Pressure Measurement MCQ Quiz - Objective Question with Answer for Pressure Measurement - Download Free PDF

Explanation:

Gauge Pressure and Its Relation to Atmospheric Pressure

Definition: Gauge pressure is the pressure measured relative to the surrounding atmospheric pressure. It is commonly used in engineering and industrial applications to measure the pressure of fluids (liquids or gases) within a system. The concept of gauge pressure is essential because it excludes the atmospheric pressure, which simplifies the measurement process in most practical scenarios.

Working Principle:

To understand the relationship between gauge pressure and atmospheric pressure, it is necessary to understand the following terms:

• Absolute Pressure: The total pressure measured relative to a perfect vacuum. It is the sum of the atmospheric pressure and the gauge pressure.
• Atmospheric Pressure: The pressure exerted by the Earth's atmosphere at a given location and altitude. At sea level, atmospheric pressure is approximately 101.325 kPa or 1 atmosphere.
• Gauge Pressure: The pressure measured relative to the atmospheric pressure. Gauge pressure can be positive (when the system pressure is above atmospheric pressure) or negative (when the system pressure is below atmospheric pressure, also called vacuum pressure).

The relationship between these pressures is given by the following equation:

Absolute Pressure = Gauge Pressure + Atmospheric Pressure

Or equivalently:

Gauge Pressure = Absolute Pressure - Atmospheric Pressure

This equation highlights that gauge pressure is the difference between the absolute pressure within a system and the atmospheric pressure. This relationship is crucial for understanding how gauge pressure behaves in different environments and under varying atmospheric conditions.

Correct Option Analysis:

The correct option is:

Option 3: Gauge pressure is the difference between absolute pressure and atmospheric pressure.

This option accurately describes the relationship between gauge pressure, absolute pressure, and atmospheric pressure. As explained above, gauge pressure is calculated by subtracting atmospheric pressure from the absolute pressure within a system. This definition aligns with the standard practices in pressure measurement and is universally accepted in engineering and scientific contexts.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Gauge pressure is independent of atmospheric pressure.

This statement is incorrect because gauge pressure is explicitly defined relative to atmospheric pressure. If atmospheric pressure changes (e.g., at higher altitudes or under different weather conditions), the gauge pressure reading for the same absolute pressure will also change. Therefore, gauge pressure is not independent of atmospheric pressure.

Option 2: Gauge pressure equals atmospheric pressure minus absolute pressure.

This option is incorrect because it reverses the actual relationship between gauge pressure, absolute pressure, and atmospheric pressure. As stated earlier, the correct formula is:

Gauge Pressure = Absolute Pressure - Atmospheric Pressure

Switching the terms, as suggested in this option, would lead to incorrect calculations and misunderstandings in practical applications.

Option 4: Gauge pressure is the sum of absolute and atmospheric pressure.

This option is incorrect because it misrepresents the relationship between the pressures. The sum of absolute pressure and atmospheric pressure would result in a value that has no physical meaning in the context of pressure measurement. Instead, the correct relationship is:

Absolute Pressure = Gauge Pressure + Atmospheric Pressure

Gauge pressure is not the sum of absolute and atmospheric pressure.

Conclusion:

Understanding the relationship between absolute pressure, atmospheric pressure, and gauge pressure is fundamental to pressure measurement in engineering and scientific applications. Gauge pressure is the difference between absolute pressure and atmospheric pressure, making it a practical and widely used parameter for measuring the pressure of fluids within a system. The correct option, Option 3, accurately describes this relationship, while the other options misrepresent the fundamental concepts involved in pressure measurement.

Concept:

Hydrostatic Pressure, P = ρgh

Where, ρ = Density, g = Acceleration due to gravity, h = Manometric head

Calculation:

Given:

h = 20 m, g = 9.81 m/s2, ρ = 1000 kg/m3 

∴ P = 1000 × 9.8 × 20 = 196000 = 196 kPa

Explanation:

Atmospheric Pressure

Definition: Atmospheric pressure is the pressure exerted by the Earth's atmosphere at any given point. It is the force per unit area exerted on a surface by the weight of the air above that surface in the atmosphere of Earth (or that of another planet). Atmospheric pressure is measured with an instrument called a barometer, and the standard atmospheric pressure at sea level is defined as 101.325 kPa (kilopascals), 1 atm (atmosphere), or 760 mmHg (millimeters of mercury).

