Water flows through the $2-$in$-$diameter pipe at $0\mathrm{.}2f\frac{t^3}{s}$ as shown in the figure. If the pressure at $A$ is $36,$ and diameter of the pipe $CD$ is changed to $1$ in$,$ then determine the pressure at $C$ and $D$ :

Solution:

According to the datum set in the diagram:

Velocity in pipe $AB$:$v_A=v_B=\frac{t}{s}$;

Velocity in pipe CD: $v_C=v_D=$ ft$/$s;

$z_A=z_B=$ t;

$z_C=z_D=$ ft;

$\frac{P_A}{}=$

$\frac{v_A^2}{2g}=$

$\frac{v_C^2}{2g}=$

Hence $\frac{P_C}{}=$

So the pressure at point $C$ is $psf=$ psi.