Answer:
Water pressure is greatest at point B where the diameter is larger
Explanation:
This is due to Bernoulli's Equation: [tex]P_{A} +\frac{pv _{A}^{2} }{2} +pgh_{A}=P_{B} +\frac{pv _{B}^{2} }{2} +pgh_{B}[/tex]
and Continuity Equation: [tex]p_{A} v_{A}A_{A}=p_{B} v_{B}A_{B}[/tex]
where...
P = Pressure of Fluid at the Center of the Pipe
ρ = Density of Fluid
v = Velocity of Fluid
g = Gravitational Constant
h = Height of Fluid at the Center of the Pipe
A = Area of Pipe Cross Section
This is the same as saying the following:
Pressure Energy ([tex]P[/tex]) + Kinetic Energy ([tex]\frac{pv _{}^{2} }{2}[/tex]) + Potential Energy ([tex]pgh_{}[/tex]) = Constant
The height of flow at the center of the pipe is the same, so we know that Potential Energy cancels out on both sides of the equation (Δ[tex]pgh_{}[/tex] = 0)
[tex]P_{A} +\frac{pv _{A}^{2} }{2} =P_{B} +\frac{pv _{B}^{2} }{2}[/tex]
Now that we've simplified Bernoulli's Equation, we need to determine which Pressure is greater using Continuity Equation.
[tex]p_{A} v_{A}A_{A}=p_{B} v_{B}A_{B}[/tex]
Density is the same, so we can cancel this out on both sides of the equation (Δρ = 0)
[tex]v_{A}A_{A}=v_{B}A_{B}[/tex]
From the problem statement, we know that [tex]A_{A} < A_{B}[/tex]
Since [tex]A_{A} < A_{B}[/tex], we know that [tex]v_{A} > v_{B}[/tex] due to the Continuity Equation.
Answer: Jumping back to Bernoulli's Equation, we know that [tex]P_{A} < P_{B}[/tex]
