A horizontal segment of pipe tapers from a cross sectional area of 50 cm2 to 0.5 cm2. The pressure at the larger end of the pipe is 1.34 105 Pa and the speed is 0.038 m/s. What is the pressure at the narrow end of the segment?
The continuity equation
Area*velocity = constant
will tell you the speed at the smaller end.
Use the Bernoulli equation for the pressure difference due to velocity change.
If you need to review that equation, it should be in your text and is easy to find online.
Additional help will be provided when work is shown.
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To solve this problem, we can use Bernoulli's equation, which states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume at any two points along a streamline is constant.
The equation can be written as:
P1 + ρgh1 + (1/2)ρv1^2 = P2 + ρgh2 + (1/2)ρv2^2
Here, P1 and P2 are the pressures at points 1 and 2 respectively, ρ is the density of the fluid (assumed to be constant), g is the acceleration due to gravity (also assumed to be constant), h1 and h2 are the heights of the fluid above a reference point at points 1 and 2 respectively, and v1 and v2 are the velocities at points 1 and 2 respectively.
Since the pipe segment is horizontal, the heights h1 and h2 are the same, so we can ignore the potential energy terms.
At the larger end of the pipe (point 1), the cross-sectional area is 50 cm^2 and the pressure is 1.34 × 10^5 Pa. The speed v1 is given as 0.038 m/s.
At the narrower end of the pipe (point 2), the cross-sectional area is 0.5 cm^2. We need to find the pressure P2.
Let's substitute the known values into the Bernoulli's equation:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Rearranging the equation, we can solve for P2:
P2 = P1 + (1/2)ρ(v1^2 - v2^2)
Since the velocities are squared in the equation, the ratio of the areas can be used instead of the velocities:
v1^2/v2^2 = A2/A1
Substituting the given values:
v1^2/v2^2 = (0.038 m/s)^2 / v2^2 = (50 cm^2) / (0.5 cm^2)
Simplifying,
v2^2 = (0.5 cm^2 / 50 cm^2) * (0.038 m/s)^2
= (0.01) * (0.001444) m^2/s^2
= 0.00001444 m^2/s^2
Now, we can substitute the value of v2^2 in the expression for P2:
P2 = P1 + (1/2)ρ(v1^2 - v2^2)
= 1.34 × 10^5 Pa + (1/2)(ρ)(v1^2 - v2^2)
Since the density of the fluid is not given, we cannot solve for the exact value of P2 without this information. The pressure at the narrow end of the segment will depend on the density of the fluid.
To find the pressure at the narrow end of the pipe, we can make use of Bernoulli's equation, which relates the pressure, velocity, and cross-sectional area of a fluid flowing through a 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 the two ends of the pipe segment.
- ρ is the density of the fluid.
- v1 and v2 are the velocities at the two ends of the pipe segment.
- g is the acceleration due to gravity.
- h1 and h2 are the heights (or elevations) at the two ends of the pipe segment (which we can assume to be the same in this case, as the pipe is horizontal).
In this scenario, we are given the following information:
- P1 = 1.34 * 10^5 Pa (pressure at the larger end)
- v1 = 0.038 m/s (velocity at the larger end)
- A1 = 50 cm^2 (cross-sectional area at the larger end)
- A2 = 0.5 cm^2 (cross-sectional area at the narrow end)
We need to find P2, the pressure at the narrow end.
First, we need to convert the units to be consistent. Let's convert the areas to square meters:
A1 = 50 cm^2 = 0.005 m^2
A2 = 0.5 cm^2 = 0.00005 m^2
Now, let's substitute the given values into Bernoulli's equation and solve for P2:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Since the segment is horizontal, the heights cancel out.
1.34 * 10^5 Pa + (1/2)ρ(0.038 m/s)^2 = P2 + (1/2)ρv2^2
Next, let's assume the fluid is an incompressible, non-viscous fluid with a constant density, and cancel out the density (ρ) terms:
1.34 * 10^5 Pa + (1/2)(0.038 m/s)^2 = P2 + (1/2)v2^2
Now, let's solve for P2:
P2 = 1.34 * 10^5 Pa + (1/2)(0.038 m/s)^2 - (1/2)v2^2
Substituting the values, we get:
P2 = 1.34 * 10^5 Pa + (1/2)(0.038 m/s)^2 - (1/2)(0.038 m/s)^2
Simplifying the equation gives us:
P2 = 1.34 * 10^5 Pa
Therefore, the pressure at the narrow end of the pipe segment is 1.34 * 10^5 Pa.