Chapter 3.8, Problem 89P

Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kg/s. CO₂ is cooled at constant pressure as it flows in the pipe, and the temperature of the CO₂ drops to 450 K at the exit. Determine the volume flow rate and the density of carbon dioxide at the inlet and the volume flow rate at the exit of the pipe using (a) the ideal-gas equation and (b) the generalized compressibility chart. Also, determine (c) the error involved in the first case.

FIGURE P3–89

To determine

The volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the ideal gas equation of state.

Answer to Problem 89P

The volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the ideal gas equation of state are $\underset{\_}{0.06297{\text{m}}^{\text{3}}/\text{kg}}$, $\underset{\_}{31.76{\text{kg/m}}^{\text{3}}}$, and $\underset{\_}{0.05667{\text{m}}^{\text{3}}/\text{kg}}$ respectively.

Explanation of Solution

Refer to Table A-1, obtain the gas constant, critical pressure, and the critical temperature of carbon dioxide.

$\begin{array}{l}R=0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\\ P_{\text{cr}}=7.39\text{MPa}\\ T_{\text{cr}}=304.2\text{K}\end{array}$

Write the equation of volume flow rate at the inlet of the pipe.

${\dot{V}}_1=\frac{\dot{m}R{T}_1}{P_1}$ (I)

Here, inlet temperature and inlet pressure are $T_1$, $P_1$, and mass flow rate of carbon dioxide is $\dot{m}$.

Calculate the density at the inlet of pipe.

${\rho }_1=\frac{P_1}{R{T}_1}$ (II)

Calculate the equation of volume flow rate at the outlet of the pipe.

${\dot{V}}_2=\frac{\dot{m}R{T}_2}{P_2}$ (III)

Here, outlet temperature and outlet pressure are $T_2$ and $P_2$ respectively.

Conclusion:

Substitute $2\text{kg/s}$ for $\dot{m}$, $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 500 K for $T_1$, and 3 MPa for $P_1$ in Equation (I).

$\begin{array}{c}{\dot{V}}_1=\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}}\\ =\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.06297{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 500 K for $T_1$, and 3 MPa for $P_1$ in Equation (II).

$\begin{array}{c}{\rho }_1=\frac{3\text{MPa}}{\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =\frac{3\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{1\text{MPa}}}{\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =31.76{\text{kg/m}}^{\text{3}}\end{array}$

Substitute $2\text{kg/s}$ for $\dot{m}$, $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 450 K for $T_2$, and 3 MPa for $P_2$ in Equation (III).

$\begin{array}{c}{\dot{V}}_2=\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}}\\ =\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.05667{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the ideal gas equation of state are $\underset{\_}{0.06297{\text{m}}^{\text{3}}/\text{kg}}$, $\underset{\_}{31.76{\text{kg/m}}^{\text{3}}}$, and $\underset{\_}{0.05667{\text{m}}^{\text{3}}/\text{kg}}$ respectively.

To determine

The volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the generalized compressibility chart.

Answer to Problem 89P

The volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the generalized compressibility chart are $\underset{\_}{0.06165{\text{m}}^{\text{3}}/\text{kg}}$, $\underset{\_}{32.44{\text{kg/m}}^{\text{3}}}$, and $\underset{\_}{0.05472{\text{m}}^{\text{3}}/\text{kg}}$ respectively.

Explanation of Solution

Calculate the equation of reduced pressure at the inlet of the pipe.

$P_R=\frac{P_1}{P_{cr}}$ (IV)

Here, the critical pressure is $P_{cr}$.

Calculate the equation of reduced temperature at the inlet of the pipe.

$T_{R,1}=\frac{T_1}{T_{cr}}$ (V)

Here, the critical temperature is $T_{cr}$.

Calculate the equation of reduced pressure at the outlet of the pipe.

$P_R=\frac{P_2}{P_{cr}}$ (VI)

Calculate the equation of reduced temperature at the outlet of the pipe.

$T_{R,2}=\frac{T_2}{T_{cr}}$ (VII)

Write the equation of volume flow rate at the inlet of the pipe.

${\dot{V}}_1=\frac{Z_1\dot{m}R{T}_1}{P_1}$ (VIII)

Here, compressibility factor at the inlet of pipe is $Z_1$.

Calculate the density at the inlet of pipe.

${\rho }_1=\frac{P_1}{Z_{1}R{T}_1}$ (IX)

Calculate the equation of volume flow rate at the outlet of the pipe.

${\dot{V}}_2=\frac{Z_2\dot{m}R{T}_2}{P_2}$ (X)

Here, compressibility factor at the outlet of pipe is $Z_2$.

Conclusion:

Substitute 3 MPa for $P_1$ and 7.39 MPa for $P_{cr}$ in equation (IV).

$\begin{array}{c}{P}_R=\frac{3\text{MPa}}{7.39\text{MPa}}\\ =0.406\end{array}$

Substitute 500 K for $T_1$ and 304.2 K for $T_{cr}$ in equation (V).

$\begin{array}{c}{T}_{R,1}=\frac{500\text{K}}{304.2\text{K}}\\ =1.64\end{array}$

Substitute 3 MPa for $P_2$ and 7.39 MPa for $P_{cr}$ in equation (VI).

$\begin{array}{c}{P}_R=\frac{3\text{MPa}}{7.39\text{MPa}}\\ =0.406\end{array}$

Substitute 450 K for $T_2$ and 304.2 K for $T_{cr}$ in equation (VII).

