Chapter 3, Problem 3.1P

(a)

Interpretation Introduction

Interpretation:

The type of flow (laminar or turbulent) for the following condition is to be determined:

Water flowing at an average velocity of 2 m/s in a 100-mm pipe at ${\text{10}}^{\text{o}}\text{ C}$ .

Concept Introduction:

In fluid flow, the Reynolds number $\left({\text{N}}_{\mathrm{Re}}\right)$ is used to determine the type of flow.

For the ${\text{N}}_{\mathrm{Re}}<2100$ , the flow is said to be laminar.

For the ${\text{2100<N}}_{\mathrm{Re}}<4000$ , the flow is said to be transient flow.

For the ${\text{N}}_{\mathrm{Re}}>4000$ , the flow is said to be turbulent.

The mathematical expression of Reynolds number is,

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$

Answer to Problem 3.1P

The type of flow for the given condition is turbulent.

Explanation of Solution

The Reynolds number is used to determine the type of flow for the given condition.

The Reynolds number is the ratio inertia forces to viscous forces and it is dimensionless.

The Reynolds number is given as:

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$ ...... (1)

Here, $\rho$ represents the density of the fluid

  $v$ represents the velocity of the fluid

  $D$ represents the diameter

  $\mu$ represents the viscosity of the fluid

The diameter of the pipe is $100\text{mm}$ .

Convert the unit of diameter from $\text{mm}$ to $\text{m}$ as follows:

  $\begin{array}{l}\text{1}\text{mm}={10}^{-3}\text{m}\\ D=100\text{mm}\times \frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}\end{array}$

  $D=0.1\text{m}$

The density of the water at $\text{10}°\text{C}$ is $\text{999}\text{.8}\text{kg}/{\text{m}}^3$ .

The viscosity of the water at $\text{10}°\text{C}$ is $1.31\times {10}^{-3}\text{Pa}\cdot \text{s}$ .

Average velocity is $\text{2}\text{m}/\text{s}$ .

Plugin the values in equation (1)

  $\begin{array}{l}{\text{N}}_{\mathrm{Re}}=\frac{\text{999}\text{.8}\text{kg}/{\text{m}}^3\times \text{2}\text{m}/\text{s}\times 0.1\text{m}}{1.31\times {10}^{-3}\text{Pa}\cdot \text{s}}\\ =\frac{199.96}{1.31\times {10}^{-3}}\\ =\frac{199.96\times {10}^3}{1.31}\\ =152.64\times {10}^3\end{array}$

  ${\text{N}}_{\mathrm{Re}}=1.53\times {10}^5$

Calculated ${\text{N}}_{\mathrm{Re}}>4000$ , therefore the flow is turbulent.

(b)

Interpretation Introduction

Interpretation:

The type of flow (laminar or turbulent) for the following condition is to be determined:

air flowing at 50 ft/s in a 12-in. duet at 2-atm and ${\text{180}}^{\text{o}}\text{ F}$ .

Concept Introduction :

In fluid flow, the Reynolds number $\left({\text{N}}_{\mathrm{Re}}\right)$ is used to determine the type of flow.

For the ${\text{N}}_{\mathrm{Re}}<2100$ , the flow is said to be laminar.

For the ${\text{2100<N}}_{\mathrm{Re}}<4000$ , the flow is said to be transient flow.

For the ${\text{N}}_{\mathrm{Re}}>4000$ , the flow is said to be turbulent.

The mathematical expression of Reynolds number is,

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$

The formula for calculating density of air is,

  $\rho =\frac{PM}{RT}$

Answer to Problem 3.1P

The type of flow for the given condition is turbulent.

Explanation of Solution

The Reynolds number is used to determine the type of flow for the given condition.

The Reynolds number is the ratio inertia forces to viscous forces and it is dimensionless.

The Reynolds number is given as:

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$ ...... (1)

Here, $\rho$ represents the density of the fluid

  $v$ represents the velocity of the fluid

  $D$ represents the diameter

  $\mu$ represents the viscosity of the fluid

The diameter of the pipe is $12\text{in}$ .

Convert the unit of diameter from $\text{in}$ to $\text{m}$ as follows:

  $\begin{array}{l}\text{1}\text{in}=0.0254\text{m}\\ D=12\text{in}\times \frac{0.0254\text{m}}{\text{1}\text{in}}\end{array}$

  $D=0.3048\text{m}$

The temperature at which air is flowing is $180°\text{F}$ .

