/*! 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 ssm The main water line enters a... [FREE SOLUTION] | 91Ó°ÊÓ

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ssm The main water line enters a house on the first floor. The line has a gauge pressure of \(1.90 \times 10^{5}\) Pa. (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) 126,235 Pa; (b) Maximum height is 19.37 m.

Step by step solution

01

Understand the Problem

The task is to compute the water pressure at a higher point (second floor) and to find the maximum height where water can no longer flow due to pressure loss. We will start by calculating how the pressure changes with height.
02

Calculate Pressure Change Due to Height

For (a), use the formula for pressure change due to elevation, which is given by \[ \Delta P = \rho g h \]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 (6.50 m). Calculate \( \Delta P \).\[ \Delta P = 1000 \times 9.81 \times 6.50 = 63765 \text{ Pa} \]
03

Calculate Gauge Pressure at the Second Floor

Subtract the pressure change (calculated in Step 2) from the initial pressure to find the gauge pressure at the second-floor faucet:\[ P_2 = P_1 - \Delta P = 1.90 \times 10^{5} - 63765 \]\[ P_2 = 126235 \text{ Pa} \]So, the gauge pressure at the second floor is approximately 126,235 Pa.
04

Find Maximum Height for Water Flow

For (b), set the gauge pressure to zero since that is the highest point where water will no longer flow. Use the formula where pressure at this height is equal to zero: \[ P_1 = \rho g h_{max} \]Substitute the known values:\[ 1.90 \times 10^5 = 1000 \times 9.81 \times h_{max} \]Solve for \( h_{max} \):\[ h_{max} = \frac{1.90 \times 10^5}{1000 \times 9.81} \approx 19.37 \text{ m} \]This is the maximum height for water flow.

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Key Concepts

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

Pressure Calculation
When dealing with fluid mechanics, calculating pressure is key. Pressure is defined as the force exerted per unit area. In the context of fluid flow, such as water flowing through pipes in a house, pressure tells us how strongly the fluid pushes against the walls of the pipe. This is crucial because it affects how water will move, especially in a vertical direction.

In the exercise, the pressure calculation begins with the initial gauge pressure at the house entry point. The formula used to determine pressure change with elevation is: \[\Delta P = \rho g h\]where:
  • \(\rho\) represents the density of the fluid. For water, this is typically \(1000\ \text{kg/m}^3\).
  • \(g\) is the acceleration due to gravity, approximately \(9.81\ \text{m/s}^2\).
  • \(h\) is the vertical height difference the fluid travels.
By understanding these components, you can calculate how pressure will vary as water moves to different heights, such as floors in a building.
Gauge Pressure
Gauge pressure is the pressure of a system above atmospheric pressure. It is often what you read on pressure gauges in various applications, because these gauges only measure pressure relative to the ambient atmospheric pressure.

In this particular exercise, the gauge pressure at the faucet on the second floor is calculated by subtracting the pressure change due to elevation \(\Delta P\) from the initial gauge pressure \(P_1\) on the first floor:\[P_2 = P_1 - \Delta P\]This formula reflects that as the water is elevated, there’s a loss of gauge pressure due to gravity pulling the water down. As a result, the gauge pressure decreases from the first floor to the second floor.

Understanding gauge pressure helps in practical situations, like ensuring that water can still flow adequately at higher elevations in a building.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth due to the weight of the fluid above. This concept is key when determining how high we can lift a fluid using its pressure alone, as seen in this exercise.

The maximum height at which water can still flow is found by setting the gauge pressure to zero, equating it to hydrostatic pressure since no additional energy is available to push it further up. The formula used in this scenario is:\[P_1 = \rho g h_{max}\]where \(h_{max}\) is the maximum height water can reach while still flowing. This is calculated by rearranging the formula to solve for \(h_{max}\):\[h_{max} = \frac{P_1}{\rho g}\]
Through understanding hydrostatic pressure, we learn why there are height limits in gravity-fed water systems. This assessment becomes important for design considerations in plumbing and civil engineering.

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Most popular questions from this chapter

Crossing a River. You and your team come to a slow flowing river that you need to cross. The nearest bridge is 20 miles to the north, too dangerous and too far to trek with your group. You explore the area down river and discover an abandoned shed with a stash of 55 -gallon drums (all empty and weighing 35.0 lb each) and a stack of 10 -foot planks: You will build a raft. The six members of your team have a combined weight of 925 pounds (assuming everyone was truthful). You also have a four-wheeler (all-terrain vehicle), which weighs 450 lb, and other gear, which adds another 315 lb. Your simple raft design is as follows: a platform of planks with barrels strapped to the bottom. You estimate that the planks weigh about \(45.0 \mathrm{lb}\) each, and you will need 20 of them to make a platform that can accommodate everything for the one-way trip. You measure the dimensions of the cylindrical drums and find they have diameter \(D=22.5\) inches and height \(h=33.5\) inches (note: not exactly \(^{455}\) gallons"). (a) What is the minimum number of barrels that you will need so that the raft will float when fully loaded? (b) What is the minimum number of barrels you will need if you want the platform to be at least 6 inches above the water when fully loaded?

A room has a volume of \(120 \mathrm{m}^{3} .\) An air-conditioning system is to replace the air in this room every twenty minutes, using ducts that have a square cross section. Assuming that air can be treated as an incompressible fluid, find the length of a side of the square if the air speed within the ducts is (a) \(3.0 \mathrm{m} / \mathrm{s}\) and (b) \(5.0 \mathrm{m} / \mathrm{s}\)

The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3},\) and the density of seawater is 1025 \(\mathrm{kg} / \mathrm{m}^{3} .\) A swimming polar bear climbs onto a piece of floating ice that has a volume of \(5.2 \mathrm{m}^{3} .\) What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance \(h,\) as the drawing shows. Using \(1.013 \times 10^{5}\) Pa for the atmospheric pressure and \(1200 \mathrm{kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is \((a) 0.15 \mathrm{m}\) and \((b) 0.10 \mathrm{m}.\)

The Mariana trench is located in the floor of the Pacific Ocean at a depth of about \(11000 \mathrm{m}\) below the surface of the water. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3} .\) (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{m}) ?\) (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{kg}.\)

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