Chapter 5, Problem 1FP

To determine

The pressure at A and the velocity of the water at B.

Explanation of Solution

Given:

The velocity at A $\left(V_A\right)$ is $6\text{ m/s}$.

The datum at A $\left(z_A\right)$ is $\text{3 m}$

Calculation:

Since the cross section of the pipe is constant throughout the length so the velocity of the pipe also constant at both end.

$V_A=V_B$ $\left(6\text{m}/\text{s}\right)$.

Thus, velocity at B is $\underset{\_}{6\text{m}/\text{s}}$.

Apply Bernoulli equation between A and B.

  $\frac{p_A}{{\rho }_w}+\frac{V_A^2}{2}+g{z}_A=\frac{p_B}{{\rho }_w}+\frac{V_B^2}{2}+z_{B}g$        (I)

Here, $p_A$ is pressure at A, $p_B$ is pressure at B, $V_A$ is velocity at A, $V_B$ is velocity at B, g is acceleration due to gravity, $z_A$ is datum at A, $z_B$ is datum at B, ${\rho }_w$ is density of water at $5°\text{C}$.

Substitute ${\rho }_w=1,000\text{kg}/{\text{m}}^{\text{3}}$, $V_A=6\text{m}/\text{s}$, $g=9.81\text{m}/{\text{s}}^{\text{2}}$, $z_A=3m$, $p_B=0$, $V_B=6\text{m}/\text{s}$, and $z_B=0$ in equation (I)

  $\begin{array}{l}\frac{p_A}{1,000}+\frac{6^2}{2\times 9.81}+\left(3\times 9.81\right)=0+\frac{6^2}{2\times 9.81}+0\\ \frac{p_A}{1,000}+29.43=0\\ p_A=-29.43\times {10}^3\text{Pa}\end{array}$

  $\begin{array}{c}{p}_A=-29.43\times {10}^3\text{Pa}\times \frac{\text{1}\text{kPa}}{1,000\text{Pa}}\\ =-29.43\text{kPa}\end{array}$

Here, Negative sign indicates partial vacuum at A.

Thus, pressure at A is $\underset{\_}{-29.43\text{kPa}}$.