Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 62

A studant needs to measure the air pressure inside a compressed air tenk but does not have ready access to a pressure gage. Using materials already in the lab, she builds a U-tube manometer using two clear 3 -ft-long plastic tubes, flexible hoses, and a tape measure. The only readily avai able liquids are water from a tap and a bottle of corn syrup. She selects the corn syrup because it has a larger density \((S G=1.4) .\) What is the maximum air pressure, in psia, that can be measured?

Short Answer

Expert verified
The maximum air pressure that can be measured is 1.84 psia.

Step by step solution

01

Calculate the Density of Corn Syrup

Using the relation for specific gravity \(\rho = SG * \rho_{water}\) and knowing SG = 1.4 and \(\rho_{water} = 1000 \ kg/m^{3}\), the density of corn syrup is calculated. It gets \(\rho = 1.4 * 1000 = 1400 \ kg/m^{3}\).
02

Convert the Height of the Manometer from Feet to Meters

To keep in SI units, convert the 3-feet-long manometer to meters. 1 foot = 0.3048 meter so 3 feet = 0.9144 meters.
03

Find the Maximum Measurable Pressure

Use the pressure formula \(P = \rho g h\) and plug in the values \(\rho = 1400 \ kg/m^{3}\), \(g = 9.81 \ m/s^{2}\), and \(h = 0.9144 \ m\). That's how one gets: \(P = 1400 * 9.81 * 0.9144 =12.7 * 10^{3} \ Pa\).
04

Convert Pressure from Pa to psia

Knowing that 1 Pa = 0.000145 psi, the maximum pressure in psi will be \(12.7 * 10^{3} * 0.000145 = 1.84 \ psia\) (rounded to two decimal places).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

Determine the pressure at the bottom of an open 5 -m-deep tank in which a chemical process is taking place that causes the density of the liquid in the tank to vary as $$\rho=\rho_{\text {surf }} \sqrt{1+\sin ^{2}\left(\frac{h}{h_{\text {bot }}} \frac{\pi}{2}\right)}$$ where \(h\) is the distance from the free surface and \(\rho_{\text {surf }}=1700 \mathrm{kg} / \mathrm{m}^{3}\).

A 30 -ft-high downspout of a house is clogged at the bottom. Find the pressure at the bottom if the downspout is filled with \(60^{\circ} \mathrm{F}\) rainwater.

A 2 -ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2 -ft- diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere.

A 5 -gal, cylindrical open container with a bottom area of 120 in. \(^{2}\) is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of \(3 \mathrm{ft} / \mathrm{s}^{2}\). (b) What resultsnt force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal \(=231\) in. \(^{3}\) )
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Kinematics Physics

Read Explanation

Radiation

Read Explanation

Quantum Physics

Read Explanation

Force

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free