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At a certain point in a horizontal pipeline, the water's speed is 2.50 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4}\) Pa. Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

Short Answer

Expert verified
The gauge pressure at the second point is approximately 20344 Pa.

Step by step solution

01

Apply the Continuity Equation

The continuity equation states that the mass flow rate must remain constant in an incompressible fluid flow. Thus, for two points in a pipeline: \[ A_1 v_1 = A_2 v_2 \]Given that \( A_2 = 2A_1 \) and the initial speed \( v_1 = 2.50 \ \mathrm{m/s} \), substitute these values to find \( v_2 \). \[ v_2 = \frac{A_1}{A_2} v_1 = \frac{1}{2} \times 2.50 = 1.25 \ \mathrm{m/s} \]
02

Apply Bernoulli's Equation

Bernoulli's equation relates the speed and pressure at two points in flowing fluid: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]Here, \( P_1 = 1.80 \times 10^4 \, \mathrm{Pa} \) and \( v_1 = 2.50 \, \mathrm{m/s} \), \( v_2 = 1.25 \, \mathrm{m/s} \). Assume water density \( \rho = 1000 \, \mathrm{kg/m^3} \). We can rearrange and solve for \( P_2 \): \[ P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) \]
03

Substitute Values into Bernoulli's Equation

Substitute the known values into the equation:\[ P_2 = 1.80 \times 10^4 + \frac{1}{2} \times 1000 \times (2.50^2 - 1.25^2) \]Compute:\[ P_2 = 1.80 \times 10^4 + 500 \times (6.25 - 1.5625) \]\[ P_2 = 1.80 \times 10^4 + 500 \times 4.6875 \]\[ P_2 = 1.80 \times 10^4 + 2343.75 \]
04

Calculate the Final Gauge Pressure

After calculating, we find:\[ P_2 = 1.80 \times 10^4 + 2343.75 = 20343.75 \, \mathrm{Pa} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation
Understanding the Continuity Equation is critical for analyzing fluid flow. This equation derives from the conservation of mass principle, stating that for any incompressible fluid, the mass flow rate must remain constant along the flow path. Essentially, it’s like ensuring that what goes into a pipe must come out, irrespective of changes in pipe size or speed.
In mathematical terms, it is expressed as:
  • \( A_1 v_1 = A_2 v_2 \)
Here, \( A \) represents the cross-sectional area, while \( v \) represents the fluid velocity. If the area at one section of the pipe increases, the fluid velocity must decrease to maintain the same mass flow rate, assuming the fluid is incompressible.
This concept prevents "piling up" of fluid at any point along the pipeline, ensuring smooth and continuous fluid travel.
Bernoulli's Equation
Bernoulli's Equation is one of the cornerstones of fluid dynamics, providing a comprehensive relationship between pressure, velocity, and height for a flowing fluid. For a constant height scenario, such as in a horizontal pipeline, it simplifies to account only for pressure and velocity changes.
This simplified version is expressed as:
  • \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \)
Here, \( P \) denotes pressure, \( \rho \) is the fluid density, and \( v \) is the velocity. The equation essentially states that an increase in fluid speed results in a decrease in pressure, and vice versa.
It is foundational in understanding how pressure and velocity interrelate, acting as a perfect tool for engineers in designing systems where pressure management is crucial.
Incompressible Fluid Flow
In the context of fluid dynamics, the term "incompressible fluid flow" indicates that a fluid’s density remains constant throughout its flow path, meaning it does not change with variations in pressure or flow conditions. This assumption simplifies many fluid dynamics problems, making it possible to apply equations like the Continuity Equation and Bernoulli's Equation more easily.
Water is a common example of an incompressible fluid in engineering due to its relatively low compressibility under typical conditions. This concept is crucial when predicting how fluids move through different environments without needing to account for volume changes due to pressure shifts.
Understanding this assumption helps in solving practical problems where fluid volumes or flow rates are critical, ensuring precise and reliable predictions in fluid system designs.
Mass Flow Rate
The mass flow rate is a pivotal concept in fluid mechanics, defined as the amount of mass passing through a given point in a fluid flow system per unit time. It is typically measured in kilograms per second (kg/s).
The calculation of mass flow rate in an incompressible fluid can be derived using:
  • \( \dot{m} = \rho A v \)
In this equation, \( \dot{m} \) represents the mass flow rate, \( \rho \) is the fluid density, \( A \) is the cross-sectional area, and \( v \) is the velocity of the fluid.
Maintaining a constant mass flow rate is essential in various applications including pipelines, where any variation could indicate leaks or blockages. Understanding this concept aids in the effective management and design of fluid transport systems, ensuring operational efficiency and safety.

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Most popular questions from this chapter

Blood pressure. Systemic blood pressure is expressed as the ratio of the systolic pressure (when the heart first ejects blood into the arteries) to the diastolic pressure (when the heart is relaxed): systemic blood pressure \(=\frac{\text { systolic pressure }}{\text { diastolic pressure }}\) Both pressures are measured at the level of the heart and are expressed in millimeters of mercury (or torr), although the units are not written. Normal systemic blood pressure is \(\frac{120}{80}\) . (a) What are the maximum and minimum forces (in newtons) that the blood exerts against each square centimeter of the heart for a person with normal blood pressure? (b) As pointed out in the text, blood pressure is normally measured on the upper arm at the same height as the heart. Due to therapy for an injury, a patient's upper arm is extended 30.0 \(\mathrm{cm}\) above his heart. In that position, what should be his systemic blood pressure reading, expressed in the standard way, if he has normal blood pressure? The density of blood is 1060 \(\mathrm{kg} / \mathrm{m}^{3}\) .

(a) Calculate the buoyant force of air (density 1.20 \(\mathrm{kg} / \mathrm{m}^{3} )\) on a spherical party balloon that has a radius of 15.0 \(\mathrm{cm}\) . (b) If the rubber of the balloon itself has a mass of 2.00 \(\mathrm{g}\) and the balloon is filled with helium (density 0.166 \(\mathrm{kg} / \mathrm{m}^{3}\) ), calculate the net upward force (the "lift") that acts on it in air.

A small circular hole 6.00 \(\mathrm{mm}\) in diameter is cut in the side of a large water tank, 14.0 \(\mathrm{m}\) below the water level in the tank. The top of the tank is open to the air. Find the speed at which the water shoots out of the tank.

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 \(\mathrm{m}\) (a) What is the gauge pressure at this depth? (You can ignore the small changes in the density of the water with depth.) (b) At the 250 \(\mathrm{m}\) depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 \(\mathrm{cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You may ignore the small variation in pressure over the surface of the window.)

Standing on your head. (a) When you stand on your head, what is the difference in pressure of the blood in your brain compared with the pressure when you stand on your feet if you are 1.85 m tall? The density of blood is 1060 \(\mathrm{kg} / \mathrm{m}^{3}\) . (b) What effect does the increased pressure have on the blood vessels in your brain?

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