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Chapter 13: Problem 27

A certain town receives its water directly from a water tower. If the top of the water in the tower is \(26.0 \mathrm{~m}\) above the water faucet in a house, what should be the water pressure at the faucet? (Neglect the effects of other water users.)

Short Answer

Expert verified
The water pressure at the faucet is 254800 Pa.

Step by step solution

01

Identify the formula

To find the water pressure at the faucet, we can use the formula for pressure due to gravity: \( P = \rho \cdot g \cdot h \), where \( P \) is the pressure, \( \rho \) is the density of water (\(1000 \, \text{kg/m}^3\)), \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \( h \) is the height of the water column (\(26.0 \, \text{m}\)).
02

Substitute known values

Substitute the known values into the formula: \( P = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 26.0 \, \text{m} \).
03

Compute the pressure

Multiply the values: \[ P = 1000 \times 9.8 \times 26.0 = 254800 \, \text{Pa} \]Thus, the water pressure at the faucet is \( 254800 \, \text{Pa} \) (Pascals).

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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 fascinating branch of physics that studies the flow of liquids and gases. In our daily lives, this concept helps us understand everything from weather patterns to the way fluids behave in various scenarios. When we talk about water flowing from a tower to a faucet, we dive deep into fluid dynamics.
Fluid dynamics involves several key principles and equations, but one of the most relevant for our water pressure example is Bernoulli鈥檚 equation. This principle tells us how the speed, pressure, and height of a fluid are related.
  • Continuity Equation: This equation ensures that the mass flow rate is constant across different cross-sections of a pipe. Hence, it relates the speed of the fluid flow with the cross-sectional area.
  • Bernoulli鈥檚 Principle: This principle is an expression of energy conservation for flowing fluids, stating that an increase in a fluid's speed results in a decrease in the fluid's potential energy or pressure.
Understanding fluid dynamics gives us the tools to predict how changes in the system, like height or pipe size, might affect water delivery and pressure.
Physics Problem Solving
Solving physics problems often involves a few methodical steps to approach the challenge effectively. It鈥檚 more than just crunching numbers; it鈥檚 about logical thinking and applying the right principles.
In physics problem solving,
  • Identify: Start by determining what you know and what you need to find out. In the case of the water tower exercise, we needed to calculate the water pressure.
  • Formulate: Choose the right formula. For this task, we used the pressure formula related to fluid height: \( P = \rho \cdot g \cdot h \). Choosing the correct formula is critical since it incorporates all known variables.
  • Substitute and Solve: Plug in the known values and calculate the solution. Completing these steps accurately provides the final result 鈥� the water pressure at the faucet.
  • Reflect: Always check if the answer makes sense and reevaluate the initial assumptions, like neglecting friction or other users in a real-life context.
Finding clear paths through structured problem solving turns even tricky physics exercises into manageable challenges.
Height and Pressure Relationship
The relationship between height and pressure is a fundamental concept in fluid dynamics. This relationship is crucial for understanding how water can flow from a high tower down to your faucet, delivering the pressure needed for water to surge out.
The principle behind this is simple: as the height (\(h\)) of a fluid column increases, so does the pressure (\(P\)) at its base. This is due to the weight of the fluid above exerting downward force.
The formula used illustrates this perfectly: \( P = \rho \cdot g \cdot h \), where:
  • \(\rho\) is the fluid density
  • \(g\) is gravitational acceleration
  • \(h\) is the height of the fluid column
This means if you increase the height of the water in a tower, the pressure at the bottom鈥攕ay at a house鈥檚 faucet鈥攁lso rises. This concept isn鈥檛 just critical for solving textbook problems but is essential in engineering and everyday applications where fluid movement is involved, like in design of plumbing systems and aqueducts.

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Most popular questions from this chapter

Atmospheric pressure is about \(1.0 \times 10^{5} \mathrm{~Pa}\). How large a force does the still air in a room exert on the inside of a window pane that is \(40 \mathrm{~cm} \times 80 \mathrm{~cm}\) ? The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and is given by $$F=P A=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.40 \times 0.80 \mathrm{~m}^{2}\right)=3.2 \times 10^{4} \mathrm{~N}$$ Of course, a nearly equal force due to the atmosphere on the outside keeps the window from breaking.

The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\). What fraction of the volume of a piece of ice will be above the liquid when floating in fresh water? The piece of ice will float in the water, since its density is less than \(1000 \mathrm{~kg} / \mathrm{m}^{3}\), the density of water. As it does, \(F_{B}=\) Weight of displaced water \(=\) Weight of piece of ice But the weight of the ice is \(\rho_{\text {ice }} g V\), where \(V\) is the volume of the piece. In addition, the weight of the displaced water is \(\rho_{w} g V^{\prime}\), where \(V^{\prime}\) is the volume of the displaced water. Substituting into the above equation $$\begin{aligned} \rho_{\text {ice }} g V &=\rho_{w} g V^{\prime} \\ V^{\prime} &=\frac{\rho_{\text {ice }}}{\rho_{w}} V=\frac{917}{1000} V=0.917 V \end{aligned}$$ The fraction of the volume that is above water is then $$ \frac{V-V^{\prime}}{V}=\frac{V-0.917 V}{V}=1-0.917=0.083 \text { or } 8.3 \% $$

A solid wooden cube, \(30.0 \mathrm{~cm}\) on each edge, can be totally submerged in water if it is pushed downward with a force of \(54.0 \mathrm{~N}\). What is the density of the wood?

A piece of pure gold \(\left(\rho=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is suspected to have a hollow center. It has a mass of \(38.25 \mathrm{~g}\) when measured in air and \(36.22 \mathrm{~g}\) in water. What is the volume of the central hole in the gold? Remember that you go from a density in \(\mathrm{g} / \mathrm{cm}^{3}\) to \(\mathrm{kg} / \mathrm{m}^{3}\) by multiplying by 1000 . From \(\rho=m / V\), Actual volume of \(38.25 \mathrm{~g}\) of gold \(=\frac{0.03825 \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}}=1.982 \times 10^{-6} \mathrm{~m}^{3}\) Volume of displaced water \(=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3}\) $$ \text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3} $$

How high would water rise in the essentially open pipes of a building if the water pressure gauge shows the pressure at the ground floor to be \(270 \mathrm{kPa}\) (about \(40 \mathrm{lb} / \mathrm{in.}^{2}\) )? Water pressure gauges read the excess pressure just due to the water, that is, the difference between the absolute pressure in the water and the pressure of the atmosphere. The water pressure at the bottom of the highest column that can be supported is \(270 \mathrm{kPa}\). Therefore, \(P=\rho_{w} g h\) gives $$ h=\frac{P}{\rho_{w} g}=\frac{2.70 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=27.5 \mathrm{~m} $$
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