/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The main water line entrs a hous... [FREE SOLUTION] | 91Ó°ÊÓ

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The main water line entrs a house on the first floor. The line has a gauge pressure of \(1.90 \times 10^{5} \mathrm{Pa}\) . \((\mathrm{a})\) A faucet on the second floor, 6.50 \(\mathrm{m}\) above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

Short Answer

Expert verified
(a) The gauge pressure at the faucet is \( 1.26235 \times 10^{5} \; \text{Pa} \). (b) Water would not flow to heights above approximately 19.4 meters.

Step by step solution

01

Understand the Pressure Drop due to Height Change

When water flows from one height to another, the pressure at the second height is decreased due to the potential energy change. For a height change \( h \), the pressure drop \( \Delta P \) can be calculated using the formula \( \Delta P = \rho g h \), where \( \rho \) is the density of water \( (1000 \; \text{kg/m}^3) \), \( g \) is the acceleration due to gravity \( (9.81 \; \text{m/s}^2) \), and \( h \) is the height change.
02

Calculate the Pressure Drop for Part (a)

Here, \( h = 6.50 \; \text{m} \). Using the formula for pressure drop, \( \Delta P = 1000 \times 9.81 \times 6.50 = 63765 \; \text{Pa} \). This pressure drop needs to be subtracted from the initial pressure at the first floor.
03

Determine the Gauge Pressure at the Second Floor

The gauge pressure at the faucet on the second floor is the initial pressure minus the pressure drop. Thus, the pressure at the faucet is \( 1.90 \times 10^{5} \;\text{Pa} - 63765 \; \text{Pa} = 1.26235 \times 10^{5} \; \text{Pa} \).
04

Calculate the Maximum Height for Part (b)

To find the maximum height at which no water would flow, we set the gauge pressure to zero and solve for height: \( 0 = 1.90 \times 10^{5} - \rho g h \). Therefore, \( h = \frac{1.90 \times 10^{5}}{1000 \times 9.81} \approx 19.4 \; \text{m} \).

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Key Concepts

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

Gauge Pressure
Gauge pressure is the pressure of a fluid measured relative to the atmospheric pressure. It helps in determining how much more pressure a fluid is exerting beyond the pressure present due to the atmosphere around us.
When thinking about gauge pressure in pipes, it’s essential because it tells us the effective pressure that is available for use. For example, in a home's water system, the gauge pressure ensures that water can travel upward to various floors of a building.
  • Gauge pressure = Measured pressure - Atmospheric pressure
  • It operates on the principle that the pressure at a point must exceed atmospheric pressure for a fluid to flow.
To calculate the gauge pressure at a faucet on a higher floor, you need to subtract the pressure lost due to the height the water travels. This helps predict the usable pressure available at those points, crucial for efficient plumbing design.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any given point within the fluid due to the force of gravity. This concept helps us understand how pressure changes with depth and height.
In the context of the original exercise, hydrostatic pressure explains why there is a decrease in pressure as water moves up to the second floor. The water column's weight above the faucet increases the downward pressure against the faucet's flow.
  • This pressure difference is calculated using the formula: \( \Delta P = \rho g h \).
  • \( \rho \) is fluid density, \( g \) is gravitational acceleration, and \( h \) is height/difference in height.
Hydrostatic pressure thus becomes a critical part of calculating the overall pressure at any elevation within a piping system.
Pressure Drop
Pressure drop refers to the loss of pressure as fluid moves through a system. It happens due to resistance from factors such as friction, elevation changes, or flow through fittings and valves.
In relation to the exercise, the height of the second floor causes a pressure drop from the first floor to the second. Calculation of this drop involves evaluating how much height has been ascended and its effect on pressure.
  • The primary formula is \( \Delta P = \rho g h \).
  • Such calculations assist in designing systems keeping pressure loss minimal or manageable, ensuring functionality even at higher elevations.
Recognizing and calculating pressure drops is crucial to maintain an effective supply of water or any fluid to its intended destination.
Fluid Mechanics
Fluid mechanics is the branch of physics concerned with the study of fluids (liquids and gases) and the forces on them. It is foundational for understanding how fluids behave under various forces and in diverse conditions.
This field covers two main branches: fluid statics and fluid dynamics. Fluid statics focus on fluids at rest, while fluid dynamics is about fluids in motion.
  • Fluid statics deals with pressure, density, and temperature as the primary factors.
  • Fluid dynamics incorporates movement and forces, involving velocity, flow rate, and some complex mathematics.
Mastery of fluid mechanics principles allows for the accurate design and analysis of systems that transport fluids, like in plumbing, hydraulics, and aerodynamics.
Bernoulli's Principle
Bernoulli's Principle is a key concept in fluid dynamics that describes how the speed of a fluid relates to its pressure. In essence, Bernoulli's principle states that an increase in the velocity of a fluid results in a decrease in pressure and vice versa.
While Bernoulli's Principle was not directly applied in this exercise, understanding it elucidates how pressure variations occur due to changes in fluid speed. For instance, in piping systems, areas where a fluid is forced through a smaller cross-section will experience faster speeds and hence lower pressures.
  • Mathematically, it can be expressed as: \( P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \).
  • This equation can be tailored to evaluate energy conservation in fluid flow.
Grasping Bernoulli's Principle aids in recognizing the interplay between velocity and pressure, a significant aspect of designing systems involving fluid movement.

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Most popular questions from this chapter

A mercury barometer reads 747.0 mm on the roof of a building and 760.0 \(\mathrm{mm}\) on the ground. Assuming a constant value of 1.29 \(\mathrm{kg} / \mathrm{m}^{3}\) for the density of air, determine the height of the building.

The aorta carries blood away from the heart at a speed of about 40 cm/s and has a radius of approximately 1.1 cm. The aorta branches eventually into a large number of tiny capillaries that distribute the blood to the various body organs. In a capillary, the blood speed is approximately 0.07 cm/s, and the radius is about \(6 \times 10^{-4} \mathrm{cm} .\) Treat the blood as an incompressible fluid, and use these data to determine the approximate number of capillaries in the human body.

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