Chapter 2, Problem 2.45P

(a)

Interpretation Introduction

Interpretation:

To determine mass flow rate ( $\stackrel{•}{m}$ ) in $\text{kg/s}$ and $\frac{\Delta P}{\Delta L}$ in $\text{kPa/m}$ for steady incompressible fluid flow in a horizontal pipe of constant cross-sectional area.

Concept Introduction:

The relation provided for $\frac{\Delta P}{\Delta L}$ during incompressible steady fluid flow in horizontal pipe:

  $\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0$

The density of liquid water is, $\rho =996{\text{kg/m}}^{\text{3}}$

Viscosity $\mu =\text{9}{\text{.0x10}}^{\text{-4}}\text{kg/m}\text{.s}$

Assumed $\frac{\epsilon }{D}=0.0001$

Answer to Problem 2.45P

Mass flow rate ( $\stackrel{•}{m}$ ) $=\text{0}\text{.3129}\text{kg/s}$

  $\frac{\Delta P}{\Delta L}=-0.632\text{kPa/m}$

Explanation of Solution

Diameter of the pipe is, $D=\text{2}\text{cm}$

Velocity, $u=\text{1}\text{m/s}$

Calculate, $R_e=\frac{Du\rho }{\mu }$

Substitute

  $\begin{array}{l}D=2\text{cm}=2{\text{x10}}^{-2}\text{m}\\ u=1\text{m/s}\\ \mu =9.0{\text{x10}}^{-4}\text{kg/m}\text{.s}\\ \rho =996{\text{kg/m}}^3\end{array}$

  $\begin{array}{c}{R}_e=\frac{Du\rho }{\mu }\\ =\frac{{\text{2x10}}^{\text{-2}}\text{x1x996}}{\text{9}{\text{.0x10}}^{\text{-4}}}\\ =\text{22133}\text{.33}\end{array}$

Calculating fanning friction factor $f_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}$

Substituting $\frac{\epsilon }{D}=0.0001$ and $R_e=22133.33$

  $\begin{array}{l}{f}_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}\\ =\text{0}\text{.3305}{\left\{\text{ln}\left[\text{0}\text{.27x0}\text{.0001+}{\left(\frac{\text{7}}{\text{22133}\text{.33}}\right)}^{\text{0}\text{.9}}\right]\right\}}^{\text{-2}}\\ =0.00635\end{array}$

Calculate mass flow rate $\stackrel{•}{m}$ as shown below:

  $\stackrel{•}{m}=\rho uA$

A=Area of pipe

  $\begin{array}{c}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left(2{\text{x10}}^{-2}\right)}^2\\ =0.00031416{\text{m}}^{\text{2}}\end{array}$

Then,

  $\begin{array}{c}\stackrel{•}{m}=996\text{x1x0}\text{.00031416}\\ =\text{0}\text{.3129}\text{kg/s}\end{array}$

Therefore, the mass flow rate of liquid water is $=\text{0}\text{.3129}\text{kg/s}$

Calculate $\frac{\Delta P}{\Delta L}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0\\ \frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\end{array}$

Substitute

  $\begin{array}{c}{f}_F=0.00635\\ \rho =996{\text{kg/m}}^{\text{3}}\\ D=2\text{cm}=2{\text{x10}}^{-2}\text{m}\\ u=1\text{m/s}\end{array}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\\ =-\frac{\text{2}}{{\text{2x10}}^{\text{-2}}}\text{x0}\text{.00635x996x(1}{)}^2\\ =-0.632\text{kPa/m}\end{array}$

Therefore $\frac{\Delta P}{\Delta L}=-0.632\text{kPa/m}$

(b)

Interpretation Introduction

Interpretation:

To determine mass flow rate ( $\stackrel{•}{m}$ ) in $\text{kg/s}$ and $\frac{\Delta P}{\Delta L}$ in $\text{kPa/m}$ for steady incompressible fluid flow in a horizontal pipe of constant cross-sectional area.

Concept Introduction:

The relation provided for $\frac{\Delta P}{\Delta L}$ during incompressible steady fluid flow in horizontal pipe

  $\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0$

The density of liquid water is, $\rho =996{\text{kg/m}}^{\text{3}}$

Viscosity $\mu =\text{9}{\text{.0x10}}^{\text{-4}}\text{kg/m}\text{.s}$

Assumed $\frac{\epsilon }{D}=0.0001$

Answer to Problem 2.45P

Mass flow rate ( $\stackrel{•}{m}$ ) $=1.9522\text{kg/s}$

  $\frac{\Delta P}{\Delta L}=-0.252\text{kPa/m}$

Explanation of Solution

Diameter of the pipe is, $D=5\text{cm}$

Velocity, $u=\text{1}\text{m/s}$

Calculate, $R_e=\frac{Du\rho }{\mu }$

Substitute

  $\begin{array}{l}D=5\text{cm}={\text{5x10}}^{-2}\text{m}\\ u=1\text{m/s}\\ \mu =9.0{\text{x10}}^{-4}\text{kg/m}\text{.s}\\ \rho =996{\text{kg/m}}^3\end{array}$

