/*! 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 13 The U-tube device connected to t... [FREE SOLUTION] | 91Ó°ÊÓ

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The U-tube device connected to the tank in Fig. 13-6 is called a manometer. As you can see, the mercury in the tube stands higher in one side than the other. What is the pressure in the tank if atmospheric pressure is \(76 \mathrm{~cm}\) of mercury? The density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\). from which \(P=95 \mathrm{kPa}\). Or, more simply perhaps, we could note that the pressure in the tank is \(5.0 \mathrm{~cm}\) of mercury lower than atmospheric. So the pressure is \(71 \mathrm{~cm}\) of mercury, which is \(94.6 \mathrm{kPa}\).

Short Answer

Expert verified
The pressure in the tank is 94.6 kPa.

Step by step solution

01

Identify Given Values

We are given the atmospheric pressure as 76 cm of mercury, the mercury's density as \( 13.6 \, \mathrm{g/cm^3} \), and that the pressure in the tank is 5 cm of mercury lower than atmospheric pressure.
02

Express Relationship between Pressures

Since the tank pressure is 5 cm less of mercury than the atmospheric pressure, we can express the tank pressure as the difference: \( P_{\text{tank}} = 76 \, \text{cm Hg} - 5 \, \text{cm Hg} \).
03

Calculate Pressure in Tank in cm Hg

Calculate the pressure in the tank by subtracting 5 cm Hg from 76 cm Hg: \( P_{\text{tank}} = 76 \, \text{cm Hg} - 5 \, \text{cm Hg} = 71 \, \text{cm Hg} \).
04

Convert Pressure from cm Hg to kPa

Use the conversion factor for mercury: \( 1 \, \text{cm Hg} = 1.33 \, \text{kPa} \). Thus, convert the pressure: \( 71 \, \text{cm Hg} \times 1.33 \, \text{kPa/cm Hg} = 94.43 \, \text{kPa} \).
05

Verify Units and Result

The calculated pressure of 94.43 kPa is in line with the provided solution (94.6 kPa after rounding).

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Key Concepts

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

Manometer
A manometer is a device used to measure pressure. Typically relying on a liquid within a U-shaped tube, it can determine the pressure of a gas or liquid in a container based on the difference in the liquid levels in the tube. This difference shows how much the pressure inside the container varies from the surrounding atmospheric pressure. Manometers can measure relative pressures with more direct visualization compared to other devices.
A simple manometer setup includes the following components:
  • An open-ended U-tube containing a liquid like mercury or water.
  • One side of the tube connected to the system whose pressure is to be measured.
  • Another side exposed to atmospheric pressure.
When there is a pressure difference, the fluid levels in the two arms of the tube aren't equal. This level difference, usually measured in units like centimeters of mercury (cm Hg), indicates the pressure balance between the atmospheric pressure and the pressure within the container.
Mercury Type Pressure Measurement
Mercury type pressure measurement uses mercury as the medium to determine pressure differences. Mercury is favored because of its high density and stable properties, enabling precise pressure measurements. This density allows even small differences in pressure to move the mercury noticeably within the manometer, providing accurate readings.
Here are some key points about mercury type manometers:
  • Due to mercury's density (13.6 g/cm³), manometers using this liquid can be quite compact yet precise.
  • Mercury's non-evaporative nature makes it particularly well-suited for consistent results over time.
  • Such devices often measure pressure in cm Hg, a natural fit given the liquid column heights they deal with.
Despite their accuracy, mercury manometers are not always practical due to the toxic nature of mercury. Hence, proper precautions are necessary during their handling and operation.
U-tube Device
The U-tube device is the core structure of a traditional manometer. Its U-shaped architecture permits comparisons between pressures at either side of its tube by observing how the liquid column adjusts.
The working principle of a U-tube device involves:
  • Equilibrating the pressure on one side with a known pressure (usually atmospheric).
  • Observing the height difference between the columns of mercury or some other fluid.
  • Using this difference to calculate the pressure in the connected system.
Thanks to the clear and straightforward operation of the U-tube, users can readily gauge pressure readings directly from the physical arrangement of liquid within the device. This simplicity makes U-tube devices particularly favorable for educational purposes and basic experiments in fluid mechanics.
Pressure Conversion
Pressure conversion involves translating pressure readings from one unit to another. This is a common requirement when working with manometers, especially when dealing with mercury, given its specific measurement units like cm Hg.
To convert the pressure reading in cm of mercury to kilopascals (kPa):
  • Recognize that the conversion rate is approximately 1 cm Hg = 1.33 kPa.
  • Multiply the pressure value in cm Hg by this conversion factor to find the equivalent kPa.
In practice, if you have a reading such as 71 cm Hg (as seen in the original problem), you would compute:\[71 \, \text{cm Hg} \times 1.33 \, \text{kPa/cm Hg} = 94.43 \, \text{kPa}\]This conversion process is crucial because different scientific and industrial contexts might use varying units to express pressure, necessitating a consistent standard for interpretation.

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Most popular questions from this chapter

The sole of a man's size-10 shoe is around \(11.0\) in. by \(4.00\) in. Determine the gauge pressure under the feet of a 200-lb man standing upright. Give your answer in both lb/in. \(^{2}\) and Pa. [Hint: \(1.00 \mathrm{lb} / \mathrm{in}^{2}=6895 \mathrm{~Pa}\). Check your work using \(1.00 \mathrm{in} .^{2}=6.45 \times\) \(10^{-4} \mathrm{~m}^{2}\) and \(\left.1.00 \mathrm{lb}=4.448 \mathrm{~N} .\right]\)

A beaker contains oil of density \(0.80 \mathrm{~g} / \mathrm{cm}^{3}\). A \(1.6-\mathrm{cm}\) cube of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) hanging vertically on a thread is submerged in the oil. Find the tension in the thread.

Suppose we have a spring scale that reads in grams and we measure the mass of a cork in air to be \(5.0 \mathrm{~g} .\) Using the same scale, it is found that a sinker has an apparent mass of 86 g when completely immersed in water. The cork is attached to the sinker, the two are completely immersed in water, and now the scale reads 71 g. Determine the density of the cork. [Hint: The buoyance of the cork is responsible for the decreased scale reading.]

A mass (or load) acting downward on a piston confines a fluid of density \(\rho\) in a closed container, as shown in Fig. \(13-2\). The combined weight of the piston and load on the right is \(200 \mathrm{~N}\), and the cross-sectional area of the piston is \(A=8.0 \mathrm{~cm}^{2}\). Find the total pressure at point- \(B\) if the fluid is mercury and \(h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13\right.\) \(600 \mathrm{~kg} / \mathrm{m}^{3}\) ). What would an ordinary pressure gauge read at \(B\) ? Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere: This added pressure is applied at all points within the fluid. Therefore, the total pressure at \(B\) is composed of three parts: Pressure of the atmosphere \(=1.0 \times 10^{5} \mathrm{~Pa}\) Pressure due to the piston and load \(=\frac{F_{w}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{P}\) Pressure due to the height \(h\) of fluid \(=h p g=0.33 \times 10^{5} \mathrm{~Pa}\) In this case, the pressure of the fluid itself is relatively small. We have Total pressure at \(B=3.8 \times 10^{5} \mathrm{~Pa}\) The gauge pressure does not in clude atmospheric pressure. Therefore, Gauge pressure at \(B=2.8 \times 10^{5} \mathrm{~Pa}\)

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, \(P=\rho_{w} p h\) 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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