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 $48,$ and diameter of the pipe $CD$ is changed to $1in,$ then determine the pressure at $C$ and $D$ :

Solution:

According to the datum set in the diagram:

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

Velocity in pipe $CD$:$v_C=v_D=f\frac{t}{s}$;

$z_A=z_B=,ft$;

$z_C=z_D=,ft$;

$\frac{P_A}{}=ft$;

$\frac{v_A^2}{2g}=,f{t}_i$

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

Hence $\frac{P_C}{}=ft$;

So the pressure at point $C$ isB