  $\begin{array}{c}{\text{R}}_{\text{e}}=\frac{\text{Duρ}}{\text{μ}}\\ =\frac{{\text{5x10}}^{\text{-2}}\text{x1x996}}{\text{9}{\text{.0x10}}^{\text{-4}}}\\ =\text{55333}\text{.33}\end{array}$

Calculating fanning friction factor $f_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}$

Substituting $\frac{\epsilon }{D}=0.0001$ and $R_e=55333.33$

  $\begin{array}{l}{f}_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}\\ =\text{0}\text{.3305}{\left\{\text{ln}\left[\text{0}\text{.27x0}\text{.0001+}{\left(\frac{\text{7}}{\text{55333}\text{.33}}\right)}^{\text{0}\text{.9}}\right]\right\}}^{\text{-2}}\\ =0.0051715\end{array}$

Calculate mass flow rate $\stackrel{•}{m}$ as shown below:

  $\stackrel{•}{m}=\rho uA$

A=Area of pipe

  $\begin{array}{c}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left({\text{5x10}}^{-2}\right)}^2\\ =0.00196{\text{m}}^{\text{2}}\end{array}$

Then,

  $\begin{array}{c}\stackrel{•}{m}=996\text{x1x0}\text{.00196}\\ =1.9522\text{kg/s}\end{array}$

Therefore, the mass flow rate of liquid water is $=1.9522\text{kg/s}$

Calculate $\frac{\Delta P}{\Delta L}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0\\ \frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\end{array}$

Substitute

  $\begin{array}{c}{f}_F=0.0051715\\ \rho =996{\text{kg/m}}^{\text{3}}\\ D=5\text{cm}={\text{5x10}}^{-2}\text{m}\\ u=1\text{m/s}\end{array}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\\ =-\frac{\text{2}}{{\text{5x10}}^{\text{-2}}}\text{x0}\text{.00635x996x(1}{)}^2\\ =-0.252\text{kPa/m}\end{array}$

Therefore $\frac{\Delta P}{\Delta L}=-0.252\text{kPa/m}$

(c)

Interpretation Introduction

Interpretation:

To determine mass flow rate ( $\stackrel{•}{m}$ ) in $\text{kg/s}$ and $\frac{\Delta P}{\Delta L}$ in $\text{kPa/m}$ for steady incompressible fluid flow in a horizontal pipe of constant cross-sectional area.

Concept Introduction:

The relation provided for $\frac{\Delta P}{\Delta L}$ during incompressible steady fluid flow in horizontal pipe

  $\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0$

The density of liquid water is, $\rho =996{\text{kg/m}}^{\text{3}}$

Viscosity $\mu =\text{9}{\text{.0x10}}^{\text{-4}}\text{kg/m}\text{.s}$

And, assume $\frac{\epsilon }{D}=0.0001$

Answer to Problem 2.45P

Mass flow rate ( $\stackrel{•}{m}$ ) $=1.565\text{kg/s}$

  $\frac{\Delta P}{\Delta L}=-15.811\text{kPa/m}$

Explanation of Solution

Diameter of the pipe is, $D=\text{2}\text{cm}$

Velocity, $u=5\text{m/s}$

Calculate, $R_e=\frac{Du\rho }{\mu }$

Substitute

  $\begin{array}{l}D=2\text{cm}=2{\text{x10}}^{-2}\text{m}\\ u=5\text{m/s}\\ \mu =9.0{\text{x10}}^{-4}\text{kg/m}\text{.s}\\ \rho =996{\text{kg/m}}^3\end{array}$

  $\begin{array}{c}{R}_e=\frac{Du\rho }{\mu }\\ =\frac{{\text{2x10}}^{\text{-2}}\text{x5x996}}{\text{9}{\text{.0x10}}^{\text{-4}}}\\ =110666.667\end{array}$

Calculating fanning friction factor $f_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}$

Substituting $\frac{\epsilon }{D}=0.0001$ and $R_e=110666.667$

  $\begin{array}{l}{f}_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}\\ =\text{0}\text{.3305}{\left\{\text{ln}\left[\text{0}\text{.27x0}\text{.0001+}{\left(\frac{\text{7}}{110666.667}\right)}^{\text{0}\text{.9}}\right]\right\}}^{\text{-2}}\\ =0.00452\end{array}$

calculate mass flow rate $\stackrel{•}{m}$ as shown below:

  $\stackrel{•}{m}=\rho uA$

A=Area of pipe

  $\begin{array}{c}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left(2{\text{x10}}^{-2}\right)}^2\\ =0.00031416{\text{m}}^{\text{2}}\end{array}$

