At a certain point in a pipeline the water's speed is 4.00 m/s and the gauge pressure is 5.00 x 10^4 Pa. Find the gauge pressure at a second point in the line, 12.0 m lower than the first, if the cross-section area at the second point is twice that at the first.
To find the gauge pressure at the second point in the pipeline, we can use Bernoulli's equation, which relates the pressure, speed, and height of a fluid in a pipe.
The equation for Bernoulli's principle is:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the gauge pressures at points 1 and 2, respectively.
v1 and v2 are the speeds of the water at points 1 and 2, respectively.
ρ is the density of water.
g is the acceleration due to gravity.
h1 and h2 are the heights of the water at points 1 and 2, respectively.
In this case, we know the following:
P1 = 5.00 x 10^4 Pa (given gauge pressure at the first point)
v1 = 4.00 m/s (given speed at the first point)
h1 = 0 m (we assume point 1 is the reference height, so it is at the same level as the first point)
h2 = -12.0 m (given that the second point is 12.0 m lower than the first point)
A2 = 2A1 (given that the cross-sectional area at the second point is twice that at the first point)
First, let's find the speed at the second point, v2:
Using the equation of continuity, we know that A1v1 = A2v2.
Since A2 = 2A1, we can substitute A2 = 2A1 into the equation:
A1v1 = 2A1v2
v2 = v1/2
v2 = 4.00 m/s / 2
v2 = 2.00 m/s
Substituting the known values into Bernoulli's equation:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
We can simplify the equation by canceling out the terms with h1 = 0:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2 + ρgh2
Substituting the known values:
5.00 x 10^4 Pa + (1/2)(ρ)(4.00 m/s)^2 = P2 + (1/2)(ρ)(2.00 m/s)^2 + (ρ)(-12.0 m)(9.8 m/s^2)
Simplifying the equation further, we need the density of water, which is typically around 1000 kg/m^3:
5.00 x 10^4 Pa + (1/2)(1000 kg/m^3)(4.00 m/s)^2 = P2 + (1/2)(1000 kg/m^3)(2.00 m/s)^2 + (1000 kg/m^3)(-12.0 m)(9.8 m/s^2)
Calculating the equation:
5.00 x 10^4 Pa + 2000 Pa = P2 + 1000 Pa + (-117600 Pa)
Simplifying further:
52000 Pa = P2 - 116600 Pa
To find P2, we rearrange the equation:
P2 = 52000 Pa + 116600 Pa
P2 = 168600 Pa
Therefore, the gauge pressure at the second point in the pipeline is 168600 Pa.
hints:
1. gauge pressure is zero-referenced against ambient atmospheric pressure, which (usually) cancels out at different points of the pipe if the distance is short.
2. For short distances, we might be able to justify assumption of no frictional loss.
3. Equation of continuity:
V1A1=V2A2
4. Bernoulli equation for incompressible flow:
v²/2+gz+p/ρ = constant