Chapter 3, Problem 3.79P

(a)

Interpretation Introduction

Interpretation:

To determine volumetric delivery rate in actual feet per day.

Concept Introduction:

Natural gas which is pure methane is delivered to city via pipeline at a volumetric flow rate of $\text{150}$ million standard feet per day.

At delivery methane gas condition are ${\text{50}}^{\text{o}}\text{F}$ and $\text{300 psia}$ .

To deliver natural gas a pipe of $\text{24}\text{in}$ schedule- $\text{40}$ steel with an inside diameter of $\text{22}\text{.624}$ in is used. Standard condition are ${\text{60}}^{\text{o}}\text{F}$ and $\text{1}\text{atm}$ .

Convert the temperature and pressure

At standard condition,

  $\begin{array}{c}\text{T}={\text{60}}^{\text{o}}\text{F}\\ =\left[\left(\text{60-32}\right)\frac{\text{5}}{\text{9}}\text{+273}\text{.15}\right]\text{K}\\ =\text{288}\text{.7}\text{K}\end{array}$

  $\begin{array}{c}\text{P}=\text{1}\text{atm}\\ =\text{1}\text{.01}\text{bar}\end{array}$

Obtain the critical temperature $\left(T_C\right)$, critical volume $\left(V_C\right)$, critical pressure $\left(P_C\right)$, critical compressibility factor $\left(Z_C\right)$, accentric factor $\left(\omega \right)$ for methane from table B.1

The obtained values are

Critical temperature $T_c=190.6\text{K}$

Critical pressure $P_c=45.99\text{bar}$

Critical volume $V_c=98.6{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$

Critical compressibility factor $Z_c=0.286$

Accentric factor $\omega =0.012$

Estimate reduced temperature $\left(T_r\right)$ and reduced pressure $\left(P_r\right)$ of ethylene

  $\begin{array}{l}{T}_r=\frac{T}{T_C}\\ =\frac{288.7\left(\text{K}\right)}{190.6\left(\text{K}\right)}\\ T_r=1.515\end{array}$

  $\begin{array}{l}{P}_r=\frac{P}{P_C}\\ \text{=}\frac{\text{1}\text{.01}\left(\text{bar}\right)}{\text{45}\text{.99}\left(\text{bar}\right)}\\ P_r=0.022\end{array}$

The governing Pitzer correlation to calculate $Z$ is,

  $Z=1+B^o\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}\mathrm{...}\left(1\right)$

Here,

$B^o=0.083-\frac{0.422}{T_r{}^{1.6}}$ and $B^1=0.139-\frac{0.172}{T_r{}^{4.2}}$

Substitute $1.515$ for $T_r$, calculate $B^o$ and $B^1$

  $\begin{array}{l}{B}^o=0.083-\frac{0.422}{{1.515}^{1.6}}\\ B^o=-0.1341\\ B^1=0.139-\frac{0.172}{{1.515}^{4.2}}\\ B^1=0.1089\end{array}$

Substitute $-0.1341$ for $B^o$, $-0.1089$ for $B^1$, $0.022$ for $P_r$, $1.515$ for $T_r$, and $0.012$ for $\omega$ in equation (1)

  $\begin{array}{l}Z=1+\left(-0.1341-\left(0.012\times 0.1089\right)\right)\frac{0.022}{1.515}\\ Z=0.998\end{array}$

Substitute $0.998$ for $Z$, $1.01\text{bar}$ for $\text{83}\text{.14}{\text{cm}}^{\text{3}}\text{.bar}/\text{K}\text{.mol}$ for $R$, and $288.7\text{K}$ for $T$ in the following equation.

  $\begin{array}{l}{V}_1=\frac{Z_{1}RT}{P}\\ =\frac{\text{0}\text{.998×83}\text{.14}\left(\frac{{\text{cm}}^{\text{3}}\text{.bar}}{\text{K}\text{.mol}}\right)\text{×288}\text{.7}\text{K}}{\text{1}\text{.01}\text{bar}}\\ =\text{23717}\text{.34}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\\ V_1=0.024\frac{{\text{m}}^{\text{3}}}{\text{mol}}\end{array}$

Answer to Problem 3.79P

Actual deliver flow rate of natural gas is $6.875\times {10}^6\text{}{\text{ft}}^{\text{3}}/\text{day}$

Explanation of Solution

The volumetric delivery rate in actual cubic feet per day.

At actual conditions,

  $\begin{array}{l}T={50}^{\text{o}}\text{F}\\ =\left[\left(50-32\right)\frac{5}{9}+273.15\right]\text{K}\\ T=283.15\text{K}\end{array}$

  $\begin{array}{l}P=\text{300}\text{psi}\\ \text{=20}\text{.68}\text{bar}\end{array}$

Calculate the reduced temperature $\left(T_r\right)$ and reduced pressure $\left(P_r\right)$ of ethylene.

