Pressure of Fluids at Rest

What do we know about the pressure of a fluid at rest? We know that if we place it inside a container, since the relationship P=F/A, holds and due to its weight, there will be pressure at every point within it, depending on the depth.

The pressure of a fluid at rest at a given point is directly proportional to its depth. This is known from the expression:

P = \rho g h

Where \rho is the fluid’s density, g is the gravitational acceleration, and h is the depth.

We can demonstrate this by submerging an imaginary cylinder with a horizontal base area A at some depth h in the fluid.

We will see that the disc will be crushed by the weight of the fluid above it and by the normal force below, which is equal and opposite to the weight (since we assume the system is at rest).

Derivation of the formula for the pressure of a fluid at rest: {P=\rho g h}

The fluid forms another cylinder over the submerged body with the same base area A, but height h, thus having a volume.

V=A h

From which we infer that

\displaystyle A=\frac{V}{h}

If the fluid has a density \rho, then the mass of the fluid pressing the disc is m=\rho V, and it exerts a weight force

F_p=m g = \rho V g

Similarly, the normal force is exerted on the lower face of the cylinder with the same magnitude but on the opposite face.

Since the weight and the normal are vertical, they do not exert pressure on the lateral sides of the cylinder.

If we consider the cylinder to be flat and light enough, its weight will not contribute anything that needs to be countered by the normal force, and any effect on the sides will be negligible compared to everything else. Thus, the total force on the body will be:

F_{total}=F_p + F_n = 2\rho V g

The “+” in this force is because the forces are oriented toward the surface.

Thus, the pressure on the submerged body will be:

\displaystyle P = \frac{F_{total}}{A_{inferior}+A_{superior}}=\frac{2 \rho V h}{2A} = \frac{\rho V g}{\frac{V}{h}} = \rho g h

This shows that the pressure exerted by a fluid (at rest) is the same at all points at the same depth. This applies to both liquids and gases (as long as they are at rest), so if we consider the column of air above us, we can also talk about “atmospheric pressure”.

The atmospheric pressure at sea level is

P_{atm} = 1[atm] = 101.325,0 [Pa] = 760 [Torr]=0.981[barr].

We must remember that 1[Pa] = 1[N/m^2].

By combining the atmospheric pressure with the pressure exerted by a fluid due to its own weight, we have the hydrostatic pressure

P = P_{atm} + \rho g h

Relative Pressure

When we measure pressure, we are usually immersed in a medium. Sometimes the pressure of the medium is relevant, and other times it is not. For example, when you measure the pressure of your car tires, you don’t worry about adding the atmospheric pressure because what really matters for their proper functioning is the pressure difference between the inside of the tire and the outside environment:

If it’s too high, it overinflates; if it’s too low, it deflates.

This is why we have different ways to talk about pressure.

Atmospheric Pressure

We already discussed this before, and it’s the pressure of the medium we are immersed in. For example, in the Himalayas, the atmospheric pressure can be as low as one-third of what it is at sea level. It is usually represented by P_{atm} or P_{0}.

Absolute Pressure

When we consider the pressure obtained from the sum of all forces acting on a body, we talk about Absolute Pressure. The hydrostatic pressure we reviewed earlier is a form of absolute pressure because it considers the sum of the pressures due to the weight of the liquid + the pressure exerted by the atmosphere. Another way to express absolute pressure is as “pressure relative to a vacuum”. It is represented by P_{abs}.

Gauge Pressure and Vacuum Pressure

When we measure the pressure of a vehicle tire, both the tire and the measuring instrument are being pressed by the surrounding atmosphere. For this reason, what the instrument measures is the pressure difference between the inside and outside. This pressure is called “gauge pressure,” represented by P_{man}, and it satisfies the relationship:

P_{man} = P_{abs} - P_{atm}

The pressure we measure considering only the weight of a fluid is an example of gauge pressure. If the absolute pressure is higher than atmospheric pressure, we measure gauge pressure; otherwise, we measure vacuum pressure P_{vac}, defined similarly:

P_{vac} = P_{atm} - P_{abs}

This occurs, for example, when you pull the air out of a syringe, seal the air inlet/outlet, and then pull the plunger; atmospheric pressure will try to push the plunger back, and the pressure inside the syringe will be, consequently, vacuum pressure.

Practical Examples

A man has just built a pool that is 39[feet] long, 26[feet] wide, and 5.2[feet] deep and has the following questions:

1. He bought a pump for the pool, but only when he got home did he think to check the manufacturer’s specifications. These say that the gauge pressure should not exceed 0.193[atm]. Can he install the pump at the bottom of his pool?
2. Not only did he forget the pump specifications, but this man has a hearing issue. His eardrums cannot withstand a force greater than 10[N]. If his eardrums have a diameter of 1[cm], almost perfectly circular, can he safely dive to the bottom of his pool?

SOLUTION:

1. 1. In this case, the gauge pressure at the bottom of the pool can be determined as the pressure exerted only by the pool’s water. Therefore, from the formula for the pressure of a fluid at rest, we get:

      P_{man} = \rho g h

      Taking the water’s density \rho=997[kg/m^3], the depth converted to meters h=5.2[feet] = 5.2\cdot 0.3048[m] and the gravitational acceleration g=9.81[m/s^2], we get that the gauge pressure at the bottom of the pool will be

      P_{man} =997[kg/m^3]\cdot 9.81[m/s^2] \cdot5.2\cdot 0.3048[m] \approx 15.501,81[Pa]

      But 1[atm] = 101.325[Pa], so

      \displaystyle P_{man} \approx \frac{15.501,81}{101.325}[atm]\approx 0.1523[atm]

      So, yes. Since the gauge pressure at the bottom of the pool is below the 0.193[atm] specified by the manufacturer as the limit for the pump’s proper function, it will work correctly, and the man hasn’t wasted his money.

   2. From the previous part, we already calculated the gauge pressure at the bottom of the pool, but now we need the total pressure. No problem with that; we just need to remember that:

P_{total} = P_{man} + P_{atm}

and we already have both. This gives us that

P_{total} \approx 15.501,81[Pa] + 101.325[Pa] = 116.286,81[Pa]

Now we need to know the area of this man’s eardrum. Since the eardrum is approximately circular, we have:

\displaystyle A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

Here I’ve expressed the area of the circle as a function of the diameter d, which is 1[cm]. Therefore:

\displaystyle A \approx \frac{3.14 \cdot 1[cm^2]}{4} = \frac{3.14 \left[\frac{m}{100}\right]^2}{4} = \frac{3.14}{4\cdot 10.000}[m^2]=0.785\cdot 10^{-4}[m^2]

Finally, since P=F/A, the total force applied by the pressure at the bottom of the pool on this man’s eardrum will be

F=PA\approx 116.826,81[Pa] \cdot 0.785\cdot 10^{-4}[m^2] \approx 9.17[N]

Since this doesn’t exceed 10[N], just barely, he can dive safely to the bottom of the pool. What a lucky man.