Water is flowing through two horizontal pipes of different diameters which are connected together. In the first pipe the speed of water 4 m/s and the pressure is `2.0xx10^(4) N//m^(2)`. Calculate the speed and pressure of water in the second pipe. The diameter of the pipes the diameter of the pipes are 3 cm and 6 cm respectively.

Explanation

If A is the area of cross section of a pipe at a point and v is the velocity of flow of water at that point. They by the principle of continuity `Av=` constant `implies`
`A_(1)v_(1)=A_(2)v_(2)impliespir_(1)^(2)v_(1)=pir_(2)^(2)v_(2)`
`impliesv_(2)=((r_(1))/(r_(2)))^(2)v_(1)=(1.5xx10^(-2))/(3xx10^(-2))xx4=1m//s`
From Bernoullis theorem:
`P_(1)+1/2rhov_(1)^(2)=P_(2)+1/2rhov_(2)^(2)`
`impliesP_(2)=P_(1)+1/2 rho (v_(1)^(2)-v_(2)^(2))`
`:.P_(2)=2xx10^(4)+1/2xx(10^(3))xx(16-1)`
`=2xx10^(4)+7.5xx10^(3)=2.75xx10^(4)N//m^(2)`