Chapter 10, Problem 1FP

To determine

The maximum velocity $\left(V\right)$ of the glycerin and the head loss over the $4\text{m}$ length.

Explanation of Solution

Given:

Temperature $\left(T\right)$ is $20°\text{C}$.

Diameter is $\left(D\right)$ is $4\text{0}\text{mm}$

Calculation:

At the temperature of $20°\text{C}$ consider the following properties:

Density of glycerin is $\left({\rho }_{gl}\right)=1260\text{kg}/{\text{m}}^3$.

Kinematic viscosity is $\left(v_{gl}\right)=1.19\times {10}^{-3}{\text{m}}^2/\text{s}$.

Calculate the Reynold’s number $\left(R_e\right)$ .

  $\begin{array}{c}{R}_e=\frac{VD}{v_{gl}}\\ R_e=\frac{V\times \left(40\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{1.19\times {10}^{-3}}\\ =33.61V\end{array}$

Consider the flow in pipe is laminar.

Calculate the Darcy’s friction factor $\left(f\right)$.

  $\begin{array}{c}f=\frac{64}{R_e}\\ =\frac{64}{33.61V}\\ =\frac{1.904}{V}\end{array}$

Calculate the head loss due to friction $\left(h_f\right)$.

  $\begin{array}{c}{h}_f=\frac{fL{V}^2}{2gD}\\ h_f=\frac{\left(\frac{1.904}{V}\right)\times 4\times V^2}{2\times 9.81\times 0.04}\\ h_f=9.704V\end{array}$        (I)

Apply the Bernoulli’s Equation between point $A$ and $B$. Consider the point $B$ as datum.

  $\begin{array}{l}\frac{P_A}{{\gamma }_{gl}}+\frac{V_A^2}{2g}+Z_A=\frac{P_B}{{\gamma }_{gl}}+\frac{V_B^2}{2g}+Z_B+h_L\\ 0+0+5.5=0+\frac{V}{2\times 9.81}+0+9.704V\\ V=0.564\text{m}/\text{s}\end{array}$

Thus, the maximum velocity $\left(V\right)$ of the glycerin is $\underset{\_}{0.564\text{m}/\text{s}}$.

Substitute $V=0.564\text{m}/\text{s}$ in Equation (I) and calculate the head loss over the $4\text{m}$ length.

  $\begin{array}{c}{h}_f=9.704V\\ =9.704\left(0.564\text{m}/\text{s}\right)\\ =5.47\text{m}\end{array}$

Thus, the head loss over the $4\text{m}$ length is $\underset{\_}{5.47\text{m}}$.