Question

A fluid, with a density of rho = 1165 kg/m3, flows in a horizontal pipe. In one segment of the pipe the flow speed is v1 = 4.53 m/s. In a second segment the flow speed is v2 = 1.77 m/s. What is the difference between the pressure in the second segment (P2) and the pressure in the first segment (P1)?

Answer 1

The pressure difference between two pipe is
`1.01 * 10^(4)` Pa

Step-by-step explanation:

Density
`\rho = 1165`
`(kg)/(m^(3) )`

Speed in one pipe
`v_(1) = 4.53`
`(m)/(s)`

Speed in second pipe
`v_(2) = 1.77`
`(m)/(s)`

According to the bernoulli equation,

The pressure difference is given by,

`P = (1)/(2) \rho v^(2)`

`P_(2) - P_(1) = (1)/(2) \rho (v_(1)^(2) - v_(2)^(2) )`

`P_(2) - P_(1) = (1)/(2) * 1165 *[ (4.53)^(2)- (1.77)^(2)]`

`P_(2) - P_(1) = 10128.51`

`P_(2) - P_(1) = 1.01 * 10^(4)` Pa

Therefore, the pressure difference between two pipe is
`1.01 * 10^(4)` Pa