Then,

  $\begin{array}{c}\stackrel{•}{m}=996\text{x5x0}\text{.00031416}\\ =1.565\text{kg/s}\end{array}$

Therefore, the mass flow rate of liquid water is $=\text{1}\text{.565}\text{kg/s}$

Calculate $\frac{\Delta P}{\Delta L}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0\\ \frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\end{array}$

Substitute

  $\begin{array}{c}{f}_F=0.00452\\ \rho =996{\text{kg/m}}^{\text{3}}\\ D=2\text{cm}=2{\text{x10}}^{-2}\text{m}\\ u=5\text{m/s}\end{array}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\\ =-\frac{\text{2}}{{\text{2x10}}^{\text{-2}}}\text{x}0.00452\text{x996x(5}{)}^2\\ =-15.811\text{kPa/m}\end{array}$

Therefore $\frac{\Delta P}{\Delta L}=-15.811\text{kPa/m}$

(d)

Interpretation Introduction

Interpretation:

To determine mass flow rate ( $\stackrel{•}{m}$ ) in $\text{kg/s}$ and $\frac{\Delta P}{\Delta L}$ in $\text{kPa/m}$ for steady incompressible fluid flow in a horizontal pipe of constant cross-sectional area.

Concept Introduction:

The relation provided for $\frac{\Delta P}{\Delta L}$ during incompressible steady fluid flow in horizontal pipe

  $\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0$

The density of liquid water is, $\rho =996{\text{kg/m}}^{\text{3}}$

Viscosity $\mu =\text{9}{\text{.0x10}}^{\text{-4}}\text{kg/m}\text{.s}$

And, assume $\frac{\epsilon }{D}=0.0001$

Answer to Problem 2.45P

Mass flow rate ( $\stackrel{•}{m}$ ) $=9.7608\text{kg/s}$

  $\frac{\Delta P}{\Delta L}=-6.324\text{kPa/m}$

Explanation of Solution

Diameter of the pipe is, $D=5\text{cm}$

Velocity, $u=5\text{m/s}$

Calculate, $R_e=\frac{Du\rho }{\mu }$

Substitute

  $\begin{array}{l}D=5\text{cm}=5{\text{x10}}^{-2}\text{m}\\ u=5\text{m/s}\\ \mu =9.0{\text{x10}}^{-4}\text{kg/m}\text{.s}\\ \rho =996{\text{kg/m}}^3\end{array}$

  $\begin{array}{c}{\text{R}}_{\text{e}}=\frac{\text{Duρ}}{\text{μ}}\\ =\frac{{\text{5x10}}^{\text{-2}}\text{x5x996}}{\text{9}{\text{.0x10}}^{\text{-4}}}\\ =276666.667\end{array}$

Calculating fanning friction factor $f_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}$

Substituting $\frac{\epsilon }{D}=0.0001$ and $R_e=276666.667$

  $\begin{array}{l}{f}_F=0.3305{\left\{\ln \left[0.27\frac{\epsilon }{D}+{\left(\frac{7}{R_e}\right)}^{0.9}\right]\right\}}^{-2}\\ =\text{0}\text{.3305}{\left\{\text{ln}\left[\text{0}\text{.27x0}\text{.0001+}{\left(\frac{\text{7}}{276666.667}\right)}^{\text{0}\text{.9}}\right]\right\}}^{\text{-2}}\\ =0.003895\end{array}$

Calculate mass flow rate $\stackrel{•}{m}$ as shown below:

  $\stackrel{•}{m}=\rho uA$

A=Area of pipe

  $\begin{array}{c}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left({\text{5x10}}^{-2}\right)}^2\\ =0.00196{\text{m}}^{\text{2}}\end{array}$

Then,

  $\begin{array}{c}\stackrel{•}{m}=996\text{x5x0}\text{.00196}\\ =9.7608\text{kg/s}\end{array}$

Therefore, the mass flow rate of liquid water is $=9.7608\text{kg/s}$

Calculate $\frac{\Delta P}{\Delta L}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}+\frac{2}{D}{f}_F\rho u^2=0\\ \frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\end{array}$

Substitute

  $\begin{array}{c}{f}_F=0.003895\\ \rho =996{\text{kg/m}}^{\text{3}}\\ D=5\text{cm}=5{\text{x10}}^{-2}\text{m}\\ u=5\text{m/s}\end{array}$

  $\begin{array}{c}\frac{\Delta P}{\Delta L}=-\frac{2}{D}{f}_F\rho u^2\\ =-\frac{\text{2}}{{\text{5x10}}^{\text{-2}}}\text{x}0.00452{\text{x996x(5)}}^2\\ =-6.324\text{kPa/m}\end{array}$

Therefore $\frac{\Delta P}{\Delta L}=-6.324\text{kPa/m}$