Chapter 8, Problem 8.1P

Fully developed conditions are known to exist for water flowing through a 25-nim-diameer tube at 0.01 kg/s and $27°\text{C}$ . What is the maximum velocity of the water in the tube? What is the pressure gradient associated with the flow?

To determine

The maximum velocity of water inside the tube and pressure gradient along the flow.

Answer to Problem 8.1P

The maximum velocity of water inside the tube is $0.0408\text{m}/\text{s}$ and pressure gradient along the flow is $-0.892\text{Pa}/\text{m}$ .

Explanation of Solution

Given:

The diameter of the tube is $25\text{mm}$ .

Mass flow rate of water is $0.01\text{kg}/\text{s}$ .

Concept used:

Write the expression for the average velocity of water flows through the pipe.

  $u_m=\frac{4\dot{m}}{\rho \pi D^2}$ ....... (1)

Here, $u_m$ is the average velocity, $\dot{m}$ is the mass flow rate of water, $\rho$ is the density of water and $D$ is the diameter of tube.

Write the expression for the maximum velocity.

  $u_{\max }=2u_m$ ....... (2)

Here, $u_{\max }$ is the maximum velocity.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{4\dot{m}}{\pi D\mu }$ ....... (3)

Here, $\mathrm{Re}$ is the Reynolds number and $\mu$ is the viscosity of water.

Write the expression for the pressure gradient.

  $\frac{dP}{dx}=-\frac{64\rho u_m{}^2}{2\mathrm{Re}D}$ ....... (4)

Here, $\frac{dP}{dx}$ is the pressure gradient.

Calculation:

Substitute $0.01\text{kg}/\text{s}$ for $\dot{m}$ , $998\text{kg}/{\text{m}}^3$ for $\rho$ and $25\text{mm}$ for $D$ in equation (1).

  $\begin{array}{c}{u}_m=\frac{4\left(0.01\right)}{\left(998\right)\left(\pi \right){\left(25\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}\\ =\frac{0.04}{\left(998\right)\left(\pi \right){\left(0.025\right)}^2}\\ =0.0204\text{m}/\text{s}\end{array}$

Substitute $0.204\text{m}/\text{s}$ for $u_m$ in equation (2).

  $\begin{array}{c}{u}_{\max }=2\left(0.0204\right)\\ =0.0408\text{m}/\text{s}\end{array}$

Substitute $0.01\text{kg}/\text{s}$ for $\dot{m}$ , $25\text{mm}$ for $D$ and $855\times {10}^{-6}\text{Pa}\text{.s}$ for $\mu$ in equation (2).

  $\begin{array}{c}\mathrm{Re}=\frac{4\left(0.01\right)}{\pi \left(25\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\left(855\times {10}^{-6}\right)}\\ =\frac{0.04\times {10}^6}{\pi \left(0.025\right)\left(855\right)}\\ =595.667\end{array}$

Substitute $998\text{kg}/{\text{m}}^3$ for $\rho$ and $25\text{mm}$ for $D$ , $0.0204\text{m}/\text{s}$ for $u_m$ and $595.667$ for $\mathrm{Re}$ in equation (3).

  $\begin{array}{c}\frac{dP}{dx}=-\frac{64\left(998\right){\left(0.0204\right)}^2}{2\left(595.667\right)\left(25\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}\\ =-\frac{2658.09}{\left(1191.334\right)\left(0.025\right)}\\ =-0.892\text{Pa}/\text{m}\end{array}$

Conclusion:

Thus, The maximum velocity of water inside the tube is $0.0408\text{m}/\text{s}$ and pressure gradient along the flow is $-0.892\text{Pa}/\text{m}$ .