2.2. FLUID STATICS

2.2A. Force, Units, and Dimensions

In a static fluid an important property is the pressure in the fluid. Pressure is familiar as a surface force exerted by a fluid against the walls of its container. Also, pressure exists at any point in a volume of a fluid.

In order to understand pressure, which is defined as force exerted per unit area, we must first discuss a basic law of Newton's. This equation for calculation of the force exerted by a mass under the influence of gravity is

Equation 2.2-1


where in SI units F is the force exerted in newtons N (kg · m/s2), m the mass in kg, and g the standard acceleration of gravity, 9.80665 m/s2.

In English units, F is in lbf, m in lbm, g is 32.1740 ft/s2, and gc (a gravitational conversion factor) is 32.174 lbm · ft/lbf · s2. The use of the conversion factor gc means that g/gc has a value of 1.0 lbf/lbm and that 1 lbm conveniently gives a force equal to 1 lbf. Often when units of pressure are given, the word “force” is omitted, as in lb/in.2 (psi) instead of lbf/in.2. When the mass m is given in g mass, F is g force, g = 980.665 cm/s2, and gc = 980.665 g mass · cm/g force · s2. However, the units g force are seldom used.

Another system of units sometimes used in Eq. (2.2-1) is that where the gc is omitted and the force (F = mg) is given as lbm · ft/s2, called poundals. Then 1 lbm acted on by gravity will give a force of 32.174 poundals (lbm · ft/s2). Or if 1 g mass is used, the force (F = mg) is expressed in terms of dynes (g · cm/s2). This is the centimeter–gram–second (cgs) systems of units.

Conversion factors for different units of force and of force per unit area (pressure) are given in Appendix A.1. Note that always in the SI system, and usually in the cgs system, the term gc is not used.

EXAMPLE 2.2-1. Units and Dimensions of Force

Calculate the force exerted by 3 lb mass in terms of the following:

  1. Lb force (English units).

  2. Dynes (cgs units).

  3. Newtons (SI units).

Solution: For part (a), using Eq. (2.2-1),


For part (b),


As an alternative method for part (b), from Appendix A.1,


To calculate newtons in part (c),


As an alternative method, using values from Appendix A.1,



2.2B. Pressure in a Fluid

Since Eq. (2.2-1) gives the force exerted by a mass under the influence of gravity, the force exerted by a mass of fluid on a supporting area, or force/unit area (pressure), also follows from this equation. In Fig. 2.2-1 a stationary column of fluid of height h2 m and constant cross-sectional area A m2, where A = A0 = A1 = A2, is shown. The pressure above the fluid is P0 N/m2; that is, this could be the pressure of the atmosphere above the fluid. The fluid at any point, say h1, must support all the fluid above it. It can be shown that the forces at any given point in a nonmoving or static fluid must be the same in all directions. Also, for a fluid at rest, the force/unit area, or pressure, is the same at all points with the same elevation. For example, at h1 m from the top, the pressure is the same at all points shown on the cross-sectional area A1.

Figure 2.2-1. Pressure in a static fluid.


The use of Eq. (2.2-1) will be shown in calculating the pressure at different vertical points in Fig. 2.2-1. The total mass of fluid for h2 m height and density ρ kg/m3 is

Equation 2.2-2


Substituting into Eq. (2.2-2), the total force F of the fluid on area A1 due to the fluid only is

Equation 2.2-3


The pressure P is defined as force/unit area:

Equation 2.2-4


This is the pressure on A2 due to the mass of the fluid above it. However, to get the total pressure P2 on A2, the pressure P0 on the top of the fluid must be added:

Equation 2.2-5


Equation (2.2-5) is the fundamental equation for calculating the pressure in a fluid at any depth. To calculate P1,

Equation 2.2-6


The pressure difference between points 2 and 1 is

Equation 2.2-7


Since it is the vertical height of a fluid that determines the pressure in a fluid, the shape of the vessel does not affect the pressure. For example, in Fig. 2.2-2, the pressure P1 at the bottom of all three vessels is the same and is equal to h1ρg + P0.

Figure 2.2-2. Pressure in vessels of various shapes.


EXAMPLE 2.2-2. Pressure in Storage Tank

A large storage tank contains oil having a density of 917 kg/m3 (0.917 g/cm3). The tank is 3.66 m (12.0 ft) tall and is vented (open) to the atmosphere of 1 atm abs at the top. The tank is filled with oil to a depth of 3.05 m (10 ft) and also contains 0.61 m (2.0 ft) of water in the bottom of the tank. Calculate the pressure in Pa and psia 3.05 m from the top of the tank and at the bottom. Also calculate the gage pressure at the tank bottom.

Solution: First a sketch is made of the tank, as shown in Fig. 2.2-3. The pressure P0 = 1 atm abs = 14.696 psia (from Appendix A.1). Also,


Figure 2.2-3. Storage tank in Example 2.2-2.