  $\begin{array}{l}{T}_r=\frac{T}{T_C}\\ \text{=}\frac{\text{283}\text{.15}\text{K}}{\text{190}\text{.6}\text{K}}\\ T_r=1.486\\ P_r=\frac{P}{P_r}\end{array}$

  $\begin{array}{l}\text{=}\frac{\text{20}\text{.68}\text{bar}}{\text{45}\text{.99}\text{bar}}\\ P_r=0.45\end{array}$

Substitute $1.486$ for $T_r$, calculate $B^o$ and $B^1$ .

  $\begin{array}{l}{B}^o=0.083-\frac{0.422}{{1.486}^{1.6}}\\ B^o=-0.141\\ B^1=0.139-\frac{0.172}{{1.486}^{4.2}}\\ B^1=0.106\end{array}$

Substitute $-0.141$ for $B^o$, $0.106$ for $B^1$, $0.45$ for $P_r$, $1.486$ for $T_r$, and $0.012$ for $\omega$ in equation

  $\begin{array}{l}{Z}_2=1+\left(-0.141-\left(0.012\times 0.106\right)\frac{0.45}{1.486}\right)\\ Z_2=0.957\end{array}$

Substitute $0.957$ for $Z$, $\text{20}\text{.68}\text{bar}$ for $P$, $\text{83}\text{.14}{\text{cm}}^{\text{3}}\text{bar}/\text{K}\text{.mol}\text{R}$, $\text{283}\text{.15}\text{K}$ for $\text{T}$ in following equation

  $\begin{array}{l}{V}_2=\frac{Z_{2}RT}{P}\\ =\frac{\text{0}\text{.957×83}\text{.14}\left(\frac{{\text{cm}}^{\text{3}}\text{.bar}}{\text{K}\text{.mol}}\right)\text{×283}\text{.15}\left(\text{K}\right)}{\text{20}\text{.68}\text{bar}}\\ =\text{1089}\text{.4}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\\ V_2=0.00\text{11}\frac{{\text{m}}^{\text{3}}}{\text{mol}}\end{array}$

Given that at standard condition volumetric flow rate $\left(q_1\right)$ is ${\text{150×10}}^{\text{6}}{\text{Ft}}^{\text{3}}/\text{day}$

Calculate the volumetric flow rate at actual conditions $\left(q_2\right)$

  $\begin{array}{l}{q}_2=q_1\frac{V_2}{V_1}\\ ={\text{150×10}}^{\text{6}}\left({\text{ft}}^{\text{3}}/\text{day}\right)\text{×}\frac{\text{0}\text{.0011}\left({\text{m}}^{\text{3}}/\text{mol}\right)}{\text{0}\text{.024}\left({\text{m}}^{\text{3}}/\text{mol}\right)}\\ q_2=6.875\times {10}^6\text{}{\text{ft}}^{\text{3}}/\text{day}\end{array}$

Therefore, actual deliver flow rate of natural gas is $6.875\times {10}^6\text{}{\text{ft}}^{\text{3}}/\text{day}$

(b)

Interpretation Introduction

Interpretation:

To determine molar delivery rate in kmol per hour

Concept Introduction:

Natural gas which is a pure methane is delivered to city via pipeline at a volumetric flow rate of $\text{150}$ million standard feet per day.

At delivery methane gas condition are ${\text{50}}^{\text{o}}\text{F}$ and $\text{300 psia}$ .

To deliver natural gas a pipe of $\text{24}\text{in}$ schedule- $\text{40}$ steel with an inside diameter of $\text{22}\text{.624}$ in is used. standard condition are ${\text{60}}^{\text{o}}\text{F}$ and $\text{1}\text{atm}$ .

Convert the temperature and pressure

At standard condition,

  $\begin{array}{l}{\text{T=60}}^{\text{o}}\text{F}\\ \text{=}\left[\left(\text{60-32}\right)\frac{\text{5}}{\text{9}}\text{+273}\text{.15}\right]\text{K}\\ \text{T=288}\text{.7}\text{K}\end{array}$

  $\begin{array}{l}\text{P=1}\text{atm}\\ \text{=1}\text{.01}\text{bar}\end{array}$

Obtain the critical temperature $\left(T_C\right)$, critical volume $\left(V_C\right)$, critical pressure $\left(P_C\right)$, critical compressibility factor $\left(Z_C\right)$, accentric factor $\left(\omega \right)$ for methane from table B.1

The obtained values are

Critical temperature $T_c=190.6\text{K}$

Critical pressure $P_c=45.99\text{bar}$

Critical volume $V_c=98.6{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$

Critical compressibility factor $Z_c=0.286$

Accentric factor $\omega =0.012$

Estimate reduced temperature $\left(T_r\right)$ and reduced pressure $\left(P_r\right)$ of ethylene

