Chapter 17, Problem 17.1PP

A cylinder $25$ mm in diameter is placed perpendicular to a fluid stream with a velocity of $0.15m/s$. If the cylinder is $1$ m long, calculate the total drag force if the fluid is (a) water at $15°C$ and (b) air at $10°C$ and atmospheric pressure,

To determine

(a)

The total drag force

Answer to Problem 17.1PP

   $\text{6}{\text{.765×10}}^{\text{-3}}\text{N}$

Explanation of Solution

Given:

Diameter, d = $\text{25}\text{mm}$

Velocity, v = $\text{0}\text{.15}\text{m/s}$

Length, l = $\text{1}\text{m}$

Formula used:

   ${\text{F}}_D=C_D\left(\frac{\rho v^2}{2}\right)A$

${\text{F}}_{\text{D}}$ is drag force

$C_D$ is drag coefficient

$\rho$ is density

v is velocity

A is area

${\mathrm{R}}_e=\frac{vd}{\upsilon }$ ${\mathrm{R}}_e$ is Reynold's number

d is diameter

   $\upsilon$ is Kinematic viscosity

Calculation:

For water $\text{υ=1}{\text{.1384×10}}^{\text{-6}}{\text{m}}^{\text{2}}\text{/s}$

   $\text{ρ=1000}{\text{kg/m}}^{\text{3}}$

   $\begin{array}{l}{\text{R}}_{\text{e}}\text{=}\frac{\text{0}{\text{.15×25×10}}^{\text{-3}}}{\text{1}{\text{.1384×10}}^{\text{-6}}}\text{=3294}\text{.097}\\ {\text{C}}_{\text{D}}\text{=1}\text{.225}\\ {\text{F}}_{\text{D}}\text{=1}\text{.225}\left(\frac{\text{1000×0}{\text{.15}}^{\text{2}}}{\text{2}}\right)\text{×}\frac{\text{π}}{\text{4}}\text{×}{\left({\text{25×10}}^{\text{-3}}\right)}^{\text{2}}\\ {\text{F}}_{\text{D}}\text{=6}{\text{.765×10}}^{\text{-3}}\text{N}\end{array}$

Conclusion:

The total drag force on a cylinder when moving through water is calculated as $\text{6}{\text{.765×10}}^{\text{-3}}\text{N}$

To determine

(b)

The total drag force

Answer to Problem 17.1PP

   $\text{5}{\text{.16×10}}^{\text{-6}}\text{N}$

Explanation of Solution

Given:

Diameter, d = $\text{25}\text{mm}$

Velocity, v = $\text{0}\text{.15}\text{m/s}$

Length, l = $\text{1}\text{m}$

Formula used:

   ${\text{F}}_D=C_D\left(\frac{\rho v^2}{2}\right)A$

   ${\text{F}}_{\text{D}}$ is drag force

   $C_D$ is drag coefficient

   $\rho$ is density

v is velocity

A is area

${\mathrm{R}}_e$ is Reynold's number

d is diameter

   $\upsilon$ is Kinematic viscosity

Calculation:

For air, $\text{υ=14}{\text{.16×10}}^{\text{-6}}{\text{m}}^{\text{2}}\text{/s}$

   $\text{ρ=1}\text{.246}{\text{kg/m}}^{\text{3}}$

   $\begin{array}{l}{\text{R}}_{\text{e}}\text{=}\frac{\text{0}{\text{.15×25×10}}^{\text{-3}}}{\text{14}{\text{.16×10}}^{\text{-6}}}\text{=264}\text{.83}\\ {\text{C}}_{\text{D}}\text{=0}\text{.75}\\ {\text{F}}_{\text{D}}\text{=0}\text{.75}\left(\frac{\text{1}\text{.246×0}{\text{.15}}^{\text{2}}}{\text{2}}\right)\text{×}\frac{\text{π}}{\text{4}}\text{×}{\left({\text{25×10}}^{\text{-3}}\right)}^{\text{2}}\\ {\text{F}}_{\text{D}}\text{=5}{\text{.16×10}}^{\text{-6}}\text{N}\end{array}$

Conclusion:

The total drag force on a cylinder when moving through water is calculated as $\text{5}{\text{.16×10}}^{\text{-6}}\text{N}$