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A barometric pressure of 29.4 in. Hg corresponds to what value of atmospheric pressure in psia, and in pascals?

Short Answer

Expert verified
The atmospheric pressure is 14.44 psia and 99,561.89 pascals.

Step by step solution

01

Understand the Conversion Factors

First, we need to know the conversion factors to convert inches of mercury (in. Hg) to both pounds per square inch absolute (psia) and pascals. The conversion factors are: 1. 1 in. Hg = 0.491154 psia 2. 1 in. Hg = 3386.39 pascals
02

Convert to PSIA

Using the conversion factor for psia, multiply the given barometric pressure by 0.491154, which is the conversion factor for transforming in. Hg to psia:\[29.4 \, ext{in. Hg} \times 0.491154 \, ext{psia/in. Hg} = 14.43952 \, ext{psia}\]
03

Convert to Pascals

Using the conversion factor for pascals, multiply the given barometric pressure by 3386.39, which is the conversion factor for converting in. Hg to pascals:\[29.4 \, ext{in. Hg} \times 3386.39 \, ext{Pa/in. Hg} = 99561.886 \, ext{Pa}\]
04

Conclude with Final Answers

The atmospheric pressure corresponding to a barometric pressure of 29.4 in. Hg is 14.44 psia and 99,561.89 pascals (rounded to two decimal places).

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Key Concepts

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

Understanding Pound Per Square Inch Absolute (PSIA)
Pound per square inch absolute (PSIA) is a measurement of pressure relative to a perfect vacuum. When you see the term "absolute," it means that it includes atmospheric pressure in the measurement. This distinguishes it from gauge pressure (PSIG), which only considers the pressure above atmospheric pressure.

For example, if you see a pressure measurement as 0 psia, it means there is a perfect vacuum with no pressure applied. On the other hand, if a pressure is given as 14.7 psia, it is at sea-level standard atmospheric pressure. This is why understanding PSIA is crucial in fields like meteorology and engineering.

To convert barometric pressure to PSIA, we use specific conversion factors. For our problem, the given conversion factor is:
  • 1 inch of mercury (in. Hg) = 0.491154 PSIA
This factor allows easy conversion by multiplying it with the barometric pressure value measured in inches of mercury. The calculation makes it simpler to understand how pressure in our atmosphere compares to a vacuum.
The Simplicity of Pascal Conversion
The Pascal (Pa) is the SI unit of pressure and is widely used for scientific calculations. It provides a straightforward measure of pressure and is named after the French mathematician Blaise Pascal. The formula to convert a pressure reading from inches of mercury (in. Hg) to pascals involves multiplying by a standard conversion factor.

Given in our steps, the conversion factor is:
  • 1 in. Hg = 3386.39 pascals
This factor means that every inch of mercury equals 3386.39 pascals. The conversion process is simple: you just need to multiply the atmospheric pressure in in. Hg by this factor to get the equivalent pressure in pascals.

This conversion is crucial in scientific settings where precise measurements are necessary and units need to be standardized globally. It helps in understanding local atmospheric conditions, latitude effects on pressure, and providing universal pressure data.
Calculating Atmospheric Pressure
Atmospheric pressure is the force per unit area exerted by the atmosphere at any given point. It plays a significant role in weather forecasting and aviation. Typically, at sea level, this pressure is about 14.7 PSIA or 101,325 Pascals, representing a standard atmosphere.

The atmospheric pressure calculation involves converting barometric readings like those in inches of mercury to useful measurements for scientific and engineering purposes. Using the conversion factors:
  • 1 in. Hg = 0.491154 psia
  • 1 in. Hg = 3386.39 pascals
You can effortlessly find equivalent pressures in other units.

Identifying atmospheric pressure helps in calibrating instruments and predicting weather patterns. Furthermore, these calculations provide engineers with crucial data when designing structures that must withstand atmospheric conditions, such as aircraft and high-altitude buildings.

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Most popular questions from this chapter

Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure at a depth of \(5 \mathrm{km}\), assuming that seawater has a constant specific weight of \(10.1 \mathrm{kN} / \mathrm{m}^{3}\) ? Express your answer in pascals and psi.

An open 1 -m-diameter tank contains water at a depth of \(0.7 \mathrm{m}\) when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank.

An open rectangular tank is \(2 \mathrm{m}\) wide and \(4 \mathrm{m}\) long. The tank contains water to a depth of \(2 \mathrm{m}\) and \(\operatorname{oil}(S G=0.8)\) on top of the water to a depth of \(1 \mathrm{m}\). Determine the magnitude and location of the resultant fluid force acting on one end of the tank.

On the suction side of a pump a Bourdon pressure gage reads 40 -kPa vacuum. What is the corresponding absolute pressure if the local atmospheric pressure is \(100 \mathrm{kPa}\) (abs)?

An area in the form of an isosceles triangle with a base width of \(6 \mathrm{ft}\) and an altitude of \(8 \mathrm{ft}\) lies in the plane forming one wall of a tank which contains a liquid having a specific weight of \(79.8 \mathrm{lb} / \mathrm{ft}^{3}\). The side slopes upward making an angle of \(60^{\circ}\) with the horizontal. The base of the triangle is horizontal and the vertex is above the base. Determine the resultant force the fluid exerts on the area when the fluid depth is \(20 \mathrm{ft}\) above the base of the triangular area. Show, with the aid of a sketch, where the center of pressure is located.

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