Chapter 2, Problem 1FP

To determine

The resultant force the water and surrounding air exert on the cap at B.

Explanation of Solution

Given:

Absolute pressure at point A $\left(p_A\right)$ is 400 kPa.

Inner diameter of the pipe $\left(d\right)$ is 50 mm.

Atmospheric pressure $\left(p_{\text{atm}}\right)$ is 101 kPa.

Pressure head of water $\left(h_w\right)$ is 0.3 m.

Density of water $\left({\rho }_w\right)$ is $1,000\text{kg}/{\text{m}}^{\text{3}}$.

Acceleration due to gravity is $\left(g\right)$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Draw the free body diagram of the pipe as in Figure (1).

Determine the pressure at point B $\left(P_B\right)$.

  $\begin{array}{l}{p}_B+p_w=p_A\\ p_B+{\gamma }_w{h}_w=p_A\\ p_B+{\rho }_{w}g{h}_w=p_A\end{array}$

  $\begin{array}{l}{p}_B+1,000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.3\text{ m}=400\times {10}^3\text{Pa}\\ p_B=400\times {10}^3-2.943\times {10}^3\\ p_B=\left(397.06\times {10}^3\right)\text{Pa}\\ p_B=397.06\text{kPa}\end{array}$

Draw the pressure distribution on the pipe as in Figure (2).

Determine the area of the pipe (A).

$A=\frac{\pi d^2}{4}$

$\begin{array}{c}A=\frac{\pi {\left(50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2}{4}\\ =1.963\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Determine the resultant force $\left(F_R\right)$ using vertical force equilibrium.

$\begin{array}{l}\Sigma F_y=0\\ F_R=p_B\times A-p_{\text{atm}}\times A\\ F_R=\left(p_B-p_{\text{atm}}\right)A\end{array}$

$\begin{array}{c}{F}_R=\left(397.06\times {10}^3\text{Pa}-101.3\times {10}^3\text{Pa}\right)1.963\times {10}^{-3}{\text{m}}^2\\ F_R=581\text{N}\end{array}$

Thus, the resultant force the water and surrounding air exert on the cap at B is $581\text{ N}$.