Ahmad Dahlan God does not play dice with the Cosmos.

Pressure and Pascal Law in Statics Fluids

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Pressure in Fluids

Ahmad Dahlan – Matter is normally classified as being in one of three states: solid, liquid, or gas. From everyday experience we know that a solid has a definite volume and shape, a liquid has a definite volume but no definite shape, and an unconfined gas has neither a definite volume nor a definite shape. These descriptions help us picture the states of matter, but they are somewhat artificial. For example, asphalt and plastics are normally considered solids, but over long time intervals they tend to flow like liquids.

Likewise, most substances can be a solid, a liquid, or a gas (or a combination of any of these three), depending on the temperature and pressure. In general, the time interval required for a particular substance to change its shape in response to an external force determines whether we treat the substance as a solid, a liquid, or a gas.

A fluid is a collection of molecules that are randomly arranged and held together by weak cohesive forces and by forces exerted by the walls of a container. Both liquids and gases are fluids.

In our treatment of the mechanics of fluids, we’ll be applying principles and analysis models that we have already discussed. First, we consider the mechanics of a fluid at rest, that is, fluid statics, and then study fluids in motion, that is, fluid dynamics.

Daftar Isi

A. Pressure

Fluids do not sustain shearing stresses or tensile stresses. therefore, the only stress that can be exerted on an object submerged in a static fluid is one that tends to compress the object from all sides. In other words, the force exerted by a static fluid on an object is always perpendicular to the surfaces of the object as shown in Figure 1.

PRessure in Fluid Mecanics Newton
Figure 1 – The forces exerted by a fluid on the surfaces of a submerged object.

The pressure in a fluid can be measured with the device pictured in Figure 2. The device consists of an evacuated cylinder that encloses a light piston connected to a spring.

Alat Ukur Tekanan Fluida
Figure 2 – A simple device for measuring the pressure exerted by a fluid.

As the device is submerged in a fluid, the fluid presses on the top of the piston and compresses the spring until the inward force exerted by the fluid is balanced by the outward force exerted by the spring. The fluid pressure can be measured directly if the spring is calibrated in advance. If F is the magnitude of the force exerted on the piston and A is the surface area of the piston, the pressure P of the fluid at the level to which the device has been submerged is defined as the ratio of the force to the area:

P = \frac{F}{A}

Pressure is a scalar quantity because it is proportional to the magnitude of the force on the piston.

If the pressure varies over an area, the infinitesimal force dF on an infinitesimal surface element of area dA is :

dF = P dA

where P is the pressure at the location of the area dA. To calculate the total force exerted on a surface of a container, we must integrate Equation 2 over the surface.

The units of pressure are newtons per square meter (N/m2) in the SI system.
Another name for the SI unit of pressure is the pascal (Pa):

1 Pa = 1 N/m2

For a tactile demonstration of the definition of pressure, hold a tack between your thumb and forefinger, with the point of the tack on your thumb and the head of the tack on your forefinger. Now gently press your thumb and forefinger together. Your thumb will begin to feel pain immediately while your forefinger will not. The tack is exerting the same force on both your thumb and forefinger, but the pressure on your thumb is much larger because of the small area over which the force is applied.

B. Variation of Pressure with Depth

As divers well know, water pressure increases with depth. Likewise, atmospheric pressure decreases with increasing altitude; for this reason, aircraft flying at high altitudes must have pressurized cabins for the comfort of the passengers. We now show how the pressure in a liquid increases with depth.

The density of a substance is defined as its mass per unit volume. The densities of various substances values vary slightly with temperature because the volume of a substance is dependent on temperature. Under standard conditions (at 0oC and at atmospheric pressure), the densities of gases are about 10-3 the densities of solids and liquids. This difference in densities implies that the average molecular spacing in a gas under these conditions is about ten times greater than that in a solid or liquid.

Now consider a liquid of density ρ at rest as shown in Figure 3. We assume ρ is uniform throughout the liquid, which means the liquid is incompressible.

Tekanan pada Fluida Statis
Figura 3 – A parcel of fluid in a larger volume of fluid is singled out.

Let us select a parcel of the liquid contained within an imaginary block of cross-sectional area A extending from depth d to depth d + h. The liquid external to our parcel exerts forces at all points on the surface of the parcel, perpendicular to the surface. The pressure exerted by the liquid on the bottom face of the parcel is P, and the pressure on the top face is P0.

Therefore, the upward force exerted by the outside fluid on the bottom of the parcel has a magnitude PA, and the downward force exerted on the top has a magnitude P0A. The mass of liquid in the parcel is Mg = ρV = ρAh; therefore, the weight of the liquid in the parcel is Mg = ρAhg. Because the parcel is at rest and remains at rest, it can be modeled as a particle in equilibrium, so that the net force acting on it must be zero. Choosing upward to be the positive y direction, we see that :

\Sigma \vec{F} = PA \hat{j} - P_oA \hat{j} - Mg \hat{j} = 0

or

PA - P_oA - \rho Ahg = 0
P=P_oA - \rho gh

That is, the pressure P at a depth h below a point in the liquid at which the pressure is P0 is greater by an amount ρgh. If the liquid is open to the atmosphere and P0 is the pressure at the surface of the liquid, then P0 is atmospheric pressure. In our calculations and working of end-of-chapter problems, we usually take atmospheric pressure to be :

P0 = 1.00 atm = 1.013 x 105 Pa

P = P0A + ρgh implies that the pressure is the same at all points having the same depth, independent of the shape of the container.

Because the pressure in a fluid depends on depth and on the value of P0, any increase in pressure at the surface must be transmitted to every other point in the fluid. This concept was first recognized by French scientist Blaise Pascal (1623 – 1662) and is called Pascal’s law: a change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

An important application of Pascal’s law is the hydraulic press illustrated
in Figure 4a. A force of magnitude F1 is applied to a small piston of surface area A1. The pressure is transmitted through an incompressible liquid to a larger piston of surface area A2. Because the pressure must be the same on both sides :

P_1=P_2
\frac{F_1}{A_1} = \frac{F_2}{A_2}

Therefore, the force F2 is greater than the force F1 by a factor of A2/A1. By designing a hydraulic press with appropriate areas A1 and A2, a large output force can be applied by means of a small input force. Hydraulic brakes, car lifts, hydraulic jacks, and forklifts all make use of this principle.

Tekanan Hidrostatis
Figure 4

Because liquid is neither added to nor removed from the system, the volume of liquid pushed down on the left in Figure 4a as the piston moves downward through a displacement Δx1 equals the volume of liquid pushed up on the right as the right piston moves upward through a displacement Δx2.

That is, A1 Δx1 = A2 Δx2 therefore A2/A1 = Δx1/Δx2. We have already shown that A2/A1 = F2/F1. Therefore F2/F1 = Δx1/Δx2, so F1Δx1 = F2Δx2.

Each side of this equation is the work done by the force on its respective piston. Therefore, the work done by F1 on the input piston equals the work done by F2 on the output piston, as it must to conserve energy. (The process
can be modeled as a special case of the non-isolated system model: the non-isolated system in steady state. There is energy transfer into and out of the system, but these energy transfers balance, so that there is no net change in the energy of the system.

Source : Serwey & Jewet. (2014). Physics for Scientist and Engineers with Modern Physics

Ahmad Dahlan God does not play dice with the Cosmos.

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