Water is flowing smoothly through a closed-pipe system. At one point the speed of the water is `3.0m//s`. While at another point `1.0m` higher the speed is `4.0m//s`. If the pressure is `20 Kpa` at the lower point, what is the pressure at the upper point? What would the pressure at the upper point be if the water were to stop flowing and the pressure at the lower point were `18 kpa`?

Options

A. the pressure at `B` when water is flowing is `6.5kPa`.
B. the pressure at `B` when water is flowing is `8.0kPa`.
C. the pressure at `B` when water is stops flowing is `10kPa`.
D. the pressure at `B` when water is stops flowing is `8.0kPa`.

Answer

A, C

Explanation

Let `p_(1),h_(1),v_(1)` and `p_(2),h_(2),v_(2)` be the presures, heights and velocities of flow at the two points, respectively. Then Beroulli's theorem gives
`p_(1)+rhoh_(1)+1/2rhov_(1)^(2)=p_(2)+rhogh_(2)+11/2rhov_(2)^(2)`…………..
Putting `v_(1)=3.0m//s, v_(2)=4.0m//s, (h_(2)-h_(1))=1m, p_(1)=20kPa,`
We get
`p_(2)=20+[10^(3)xx10(-1)+(10^(3))/2[9-16]]xx10^(-3)`
`=20-10-3.5=6.5kPa`
Also when the flow stops, `v_(1)=v_(2)=0` and then from eqn i we have `P_(2)=20-10=0kPa`