Working Principle: The atmospheric pressure at a given point is the result of the weight of the column of air above that point. As altitude increases, the atmospheric pressure decreases because there is less air above the point exerting pressure. Conversely, as altitude decreases (for example, going below sea level), atmospheric pressure increases.

Atmospheric pressure is crucial for various natural and man-made processes. For example, it influences weather patterns and wind currents. In addition, it plays a significant role in the boiling point of liquids, as lower atmospheric pressure at higher altitudes results in a lower boiling point.

Measurement: Atmospheric pressure can be measured using different types of barometers:

• Mercury Barometer: This device uses a column of mercury in a glass tube. The height of the mercury column changes in response to the atmospheric pressure. The standard atmospheric pressure supports a mercury column 760 mm high.
• Aneroid Barometer: This instrument uses a small, flexible metal box called an aneroid cell. The cell expands or contracts with changes in atmospheric pressure, and this movement is translated into a pressure reading.

Importance in Everyday Life: Atmospheric pressure affects various aspects of daily life, including weather conditions, aviation, and even the human body. For instance:

• Weather Forecasting: Changes in atmospheric pressure are closely monitored by meteorologists to predict weather changes. A falling atmospheric pressure often indicates stormy weather, while rising pressure suggests fair weather.
• Aviation: Pilots need to be aware of atmospheric pressure to determine altitude and to ensure safe takeoff and landing procedures. Aircraft altimeters are calibrated based on atmospheric pressure.
• Human Health: Rapid changes in atmospheric pressure can affect the human body, leading to conditions such as headaches or joint pains in some individuals.

Applications: Atmospheric pressure is a fundamental concept in various scientific and engineering fields. Some applications include:

• Vacuum Systems: Understanding atmospheric pressure is essential for designing and operating vacuum systems used in manufacturing, scientific research, and space exploration.
• Hydraulics and Pneumatics: Atmospheric pressure is considered when designing systems that rely on fluid or air pressure for operation.
• Environmental Science: Studying atmospheric pressure helps in understanding climate change and environmental phenomena.

Correct Option Analysis: Option 1 is correct because it accurately defines atmospheric pressure as the pressure exerted by the Earth's atmosphere at any given point. This definition encompasses the fundamental concept of atmospheric pressure and its implications in various fields.

Analysis of Other Options:

Option 2: This option describes the pressure difference between two fluids, which is not the same as atmospheric pressure. The pressure difference between two fluids is a concept used in fluid dynamics and is typically measured using devices such as differential pressure gauges. This option does not accurately represent the definition of atmospheric pressure.

Option 3: This option refers to the pressure inside a fluid container, which is again a different concept from atmospheric pressure. The pressure inside a fluid container can be influenced by various factors, including the type of fluid, the volume of the container, and the temperature. This option does not capture the essence of atmospheric pressure exerted by the Earth's atmosphere.

Option 4: This option mentions the pressure of a vacuum, which is essentially the absence of pressure or a very low-pressure environment. Vacuum pressure is the pressure below the atmospheric pressure and is measured in terms of absolute pressure or vacuum levels. This option does not define atmospheric pressure but rather its opposite.

In conclusion, atmospheric pressure is a fundamental concept that plays a crucial role in various natural and engineered systems. Understanding atmospheric pressure and its measurement is essential for weather forecasting, aviation, environmental science, and many other fields. The correct option accurately defines atmospheric pressure as the pressure exerted by the Earth's atmosphere at any given point, distinguishing it from other types of pressure described in the incorrect options.

Concept:

Pressure head:

• Pressure head is the vertical height of a fluid column that corresponds to a specific pressure exerted by the fluid.

Mathematically, it is given as: \(h = \frac{P}{ρ g}\)

Where, P is pressure, ρ is fluid density, and g is acceleration due to gravity.

Hence, pressure head is best described as the height of a fluid column equivalent to the pressure exerted by the fluid.

Concept:

For constant volume: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \Rightarrow P_2 = P_1 \cdot \frac{T_2}{T_1} \)

Calculation:

Given:

• \( P_1^{gauge} = 1.75 \, \text{bar} \), so \( P_1^{abs} = 1.75 + 1.01 = 2.76 \, \text{bar} \)
• \( T_1 = 27^\circ C = 300 \, \text{K}, ~T_2 = 87^\circ C = 360 \, \text{K} \)

\( P_2^{abs} = 2.76 \cdot \frac{360}{300} = 3.312 \, \text{bar} \)

\( P_2^{gauge} = 3.312 - 1.01 = {2.302 \, \text{bar}} \)

The correct answer is Pascal's law.

