Oil enters the bend of a pipe in the horizontal plane with velocity `4ms^(-1)` and pressure `280xx10^(3)Nm^(-2)` as shown in the figure. Find the pressure of oil at the point `Q` in `(kNm^(2))`
(Take specific gravity of oil as 0.9 and `sin37^@=0.6.`)

Answer

`172(kNm^(-2))`

Explanation

Let `A_(1)` and `A_(2)` be the cross sectional areas of the pipe at points `P` and `Q`, respectively.
Let `v_(1)` and `v_(2)` be velocities of oil at the points `P` and `Q`, respectively. By conservation of mass.
`Q=A_(1)v_(1)=A_(2)v_(2)impliesv_(2)=((A_(1))/(A_(2)))(v_(1))=4v_(1)`
Applying Bernoulli's equation between points `P` and `Q` we have
`P_(1)+1/2rhov_(1)^(2)=P_(2)+1/2rhov_(2)^(2)`
`P_(2)=P_(1)+1/2rho(v_(1)^(2)-v_(2)^(2))`
`=P_(1)+1/2xx0.9[16-256]xx10^(3)`
`=280xx10^(3)+(-108)xx10^(3)`
`P_(2)=172xx10^(3)Nm^(-2)`
`P_(2)=172(kNm^(-2))`