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Chapter 3: Problem 5

Knowing the density of mercury is based on the specific gravity \(13.6\), find the height of some column of liquid in meters that would be equivalent to 1 atmosphere.

Short Answer

Expert verified
The height of the mercury column is approximately 0.760 meters.

Step by step solution

01

Understanding the Problem

First, let's understand that we need to find the height of a liquid column of mercury that exerts the same pressure as one atmosphere. We know that one atmosphere is equivalent to 101325 Pa.
02

Using the Formula for Pressure

The pressure exerted by a liquid column is given by the formula: \[ P = h \cdot \rho \cdot g \]where \(P\) is the pressure, \(h\) is the height of the liquid column, \(\rho\) is the density of the liquid, and \(g\) is the acceleration due to gravity (approximately 9.81 m/s^2 on Earth).
03

Calculate the Density of Mercury

Specific gravity is the ratio of the density of a substance to the density of water. Given that the specific gravity of mercury is 13.6, and the density of water is 1000 kg/m^3, the density of mercury \(\rho\) is:\[ \rho = 13.6 \times 1000 = 13600 \text{ kg/m}^3 \]
04

Rearranging for Height

Rearrange the pressure formula to solve for the height \(h\):\[ h = \frac{P}{\rho \cdot g} \]
05

Substitute Known Values

Substitute the known values into the equation:\[ h = \frac{101325}{13600 \times 9.81} \approx 0.760 \text{ meters} \]
06

Conclusion

The height of the mercury column that is equivalent to one atmosphere is approximately 0.760 meters.

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Key Concepts

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

Specific Gravity
Specific gravity is a very useful concept in fluid mechanics that helps us understand how dense a substance is compared to water. It is essentially a ratio, calculated by dividing the density of a substance by the density of water.

For example, if a substance has a specific gravity of 1, that means it has the same density as water. When specific gravity is greater than 1, the substance is denser than water. Conversely, if it's less than 1, the substance is lighter than water.

In our exercise, the specific gravity of mercury is given as 13.6. To find its density, we use the known density of water, which is 1000 kg/m鲁. So, the density of mercury becomes:
  • Density of mercury = Specific gravity of mercury 脳 Density of water
This translates to 13.6 脳 1000, giving us a density of 13600 kg/m鲁 for mercury.
Density
Density is a fundamental property of materials and plays a critical role in fluid mechanics. It refers to the mass contained in a given unit of volume and is usually expressed in kilograms per cubic meter (kg/m鲁).

Understanding density helps us determine how substances will behave in different conditions. For example, a dense fluid like mercury is much heavier than water for the same volume.

In our context, the density of mercury was computed using its specific gravity. Knowing the density allows us to calculate other essential parameters, such as pressure and buoyancy, which are pivotal in understanding fluid behavior.
Pressure
Pressure in the context of fluid mechanics is the force exerted by the fluid per unit area. It is usually measured in Pascals (Pa).

The pressure exerted by a liquid column can be calculated using the formula:
  • \[P = h \cdot \rho \cdot g\]
where \(P\) is the pressure, \(h\) is the height of the liquid column, \(\rho\) is the density of the fluid, and \(g\) is acceleration due to gravity, approximately 9.81 m/s虏.

This formula allows us to derive the height of a fluid column that corresponds to a specific pressure, as demonstrated in our exercise where the height of mercury needed to equal the atmospheric pressure was calculated.
Atmosphere
The term "atmosphere" has a specific meaning in fluid mechanics. It refers to the pressure exerted by the weight of the air above a particular point. Standard atmosphere (atm) is a unit of pressure defined as 101325 Pascals.

Understanding atmospheric pressure is crucial since it's a baseline for measuring the pressure exerted by other fluids. When we say a liquid exerts 1 atmosphere of pressure, it means the pressure exerted by the fluid column equates to the atmospheric pressure at sea level.

In practical terms, knowing atmospheric conditions can help in designing accurate systems for pressure measurement and fluid movement, understanding phenomena like boiling points and weather patterns.

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Most popular questions from this chapter

Why would a differential manometer be unsuitable for measuring pressure drops in hydraulic systems aboard an orbiting spacecraft?

A sunken ship of original displacement of 5 tons is to be raised from \(70 \mathrm{ft}\) of seawater by means of a rubber flotation device. How many \(\mathrm{ft}^{3}\) of STD sea level air \(\left(60^{\circ} \mathrm{F}, 14.7 \mathrm{psia}\right)\) must be pumped into the device to just lift the boat off the bottom?

The profile of the inner face of a dam is described by a parabola having the form \(x^{2}=3 z\), where \(z\) is the height above the base and \(x\) is the horizontal distance of the face from the vertical reference line. If the water level is \(3 \mathrm{~m}\) above the base, find the thrust on the dam per unit width due to the water pressure, and its line-of-action.

A piece of ore weighing \(10 \mathrm{lbf}\) in air was found to raise the level of water in a square \(\operatorname{tank}(1 \mathrm{ft} \times 1 \mathrm{ft} \times 1 \mathrm{ft})\) by \(0.03 \mathrm{ft}\) when submerged in the tank. Find (a) the volume of the piece of ore \(\left(f t^{3}\right)\), (b) the buoyant force acting on the ore (lbf), and (c) the specific gravity \((S)\) of the ore (neglect the buoyant force of air).

Given: Liquids \(A\) and \(B\) are water; \(S_{2}=\) \(0.8, h_{1}=1 \mathrm{ft}, h_{2}=0.5 \mathrm{ft}\), and \(h_{3}=2 \mathrm{ft}\). (a) Find \(p_{A}-p_{B}\) in psi. (b) If \(p_{B}=10\) psia and the barometer reading is \(29.5\) in. of mercury, find the gauge pressure at \(A\) in \(1 \mathrm{bf} / \mathrm{ft}^{2}\). $$ \text { (a) } \begin{aligned} h_{A}-h_{1} S_{1}-h_{2} S_{2}+h_{3} S_{3}=h_{B} \\ h_{A}-1(1)-0.5(0.8)+2(1)=h_{B} \\ h_{A}-h_{B}=1+0.4-2 \\ =-0.6 \mathrm{ft} \text { water } \end{aligned} $$ \(\begin{aligned} p_{A}-p_{B}=&-0.6 \mathrm{ft} \text { water } \\ &(14.7 \mathrm{psi} / 33.9 \mathrm{ft} \text { water }) \\ &=-0.26 \mathrm{psi} \\\ \text { (b) } p_{A}=p_{B}-0.26 \mathrm{psi} \\\=10-0.26=9.74 \mathrm{psia} \\\ \text { Atm pressure }=29.5 \mathrm{in} . \mathrm{Hg} \\ \quad(14.7 \mathrm{psi} / 30 \text { in. } \mathrm{Hg}) \\\=14.47 \mathrm{psi} \\\ p_{A}=9.74-14.47=-4.73 \mathrm{psi} \\\=-4.73 \mathrm{lbf} / \mathrm{in} .{ }^{2}\left(144 \mathrm{in} .^{2} / \mathrm{ft}^{2}\right) \\\=681 \mathrm{lbf} / \mathrm{ft}^{2} \text { vacuum } \end{aligned}\)
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