Incomplete question.The complete question is here
At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is 5.00×10⁴Pa.Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.
Answer:
P₂=1.62×10⁵ Pa
Explanation:
Given Data
Pressure=5.00×10⁴Pa
Water Speed= 3.0 m/s
Length= 11.0 m
To find
Pressure at a second point in the line
Solution
Apply Bernoulli equation at two points
Continuity equation P₁A₁=P₂A₂
As we have to find P₂.So
P₂=(P₁A₁)/A₂
And
V₁=1/A
So
P₂=P₁+1/2ρ(V₁²-V₂²)+ρg(y₁-y₂)
P₂=P₁+ρ{15/32V₁²+g(y₁-y₂)}
P₂=5.00×10⁴Pa+(1.00×10³ kg/m³){15/(3.00)²+9.8(11 m)}
P₂=1.62×10⁵ Pa
