Chapter 9, Problem 9.105P

To determine

(a)

To find: the length of the pipe by using given information.

Answer to Problem 9.105P

The value of length of the pipe is determined below.

Explanation of Solution

Given information:

First convert the pressure we get

$\begin{array}{l}1Ib/i{n}^2=6.8948\times {10}^3\times 1Pa\\ =6.8948\times {10}^{3}Pa\end{array}$

$\begin{array}{l}2500Ib/i{n}^2=2500\times 6.8948\times {10}^{3}Pa\\ =17.237\times {10}^{6}Pa\\ =17.237\times {10}^{6}N/m^2\end{array}$

Then convert the temperature we get

$T_c=\frac{5}{9}\left(T_F-{32}^0\right)$

Convert 140⁰ to°C we get

$\begin{array}{l}{T}_c=\frac{5}{9}\left(T_F-{32}^0\right)\\ T_c=\frac{5}{9}\left(140-32\right)\\ ={60}^{0}C\end{array}$

Then

$\begin{array}{l}{T}_K=T_C+273\\ T_K=60+273\\ =333K\end{array}$

Find the density of natural gas which is mostly comprised of methane

${\rho }_1=\frac{P_1}{R{T}_1}$

Substitute above values we get

$\begin{array}{l}{\rho }_1=\frac{17.2327\times {10}^{6}N/m^2}{518m^2/s^2.K\times 333K}\\ =99.93kg/m^3\\ =\frac{17237000}{172571.7}\\ =99.88kg/m^3\end{array}$

Convert diameter from in to m

$1in=0.0254m$

Then

$\begin{array}{l}52in=52\times 0.0254m\\ =1.321m\end{array}$

Find the area of the pipe

$\begin{array}{l}{A}_1=\frac{\pi }{4}{D}_1^2\\ A_1=\frac{\pi }{4}\times {1.321}^2\\ =1.371m^2\end{array}$

Find the velocity of natural gas in the pipe

$\begin{array}{l}{V}_1=\frac{\dot{m}}{{\rho }_1{A}_1}\\ V_1=\frac{890}{99.88*1.371}\\ =6.5m/s\end{array}$

We know that

$\mu =1.03*{10}^{-5}kg/m.s$

And then formula for Reynolds number for flow of natural gas

$\begin{array}{l}\mathrm{Re}=\frac{{\rho }_1{V}_{1}D}{\mu }\\ \mathrm{Re}=\frac{99.88\times 6.5\times 1.321}{1.03\times {10}^{-5}}\\ =\frac{857.62}{1.03\times {10}^{-5}}\\ =83264077.67\\ =8.33\times {10}^7\end{array}$

Find the friction factor of the pipe

$\begin{array}{l}\frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{\epsilon /D_1}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{0/1.321}{3.7}+\frac{2.51}{8.33\times {10}^7\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{2.51}{8.33\times {10}^7\times \sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{3.01\times {10}^{-8}}{\sqrt{f}}\right)\\ f_{smooth}=0.00608\end{array}$

Find the mach number

$\begin{array}{l}M{a}_1=\frac{V_1}{V_a}\\ M{a}_1=\frac{6.5}{477}\\ =0.0136\end{array}$

Find the value of $\frac{f{L}_1^{*}}{D}$

$\begin{array}{l}\frac{f{L}_1^{*}}{D}=\frac{1-M{a}_1^2}{kM{a}_1^2}+\frac{k+1}{2k}\ln \left(\frac{\left(k+1\right)M{a}_1^2}{2+\left(k-1\right)M{a}_1^2}\right)\\ \frac{f{L}_1^{*}}{D}=\frac{1-{0.0136}^2}{1.32\times {0.0136}^2}+\frac{1.32+1}{2\times 1.32}\ln \left(\frac{\left(1.32+1\right){0.0136}^2}{2+\left(1.32-1\right){0.0136}^2}\right)\\ =4095.132+0.88\times \ln \left(2.145\times {10}^{-4}\right)\\ =4095.132-7.433\\ =4087.7\end{array}$

Find the pressure ratio by using below formula

$\begin{array}{l}\frac{P_1}{p^{*}}=\frac{1}{M{a}_1}{\left[\frac{k+1}{2+\left(k-1\right)M{a}_1^2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \frac{P_1}{p^{*}}=\frac{1}{0.0136}{\left[\frac{1.32+1}{2+\left(1.32-1\right){0.0136}^2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{0.0136}{\left(1.16\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =79.1924\end{array}$

And then find the temperature ratio we get

$\begin{array}{l}\frac{T_1}{T^{*}}=\frac{k+1}{2+\left(k-1\right)M{a}_1^2}\\ \frac{T_1}{T^{*}}=\frac{1.32+1}{2+\left(\left(1.32-1\right)\times {0.0136}^2\right)}\\ =1.16\end{array}$

Find the new pressure ratio we get

$\begin{array}{l}\frac{p_2}{p^{*}}=\frac{p_2}{p_1}\frac{p_1}{p^{*}}\\ \frac{p_2}{p^{*}}=\frac{2000}{2500}\times 79.1924\\ =63.35\end{array}$

Find the new mach number by using below formula

$\begin{array}{l}\frac{P_2}{p^{*}}=\frac{1}{M{a}_2}{\left[\frac{k+1}{2+\left(k-1\right)M{a}_2^2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ 63.35=\frac{1}{M{a}_2}{\left[\frac{1.32+1}{2+\left(1.32-1\right)M{a}_2{}^2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ M{a}_2=0.017\end{array}$

Find the value of $\frac{f{L}_2^{*}}{D}$

$\begin{array}{l}\frac{f{L}_2^{*}}{D}=\frac{1-M{a}_2^2}{kM{a}_2^2}+\frac{k+1}{2k}\ln \left(\frac{\left(k+1\right)M{a}_2^2}{2+\left(k-1\right)M{a}_2^2}\right)\\ \frac{f{L}_1^{*}}{D}=\frac{1-{0.017}^2}{1.32*{0.017}^2}+\frac{1.32+1}{2*1.32}\ln \left(\frac{\left(1.32+1\right){0.017}^2}{2+\left(1.32-1\right){0.017}^2}\right)\\ =2620.612+0.88*\ln \left(3.35*{10}^{-4}\right)\\ =262.0612-7.041\\ =2613.57\end{array}$

Find the change in length of the pipe

$\begin{array}{l}\frac{f\delta L^{*}}{D}=\frac{f{L}_1^{*}}{D}-\frac{f{L}_2^{*}}{D}\\ \frac{f\delta L^{*}}{D}=4087.7-2613.57\\ =1474.13\end{array}$

Substitute above values we get

$\begin{array}{l}\frac{0.00608*\delta L^{*}}{1.321}=1474.13\\ \delta L^{*}=320283.84m\end{array}$

Convert the units we get

$\begin{array}{l}320283.84in=320283.84*0.6214*{10}^{-3}mi\\ =199mi\end{array}$.

To determine

(b)

Find the temperature at that point by using given information.

Answer to Problem 9.105P

The value of temperature is determined below.

Explanation of Solution

Find the temperature ratio we get

$\begin{array}{l}\frac{T_2}{T^{*}}=\frac{k+1}{2+\left(k-1\right)M{a}_2^2}\\ \frac{T_2}{T^{*}}=\frac{1.32+1}{2+\left(\left(1.32-1\right)*{0.017}^2\right)}\\ =1.16\end{array}$

Find the final temperature we get

$\begin{array}{l}\frac{T_2}{T^{*}}=\frac{T_2}{333}1.16\\ T_2=333K\\ ={60}^{0}C\\ ={140}^{0}F\end{array}$.