Chapter 8, Problem 1P

Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 10 mm. Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.

To determine

The general expression for Reynolds number in terms of volume flow rate and tube diameter.

Explanation of Solution

Given:

Tube diameter 1$\left(D_1\right)$ is $10\text{mm}$.

Tube diameter 2$\left(D_2\right)$ is $6\text{mm}$.

Reynolds number 1$\left({\mathrm{Re}}_1\right)$ is $1800$.

Calculation:

Write the equation for the volume flow rate $\left(Q\right)$.

  $Q=A\overline{V}$

Write the equation for the mass flow rate $\left(\dot{m}\right)$.

  $\dot{m}=\rho A\overline{V}$

Write the equation for the cross sectional area $\left(A\right)$.

  $A=\frac{\pi D^2}{4}$

Calculate the Reynolds number in terms of volume flow rate and tube diameter $\left(\mathrm{Re}\right)$.

  $\begin{array}{c}\mathrm{Re}=\frac{\rho D\overline{V}}{\mu }\\ =\frac{\rho D}{\mu }\left(\frac{Q}{A}\right)\\ =\frac{\rho D}{\mu }\left(\frac{Q}{\frac{\pi D^2}{4}}\right)\\ =\frac{4Q}{\pi D}\frac{\rho }{\mu }\end{array}$

  $\begin{array}{c}=\frac{4Q}{\pi D}\left(\frac{1}{\overline{V}}\right)\\ =\frac{4Q}{\pi D\overline{V}}\end{array}$

Thus, the Reynolds number in terms of volume flow rate and tube diameter is $\frac{4Q}{\pi D\overline{V}}$.

To determine

The general expression for Reynolds number in terms of mass flow rate and tube diameter and the Reynolds number for the same flow rate in the section.

Explanation of Solution

Calculate the Reynolds number in terms of mass flow rate and tube diameter $\left(\mathrm{Re}\right)$.

  $\begin{array}{c}\mathrm{Re}=\frac{\rho D\overline{V}}{\mu }\\ =\frac{D}{\mu }\left(\frac{\rho \overline{V}A}{A}\right)\\ =\frac{D}{\mu }\left(\frac{\dot{m}}{\frac{\pi D^2}{4}}\right)\\ =\frac{4\dot{m}}{\pi D\mu }\end{array}$

Thus, the Reynolds number in terms of mass flow rate and tube diameter is $\frac{4\dot{m}}{\pi D\mu }$.

Calculate the Reynolds number for the same flow rate in the section $\left({\mathrm{Re}}_2\right)$.

  $D_1{\mathrm{Re}}_1=D_2{\mathrm{Re}}_2$

  $\begin{array}{c}{\mathrm{Re}}_2=\frac{D_1}{D_2}{\mathrm{Re}}_1\\ =\frac{\left(10\text{mm}\right)}{\left(6\text{mm}\right)}\left(1800\right)\\ =3000\end{array}$

Thus, the Reynolds number for the same flow rate in the section is $3000$.