Chapter 4, Problem 91P

(a)

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.

Explanation of Solution

Given:

The inlet temperature $\left(T_1\right)$ is 500 K

The inlet pressure $\left(P_1\right)$ is 3 MPa.

The mass flow rate of carbon dioxide $\left(\dot{m}\right)$ is $2\text{kg/s}$.

The outlet temperature  $\left(T_2\right)$ is 450 K.

The outlet pressure $\left(P_2\right)$ is 3 MPa.

Calculation:

Refer to Table A-1, obtain the gas constant $\left(R\right)$, critical pressure $\left(P_{\text{cr}}\right)$, and the critical temperature $\left(T_{\text{cr}}\right)$ 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.

  $\begin{array}{c}{\dot{V}}_1=\frac{\dot{m}R{T}_1}{P_1}\\ {\dot{V}}_1=\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}}\\ =\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.06297{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, the volume flow rate at the inlet of the pipe using the ideal gas equation is $\underset{\_}{0.06297{\text{m}}^{\text{3}}/\text{kg}}$.

Calculate the density at the inlet of pipe.

  $\begin{array}{c}{\rho }_1=\frac{P_1}{R{T}_1}\\ {\rho }_1=\frac{3\text{MPa}}{\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =\frac{3\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{1\text{MPa}}}{\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =31.76{\text{kg/m}}^{\text{3}}\end{array}$

Thus, The density of carbon dioxide at the inlet using the ideal gas equation is $\underset{\_}{31.76{\text{kg/m}}^{\text{3}}}$.

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

  $\begin{array}{c}{\dot{V}}_2=\frac{\dot{m}R{T}_2}{P_2}\\ {\dot{V}}_2=\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}}\\ =\frac{\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.05667{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Thus, The volume flow rate at the exit of the pipe using the ideal gas equation of state is,, and $\underset{\_}{0.05667{\text{m}}^{\text{3}}/\text{kg}}$.

(b)

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.

Explanation of Solution

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

  $\begin{array}{c}{P}_R=\frac{P_1}{P_{cr}}\\ P_R=\frac{3\text{MPa}}{7.39\text{MPa}}\\ =0.406\end{array}$

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

  $\begin{array}{c}{T}_{R,1}=\frac{T_1}{T_{cr}}\\ T_{R,1}=\frac{500\text{K}}{304.2\text{K}}\\ =1.64\end{array}$

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

  $\begin{array}{c}{P}_R=\frac{P_2}{P_{cr}}\\ P_R=\frac{3\text{MPa}}{7.39\text{MPa}}\\ =0.406\end{array}$

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

  $\begin{array}{c}{T}_{R,2}=\frac{T_2}{T_{cr}}\\ T_{R,2}=\frac{450\text{K}}{304.2\text{K}}\\ =1.48\end{array}$

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.

  $Z_2=0.9656$.

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

  $\begin{array}{c}{\dot{V}}_1=\frac{Z_1\dot{m}R{T}_1}{P_1}\\ {\dot{V}}_1=\frac{0.9791\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}}\\ =\frac{0.9791\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)500\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.06165{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Calculate the density at the inlet of pipe.

  $\begin{array}{c}{\rho }_1=\frac{P_1}{Z_{1}R{T}_1}\\ {\rho }_1=\frac{3\text{MPa}}{0.9791\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =\frac{3\text{MPa}\times {\text{10}}^3\frac{\text{kPa}}{1\text{MPa}}}{0.9791\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(500\text{K}\right)}\\ =32.44{\text{kg/m}}^{\text{3}}\end{array}$

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

  $\begin{array}{c}{\dot{V}}_2=\frac{Z_2\dot{m}R{T}_2}{P_2}\\ {\dot{V}}_2=\frac{0.9656\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}}\\ =\frac{0.9656\left(2\text{kg/s}\right)\left(0.1889\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)450\text{K}}{3\text{MPa}\times \frac{\text{1000}\text{kPa}}{1\text{MPa}}}\\ =0.05472{\text{m}}^{\text{3}}/\text{kg}\end{array}$

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.

(c)

To determine

The error involved in the first case.

Answer to Problem 91P

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.

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

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 ${\dot{V}}_{1,\exp }$.

Substitute ${\dot{V}}_{\text{1,}\text{calculated}}=0.06297{\text{m}}^{\text{3}}/\text{kg}$ and  ${\dot{V}}_{1,\exp }=0.06165{\text{m}}^{\text{3}}/\text{kg}$ in the above Equation.

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

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 ${\rho }_{1,\exp }$.

Substitute ${\rho }_{\text{1,}\text{calculated}}=\text{31}\text{.76}{\text{kg/m}}^{\text{3}}$ and ${\rho }_{1,\exp }=\text{32}\text{.44}{\text{kg/m}}^{\text{3}}$ in the above Equation.

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

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 ${\dot{V}}_{2,\exp }$.

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

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

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