Chapter 8, Problem 8.1P

(a)

Interpretation Introduction

Interpretation:

The pipe dimensions are to be determined for flow of gas through pipeline.

Concept Introduction:

The fundamental concept of mass conservation is used to describe the fluid flow and determine the pipe dimensions. Here the material transporting through pipeline is natural gas and obeys the ideal-gas law.

Answer to Problem 8.1P

The cross-sectional radius of pipe is $9.64\text{in}$ that is close to acceptable standard of $10\text{in}$

Explanation of Solution

Given information:

The operating pressure of natural gas is $20\text{atm}$

The operating temperature of natural gas is $20{}^{o}C$

Operational flow rate of natural gas is $250000\frac{{\text{m}}^{\text{3}}}{\text{h}}$

Say, reference pressure of gas flow is $1\text{atm}$ and reference temperature of gas is $273\text{K}$

Calculation:

The actual flow rate of gas with respect to reference conditions is given by the following equation.

  $q_{act}=q_{op}\times \frac{P^{ref}}{P^{op}}\times \frac{T^{op}}{T^{ref}}$

Here, the volumetric flow rate is given by $q$ , pressure of gas is given by $P$ and temperature of gas is given by $T$ . The subscripts $op$ and $ref$ denotes operational and reference conditions respectively.

Substitute the value of known variables in the above expression.

  $\begin{array}{l}{q}_{act}=250000\frac{{\text{m}}^{\text{3}}}{\text{h}}\times \frac{1\text{h}}{3600\text{s}}\times \frac{1\text{atm}}{20\text{atm}}\times \frac{293.15\text{K}}{273.15\text{K}}\\ \text{= 3}\text{.726}\frac{{\text{m}}^{\text{3}}}{\text{s}}\end{array}$

Thus, the actual flow rate of natural gas with respect to given conditions is $3.726\frac{{\text{m}}^{\text{3}}}{\text{s}}$

Furthermore, the volumetric flow rate of fluids is given by the following equation.

  $q_{act}=A\times v$

Here, $A$ is the cross-sectional area of pipeline and $v$ is the velocity of natural gas in pipeline. As per international recommendations, assume that the gas flow velocity is $20\text{m/s}$

Hence, the cross-sectional area is calculated as below.

  $\begin{array}{l}A=\frac{q_{act}}{v}\\ \text{=}\frac{3.726\frac{{\text{m}}^{\text{3}}}{\text{s}}}{20\frac{\text{m}}{\text{s}}}\\ \text{=}0.1863{\text{m}}^{\text{2}}\\ \text{=}2.03{\text{ft}}^{\text{2}}\end{array}$

The cross-section of pipeline is circular in appearance; hence, the area is that of a circle as shown below.

  $A=\pi r^2$

Here, $r$ is the cross-sectional radius.

By simplifying the above equation, radius of pipeline can be simply calculated as below.

  $\begin{array}{l}r=\sqrt{\frac{A}{\pi }}\\ \text{=}\sqrt{\frac{2.03}{\pi }}\\ \text{=}0.8036\text{ft}\\ \text{=}9.64\text{in}\end{array}$

From the data available on dimensions, specifications and weights of schedule-40 steel pipes it is observed that the closest standard size available is 10 inch, which is also agreeable with the answer obtained above. Therefore, the inner radius of pipeline is $9.64\text{in}$

(b)

Interpretation Introduction

Interpretation:

The pipe dimensions are to be determined for transporting slurry into a centrifuge separator.

Concept Introduction:

Area of pipe depends on the volumetric flow rate of material. Conversely, volumetric flow rate depends on the mass flow rate and density of the material. In case of slurries, it is significant to have prior information about the mass fraction solid particles and solvents (such as water) to determine the physical properties.

