/*! 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 5 How high would water rise in the... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ h=\frac{P}{\rho_{w} g}=\frac{2.70 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=27.5 \mathrm{~m} $$

Short Answer

Expert verified
The water will rise to a height of 27.5 meters.

Step by step solution

01

Given Information

We know that the water pressure at the ground floor is given as \(270\ \text{kPa}\). This pressure is the excess pressure due to water alone.
02

Understanding the Formula

The water pressure equation is given by \(P = \rho_w g h\), where \(P\) is the pressure, \(\rho_w\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the height of the water column. We want to solve for \(h\).
03

Rearranging the Formula

We rearrange the equation \(P = \rho_w g h\) to solve for \(h\). This gives us: \[ h = \frac{P}{\rho_w g} \]
04

Substitute Known Values

Substitute the given values into the formula:- \(P = 270\ \text{kPa} = 2.70 \times 10^5\ \text{N/m}^2\)- \(\rho_w = 1000\ \text{kg/m}^3\), the density of water- \(g = 9.81\ \text{m/s}^2\), the acceleration due to gravity\[ h = \frac{2.70 \times 10^5\ \text{N/m}^2}{1000\ \text{kg/m}^3 \times 9.81\ \text{m/s}^2} \]
05

Calculate the Height

Perform the calculation:\[ h = \frac{2.70 \times 10^5}{1000 \times 9.81} \]\[ h = 27.5\ \text{m} \]This means the water will rise to a height of 27.5 meters.

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Key Concepts

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

Water Pressure Calculation
Water pressure calculation is an essential concept in hydrostatics that helps determine how high water can ascend in pipes or similar structures. The formula used in this context is:
  • \( P = \rho_w g h \)
This equation shows the relationship between pressure \( P \), the density of water \( \rho_w \), the gravitational acceleration \( g \), and the height \( h \) of the water column. To find the height to which water can rise, we rearrange the formula to solve for \( h \):
  • \( h = \frac{P}{\rho_w g} \)
Understanding this calculation means recognizing that the height is directly proportional to the pressure and inversely proportional to both the density of water and the gravitational acceleration. It provides insight into how pressure changes can affect water's ability to move upwards through pipes and other vertical structures.
Density of Water
Density is a critical factor in calculating water pressure. For liquid water at typical conditions, the density \( \rho_w \) is approximately \( 1000 \ \text{kg/m}^3 \). This value indicates how much mass of water is contained in a cubic meter volume.
  • High density means more mass per unit volume.
  • The density of water remains relatively constant under standard atmospheric conditions.
  • Variations can occur with temperature changes but are typically negligible for most pressure calculations.
This consistent density helps simplify calculations in hydrostatics, as seen in the exercise, because it remains a steady and predictable factor. Understanding water density is fundamental to mastering concepts of buoyancy and fluid dynamics.
Acceleration Due to Gravity
Acceleration due to gravity \( g \) is a constant that affects how fluids exert pressure. On Earth, this value is approximately \( 9.81 \ \text{m/s}^2 \). It plays a crucial role in determining the pressure exerted by fluids like water in a column:
  • Gravity's constant pull defines the potential energy of fluids in motion.
  • This acceleration influences how high water can rise in open columns based on pressure.
  • Without gravity, pressure calculations would be drastically different.
In pressure-height relationships, gravity helps translate pressure measurements into understandable height equivalents. Understanding \( g \) allows us to predict and manage fluid systems effectively.
Pressure-Height Relationship
The pressure-height relationship in hydrostatics is a direct connection between the pressure at a given point and the height of a fluid column it can support. This relationship is crucial in many real-world applications, such as designing water supply systems or calculating the capacity of hydraulic machines:
  • Pressure increases with height due to the weight of the fluid above.
  • In vertical columns, higher pressure implies the potential for greater height.
  • This relationship guides engineers in planning efficient fluid distribution systems.
In our exercise, this relationship is exploited to determine that a pressure of \( 270 \ \text{kPa} \) can support a column of water up to \( 27.5 \ \text{meters} \). By mastering these calculations and relationships, we gain the ability to design and analyze fluid systems more effectively.

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Most popular questions from this chapter

Find the pressure due to the fluid at a depth of \(76 \mathrm{~cm}\) in still \((a)\) water \(\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and (b) mercury ( \(\left.\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)\). (a) \(P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=7.5 \mathrm{kPa}\) (b) \(P=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx 1.0 \mathrm{~atm}\)

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 \(=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3}\) $$ \text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3} $$

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.

A barrel will rupture when the gauge pressure within it reaches \(350 \mathrm{kPa}\). It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil \(\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)\). How long can the pipe be if the barrel is not to rupture? From \(P=\rho g h\) we have $$ h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m} $$

A \(2.0-\mathrm{cm}\) cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of \(47.3 \mathrm{~g}\) when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp \(\mathrm{gr}=1.26 ?[\) Hint \(:\) Find \(\rho\) too. \(]\)

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