The 3.0 -diameter water line in the figure splits into two 1.0 -diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 and the gauge pressure is 50 .What is the gauge pressure at point B?
I know we have to find P in kPa, but am unsure how to set this problem up or what to plug in where. Any help would be great
To solve this problem, we can use the principle of continuity and Bernoulli's equation. Let's break it down step by step:
Step 1: Applying the principle of continuity
The principle of continuity states that the volume flow rate of an incompressible fluid remains constant in a pipe.
In this case, water is flowing through pipes of different diameters. According to the principle of continuity, the volume flow rate should remain the same at any given point. Mathematically, this can be expressed as:
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas of the pipes, and v1 and v2 are the velocities of water at points A and B, respectively.
Step 2: Finding the cross-sectional areas
The cross-sectional area of a pipe can be calculated using the formula:
A = πr^2
where r is the radius of the pipe.
In this problem, we are given that the diameter of the 3.0-diameter pipe is used, so its radius (r1) would be 1.5. The diameter of the 1.0-diameter pipe is half of that, so its radius (r2) would be 0.5.
Step 3: Finding the velocity at point B
Using the principle of continuity and the values we obtained for the cross-sectional areas, we can rewrite the equation as:
A1v1 = A2v2
πr1^2v1 = πr2^2v2
(1.5)^2v1 = (0.5)^2v2
2.25v1 = 0.25v2
v2 = 9v1
This equation tells us that the velocity at point B is nine times the velocity at point A.
Step 4: Applying Bernoulli's equation
Bernoulli's equation relates the pressure, velocity, and elevation of a fluid in a pipe. It can be written as:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2
where P1 and P2 are the pressures at points A and B, ρ is the density of water, v1 and v2 are the velocities at points A and B, g is the acceleration due to gravity, and h1 and h2 are the elevations at points A and B, respectively.
In this problem, the elevations are given to be the same for points A and B. Hence, the equation simplifies to:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
Step 5: Solving for the gauge pressure at point B (P2)
We have the values for the velocity at point A (v1 = 2.0 m/s) and the velocity at point B (v2 = 9v1). We are also given the gauge pressure at point A (P1 = 50 kPa). Now, we need to solve the equation above for P2.
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
Substituting the given values:
50 + 0.5ρ(2.0)^2 = P2 + 0.5ρ(9(2.0))^2
Simplifying and solving the equation will give you the gauge pressure at point B (P2) in kPa.
Note: The density of water (ρ) can be taken as 1000 kg/m^3.
I hope this explanation helps you set up the problem and solve it correctly!