  $\begin{array}{l}{T}_r=\frac{T}{T_C}\\ =\frac{288.7\left(\text{K}\right)}{190.6\left(\text{K}\right)}\\ T_r=1.515\end{array}$

  $\begin{array}{l}{P}_r=\frac{P}{P_C}\\ \text{=}\frac{\text{1}\text{.01}\left(\text{bar}\right)}{\text{45}\text{.99}\left(\text{bar}\right)}\\ P_r=0.022\end{array}$

The governing Pitzer correlation to calculate $Z$ is,

  $Z=1+B^o\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}\mathrm{...}\left(1\right)$

Here,

$B^o=0.083-\frac{0.422}{T_r{}^{1.6}}$ and $B^1=0.139-\frac{0.172}{T_r{}^{4.2}}$

Substitute $1.515$ for $T_r$, calculate $B^o$ and $B^1$

  $\begin{array}{l}{B}^o=0.083-\frac{0.422}{{1.515}^{1.6}}\\ B^o=-0.1341\\ B^1=0.139-\frac{0.172}{{1.515}^{4.2}}\\ B^1=0.1089\end{array}$

Substitute $-0.1341$ for $B^o$, $-0.1089$ for $B^1$, $0.022$ for $P_r$, $1.515$ for $T_r$, and $0.012$ for $\omega$ in equation (1)

  $\begin{array}{l}Z=1+\left(-0.1341-\left(0.012\times 0.1089\right)\right)\frac{0.022}{1.515}\\ Z=0.998\end{array}$

Substitute $0.998$ for $Z$, $1.01\text{bar}$ for $\text{83}\text{.14}{\text{cm}}^{\text{3}}\text{.bar}/\text{K}\text{.mol}$ for $R$, and $288.7\text{K}$ for $T$ in the following equation.

  $\begin{array}{l}{V}_1=\frac{Z_{1}RT}{P}\\ =\frac{\text{0}\text{.998×83}\text{.14}\left(\frac{{\text{cm}}^{\text{3}}\text{.bar}}{\text{K}\text{.mol}}\right)\text{×288}\text{.7}\text{K}}{\text{1}\text{.01}\text{bar}}\\ =\text{23717}\text{.34}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\\ V_1=0.024\frac{{\text{m}}^{\text{3}}}{\text{mol}}\end{array}$

Answer to Problem 3.79P

Molar flow rate is $7291.67\text{kmol}/\text{hour}$

Explanation of Solution

The molar delivery rate in $\text{Kmol}$ per hour.

  $n_1=\frac{q_1}{V_1}$

Here $n_1$ is molar flow rate, $q_1$ is volumetric flow rate and $V_1$ is molar volume

  $n_1=\frac{q_1}{V_1}$

  $\begin{array}{l}=\frac{{\text{150×10}}^{\text{6}}\left({\text{ft}}^{\text{3}}/\text{day}\right)\text{×0}\text{.028}\left({\text{m}}^{\text{3}}/{\text{ft}}^{\text{3}}\right)\text{×}\frac{\text{1}\text{day}}{\text{24}\text{hour}}}{\text{0}\text{.024}\left({\text{m}}^{\text{3}}/\text{mol}\right)\text{×}\frac{\text{1}\left(\text{mol}\right)}{{\text{10}}^{\text{-3}}\left(\text{Kmol}\right)}}\\ =\frac{150\times {10}^6\times 0.028\times \frac{1}{24}}{0.024\times \frac{1}{{10}^{-3}}}\\ n_1=7291.67\text{Kmol}/\text{hour}\end{array}$

Therefore molar flow rate is $7291.67\text{kmol}/\text{hour}$

(c)

Interpretation Introduction

Interpretation:

To determine velocity at delivery conditions in $\text{m/s}$ .

Concept Introduction:

Natural gas which is a pure methane is delivered to city via pipeline at a volumetric flow rate of $\text{150}$ million standard feet per day.

At delivery methane gas condition are ${\text{50}}^{\text{o}}\text{F}$ and $\text{300 psia}$ .

To deliver natural gas a pipe of $\text{24}\text{in}$ schedule- $\text{40}$ steel with an inside diameter of $\text{22}\text{.624}$ in is used. standard condition are ${\text{60}}^{\text{o}}\text{F}$ and $\text{1}\text{atm}$ .