Convert the unit of temperature from $°\text{F}\text{to}°\text{K}$ as follows:

  $\begin{array}{l}°\text{C}=\frac{1}{1.8}\left(°\text{F}-\text{32}\right)\\ T=\frac{1}{1.8}\left(180-\text{32}\right)°\text{C}\\ T=82.22°\text{C}\\ \text{K}\text{=}°\text{C}+273.15\\ T=82.22+273.15\\ T=355.37\text{K}\end{array}$

The density of the air is calculated as follows:

The value of universal gas constant is $0.08314\text{atm}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}$

The value of pressure is $2\text{atm}$

The molecular weight of air is $29\text{kg}/\text{kmol}$ .

The formula for calculating the density of air is,

  $\rho =\frac{PM}{RT}$ ….. (2)

Plugin the values in equation (2)

  $\begin{array}{l}\rho =\frac{2\text{atm}\times 29\text{kg}/\text{kmol}}{0.08314\text{atm}\cdot {\text{m}}^3/\text{kmol}\cdot \text{K}\times 355.37\text{K}}\\ =\frac{58}{29.54}\end{array}$

  $\rho =1.96\text{kg}/{\text{m}}^3$

The viscosity of the water at $355.37\text{K}$ is $0.3478\times {10}^{-3}\text{Pa}\cdot \text{s}$ .

The velocity is $50\text{ft}/\text{s}$ .

Convert the unit of velocity from $\text{ft}/\text{s}\text{to}\text{m}/\text{s}$

  $\begin{array}{l}1\text{ft}=0.3048\text{m}\\ v\text{=}50\text{ft}/\text{s}\times \frac{0.3048\text{m}}{\text{1ft}}\end{array}$

  $v=15.24\text{m}/\text{s}$

Plugin the values in equation (1)

  $\begin{array}{l}{\text{N}}_{\mathrm{Re}}=\frac{1.96\text{kg}/{\text{m}}^3\times 15.24\text{m}/\text{s}\times 0.3048\text{m}}{0.3478\times {10}^{-3}\text{Pa}\cdot \text{s}}\\ =\frac{9.1045\times {10}^3}{0.3478}\end{array}$

  ${\text{N}}_{\mathrm{Re}}=26177.4$ .

The calculated ${\text{N}}_{\mathrm{Re}}>4000$ , therefore the flow is turbulent.

(c)

Interpretation Introduction

Interpretation:

The type of flow (laminar or turbulent) for the following condition is to be determined:

Oil with viscosity of 20 cP flowing at 5 ft/s in a 2-in. pipe with a specific gravity of 0.78.

Concept Introduction :

In fluid flow, the Reynolds number $\left({\text{N}}_{\mathrm{Re}}\right)$ is used to determine the type of flow.

For the ${\text{N}}_{\mathrm{Re}}<2100$ , the flow is said to be laminar.

For the ${\text{2100<N}}_{\mathrm{Re}}<4000$ , the flow is said to be transient flow.

For the ${\text{N}}_{\mathrm{Re}}>4000$ , the flow is said to be turbulent.

The mathematical expression of Reynolds number is,

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$

The specific gravity of the oil is given. The formula for calculating the density of oil is,

  $\text{Specific}\text{gravity}\text{=}\frac{\text{Density}\text{of}\text{liquid}}{\text{Density}\text{of}\text{water}}$

  $\text{Density}\text{of}\text{liquid}=\text{Density}\text{of}\text{water}\times \text{Specific}\text{gravity}$

Answer to Problem 3.1P

The type of flow for the given condition is transient.

Explanation of Solution

The Reynolds number is used to determine the type of flow for the given condition.

The Reynolds number is the ratio inertia forces to viscous forces and it is dimensionless.

The Reynolds number is given as:

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$ ...... (1)

Here, $\rho$ represents the density of the fluid

  $v$ represents the velocity of the fluid

  $D$ represents the diameter

  $\mu$ represents the viscosity of the fluid

The diameter of the pipe is $2\text{in}$ .

Convert the unit of diameter from $\text{in}$ to $\text{m}$ as follows:

  $\begin{array}{l}\text{1}\text{in}=0.0254\text{m}\\ D=2\text{in}\times \frac{0.0254\text{m}}{\text{1}\text{in}}\end{array}$

  $D=0.0508\text{m}$

The specific gravity is the ratio of density of liquid to the density of water.

