Chapter 8, Problem 8.1PP

A 4-in-ductile iron pipe carries $0.20f{t}^3/s$ of glycerin (sg = $1.26$ ) at $100°F$. Is the flow laminar or turbulent?

To determine

The flow is laminar or turbulent.

Answer to Problem 8.1PP

The flow is laminar

Explanation of Solution

Given information:

The diameter of the ductile iron pipe is $4\text{in}$and the volume flow rate is $0.20{\text{ft}}^{\text{3}}/\text{s}$.

Write the expression for Reynolds number.

$N_R=\frac{VD}{v}$.......(I)

Here, Reynolds number is $N_R$, velocity of flow is $V$., diameter of the pipe is $D$and the kinematic viscosity is $v$.

Write the expression for the area of the pipe.

$A=\frac{\pi D^2}{4}$.......(II)

Here, area of the pipe is $A$.

Write the expression for velocity..

$V=\frac{Q}{A}$.......(III)

Here, volume flow rate is $Q$.

Calculation:

Substitute $4\text{in}$for $D$on Equation (II).

$\begin{array}{c}A=\frac{\pi {\left(4\text{in}\right)}^2}{4}\\ =\frac{\pi {\left(\left(4\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}{4}\\ =0.0855{\text{ft}}^{\text{2}}\end{array}$

Substitute $0.0855{\text{ft}}^{\text{2}}$for $A$, $0.20{\text{ft}}^{\text{3}}/\text{s}$for $Q$in Equation (III).

$\begin{array}{c}V=\frac{0.20{\text{ft}}^{\text{3}}/\text{s}}{0.0855{\text{ft}}^{\text{2}}}\\ =2.34\text{ft}/\text{s}\end{array}$

Refer to Appendix-A, "Properties of Liquid" at $100\text{°F}$to obtain the kinematic viscosity as $3.0\times {10}^{-3}{\text{ft}}^2/\text{s}$.

Substitute $2.34\text{ft}/\text{s}$for $V$, $4\text{in}$for $D$and $3.0\times {10}^{-3}{\text{ft}}^2/\text{s}$for $v$in Equation (I).

$\begin{array}{c}{N}_R=\frac{\left(2.34\text{ft}/\text{s}\right)\left(4\text{in}\right)}{3.0\times {10}^{-3}{\text{ft}}^2/\text{s}}\\ =\frac{\left(2.34\text{ft}/\text{s}\right)\left(4\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)}{3.0\times {10}^{-3}{\text{ft}}^2/\text{s}}\\ =260\end{array}$

Since Reynolds number is $260$and it is less than $2000$so the flow is laminar.

Conclusion:

The flow is laminar.