Convert the temperature and pressure

At standard condition,

  $\begin{array}{l}{\text{T=60}}^{\text{o}}\text{F}\\ \text{=}\left[\left(\text{60-32}\right)\frac{\text{5}}{\text{9}}\text{+273}\text{.15}\right]\text{K}\\ \text{T=288}\text{.7}\text{K}\end{array}$

  $\begin{array}{l}\text{P=1}\text{atm}\\ \text{=1}\text{.01}\text{bar}\end{array}$

Obtain the critical temperature $\left(T_C\right)$, critical volume $\left(V_C\right)$, critical pressure $\left(P_C\right)$, critical compressibility factor $\left(Z_C\right)$, accentric factor $\left(\omega \right)$ for methane from table B.1

The obtained values are

Critical temperature $T_c=190.6\text{K}$

Critical pressure $P_c=45.99\text{bar}$

Critical volume $V_c=98.6{\text{cm}}^{\text{3}}{\text{.mol}}^{\text{-1}}$

Critical compressibility factor $Z_c=0.286$

Accentric factor $\omega =0.012$

Estimate reduced temperature $\left(T_r\right)$ and reduced pressure $\left(P_r\right)$ of ethylene

  $\begin{array}{l}{T}_r=\frac{T}{T_C}\\ =\frac{288.7\left(\text{K}\right)}{190.6\left(\text{K}\right)}\\ T_r=1.515\end{array}$

  $\begin{array}{l}{P}_r=\frac{P}{P_C}\\ \text{=}\frac{\text{1}\text{.01}\left(\text{bar}\right)}{\text{45}\text{.99}\left(\text{bar}\right)}\\ P_r=0.022\end{array}$

The governing Pitzer correlation to calculate $Z$ is,

  $Z=1+B^o\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}\mathrm{...}\left(1\right)$

Here,

$B^o=0.083-\frac{0.422}{T_r{}^{1.6}}$ and $B^1=0.139-\frac{0.172}{T_r{}^{4.2}}$

Substitute $1.515$ for $T_r$, calculate $B^o$ and $B^1$

  $\begin{array}{l}{B}^o=0.083-\frac{0.422}{{1.515}^{1.6}}\\ B^o=-0.1341\\ B^1=0.139-\frac{0.172}{{1.515}^{4.2}}\\ B^1=0.1089\end{array}$

Substitute $-0.1341$ for $B^o$, $-0.1089$ for $B^1$, $0.022$ for $P_r$, $1.515$ for $T_r$, and $0.012$ for $\omega$ in equation (1)

  $\begin{array}{l}Z=1+\left(-0.1341-\left(0.012\times 0.1089\right)\right)\frac{0.022}{1.515}\\ Z=0.998\end{array}$

Substitute $0.998$ for $Z$, $1.01\text{bar}$ for $\text{83}\text{.14}{\text{cm}}^{\text{3}}\text{.bar}/\text{K}\text{.mol}$ for $R$, and $288.7\text{K}$ for $T$ in the following equation.

  $\begin{array}{l}{V}_1=\frac{Z_{1}RT}{P}\\ =\frac{\text{0}\text{.998×83}\text{.14}\left(\frac{{\text{cm}}^{\text{3}}\text{.bar}}{\text{K}\text{.mol}}\right)\text{×288}\text{.7}\text{K}}{\text{1}\text{.01}\text{bar}}\\ =\text{23717}\text{.34}\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\\ V_1=0.024\frac{{\text{m}}^{\text{3}}}{\text{mol}}\end{array}$

Answer to Problem 3.79P

Velocity at delivery condition is $8.57\text{m}/\text{sec}$

Explanation of Solution

Gas velocity at delivery condition in $\text{m}{\text{.s}}^{\text{-1}}$ .

  $u_2=\frac{q_2}{A}$

Here, $u_2$ is gas velocity at delivery condition $q_2$ is volumetric flow rate and $A$ is pipe cross section area

Calculate volumetric flow rate in ${\text{m}}^{\text{3}}/\text{sec}$

  $\begin{array}{l}{q}_2=6.875\times {10}^6{\text{ft}}^{\text{3}}/\text{day}\\ =6.875\times {10}^6\left(\frac{{\text{ft}}^{\text{3}}}{\text{day}}\right)\text{×0}\text{.028}\left(\frac{{\text{m}}^{\text{3}}}{{\text{ft}}^{\text{3}}}\right)\text{×}\frac{\text{1}\text{day}}{\text{24×60×60}\text{sec}}\\ q_2=2.228{\text{m}}^{\text{3}}/\text{sec}\end{array}$

Calculate pipe cross section area

  $\begin{array}{l}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}\times {\left(\text{22}\text{.624}\left(\text{in}\right)\text{×0}\text{.0254}\left(\text{m}/\text{in}\right)\right)}^{\text{2}}\\ A=0.26{\text{m}}^{\text{2}}\end{array}$

Calculate velocity

  $\begin{array}{l}{u}_2=\frac{\text{2}\text{.228}\left({\text{m}}^{\text{3}}/\text{sec}\right)}{\text{0}\text{.26}\left({\text{m}}^{\text{2}}\right)}\\ u_2=8.57\text{m}/\text{sec}\end{array}$

Therefore, velocity at delivery condition is $8.57\text{m}/\text{sec}$