• The Pascal's law states that in a fluid which is at rest in a container, the pressure applied to one part of the fluid is uniformly transmitted to all the parts of the fluid.

Key Points

• A hydraulic lift employs this principle to lift heavy objects.
• When pressure is applied to a fluid through one piston, it results in an equivalent pressure on another piston in the system which is then able to lift objects.
• With the increase in the area of the second piston, the force exerted by it also increases thus enabling lifting of heavier objects.

Additional Information

• Hooke's law states that force needed to extend or compress a spring by some distance is directly proportional to that distance.
• Newton's first law of motion - A body at rest remains at rest, or if in motion, remains in motion at constant velocity unless acted on by a net external force. 
• Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

Explanation:

The total energy of a flowing fluid can be represented in terms of head, which is given by

\(\frac{p}{ρ{g}}\;+\;\frac{V^2}{2g}\;+\;z\; \)

The sum of the pressure head and hydrostatic pressure head is called the piezometric head. It is given by 

Piezometric head = \(\mathbf{\frac{P}{ρ{g}}+z}\)

where \(P\over\gamma \)= pressure energy per unit weight or pressure head

\(V^2\over{2g}\)= kinetic energy per unit weight or kinetic energy head

z = potential energy per unit weight or elevation head

The pressure at any point in a static fluid is obtained by Hydro-static law which is given by -

\(\frac{dP}{dz}=-ρ{g}\)

∴ P = -ρgz

∴ P = ρgh

where P = pressure above atmospheric pressure and h = height of the point from the free surface.

At point A, pressure head = \(P_A\over\gamma\) = hA  and datum head = zA

At point B, pressure head = \(P_B\over\gamma\) = hB  and datum head = zB

Piezometric head at point A = \(\frac{P}{ρ{g}}+z\) = hA + zA = H

Piezometric head at point B = \(\frac{P}{ρ{g}}+z\) = hB + z₀ = H

∴ piezometric head remains constant at all points in the liquid.

Explanation:

A manometer is a device that measures pressure by balancing a column of liquid against a column of gas or another liquid. The liquid used in a manometer should have the following characteristics:

• The liquid should stick on the walls: The liquid used in a manometer should stick to the walls of the tube to prevent it from flowing back and forth due to vibration or turbulence.
• Low surface tension: The liquid used in a manometer should have low surface tension to ensure that the meniscus does not significantly affect the pressure reading.
• It should be immiscible: The liquid used in a manometer should be immiscible with the gas or liquid being measured, to prevent the two fluids from mixing and affecting the accuracy of the pressure measurement.
•  High viscosity is NOT a desirable characteristic for a manometer liquid. A highly viscous liquid will not respond quickly to changes in pressure, leading to slow and inaccurate readings. Thus, manometer liquids are typically chosen to have relatively low viscosity.

Explanation:

From hydrostatic law:

Rate of increase of pressure in a vertical direction equal to the weight density of the fluid at that point.

\(\frac{{\partial p}}{{\partial x}} = - ρ g\)    ....eq (1)

For a compressible fluid, density (ρ) changes with the change of pressure and temperature. Thus, eq (1) cannot be integrated directly.

∵ Air is an ideal gas so,

ρ = p/RT (∵ PV = mRT)

\( \Rightarrow \frac{{dp}}{{dx}} = -\frac{p}{{RT}}g \Rightarrow \frac{{dp}}{p} = -\frac{g}{{RT}}\;dx\)

\( \Rightarrow \smallint \frac{{dp}}{p} = \smallint \frac{g}{{RT}}dx \Rightarrow \ln\;p = - \frac{{gh}}{{RT}}\;\)

∴ p = e^-gh/RT i.e. the atmospheric pressure varies exponentially with height.

Explanation:-

Manometer - 

A manometer is an instrument that uses a column of liquid to measure pressure, although the term is currently often used to mean any pressure instrument.

