§2.3 Fluid hydrostatic pressure distribution under the action of gravity

Mass force includes gravity and inertia force.The distribution law of hydrostatic pressure is different under the action of different mass force.In practical engineering or daily life, fluid equilibrium means the fluids are motionless relative to the earth, or the mass force exerted on the fluids is only gravity.So this section focuses on hydrostatic pressure distribution under the action of gravity.

2.3.1 Basic equations of hydrostatics

1．Three expressions for the basic equations of hydrostatics

A stationary fluid, for which the mass force only includes gravity, was considered as the obj ect of study.In a rectangular coordinate system, the z axis is vertically upward as shown in Fig.2-5.The pressure on the liquid surface is p0and the pressure at an arbitrary point A was is p.And unit mass force is:fx=fy=0, fz=-g.Thus, integrated form of the differential balance equations of fluids (equation(2-2))can be rewritten as:

dp=-ρgdz

After this equation is integrated, equation(2-4)can be obtained as:

On the free surface, z=H, p=p0.Then, we can write

C＇=p0+ρgH

And h=H-z.By substituting these two equations above into equation(2-4), equation(2-4)can thus be written as:

After each term being divided by ρg, equation(2-4)becomes

where p is the pressure at a point in the stationary fluid, and p0is liquid surface pressure.For the liquid in an open-atmospheric container, p0is atmospheric pressure, which is denoted as pa ·h is submerged depth and z is the height of this point above the coordinate plane.

Equation(2-5)and equation(2-6)are two expressions for basic equations of hydrostatics.equation(2-5)indicates:(1)For a stationary fluid, the pressure increases linearly with depth and the hydrostatic pressure of each point on the same horizontal plane is equal to each other, which is independent of the shape of container; (2)The pressure p at an arbitrary point in the fluid is equal to the addition of liquid surface pressure p 0 and the weight ρg h of vertical liquid column per an unit area from this point to the liquid surface.

Cis a constant in equation(2-6), the magnitude of which can be determined on the basis of boundary conditions.As shown in Fig.2-6, a reference plane 0-0 was taken at random.The height of the two points (1)and (2)away from the reference plane 0-0 is z1 and z2, respectively.The hydrostatic pressure of these two points is p1 and p2, respectively.According to equation(2-6), equation(2-7)can be deduced as:

Equation(2-7)is the third expression for basic equations of hydrostatics.This equation is applicable to stationary, continuous and homogeneous fluids, in which mass force is only gravity.

2．Physical meaning of each term in the basic equations of hydrostatics and basic concepts

A few important basic concepts were derived from equation(2-6), which are described as follows:

Piezometric tube When the absolute pressure at the measuring point is greater than the local atmospheric pressure, a transparent tube with an upward opening was set up at this measuring point.This transparent tube is called piezometric tube.

Elevation head z is the position height at an arbitrary point away from reference plane 0-0.z also indicates the position potential energy of a unit weight fluid from a datum plane.

Height of piezometric tube is rising height of liquid in the piezometric tube, which is also called pressure head, indicating the pressure potential energy (pressure energy)of the unit weight fluid from atmosphere pressure.

Piezometric headis the total potential energy of a unit weight fluid from a reference plane.

The physical meaning of equation(2-6)can be summarized as follows:Under the action of gravity, total potential energy of unit weight fluid at an arbitrary point in a continuous and homogeneous fluid at rest is same.In other words, the piezometric head is equal everywhere.This is the law of energy distribution in a stationary fluid.

3．Inference

(1)Discriminance of equipressure surface

According to equation(2-7), the horizontal surface of a stationary and continuous fluid, in which mass force is only gravity, is equipressure surface.In other words, it is necessary to analyze whether the fluid in the horizontal plane satisfies the following four conditions:stationary, continuous, homogeneous and mass force is only gravity, in order to distinguish whether a horizontal surface is an equipressure surface.If the fluids are discontinuous, or two or more fluids coexist, or the fluids are in a magnetic field, or the fluids are in relative balance (refer to section 2.4), the horizontal surface is not necessarily equipressure surface.For example, the horizontal surface B-B＇ is an equipressure surface while surface C-C＇is not an equipressure surface as illustrated in Fig.2-7.

(2)For the pressure distribution of gas in limited space, the pressure at an arbitrary point in the gas is equal (pgas=C0)because the density of gas is small, the height z is limited and the influence gravity exerts on gas pressure can be neglected.

2.3.2 Measurement of pressure

1．The representation of pressure

The pressure can be calculated from different reference plane.Thus, there are different expression ways such as:absolute pressure, relative pressure and vacuum degree.

Absolute pressure is zero-referenced against a perfect vacuum, using an absolute scale, so it is equal to gauge pressure plus atmospheric pressure.

