The 3.0cm-diameter water line in the figure splits into two 1.0cm -diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa. What is the gauge pressure at point B.
I keep getting an answer that I am being told is wrong.
I use the continuity equation to solve for the speed at point B then use Bernoulli's principle to determine the pressure at that point.
To solve this problem, you can use the principles of fluid dynamics, specifically the continuity equation and Bernoulli's principle. Let's go through the steps to find the gauge pressure at point B.
Step 1: Apply the continuity equation.
The continuity equation states that the mass flow rate at any given point in an incompressible fluid is constant. It can be expressed as:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the pipes at points A and B respectively, and v1 and v2 are the velocities of the water at those points.
Given:
- Diameter of the water line at point A (D1) = 3.0 cm
- Diameter of the two pipes at point B (D2) = 1.0 cm
- Velocity of water at point A (v1) = 2.0 m/s
First, we need to calculate the cross-sectional area at each point using the formula for the area of a circle:
A = πr^2
where r is the radius. For convenience, let's convert the diameters to radii (in meters):
r1 = D1/2 = 0.03 m/2 = 0.015 m
r2 = D2/2 = 0.01 m/2 = 0.005 m
Now we can calculate the cross-sectional areas:
A1 = πr1^2 = π(0.015^2) ≈ 0.00070685 m^2
A2 = πr2^2 = π(0.005^2) ≈ 0.00007854 m^2
Using the continuity equation, we can solve for v2:
A1v1 = A2v2
v2 = (A1v1) / A2
v2 = (0.00070685 m^2 * 2.0 m/s) / 0.00007854 m^2
v2 ≈ 14.263 m/s
Step 2: Apply Bernoulli's principle.
Bernoulli's principle states that in steady, idealized flow of an incompressible fluid, the sum of the fluid's potential energy, kinetic energy, and pressure energy per unit volume remains constant.
P1/ρ + v1^2/2 + gh1 = P2/ρ + v2^2/2 + gh2
Where P1 and P2 are the gauge pressures at points A and B respectively, ρ is the density (assumed constant), g is the acceleration due to gravity, h1 and h2 are the elevations of the two points from a reference level, and v1 and v2 are the velocities at those points.
Given:
- Velocity of water at point B (v2) = 14.263 m/s (from step 1)
- Gauge pressure at point A (P1) = 50 kPa (convert to Pa: 50,000 Pa)
- Height of point A and point B (h1 and h2) are the same, so they can be ignored in this case.
P1/ρ + v1^2/2 = P2/ρ + v2^2/2
Now we need to rearrange the equation to solve for P2 (the gauge pressure at point B):
P2 = P1 + ρ(v1^2 - v2^2)/2
Plugging in the known values:
P2 = 50,000 Pa + ρ(2.0 m/s)^2 - (14.263 m/s)^2) / 2
Step 3: Calculate the gauge pressure at point B.
To calculate the gauge pressure at point B, we need to know the density of water, which is approximately 1000 kg/m^3.
P2 = 50,000 Pa + 1000 kg/m^3 * (2.0 m/s)^2 - (14.263 m/s)^2) / 2
P2 = 50,000 Pa + 1000 * (4 - 203.505158) / 2
P2 ≈ 50,000 Pa - 99,505.158 / 2
P2 ≈ 0.497 mmHg
Therefore, the gauge pressure at point B is approximately 0.497 mmHg. Note that this value is very small, so it may be possible that there is a mistake in the calculations or problem statement.