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Chapter 11: Problem 60

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{~Pa}\) and a speed of \(2.1 \mathrm{~m} / \mathrm{s}\). However, on the second floor, which is \(4.0 \mathrm{~m}\) higher, the speed of the water is \(3.7 \mathrm{~m} / \mathrm{s}\). The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

Short Answer

Expert verified
The gauge pressure on the second floor is \(2.9576 \times 10^5 \mathrm{~Pa}\).

Step by step solution

01

Understand the Bernoulli's Equation

We鈥檒l use Bernoulli's equation which is \( P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \) where \( P \) is pressure, \( \rho \) is the water density, \( v \) is the speed, \( h \) is the height, and \( g \) is the gravitational acceleration.
02

Substitute the Known Values

Set \( P_1 = 3.4 \times 10^5 \mathrm{~Pa} \), \( v_1 = 2.1 \mathrm{~m/s} \), \( v_2 = 3.7 \mathrm{~m/s} \), \( h_1 = 0 \), \( h_2 = 4 \mathrm{~m} \), and \( g = 9.81 \mathrm{~m/s^2} \). Assume the density of water \( \rho = 1000 \mathrm{~kg/m^3} \).
03

Calculate the Pressure Difference Due to Speed Change

Apply the terms related to velocity: \( \frac{1}{2} \rho v_1^2 = \frac{1}{2} \times 1000 \times (2.1)^2 = 2205 \mathrm{~Pa} \) and similarly \( \frac{1}{2} \rho v_2^2 = 6845 \mathrm{~Pa} \). The difference due to speed change is \( 6845 - 2205 = 4640 \mathrm{~Pa} \).
04

Calculate the Pressure Difference Due to Height Change

Include the gravitational potential term: \( \rho g h = 1000 \times 9.81 \times 4 = 39240 \mathrm{~Pa} \).
05

Calculate the Gauge Pressure on the Second Floor

Using the Bernoulli equation, calculate \( P_2 = P_1 + (\frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2) - \rho g h \). Substitute values: \( P_2 = 3.4 \times 10^5 - 4640 - 39240 = 2.9576 \times 10^5 \mathrm{~Pa} \).

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a field of physics concerned with the movement of liquids and gases. In this context, we focus on how water moves through pipes within a building.
Bernoulli鈥檚 Equation is a fundamental principle used in fluid dynamics to describe the conservation of energy in a fluid flow.
  • It states that the sum of pressure energy, kinetic energy, and potential energy in a flowing fluid remains constant, provided the fluid flow is steady and frictionless.
  • This principle is especially useful for understanding behaviors like how fluid speeds increase as the diameter of a pipe decreases.
With Bernoulli鈥檚 Equation, it becomes easier to predict how changes in fluid speed and height affect pressure, which is crucial when dealing with multi-level systems like the apartment scenario outlined above.
Gauge Pressure
Gauge pressure is the pressure within a system above atmospheric pressure. It is essential in assessing the operational efficiency of a fluid system.
  • Unlike absolute pressure, which includes atmospheric pressure in its measurement, gauge pressure zeroes in solely on the pressure exerted by a fluid.
  • In the given exercise, understanding gauge pressure helps us determine how pressure changes from one floor to another.
This understanding is critical as it highlights how pressure can drop due to increased height or speed, affecting the flow dynamics within the system.
Pipe Flow
Pipe flow refers to the movement of fluids through a closed conduit, typically round in cross-section. In our exercise, water moves through the apartment's pipe system, subject to changes in diameter.
  • Variation in pipe diameters causes fluid velocities to change, due to the principle of continuity, which states that the mass flow rate must be constant along the pipe.
  • As a result, smaller diameter sections lead to higher velocities, and larger diameter sections lead to lower velocities.
Understanding pipe flow is vital in designing piping systems that maintain optimal fluid delivery across various floors without unnecessary pressure losses.
Water Pressure Calculation
Water pressure calculation involves applying Bernoulli鈥檚 Equation to determine pressure changes in a fluid network. To solve the exercise provided, apply these steps:
  • Identify the initial and final points in the pipe system, considering changes in height and velocity.
  • Substitute given values such as initial gauge pressure, fluid density, gravity, and height difference.
By carefully following these calculations, one can determine the new pressure at any point in a piping system. Correctly calculating water pressure ensures the system operates within its designed parameters, avoiding potential disruptions and maintaining efficient water flow.

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Most popular questions from this chapter

An airplane has an effective wing surface area of \(16 \mathrm{~m}^{2}\) that is generating the lift force. In level flight the air speed over the top of the wings is \(62.0 \mathrm{~m} / \mathrm{s},\) while the air speed beneath the wings is \(54.0 \mathrm{~m} / \mathrm{s}\). What is the weight of the plane?

A fountain sends a stream of water straight up into the air to a maximum height of 5.00 \(\mathrm{m} .\) The effective cross-sectional area of the pipe feeding the fountain is \(5.00 \times 10^{-4} \mathrm{~m}^{2}\) Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (Note: \(1 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\).)

A \(0.10-m \times 0.20-m \times 0.30-m\) block is suspended from a wire and is completely under water. What buoyant force acts on the block?

A 1.3-m length of horizontal pipe has a radius of \(6.4 \times 10^{-3} \mathrm{~m}\). Water within the pipe flows with a volume flow rate of \(9.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) out of the right end of the pipe and into the air. What is the pressure in the flowing water at the left end of the pipe if the water behaves as (a) an ideal fluid and (b) a viscous fluid \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) ?

A person who weighs \(625 \mathrm{~N}\) is riding a \(98-\mathrm{N}\) mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is \(7.60 \times 10^{5} \mathrm{~Pa}\), what is the area of contact between each tire and the ground?
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