Pressure in a fluid
To measure the pressure inside a fluid, imagine introducing a sensor in the form of a tiny evacuated cylinder, with a piston at one end supported by a spring.
This is diagram of a sensor for measuring pressure inside a fluid. The fluid exerts a force of magnitude along the axis of the piston, causing it to compress. The ratio of to the area A of the piston defines the pressure.
The compression of the spring indicates the magnitude of the force exerted along the axis of the piston by the fluid. Then, if the cross-sectional area of the piston is A, and the force acting along the piston’s axis has magnitude , the pressure P is defined to be
It is assumed that the act of introducing the sensor has no significant effect on the pressure in the fluid. If the fluid is static this is a fair assumption, but if the fluid is flowing, like a river in spate, we should be more careful. It is essential for the sensor to drift with the fluid, so that it experiences exactly the same pressure as the fluid itself. We can imagine introducing a large number of sensors and letting them drift with the flow, allowing the pressure to be measured at different places and times throughout the fluid. The expense and practical difficulty of putting this plan into action need not concern us — the important point is simply that pressure at any point in the fluid can be given a precise definition.
Our definition of pressure is based on an important observation. In a fluid, the reading on a pressure sensor does not depend on the orientation of the sensor. If the sensor is turned around so that it points in a different direction, the same pressure will be recorded. In mathematical terms we can say that pressure is a scalar quantity, with no associated direction, in contrast to force which is a vector and acts in a definite direction. Both the magnitude of and the cross-sectional area are positive quantities, so pressure is itself a positive quantity. Zero pressure corresponds to a perfect vacuum.
In this section, we will consider the pressure inside a fluid that is at rest. Two rather different examples will be discussed: the pressure experienced by a diver and the pressure experienced by a mountaineer.
The pressure experienced by a diver
One of the main problems confronting a diver is the rapid increase in pressure that occurs with increasing depth. This is noticeable even for shallow dives, such as those to the bottom of a swimming pool. A serious problem arises because the human body is not completely solid: internal spaces like the lungs are filled with air, so deep diving introduces the danger of being crushed by the pressure of the surrounding water, like an empty aluminium can being crushed underfoot. Breathing special mixtures of high-pressure gases, scuba divers have dived to 300 metres; but unfortunately, air becomes toxic at very high pressures, so special diving suits or submarines become essential beyond this depth.
In order to learn about the increase of pressure with depth, we will consider a fluid that is at rest, with no currents flowing. We will also assume that the fluid is incompressible, so that it has a fixed density, independent of the pressure. Both these assumptions are reasonable because currents have almost no effect on the pressure experienced by a diver, and water is practically incompressible.
Suppose a cube of water is at rest somewhere within a lake. If you like, you can imagine the cube to be surrounded by a thin plastic membrane, so that it is clearly separated from the rest of the water, although this is not really necessary. What are the forces acting on the cube? There is the weight of the cube acting downwards and there are inward forces acting on each face of the cube due to the pressure of the surrounding water.
First, consider the forces acting along the horizontal x-axis. These are due solely to the external pressure on the two shaded faces in shown here.
Forces on an imaginary cube of water in a lake.
Suppose the area of one face of the cube is A, the pressure on the left-hand face is PL, and the pressure on the right-hand face is PR. Then the x-component of the total external force on the cube is
Fx = PLA1-1PRA1.
Notice how the signs appear here: the force on the left-hand face is positive because it acts along the x-axis, while that on the right-hand face is negative because it acts in the opposite direction. Finally, remember that the cube is supposed to be at rest so Fx must be zero, leading to the conclusion that PL = PR. More generally, we can say that any two points at the same depth in a static fluid must be at the same pressure — if they were not, currents would flow, driven by the pressure difference.
In order to see how pressure depends on depth, we must consider the vertical forces acting on the cube.
Forces on an imaginary cube of water in a lake
It is convenient to choose a z-axis that points vertically downwards, with its origin at the surface of the water. Suppose the cube has mass M, the pressure on the upper face is P1 and the pressure on the lower face is P2. Then the z-component of the total external force on the cube is
Fz = Mg + P1A - P2A1.
Again, we can argue that Fz must be zero in order for the cube to remain at rest, so it follows that.
The pressure on the lower face is greater than the pressure on the upper face, allowing the weight of the cube to be supported. It is worth expressing this result in a slightly different way by noting that the mass of the cube is given by
where is the density of the water and V is the volume of the cube. Since V = A(z2 - z1), we have
This allows us to write
If we now place the cube with its upper face at the surface of the water, z1 will be zero and P1 will be equal to atmospheric pressure, PA. On the lower face of the cube, at depth z2, the pressure is then
Now there was nothing special about z2. It could have been any point below the surface, so the subscript 2 can be dropped, giving a simple formula for the pressure P at depth z below the surface:
The quantity , which is the contribution of the liquid to the total pressure, is known as the gauge pressure, since it is the pressure that would be registered on a diver’s pressure gauge, normally calibrated to read zero at the surface.
It is interesting to calculate the rate of increase of pressure with depth. This can be done by rearranging Equation
The left-hand side can be written as P / z so, taking the limit as z becomes very small.
The right-hand side of this equation, , is the force magnitude per unit volume due to gravity. The left-hand side of the equation is the pressure gradient, which can be interpreted as the force magnitude per unit volume due to pressure.
Notice that the pressure gradient, dP / dz, has the constant value . Inserting appropriate values for , the density of pure water, and g, the magnitude of the acceleration due to gravity close to the Earth’s surface, we obtain
Comparing this rate of increase of pressure with standard atmospheric pressure (1051Pa), we see that each metre of depth corresponds to about 10% of atmospheric pressure. So, if you dive to a depth of 10 metres, you will experience roughly twice the pressure as at the surface.