =0.14602{\text{m}}^3\text{/kg}\end{array}$

Substitute $6000\text{kg/h}$ for $\dot{m}$, $0.14602{\text{m}}^3\text{/kg}$ for $v$ and $40\text{}{\text{cm}}^2$ for $A$ in Equation (III).

$\begin{array}{c}{V}_1=\frac{\left(6000\text{kg/h}\right)\left(0.14602{\text{m}}^3\text{/kg}\right)}{40\text{}{\text{cm}}^2}\\ =\frac{\left(6000\text{kg/h}\times \frac{1\text{h}}{3600\text{s}}\right)\left(0.14602{\text{m}}^3\text{/kg}\right)}{\left(40\text{}{\text{cm}}^2\times \frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)}\\ =\frac{0.24336}{40\times {10}^{-4}}\\ =60.8416\text{m/s}\end{array}$

Thus, the inlet velocity is $60.8416\text{m/s}$.

To determine

The exit temperature.

Answer to Problem 32P

The exit temperature is $685.7696\text{K}$.

Explanation of Solution

Write the energy rate balance equation.

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

Here, the rate of energy transfer in by heat, work and mass is ${\dot{E}}_{\text{in}}$, the rate of energy transfer out by heat, work and mass is ${\dot{E}}_{\text{out}}$, and the rate of change in internal, kinetic, potential, etc. energies is $\Delta {\dot{E}}_{\text{system}}$.

Here, the nozzle operates at steady state. Hence, the rate of change in internal, kinetic, potential, etc. energies becomes zero.

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

The rate of energy transfer in $\left({\dot{E}}_{\text{in}}\right)$ by heat, work and mass is expressed as follows.

${\dot{E}}_{\text{in}}={\dot{Q}}_{\text{in}}+{\dot{W}}_{\text{in}}+{\displaystyle \sum _{\text{in}}\dot{m}\left(h+\frac{V^2}{2}+gz\right)}$

The rate of energy transfer out $\left({\dot{E}}_{\text{out}}\right)$  by heat, work and mass is expressed as follows.

${\dot{E}}_{\text{out}}={\dot{Q}}_{\text{out}}+{\dot{W}}_{\text{out}}+{\displaystyle \sum _{\text{out}}\dot{m}\left(h+\frac{V^2}{2}+gz\right)}$

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$ and the elevation from the datum is $z$; the suffixes in and out indicates the inlet and outlet of the system.

Here, the nozzle has one inlet and one outlet. Say inlet condition as 1 and outlet condition as 2 as follows.

${\dot{E}}_{\text{1}}={\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)$ (V)

${\dot{E}}_{\text{2}}={\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)$ (VI)

Since, the nozzle is adiabatic nozzle, the heat transfer rate and work transfer rates are negligible i.e. $\dot{Q}=0$ and $\dot{W}=0$. The inlet, outlet are at same elevation, the potential energy becomes negligible i.e. $gz=0$.

The Equations (V) and (VI) are reduced to as follows.

$\begin{array}{c}{\dot{E}}_{\text{1}}=0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\\ =\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)\end{array}$

$\begin{array}{c}{\dot{E}}_{\text{2}}=0+0+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\\ =\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\end{array}$

Substitute $0$ for $\Delta {\dot{E}}_{\text{system}}$, $\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)$ for ${\dot{E}}_{\text{in}}$ and $\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)$ for ${\dot{E}}_{\text{out}}$ in

Equation (IV).

$\begin{array}{l}\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ h_1+\frac{V_1{}^2}{2}=h_2+\frac{V_2{}^2}{2}\\ h_2=h_1+\frac{V_1{}^2}{2}-\frac{V_2{}^2}{2}\end{array}$

$h_2=h_1+\frac{\left(V_1{}^2-V_2{}^2\right)}{2}$ (VII)

Refer Table A-20, “Ideal-gas properties of carbon dioxide, ${\text{CO}}_2$”, the properties are given in molar basis.

The enthalpy in molar basis is as follows,

$\begin{array}{l}{h}_1=\frac{\overline{h_1}}{M}\\ h_2=\frac{\overline{h_2}}{M}\end{array}$

Here, the molar mass of carbon dioxide is $M$.

Substitute $\frac{\overline{h_1}}{M}$ for $h_1$ and $\frac{\overline{h_2}}{M}$ for $h_2$ in Equation (VII).

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

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The molar mass of carbon dioxide is, $M=44.01\text{kg/kmol}$.

Refer Table A-20, “Ideal-gas properties of carbon dioxide, ${\text{CO}}_2$”.

The inlet enthalpy $\left(h_1\right)$ corresponding to the temperature of $500\text{}°\text{C }\left(773\text{K}\right)$- using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (IX)

Show the temperature and enthalpy values from the Table A-20 as in below table.

S.No.	x	y
	Temperature $\left(T\right)$,in $\text{K}$	Enthalpy $\left(\overline{h}\right)$, in $\text{kJ/kmol}$
1	770	30644
2	773	?
3	780	31154

Substitute $770$ for $x_1$, $773$ for $x_2$, $780$ for $x_3$, $30644$ for $y_1$, and $31154$ for $y_3$ in Equation (IX).

$\begin{array}{c}{y}_2=\frac{\left(773-770\right)\left(31154-30644\right)}{\left(780-770\right)}+30644\\ =30797\text{kJ/kmol}\end{array}$

Thus, the enthalpy $\left(h_2\right)$ corresponding to the temperature of $500\text{}°\text{C }\left(773\text{K}\right)$ is $30797\text{kJ/kmol}$.

Conclusion:

Substitute $30797\text{kJ/kmol}$ for $\overline{h_1}$,$44.01\text{kg/kmol}$ for $M$, $60.8416\text{m/s}$ for $V_1$ and $450\text{m/s}$ for $V_2$ in Equation (IX).

$\begin{array}{c}\overline{h_2}=30797\text{kJ/kmol}+\frac{44.01\text{kg/kmol}\left[{\left(60.8416\text{m/s}\right)}^2-{\left(450\text{m/s}\right)}^2\right]}{2}\\ =30797\text{kJ/kmol}+44.01\text{kg/kmol}\left(-99399.1458{\text{m}}^2{\text{/s}}^2\times \frac{1\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}\right)\\ =30797\text{kJ/kmol}-4374.5564\text{kJ/kmol}\\ \text{=}\text{26422}\text{.4436}\text{kJ/kmol}\end{array}$

Refer Table A-20, “Ideal-gas properties of carbon dioxide, ${\text{CO}}_2$”.

The temperature corresponding to exit enthalpy of $\text{26422}\text{.4436}\text{kJ/kmol}$-using interpolation method.

Show the enthalpy and temperature values from the Table A-20 as in below table.

S.No.	x	y
	Enthalpy $\left(\overline{h}\right)$, in $\text{kJ/kmol}$	Temperature $\left(T\right)$,in $\text{K}$
1	26138	680
2	26422.4436	?
3	26631	690

Substitute $26138$ for $x_1$, $26422.4436$ for $x_2$, $26631$ for $x_3$, $680$ for $y_1$, and $690$ for $y_3$ in Equation (IX).

$\begin{array}{c}{y}_2=\frac{\left(26422.4436-26138\right)\left(690-680\right)}{\left(26631-26138\right)}+680\\ =685.7696\text{K}\end{array}$

Thus, the temperature corresponding to exit enthalpy of $\text{26422}\text{.4436}\text{kJ/kmol}$ is $685.7696\text{K}$.

Thus, the exit temperature is $685.7696\text{K}$.