Two types of manometer, such as

1. Simple manometer

A simple manometer consists of a glass tube having one of its ends connected to a point where pressure is to be measured and the other end remains open to the atmosphere. Common types of simple manometers are:

• Piezometer
• U tube manometer
• Single Column manometer

2. Differential manometer

Differential Manometers are devices used for measuring the difference of pressure between two points in a pipe or in two different pipes. A differential manometer consists of a U-tube, containing a heavy liquid, whose two ends are connected to the points, which difference in pressure is to be measured.

The most common types of differential manometers are:

• U-tube differential manometer.
• Inverted U-tube differential manometer

3. Inclined U-tube manometer - 

If the pressure to be measured is very small. It is more accurate than a U-tube manometer.

Then tilting the arm provides a convenient way of obtaining a larger (more easily read) movement of the manometer.

The pressure difference is still given by the height change of the manometric fluid(z₂).

The sensitivity to pressure change can be increased further by a greater inclination of the manometer arm.

An alternative solution to increase sensitivity is to reduce the density of the manometric fluid.

Concept:

We know that; P = ρgh

Notice that all the options are given in terms of water column hence we will calculate on the basis of properties of water.

When Pressure is equivalent for two different liquids,

hHg  × SHg = hwater × Swater

Calculation:

Given:

S_Hg = 13.6

hHg  = 70 cm = 0.7 m

S_water= 1

hHg  × SHg = hwater × Swater

hwater = 13.6 × 0.7 = 9.52

Concept:

Mathematically, it can be represented as:

Absolute Pressure = Atmospheric pressure + Gauge Pressure.

Pabs = Pgauge + Patm

Calculation:

Given:

h = 5 m

P_atm = 0.75 m of mercury = ρ_Hg × g × 0.75 = 13600 × 9.81 × 0.75 = 100062 Pa

P_gauge = ρ_water × g × h

P_gauge = 1000 × 9.81 × 5 = 49050 Pa

P_abs = P_gauge + P_atm

P_abs = 49050 + 100062

P_abs = 149112 Pa = 149112 N/m² = 0.149 N/mm² ≃ 0.15 N/mm²

Concept:

In an open tube manometer

• The pressure at both the open ends is atmospheric.
• The pressure at any point inside the column can be calculated from either side.

Calculation:

Given:

ρ_water = 1000 kg/m3, l = 80 mm and d = 20 mm

So, the pressure at the bottom of the oil column can be equated from either end to find the required value of ρ_oil.

ρ_oil × g × (d + l) = ρ_water × g × l

ρ_oil × (20 + 80) = 1000 × 80

ρ_oil = 800 kg/m3

Hence the required density of oil is 800 kg/m3.

Concept:

Pabsolute = Patmospheric + Pgauge

The pressure at any point in a static fluid is obtained by Hydro-static law which is given by-

\(\frac{dP}{dz}=-ρ{g}\)

∴ P = -ρgz

P increases when we go down (z negative) and decreases when we go up (z positive).

where P = pressure above atmospheric pressure and h = height of the point from the free surface.

The difference in barometric (Atmospheric) pressure at sea level and that on the mountain is due to the elevation difference i.e. a pressure equivalent to the extra height of air column equal to the elevation of the mountain acting at the sea level as compared to on the mountain.

P_s = P_h + (ρ_a × g × h)

P_s = Pressure at the sea level, P_h = Pressure at the mountain top, ρ_a = Density of air, h = Height of mountain

Calculation:

Given:

P_s = 760 mm of Mercury, P_h = 724 mm of Mercury, ρ_a = 1.2 Kg/m³

∴ Ps = Ph + (ρa × g × h)

⇒ (ρa × g × h) = Ps - Ph

So,

(ρa × g × h) = (760 - 724) × 10^-3 × g × 13600

⇒ h = \(\frac{13.6~\times~36}{1.2}\) = 408 m

So, Height of the mountain is 410 meters.

Explanation:

Hydrostatic law:

The pressure at any point in a static fluid is obtained by Hydro-static law which is given by -

\(\frac{dP}{dz}=-ρ{g}\)

∴ P = -ρgz

∴ P = ρgh

where P = pressure above atmospheric pressure and h = height of the point from the free surface.

Specific Weight is the weight of a substance per unit volume.

Specific weight, w = \(\frac{{{\rm{\;mg}}}}{{\rm{V}}}{\rm{\;or\;\rho g\;}}\)

where, m = mass, V = Volume

∴ the correct answer is specific weight.