Water is floating smoothly through a closed-pipe system. At one point A, the speed of the water is 3.0 m/s while at another point B, 1.0 m higher, the speed is 4.0 m/s. The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. Then

Let p₁, h₁, v₁ and p₂, h₂, v₂ be the pressures, heights and velocities of flow at the two points, respectively. Then Bernoulli's theorem gives

           ${\mathrm{p}}_1+{\mathrm{\rho gh}}_1+\frac{1}{2}{\mathrm{\rho v}}_1^2={\mathrm{p}}_2+{\mathrm{\rho gh}}_2+\frac{1}{2}{\mathrm{\rho v}}_2^2$                …(i)

Putting ${\mathrm{v}}_1=3.0\mathrm{m}/\mathrm{s}, {\mathrm{v}}_2=4.0\mathrm{m}/\mathrm{s}, \left({\mathrm{h}}_2-{\mathrm{h}}_1\right)=1\mathrm{m}, {\mathrm{p}}_1=20\mathrm{kPa}$, we get

         $\begin{array}{l}{\mathrm{p}}_2=20+\left[{10}^3\times 10(-1)+\frac{{10}^3}{2}[9-16]\right]\times {10}^{-3}\\ =20-10-3.5=6.5\mathrm{kPa}\end{array}$

Also when the flow stops, ${\mathrm{v}}_1={\mathrm{v}}_2=0$ and then from Eq. (i), we have ${\mathrm{P}}_2=20-10=10\mathrm{kPa}$