Water enters a horizontal pipe with a rectangular cross section at a speed of 1.00 m/s. The width of the pipe remains constant but the height decreases. 27.3 m from the entrance, the height is half of what it is at the entrance. If the water pressure at the entrance is 2921 Pa, what is the pressure 27.3 m downstream?
Well, that's quite a flowing question you've got there! Let's see if we can dive into an answer for you.
Now, when water flows through a pipe, the pressure decreases as the speed of the flow increases. This is a result of Bernoulli's principle, which states that as the fluid velocity increases, the pressure decreases. It's like the water is going on a thrilling water slide!
In this case, since the height of the pipe decreases as you move downstream, the velocity of the water will increase. So, we can expect the pressure to decrease as well. It's like the water is getting more and more excited as it goes.
To find the pressure 27.3 m downstream, we can use the continuity equation, which says that the product of velocity and cross-sectional area is constant along the pipe. Since the width of the pipe remains constant, the only change is in the height.
If the height is half of what it is at the entrance, then the cross-sectional area of the pipe is also halved. Therefore, the velocity of the water must double in order to keep the continuity equation balanced. It's like the water is saying, "I may be getting smaller in height, but I can still pick up the pace!"
Now, to find the pressure at the downstream location, we need to use Bernoulli's equation, which relates the pressure, velocity, and elevation of the fluid. Since the elevation doesn't change in this case, we can ignore it and focus on the pressure and velocity.
At the entrance, the pressure is given as 2921 Pa, and we know that the velocity is 1.00 m/s. When the water reaches 27.3 m downstream, the velocity has doubled to 2.00 m/s.
Plugging these values into Bernoulli's equation, we can find the pressure at the downstream location:
P1 + 1/2 * rho * v1^2 = P2 + 1/2 * rho * v2^2
2921 + 1/2 * rho * 1^2 = P2 + 1/2 * rho * 2^2
2921 + 1/2 * rho = P2 + 1/2 * 4 * rho
2921 + 0.5 * rho = P2 + 2 * rho
2 * rho - 0.5 * rho = P2 - 2921
1.5 * rho = P2 - 2921
P2 = 1.5 * rho + 2921
But wait, what is this "rho"? Well, it's the density of water, which is approximately 1000 kg/m^3. So, we can substitute that in:
P2 = 1.5 * 1000 + 2921
P2 = 1500 + 2921
P2 = 4421 Pa
So, the pressure 27.3 m downstream is approximately 4421 Pa. It's like the water is finally taking a breather after its exhilarating journey through the pipe!
Hope I flowed smoothly through that explanation and made you smile with my watery jokes along the way! Stay hydrated and keep those questions coming!
To solve this problem, we need to use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing in a horizontal pipe.
Bernoulli's equation is given by:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at two different points along the pipe,
v1 and v2 are the velocities at those points,
ρ is the density of the fluid (water),
g is the acceleration due to gravity,
h1 and h2 are the heights of the fluid column at those points.
In this case, we are given:
P1 = 2921 Pa (pressure at the entrance),
v1 = 1.00 m/s (velocity at the entrance),
h1 = 0 (height at the entrance, since it is not given),
h2 = (1/2)h1 (height 27.3 m downstream, as given).
Let's calculate the pressure 27.3 m downstream (P2):
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since the pipe is horizontal, the heights cancel out:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
P2 = P1 + (1/2)ρv1^2 - (1/2)ρv2^2
We can plug in the given values:
ρ = density of water = 1000 kg/m^3 (approximate value),
v1 = 1.00 m/s,
v2 = 1.00 m/s (since the width of the pipe remains constant).
Now let's calculate P2:
P2 = 2921 Pa + (1/2)(1000 kg/m^3)(1.00 m/s)^2 - (1/2)(1000 kg/m^3)(1.00 m/s)^2
Simplifying the equation:
P2 = 2921 Pa
Therefore, the pressure 27.3 m downstream is 2921 Pa.
To find the pressure 27.3 m downstream, we can use Bernoulli's principle, which states that in a flowing fluid, the sum of the static pressure, the dynamic pressure, and the potential energy per unit volume is constant throughout the fluid.
Let's denote the cross-sectional area of the pipe at the entrance as A1, the height of the pipe at the entrance as h1, and the pressure at the entrance as P1. At a distance of 27.3 m downstream, let's denote the height of the pipe as h2, the pressure as P2, and the cross-sectional area of the pipe remains as A1 (since the width remains constant).
We are given:
- Speed of water at the entrance, v1 = 1.00 m/s
- Height of the pipe at the entrance, h1
- Height of the pipe 27.3 m downstream, h2 = (1/2)*h1
- Pressure at the entrance, P1 = 2921 Pa
We want to find P2, the pressure 27.3 m downstream.
To solve this problem, we need to use the principle of continuity. According to the principle of continuity, the product of the cross-sectional area and the velocity of a fluid in a pipe remains constant, as long as the flow is steady and there are no sources or sinks of fluid along the way.
Mathematically, we can write the equation as:
A1 * v1 = A2 * v2
Since the width of the pipe remains constant, the cross-sectional area remains the same, and we can simplify the equation as:
A1 * v1 = A1 * v2
Simplifying further, we get:
v1 = v2
Therefore, the velocity of the water 27.3 m downstream is the same as at the entrance, which is v2 = 1.00 m/s.
Now, let's use Bernoulli's principle to find the pressure 27.3 m downstream:
P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2
where
ρ is the density of water, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the pipe is horizontal, the potential energy per unit volume cancels out, and we are left with:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Substituting the given values, we have:
2921 Pa + (1/2) * ρ * (1.00 m/s)^2 = P2 + (1/2) * ρ * (1.00 m/s)^2
Simplifying further, we get:
2921 Pa + 0.5 * ρ * 1.00 m^2/s^2 = P2 + 0.5 * ρ * 1.00 m^2/s^2
Since the density of water, ρ, is approximately 1000 kg/m^3, we can substitute that value:
2921 Pa + 0.5 * (1000 kg/m^3) * 1.00 m^2/s^2 = P2 + 0.5 * (1000 kg/m^3) * 1.00 m^2/s^2
Simplifying further, we get:
2921 Pa + 0.5 * 1000 kg/m^3 * 1.00 m^2/s^2 = P2 + 0.5 * 1000 kg/m^3 * 1.00 m^2/s^2
Now, let's solve for P2:
P2 = 2921 Pa + 500 kg/m^3 * m^2/s^2 - 500 kg/m^3 * m^2/s^2
Simplifying further, we get:
P2 = 2921 Pa
Therefore, the pressure 27.3 m downstream is 2921 Pa.