From Eq. (2.2-6), using English and then SI units,


To calculate P2 at the bottom of the tank, ρwater = 1.00 g/cm3 and


The gage pressure at the bottom is equal to the absolute pressure P2 minus 1 atm pressure:



2.2C. Head of a Fluid

Pressures are given in many different sets of units, such as psia, dyn/cm2, and newtons/m2, as given in Appendix A.1. However, a common method of expressing pressures is in terms of head in m or feet of a particular fluid. This height or head in m or feet of the given fluid will exert the same pressure as the pressures it represents. Using Eq. (2.2-4), which relates pressure P and height h of a fluid, and solving for h, which is the head in m,

Equation 2.2-8


EXAMPLE 2.2-3. Conversion of Pressure to Head of a Fluid

Given the pressure of 1 standard atm as 101.325 kN/m2 (Appendix A.1), do as follows:

  1. Convert this pressure to head in m water at 4°C.

  2. Convert this pressure to head in m Hg at 0°C.

Solution: For part (a), the density of water at 4°C in Appendix A.2 is 1.000 g/cm3. From A.1, a density of 1.000 g/cm3 equals 1000 kg/m3. Substituting these values into Eq. (2.2-8),


For part (b), the density of Hg in Appendix A.1 is 13.5955 g/cm3. For equal pressures P from different fluids, Eq. (2.2-8) can be rewritten as

Equation 2.2-9


Solving for hHg in Eq. (2.2-9) and substituting known values,



2.2D. Devices to Measure Pressure and Pressure Differences

In chemical and other industrial processing plants, it is often important to measure and control the pressure in a vessel or process and/or the liquid level in a vessel. Also, since many fluids are flowing in a pipe or conduit, it is necessary to measure the rate at which the fluid is flowing. Many of these flow meters depend upon devices for measuring a pressure or pressure difference. Some common devices are considered in the following paragraphs.

1. Simple U-tube manometer

The U-tube manometer is shown in Fig. 2.2-4a. The pressure pa N/m2 is exerted on one arm of the U tube and pb on the other arm. Both pressures pa and pb could be pressure taps from a fluid meter, or pa could be a pressure tap and pb the atmospheric pressure. The top of the manometer is filled with liquid B, having a density of ρB kg/m3, and the bottom with a more dense fluid A, having a density of ρA kg/m3. Liquid A is immiscible with B. To derive the relationship between pa and pb, pa is the pressure at point 1 and pb at point 5. The pressure at point 2 is

Equation 2.2-10


Figure 2.2-4. Manometers to measure pressure differences: (a) U tube; (b) two-fluid U tube.


where R is the reading of the manometer in m. The pressure at point 3 must be equal to that at 2 by the principles of hydrostatics:

Equation 2.2-11


The pressure at point 3 also equals the following:

Equation 2.2-12


Equating Eq. (2.2-10) to (2.2-12) and solving,

Equation 2.2-13


Equation 2.2-14


The reader should note that the distance Z does not enter into the final result nor do the tube dimensions, provided that pa and pb are measured in the same horizontal plane.

EXAMPLE 2.2-4. Pressure Difference in a Manometer

A manometer, as shown in Fig. 2.2-4a, is being used to measure the head or pressure drop across a flow meter. The heavier fluid is mercury, with a density of 13.6 g/cm3, and the top fluid is water, with a density of 1.00 g/cm3. The reading on the manometer is R = 32.7 cm. Calculate the pressure difference in N/m2 using SI units.

Solution: Converting R to m,


Also converting ρA and ρB to kg/m3 and substituting into Eq. (2.2-14),



2. Two-fluid U tube

In Fig. 2.2-4b a two-fluid U tube is shown, which is a sensitive device for measuring small heads or pressure differences. Let A m2 be the cross-sectional area of each of the large reservoirs and a m2 be the cross-sectional area of each of the tubes forming the U. Proceeding and making a pressure balance as for the U tube,

Equation 2.2-15


where R0 is the reading when pa = pb, R is the actual reading, ρA is the density of the heavier fluid, and ρB is the density of the lighter fluid. Usually, a/A is made sufficiently small as to be negligible, and also R0 is often adjusted to zero; then

Equation 2.2-16


If ρA and ρB are close to each other, the reading R is magnified.

EXAMPLE 2.2-5. Pressure Measurement in a Vessel

The U-tube manometer in Fig. 2.2-5a is used to measure the pressure pA in a vessel containing a liquid with a density ρA. Derive the equation relating the pressure pA and the reading on the manometer as shown.

Figure 2.2-5. Measurements of pressure in vessels: (a) measurement of pressure in a vessel, (b) measurement of differential pressure.


Solution: At point 2 the pressure is

Equation 2.2-17


At point 1 the pressure is

Equation 2.2-18


Equating p1 = p2 by the principles of hydrostatics and rearranging,

Equation 2.2-19



Another example of a U-tube manometer is shown in Fig. 2.2-5b. In this case the device is used to measure the pressure difference between two vessels.

3. Bourdon pressure gage

Although manometers are used to measure pressures, the most common pressure-measuring device is the mechanical Bourdon-tube pressure gage. A coiled hollow tube in the gage tends to straighten out when subjected to internal pressure, and the degree of straightening depends on the pressure difference between the inside and outside pressures. The tube is connected to a pointer on a calibrated dial.

4. Gravity separator for two immiscible liquids

In Fig. 2.2-6 a continuous gravity separator (decanter) is shown for the separation of two immiscible liquids A (heavy liquid) and B (light liquid). The feed mixture of the two liquids enters at one end of the separator vessel, and the liquids flow slowly to the other end and separate into two distinct layers. Each liquid flows through a separate overflow line as shown. Assuming the frictional resistance to the flow of the liquids is essentially negligible, the principles of fluid statics can be used to analyze the performance.

Figure 2.2-6. Continuous atmospheric gravity separator for immiscible liquids.


In Fig. 2.2-6, the depth of the layer of heavy liquid A is hA1 m and that of B is hB. The total depth hT = hA1 + hB and is fixed by position of the overflow line for B. The heavy liquid A discharges through an overflow leg hA2 m above the vessel bottom. The vessel and the overflow lines are vented to the atmosphere. A hydrostatic balance gives

Equation 2.2-20


Substituting hB = hThA1 into Eq. (2.2-20) and solving for hA1,

Equation 2.2-21


This shows that the position of the interface or height hA1 depends on the ratio of the densities of the two liquids and on the elevations hA2 and hT of the two overflow lines. Usually, the height hA2 is movable so that the interface level can be adjusted.

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