por The gauge pressure Pm is that which is measured in relation to a reference pressure, which in most cases is chosen as the atmospheric pressure Patm at sea level. It is then a relative pressureanother term by which it is also known.
The other way in which pressure is usually measured is by comparing it to absolute vacuum, whose pressure is always zero. In such a case, one speaks of the absolute pressurewhich we will denote as Pa.
The mathematical relationship between these three quantities is:
Pa = Patm + Pm
Therefore:
Pm = Pa – Patm
Figure 1 conveniently illustrates this relationship. Since the vacuum pressure is 0, the absolute pressure is always positive and the same is true for the atmospheric pressure Patm.
Gauge pressure is usually used to denote pressures above atmospheric pressure, such as that carried by tires or the one at the bottom of the sea or a swimming pool, which is exerted by the weight of the water column. . In these cases Pm > 0, since Pa > Patm.
However, there are absolute pressures below Patm. In these cases Pm < 0 and is called vacuum pressure and it should not be confused with the vacuum pressure already described, which is the absence of particles capable of exerting pressure.
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Formulas and equations
The pressure in a fluid -liquid or gas- is one of the most significant variables in his 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 pipes is caused by changes in pressure.
Average pressure is defined as the ratio between the force perpendicular to a surface F⊥ and the area of said surface A, which is expressed mathematically as follows:
P = F⊥ /A
Pressure is a scalar quantity, whose dimensions are force per unit area. The units of its measurement in the International System of Units (SI) are newton/m2, called Pascal and abbreviated as Pa, in honor of Blaise Pascal (1623-1662).
The multiples like kilo (103) and mega (106) are frequently 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 mega pascals are not uncommon, so it is important to become familiar with prefixes.
In Anglo-Saxon units, pressure is measured in pounds/foot2, however, it is commonly done in pounds/inch2 or psi (pounds-force per square inch).
pressure variation with depth
The more we submerge ourselves in the water of a swimming pool or in the sea, the more pressure we experience. On the contrary, as the height increases, the atmospheric pressure decreases.
The average 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 greatest depth known – 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 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 cross sectional area TOweight dW and height dy.
we will call P to the pressure that exists at depth «and» and P + dP to 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 dVyou have to:
ρ = dm/dV ⇒ dm= ρ.dV
So the weight dW of the element is:
dW = g. dm = ρ.g.dV
And now Newton’s second law applies:
ΣFy = F2 – F1 – dW = 0
(P + dP).A – PA – ρ.g.dV = 0
(P + dP).A – PA – ρ.g. A.dy = 0
dP = ρ.g.dy
Solution of the differential equation
Integrating both sides and considering that the density ρas well as gravity g are constants, the searched expression is found:
P2 – P1 =ΔP = ρ.g.(y2 – y1)
ΔP = ρ.g. Δand
If in the previous expression we choose P1 like atmospheric pressure and y1 as the surface of the liquid, then y2 It is located at a depth h and ΔP= P2 – Patm is the gauge pressure as a function of depth:
pm = ρ.gh
In case of needing the value of the absolute pressure, simply add the atmospheric pressure to the previous result.
examples
A device called a gauge is used to measure gauge pressure. pressure gauge, which generally offer pressure differences. At the end, the principle of operation of a U-tube manometer will be described, but now let’s see some examples and important consequences of the previously derived equation.
Pascal’s principle
The equation ΔP = ρ.g.(y2 – y1) can be written as P = Po + ρ.gh, where P is the pressure at depth hwhile Po is the pressure at the surface of the fluid, usually patm.
Of course, every time you increase Poincreases P in 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 place it outside the integral resolved 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 in said fluid. Through this property, it is possible to multiply the force F1 applied to the small plunger on the left, and obtain F2 in the one on the right.
Car brakes work on this principle: a relatively small force is applied to the pedal, which is converted into a larger 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 than, 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 that of 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 was deduced in the previous section.
Let’s look at some different containers. Being communicated, when they are filled with liquid they all reach the same height h. The prominent points 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 the reasons for the differences, if any.
container 1
In this container the area of the base is A, therefore:
fluid weight: mg = ρ.Vg = ρ . A.h. g
bottom pressure: ρ. g. h
force due to pressure: F = PA = ρ. gh A
Weight and force due to pressure are equal.
container 2
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 A2 is external to the container, h2 is the height of the narrow part, h1 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:
fluid weight:m. g = ρ . g. V = ρ . g. [A1 .h1+ (A1 -A2) .h2]=
= ρ . g (A1.h –A2h2)= ρ . g. A1.h – ρ . g. A.. h2 (Use has been made of h=h1 +h2)
bottom pressure:P=ρ. g. h
Force on bottom due to pressure: F = P. A1 =ρ. gh A1
Comparing the weight of the fluid with the force due to the pressure, it is noticed 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 red arrows in the figure) that are included in the above calculation. This upward force counteracts the downward force 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. A1.h – ρ . g. A..h2
Exercise 2
An open tube manometer is shown in the figure. It consists of a U-shaped tube, in which one end is at atmospheric pressure and the other is connected to S, the system whose pressure is to be measured.
The liquid in the tube (in yellow in the figure) can 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.)
You are asked to find the gauge pressure P.m in system S, as a function of the height H of the liquid column.
Solution
The bottom pressure for both tube branches is the same, since they are at the same depth. Let PA be the pressure at point A, located at y1, and PB the pressure at point B, located at height y2. Since point B is at the interface of liquid and air, the pressure there is Po. In this branch of the manometer, the pressure at the bottom is:
Po + ρ.g.y2
On the other hand, the bottom pressure for the left branch is:
P + ρ.g.y1
where P is the absolute pressure of the system and ρ is the density of the fluid. Equating both pressures:
Po + ρ.g.y2 = P + ρ.g.y1
clearing P:
P = Po + ρ.g.y2 – ρ.g.y1 = Po + ρ.g(y2 – y1)= Po + ρ.g. h
Therefore, the gauge pressure P.m is given by P – Po = 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
Cimbala, C. 2006. Fluid Mechanics, Fundamentals and Applications. Mc. Graw Hill. 66-74.
Figueroa, D. 2005. Series: Physics for Science and Engineering. Volume 4. Fluids and Thermodynamics. Edited by Douglas Figueroa (USB). 3-25.
Mott, R. 2006. Fluid Mechanics. 4th. Edition. Pearson Education. 53-70.
Shaugnessy, E. 2005. Introduction to Fluid Mechanics. Oxford University Press. 51-60.
Stylianos, V. 2016. A simple explanation of the classic hydrostatic paradox. Retrieved from: haimgaifman.files.wordpress.com