$\begin{array}{c}{T}_{R,2}=\frac{450\text{K}}{304.2\text{K}}\\ =1.48\end{array}$

Refer to Figure 3-48, obtain the compressibility factor at inlet state $\left(Z_1\right)$ by reading the values of reduced pressure and reduce temperature at inlet conditions of 0.406 and 1.64.

$Z_1=0.9791$.

Refer to Figure 3-48, obtain the compressibility factor at outlet state $\left(Z_2\right)$ by reading the values of reduced pressure and reduce temperature at outlet conditions of 0.406 and 1.48.

$Z_2=0.9656$.

Substitute 0.9791 for $Z_1$, $2\text{kg/s}$ for $\dot{m}$, $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 500 K for $T_1$, and 3 MPa for $P_1$ in Equation (VIII).

$\begin{array}{c}{\dot{V}}_1=\frac{0.9791\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}}\\ =\frac{0.9791\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.06165{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute 0.9791 for $Z_1$, $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 500 K for $T_1$, and 3 MPa for $P_1$ in Equation (IX).

$\begin{array}{c}{\rho }_1=\frac{3\text{MPa}}{0.9791\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =\frac{3\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{1\text{MPa}}}{0.9791\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =32.44{\text{kg/m}}^{\text{3}}\end{array}$

Substitute 0.9656 for $Z_2$, $2\text{kg/s}$ for $\dot{m}$, $0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, 450 K for $T_2$, and 3 MPa for $P_2$ in Equation (X).

$\begin{array}{c}{\dot{V}}_2=\frac{0.9656\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}}\\ =\frac{0.9656\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.05472{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the volume flow rate, density of carbon dioxide at the inlet, and the volume flow rate at the exit of the pipe using the generalized compressibility chart are $\underset{\_}{0.06165{\text{m}}^{\text{3}}/\text{kg}}$, $\underset{\_}{32.44{\text{kg/m}}^{\text{3}}}$, and $\underset{\_}{0.05472{\text{m}}^{\text{3}}/\text{kg}}$ respectively.

To determine

The error involved in the first case.

Answer to Problem 89P

The error involved in the first case are $\underset{\_}{2.1\%}$, $\underset{\_}{2.1\%}$, and $\underset{\_}{3.6\%}$ respectively.

Explanation of Solution

Calculate the percentage of error involved in the first case of volume flow rate at the inlet condition.

$\text{Error}=\frac{{\dot{V}}_{\text{1,}\text{calculated}}-{\dot{V}}_{1,\exp }}{{\dot{V}}_{1,\exp }}\times 100\%$ (XI)

Here, calculated volume flow rate at inlet state from EOS is ${\dot{V}}_{1,\text{calculated}}$ and expected volume flow rate at inlet state from compressibility chart is ${\dot{V}}_{1,\exp }$.

Calculate the percentage of error involved in the first case of density at the inlet condition.

$\text{Error}=\frac{{\rho }_{\text{1,}\text{calculated}}-{\rho }_{1,\exp }}{{\rho }_{1,\exp }}\times 100\%$ (XII)

Here, calculated density at inlet state from EOS is ${\rho }_{1,\text{calculated}}$ and expected density at inlet state from compressibility chart is ${\rho }_{1,\exp }$.

Calculate the percentage of error involved in the first case of volume flow rate at the outlet condition.

$\text{Error}=\frac{{\dot{V}}_{\text{2,}\text{calculated}}-{\dot{V}}_{2,\exp }}{{\dot{V}}_{2,\exp }}\times 100\%$ (XIII)

Here, calculated volume flow rate at outlet state from EOS is ${\dot{V}}_{2,\text{calculated}}$ and expected volume flow rate at outlet state from compressibility chart is ${\dot{V}}_{2,\exp }$.

Conclusion:

Substitute $0.06297{\text{m}}^{\text{3}}/\text{kg}$ for ${\dot{V}}_{\text{1,}\text{calculated}}$ and $0.06165{\text{m}}^{\text{3}}/\text{kg}$ for ${\dot{V}}_{1,\exp }$ in equation (XI).

$\begin{array}{c}\text{Error}=\frac{0.06297{\text{m}}^{\text{3}}/\text{kg}-0.06165{\text{m}}^{\text{3}}/\text{kg}}{0.06165{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =2.1\%\end{array}$

Substitute $\text{31}\text{.76}{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{1,}\text{calculated}}$ and $\text{32}\text{.44}{\text{kg/m}}^{\text{3}}$ for ${\rho }_{1,\exp }$ in equation (XII).

$\begin{array}{c}\text{Error}=\frac{\text{31}\text{.76}{\text{kg/m}}^{\text{3}}-\text{32}\text{.44}{\text{kg/m}}^{\text{3}}}{\text{32}\text{.44}{\text{kg/m}}^{\text{3}}}\times 100\%\\ =-2.09\%\\ \simeq 2.1\%\end{array}$

Substitute $0.05667{\text{m}}^{\text{3}}/\text{kg}$ for ${\dot{V}}_{\text{2,}\text{calculated}}$ and $0.05472{\text{m}}^{\text{3}}/\text{kg}$ for ${\dot{V}}_{2,\exp }$ in equation (XIII).

$\begin{array}{c}\text{Error}=\frac{0.05667{\text{m}}^{\text{3}}/\text{kg}-0.05472{\text{m}}^{\text{3}}/\text{kg}}{0.05472{\text{m}}^{\text{3}}/\text{kg}}\times 100\%\\ =3.6\%\end{array}$

Thus, the error involved in the first case are $\underset{\_}{2.1\%}$, $\underset{\_}{2.1\%}$, and $\underset{\_}{3.6\%}$ respectively.