In the Hydraulic Engineering, the pressure on the surface of water flow and building is ambient air pressure.As a result, ambient air pressure is considered as a reference.The pressure which is calculated with the ambient air pressure as a reference is called relative pressure denoted as p.Relative pressure is also called gauge pressure. The pressure measured by a pressure gauge is the relative pressure since ambient air pressure is assumed to be zero in the pressure gauge.Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure.Obviously, absolute pressure is greater than or equal to zero while relative pressure can be positive, negative or zero.The relationship between absolute pressure, relative pressure and ambient air pressure can be determined as:

As shown in Fig.2-8, the pressure at the measuring point 1 is negative.The value of pressure may be appended with the word“vacuum”, indicating absolute pressure is smaller than ambient air pressure.Vacuum degree is usually represented by which can be calculated according to equation(2-9):

As illustrated in Fig.2-8, at the measuring point, the vacuum degree can be determined as:pv2=pa-pabs2

Vacuum degree can be expressed by the water column height as:

where hv is called as vacuum height.

The relationship between absolute pressure, relative pressure and vacuum degree is described in Fig.2-8.The reference for absolute pressure and the reference for relative pressure differ by a local atmospheric pressure pa.It should be noted the pressure mentioned in this book means relative pressure unless otherwise stated.

2．Measurement unit of pressure

(1)The SI unit of pressure is Pa (N/m2).

(2)Expression methods of atmospheric pressure include standard atmosphere and engineering atmosphere.Engineering atmosphere is slightly different from standard atmosphere.A standard atmosphere (atm)equals 101.325 kPa while an engineering atmosphere (at)is 98 kPa.Engineering sciences usually use engineering atmosphere in place of standard atmosphere.Atmospheric pressure slightly varies in different locations.

(3)The height of water column or mercury column is usually used to represent the height of liquid column.The common unit of the height of liquid column is mH2O or mmHg.

Example 2-1 Try to mark the elevation head, the height of piezometric tube and piezometric head at points A, B and C in the container filled with liquid as shown in Fig. 2-9.The plane 0-0 was set as reference plane.

Solution:The elevation head, the height of piezometric tube and the piezometric head at point A is 3 m,2 m and 5 m, respectively (see Fig.2-10).

Because, according to the piezometric head at point A, the elevation head and the height of piezometric tube at point B can be determined to be 2 m and 3 m, respectively.

At point C, the piezometric head is .The elevation head is zC=6m.The height of piezometric tube is .pC<0, indicating point C is in a vacuum.

Example 2-2 As shown in Fig.2-11, hv=2 m, the liquid in the container B is water.Calculate the vacuum degree in the sealed container A.If the vacuum degree is fixed and oil with a densityρ＇of 820 kg/m3replaces water, calculate the height h＇vof oil column in the piezometric tube.

Solution:(1)Calculate the vacuum degree pv in the sealed container A.

According to equation(2-10), p v=ρg h v=9800 N/m3×2 m=19.6 kN/m2

The vacuum degree in container A is 19.6 kN/m2.

(2)Calculate the height of oil column.

Example 2-3 There is a sealed water tank as shown in Fig.2-12.If the relative pressure on the water surface is -44.5 kPa, calculate (1)h; (2)Relative pressure, absolute pressure and vacuum degree with the unit of at and water column height at point M underwater 0.3 m; (3)the piezometric head at point M with plane 0-0 as a reference.

Solution:(1)Calculate h

On the equipressure surface 1-1, pN=pR=pa.Relative pressure was used in the following calculation.

p0+ρgh=0

-44.5×103Pa+9800 N/m3×h=0

h=4.54 m

(2)Calculate pM

Relative pressure:Absolute pressure:

pM=p0+ρghM=-44.5×103Pa+9800N/m3×0.3m=-41.56kPa

or

or

pMabs=pM+pa=-41.56kPa+98kPa=56.44kPa=0.576 at=5.76 mH2 O

Vacuum degree:

pv=-p=41.56kPa=0.424 atm=4.24mH2 O

Vacuum degree expressed by the height of water column:

(3)The piezometric head at point M with plane 0-0 as a reference is:

Example 2-4 As shown in Fig.2-13, the difference of piezometric head between point A and B was measured by a mercury pressure gauge.Try to write its expression.

Solution:The horizontal plane 0-0 in which upper liquid surface of mercury column locates was taken as reference.The plane where the lower liquid surface locates was taken as equipressure plane.Then,

pM=pA+ρwater gzA+ρwater gΔhp

pN=pB+ρwater gzB+ρpgΔhp

Because pM=pN, we can write

Because the ratio of mercury density to water density is 13.6, the following expression can be obtained: