GAUGE PRESSURE: EXPLANATION, FORMULAS, EQUATIONS, EXAMPLES - PHYSICAL - 2025

The gauge pressure P _m is that which is measured in relation to a reference pressure, which in most cases is chosen as the atmospheric pressure P _atm at sea level. It is then a relative pressure, another term by which it is also known.

The other way in which pressure is usually measured is by comparing it with absolute vacuum, whose pressure is always zero. In this case we speak of the absolute pressure, which we will denote as P _a.

Figure 1. Absolute pressure and gauge pressure. Source: F. Zapata.

The mathematical relationship between these three quantities is:

Thus:

Figure 1 conveniently illustrates this relationship. Since the vacuum pressure is 0, the absolute pressure is always positive and so is the atmospheric pressure P _atm.

Manometric pressure is usually used to denote pressures above atmospheric pressure, such as that found in tires or that found at the bottom of the sea or a swimming pool, which is exerted by the weight of the water column.. In these cases P _m > 0, since P _a > P _atm.

However, there are absolute pressures below P _atm. In these cases, P _m <0 and is called the vacuum pressure and should not be confused with the vacuum pressure already described, which is the absence of particles capable of exerting pressure.

Formulas and equations

The pressure in a fluid -liquid or gas- is one of the most significant variables in its study. In a stationary fluid, the pressure is the same at all points at the same depth regardless of orientation, while the movement of fluids in the pipes is caused by changes in pressure.

The mean pressure is defined as the quotient between the force perpendicular to a surface F _⊥ and the area of ​​said surface A, which is expressed mathematically as follows:

Pressure is a scalar quantity, the dimensions of which are force per unit area. The units of its measurement in the International System of Units (SI) are newton / m ², called Pascal and abbreviated as Pa, in honor of Blaise Pascal (1623-1662).

Multiples such as kilo (10 ³) and mega (10 ⁶) are often used, since atmospheric pressure is usually in the range of 90,000 - 102,000 Pa, which is equal to: 90 - 102 kPa. Pressures on the order of megapascals are not uncommon, so it is important to become familiar with the prefixes.

In Anglo-Saxon units, pressure is measured in pounds / foot ², however, it is common to measure it in pounds / inch ² or psi (pounds-force per square inch).

Variation of pressure with depth

The more we immerse ourselves in the water in a pool or in the sea, the more pressure we experience. On the contrary, as the height increases, the atmospheric pressure decreases.

The mean atmospheric pressure at sea level is established at 101,300 Pa or 101.3 kPa, while in the Mariana Trench in the Western Pacific - the deepest known depth - it is about 1000 times greater and at the top of Everest it is just 34 kPa.

It is clear that pressure and depth (or height) are related. To find this out, in the case of a fluid at rest (static equilibrium), a disk-shaped portion of fluid is considered, confined in a container, (see figure 2). The disk has a cross section of area A, weight dW, and height dy.

Figure 2. Differential element of fluid in static equilibrium. Source: Fanny Zapata.

We will call P the pressure that exists at depth “y” and P + dP the pressure that exists at depth (y + dy). Since the density ρ of the fluid is the ratio between its mass dm and its volume dV, we have:

Therefore the weight dW of the element is:

And now Newton's second law applies:

Solution of the differential equation

Integrating both sides and considering that the density ρ, as well as the gravity g are constant, the expression sought is found:

If in the previous expression P ₁ is chosen as the atmospheric pressure and y ₁ as the surface of the liquid, then y ₂ is located at a depth h and ΔP = P ₂ - P _atm is the gauge pressure as a function of depth:

In case you need the absolute pressure value, simply add the atmospheric pressure to the previous result.

Examples

A device called a manometer is used to measure gage pressure, which generally offer pressure differences. At the end the working principle of a U-tube manometer will be described, but now let's look at some important examples and consequences of the previously derived equation.

Pascal's principle

The equation Δ P = ρ.g. (Y ₂ - y ₁) can be written as P = Po + ρ.gh, where P is the pressure at depth h, while P _o is the pressure at the surface of the fluid, usually P _atm.

Obviously, each time Po increases, P increases by the same amount, as long as it is a fluid whose density is constant. This is precisely what was assumed when considering ρ constant and placing it outside the integral solved in the previous section.

Pascal's principle states that any increase in the pressure of a confined fluid in equilibrium is transmitted without any variation to all points of said fluid. Using this property, it is possible to multiply the force F ₁ applied to the small piston on the left, and obtain F ₂ on the one on the right.

Figure 3. Pascal's principle is applied in the hydraulic press. Source: Wikimedia Commons.

Car brakes work on this principle: a relatively small force is applied on the pedal, which is converted into a greater force on the brake cylinder at each wheel, thanks to the fluid used in the system.

Stevin's hydrostatic paradox

The hydrostatic paradox states that the force due to the pressure of a fluid at the bottom of a container can be equal to, greater or less than the weight of the fluid itself. But when you put the container on top of the scale, it will normally register the weight of the fluid (plus the container of course). How to explain this paradox?

We start from the fact that the pressure at the bottom of the container depends exclusively on the depth and is independent of the shape, as it was deduced in the previous section.

Figure 4. The liquid reaches the same height in all the containers and the pressure at the bottom is the same. Source: F. Zapata.

Let's look at a few different containers. Being communicated, when they are filled with liquid they all reach the same height h. The highlights are at the same pressure, since they are at the same depth. However, the force due to pressure at each point may differ from the weight, (see example 1 below).

Exercises

Exercise 1

Compare the force exerted by the pressure on the bottom of each of the containers with the weight of the fluid, and explain why the differences, if any.

Container 1

Figure 5. The pressure at the bottom is equal in magnitude to the weight of the fluid. Source: Fanny Zapata.

In this container the area of ​​the base is A, therefore:

The weight and the force due to pressure are equal.

Container 2

Figure 6. The force due to pressure in this container is greater than the weight. Source: F. Zapata.

The container has a narrow part and a wide part. In the diagram on the right it has been divided into two parts and geometry will be used to find the total volume. The area A ₂ is external to the container, h ₂ is the height of the narrow part, h ₁ is the height of the wide part (base).

The full volume is the volume of the base + the volume of the narrow part. With these data we have:

Comparing the weight of the fluid with the force due to pressure, it is found that this is greater than the weight.

What happens is that the fluid also exerts force on the part of the step in the container (see the arrows in red in the figure) that are included in the above calculation. This upwards force counteracts those exerted downwards and the weight registered by the scale is the result of these. According to this, the magnitude of the weight is:

W = Force on the bottom - Force on the stepped part = ρ. g. At ₁.h - ρ. g. A _.. h ₂

Exercise 2

The figure shows an open tube manometer. It consists of a U tube, in which one end is at atmospheric pressure and the other is connected to S, the system whose pressure is to be measured.

Figure 7. Open tube manometer. Source: F. Zapata.

The liquid in the tube (yellow in the figure) may be water, although mercury is preferably used to reduce the size of the device. (A difference of 1 atmosphere or 101.3 kPa requires a 10.3 meter water column, nothing portable).

It is asked to find the gauge pressure P _m in the system S, as a function of the height H of the liquid column.

Solution

The pressure at the bottom for both branches of the tube is the same, as they are at the same depth. Let P _{A be} the pressure at point A, located at y ₁ and P _B those at point B at y ₂. Since point B is at the interface of liquid and air, the pressure there is P _o. In this branch of the manometer, the pressure at the bottom is:

For its part, the pressure at the bottom for the branch on the left is:

Where P is the absolute pressure of the system and ρ is the density of the fluid. Equalizing both pressures:

Solving for P:

Therefore, the gauge pressure P _m is given by P - P _o = ρ.g. H and to have its value, it is enough to measure the height to which the manometric liquid rises and multiply it by the value of g and the density of the fluid.

References

1. Cimbala, C. 2006. Fluid Mechanics, Fundamentals and Applications. Mc. Graw Hill. 66-74.
2. Figueroa, D. 2005. Series: Physics for Sciences and Engineering. Volume 4. Fluids and Thermodynamics. Edited by Douglas Figueroa (USB). 3-25.
3. Mott, R. 2006. Fluid Mechanics. 4th. Edition. Pearson Education. 53-70.
4. Shaugnessy, E. 2005. Introduction to Fluid Mechanics. Oxford University Press. 51 - 60.
5. Stylianos, V. 2016. A simple explanation of the classic hydrostatic paradox. Recovered from: haimgaifman.files.wordpress.com