The density of the oil is calculated as follows:

  $\begin{array}{l}\text{Specific}\text{gravity}\text{=}\frac{{\rho }_{oil}}{{\rho }_{water}}\\ 0.78=\frac{{\rho }_{oil}}{1000\text{kg}/{\text{m}}^3}\\ {\rho }_{oil}=0.78\times 1000\text{kg}/{\text{m}}^3\end{array}$

  ${\rho }_{oil}=780\text{kg}/{\text{m}}^3$

The viscosity of the oil is $20\text{cP}$ .

Convert the unit of viscosity from $\text{cP}\text{to Pa}\cdot \text{s}$

  $\begin{array}{l}\text{1}\text{cP}=0.001\text{Pa}\cdot \text{s}\\ \mu =20\text{cP}\times \frac{0.001\text{Pa}\cdot \text{s}}{\text{1}\text{cP}}\end{array}$

  $\mu =0.02\text{Pa}\cdot \text{s}$

The velocity is $5\text{ft}/\text{s}$ .

Convert the unit of velocity from $\text{ft}/\text{s}\text{to}\text{m}/\text{s}$

  $\begin{array}{l}1\text{ft}=0.3048\text{m}\\ v\text{=}5\text{ft}/\text{s}\times \frac{0.3048\text{m}}{\text{1ft}}\end{array}$

  $v=1.524\text{m}/\text{s}$

Plugin the values in equation (1)

  $\begin{array}{l}{\text{N}}_{\mathrm{Re}}=\frac{780\text{kg}/{\text{m}}^3\times 1.524\text{m}/\text{s}\times 0.0508\text{m}}{0.02\text{Pa}\cdot \text{s}}\\ =\frac{60.387}{0.02}\end{array}$

  ${\text{N}}_{\mathrm{Re}}=3019.35$ .

The calculated ${\text{N}}_{\mathrm{Re}}>2100\text{and}<4000$ , therefore the flow is transient.

(d)

Interpretation Introduction

Interpretation:

The type of flow (laminar or turbulent) for the following condition is to be determined:

Polymer (density = 900 kg/m³and viscosity 1 Pa.s) melts and flowing at 0.2 m/s in a 15-mm tube.

Concept Introduction:

In fluid flow, the Reynolds number $\left({\text{N}}_{\mathrm{Re}}\right)$ is used to determine the type of flow.

For the ${\text{N}}_{\mathrm{Re}}<2100$ , the flow is said to be laminar.

For the ${\text{2100<N}}_{\mathrm{Re}}<4000$ , the flow is said to be transient flow.

For the ${\text{N}}_{\mathrm{Re}}>4000$ , the flow is said to be turbulent.

The mathematical expression of Reynolds number is,

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$

Answer to Problem 3.1P

The type of flow for the given condition is laminar.

Explanation of Solution

The Reynolds number is used to determine the type of flow for the given condition.

The Reynolds number is the ratio inertia forces to viscous forces and it is dimensionless.

The Reynolds number is given as:

  ${\text{N}}_{\mathrm{Re}}=\frac{\rho vD}{\mu }$ ...... (1)

Here, $\rho$ represents the density of the fluid

  $v$ represents the velocity of the fluid

  $D$ represents the diameter

  $\mu$ represents the viscosity of the fluid

The diameter of the pipe is $15\text{mm}$ .

Convert the unit of diameter from $\text{mm}$ to $\text{m}$ as follows:

  $\begin{array}{l}\text{1}\text{mm}={10}^{-3}\text{m}\\ D=15\text{mm}\times \frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}\end{array}$

  $D=0.015\text{m}$

The density of the polymer melt is $\text{900}\text{kg}/{\text{m}}^3$ .

The viscosity of the polymer melt is $1\text{Pa}\cdot \text{s}$ .

The velocity is $\text{0}\text{.2}\text{m}/\text{s}$ .

Plugin the values in equation (1)

  $\begin{array}{l}{\text{N}}_{\mathrm{Re}}=\frac{\text{900}\text{kg}/{\text{m}}^3\times 0.\text{2}\text{m}/\text{s}\times 0.015\text{m}}{1\text{Pa}\cdot \text{s}}\\ =\frac{2.7}{1}\end{array}$

  ${\text{N}}_{\mathrm{Re}}=2.7$

Calculated ${\text{N}}_{\mathrm{Re}}<2100$ , therefore the flow is laminar.