Answer to Problem 8.1P

The radius of pipe under given condition is $0.36\text{in}$ and the acceptable and available dimension is of schedule-80 pipe with inner diameter as $0.75\text{in}$

Explanation of Solution

Given information:

The slurry transported through the pipeline contains solid crystals of p-nitrophenol and water. Let the total mass of slurry pushed into the centrifuge is $100\text{kg}$

Mass of solid in slurry is $45\%$ i.e. $\frac{45\text{kg-solid}}{100\text{kg-mix}}$

Density of p-Nitrophenol $\left({\rho }_s\right)$ is $1475\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Density of water $\left({\rho }_w\right)$ is $1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Mass flow rate of p-Nitrophenol $\left(m_s\right)$ is $1000\frac{\text{kg}}{\text{h}}$

Calculation:

Mass flow rate of slurry $\left(m_y\right)$ is given by the following expression.

  $\begin{array}{l}{m}_y=\frac{m_s}{f}\\ =\frac{1000\frac{\text{kg-solid}}{\text{h}}}{0.45\frac{\text{kg-solid}}{\text{kg-mix}}}\\ =2222.22\frac{\text{kg-mix}}{\text{h}}\end{array}$

Here, $m_y$ is the mass flow rate of total slurry, $m_s$ is the mass flow rate of solid, and $f$ is the mass of solid in slurry as given in the information above.

Mass flow rate of water $\left(m_w\right)$ is given by the following expression.

  $\begin{array}{l}{m}_w=m_y-m_s\\ =2222.22-1000\\ =1222.22\frac{\text{kg}}{\text{h}}\end{array}$

Volumetric flow rate $\left(q_s\right)$ of p-Nitrophenol is given as below.

  $\begin{array}{l}q=\frac{m_s}{{\rho }_s}\\ =\frac{1000\frac{\text{kg}}{\text{h}}}{1475\frac{\text{kg}}{{\text{m}}^{\text{3}}}}\\ =0.678\frac{{\text{m}}^{\text{3}}}{\text{h}}\times \frac{1\text{h}}{\text{3600 s}}\\ =1.883\times {10}^{-4}\frac{{\text{m}}^{\text{3}}}{\text{s}}\end{array}$

Volumetric flow rate $\left(q_w\right)$ of water is given as below.

  $\begin{array}{l}q=\frac{m_w}{{\rho }_w}\\ =\frac{1222.22\frac{\text{kg}}{\text{h}}}{998\frac{\text{kg}}{{\text{m}}^{\text{3}}}}\\ =1.225\frac{{\text{m}}^{\text{3}}}{\text{h}}\times \frac{1\text{h}}{\text{3600 s}}\\ =3.40\times {10}^{-4}\frac{{\text{m}}^{\text{3}}}{\text{s}}\end{array}$

Therefore, the total volumetric mass flow rate of slurry $\left(q_y\right)$ is as below expression.

  $\begin{array}{l}q=q_s+q_w\\ =5.283\times {10}^{-4}\frac{{\text{m}}^{\text{3}}}{\text{s}}\end{array}$

Thus, the actual flow rate of slurry is $5.085\times {10}^{-4}\frac{{\text{m}}^{\text{3}}}{\text{s}}$

Assume that the slurry velocity is $2\text{m/s}$

The, the area of pipeline can be obtained by the following relationship.

  $\begin{array}{l}A=\frac{q}{v}\\ \text{=}\frac{5.283\times {10}^{-4}\frac{{\text{m}}^{\text{3}}}{\text{s}}}{2\frac{\text{m}}{\text{s}}}\\ \text{=}2.6415\times {10}^{-4}{\text{m}}^{\text{2}}\\ \text{=}2.87\times {10}^{-3}{\text{ft}}^{\text{2}}\end{array}$

Here, $A$ is the cross-sectional area, and $v$ is the maximum permissible velocity.

The radius $\left(r\right)$ of pipeline can be obtained by the following relationship.

  $\begin{array}{l}r=\sqrt{\frac{A}{\pi }}\\ \text{=}\sqrt{\frac{2.87\times {10}^{-3}}{\pi }}\\ \text{=}0.03026\text{ft}\\ \text{=}0.363\text{in}\end{array}$

The radius of pipe to carry slurry of p-Nitrophenol and water under given condition is $0.363\text{in}$ whereas the standard pipe of schedule-80 has acceptable inner diameter of $0.75\text{in}$