Question

At one point in a pipeline, the water's speed is 4.37 m/s and the gauge pressure is 46.4 kPa. Find the gauge pressure at a second point in the line, 11.2 m lower than the first, if the pipe diameter at the second point is twice that at the first. Remember that the density of water is 1000 kg/m^3. Please give your answer in units of kPa.

Answer 1

165.11 K Pa

Step-by-step explanation:

Applying Bernoulli theorem ,the problem which deals with the conservation of energy of a flowing fluid , we have

P + 1/2ρ v² + ρgh = constant

P represents pressure , ρ represents density of liquid flowing v is velocity, h is the height from the ground.

P₁ + 1/2ρ v₁² + ρgh₁ =P₂ + 1/2ρ v₂² + ρgh₂

Substituting the values we get

46.4 x 10³ + .5 x 1000 x (4.37)²+ρg( h₁ - h₂) = P₂ + .5 X 1000x(4.37/4)²

46400+ 9548.45 +1000x 9.8 x 11.2 = P₂ +596.78

P₂ = 165.11 K Pa