Chapter 3

Hydrostatics

3.1 Review

• Vapor pressure: High velocity ⇒ low pressure ⇒ cavitation.

• Surface tension: ∆P is inversely related to R - the surface curvature (e.g., for

bubble we found ∆P = 2π/R).

• Contact angle: defined as the angle between the tangent to the free surface directed

from the solid-liquid-gas interface to the solid boundary through the liquid.

3.2 Hydrostatics and Pressure

In many fluid problems the velocity is zero or the velocity is constant ⇒ τ = 0.

If the tangential stresses are zero this leaves only the normal stresses, which in the ab-

sence of acceleration arise only from the pressure. Then our question is one of statics,

meaning we have no motion (or relative motion), and is how does the pressure vary?

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Let’s look at a small fluid element:

Our starting point is F = ma = 0 hence we write in component form that∑

Fx =∑

Fz = 0.

∑

Fx = Px∆z∆y − Ps sin θ∆s∆y =∆x∆z

2∆y ρax

∑

Fz = Pz∆x∆y − Ps cos θ∆s∆y − ρg∆x∆z

2∆y =

∆x∆z

2∆y ρaz

But from the trigonometry we have

∆s cos θ = ∆x; ∆s sin θ = ∆z

and hence we can write

∑

Fx = Px − Ps = ρax

∆x

2∑

Fz = Pz − Ps = ρaz

∆z

2(g + az)

Now, we take the limit as ∆x, ∆y, ∆z → 0. Aha!

∑

Fx = Px − Ps = 0 ⇒ Px = Ps

∑

Fz = Pz − Ps = 0 ⇒ Pz = Ps = Px

Now, since θ is arbitrary this is generically true. In other words the pressure at a point

is independent of direction! Hence we speak of pressure as a scalar quantity as it is a

quantity with no dependence on direction (as opposed to velocity, which is a vector).

This result is known as Pascal’s law and says that if τ = 0 the pressure at a point is the

same in all directions.

CEE 3310 – Hydrostatics, Sept. 9, 2013 23

3.3 Equation of Motion in absence of Shear Stresses

Let’s investigate the force balance on a small volume of fluid to determine the equation

of motion in the absence of shear stresses

Surface Forces

d~Fs

d∀= −∇P =

Force

Volume

Body Forces

What body forces might act on the fluid element? Gravity, electromagnetic, ...

∆~Fg = ρ~g∆x∆y∆z

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Taking the limit as ∆x, ∆y, ∆z → 0 we have

d~Fg

d∀= −γ~k

Now, putting it all together we have

~F = m~a

d~Fs

d∀+

d~Fg

d∀= ρ~a

−∇P − γ~k = ρ~a

3.4 Hydrostatic Pressure

We found the equation of motion for a fluid with τ = 0:

−∇P − γ~k = ρ~a

Now, if the fluid is at rest (or at least moving at a constant velocity):

~a = 0 ⇒ ∇P = −γ~k

Hence we can write:

∂P

∂x= 0,

∂P

∂y= 0

∂P

∂z= −γ

What does this tell us? Well, we see that P = P (z) only, hence the pressure at a given

elevation (z position) is constant. In the vertical we have

dP

dz= −γ where we have replaced ∂ with d now.

Hence as z ↑ P ↓

CEE 3310 – Hydrostatics, Sept. 9, 2013 25

3.4.1 Incompressible Fluids

dP = −γ dz ⇒

∫ P2

P1

dP = −

∫ z2

z1

γ dz ⇒ P2 − P1 = −

∫ z2

z1

γ dz

Now let’s consider particular solutions:

Constant γ

It is usually reasonable to assume that all liquids essentially have a constant γ, certainly

true for all fluids at constant temperature and pressure. Under this condition we have

P2 − P1 = −γ

∫ z2

z1

dz = −γ(z2 − z1) = −γh

Therefore

P1 = γh + P2 or h =P1 − P2

γ

where we refer to h as the pressure head as it is a pressure measured in units of length.

Liquids with a Free Surface

Examples include lakes, rivers, oceans, reservoirs, your pint glass of water, ...

The surface is the datum (at this point the pressure is the pressure of the atmosphere or

surrounding gas). Therefore we have z = 0, P = Patm.

P1 = γh + Patm but h = z2 − z1 = −z1 ⇒ P = Patm − γz

where z < 0 in liquid (water) and z > 0 in gas (atmosphere).

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3.4.2 Compressible Gases

dP

dz= −γ = −ρg = −

P

RΘg

How do we solve this? ⇒ separation of variables!

∫ P2

P1

dp

P= −

∫ z2

z1

g dz

RΘ

we assume g is a constant but that Θ may vary (e.g., the atmosphere). Then we can

write the above

lnP2

P1

= −g

R

∫ z2

z1

dz

Θ

If we assume that the temperature is constant within the elevation range of interest, an

assumption known as isothermal conditions (i.e., Θ = Θ0 = a constant), we can write

lnP2

P1

= −g

RΘ0

(z2 − z1) or P2 = P1 exp

(

−g(z2 − z1)

Rθ0

)

3.4.3 The Standard Atmosphere

In the troposphere (first 11.0 km of atmosphere) the temperature drops linearly with

elevation. The rate of temperature drop is known as the lapse rate and is given by

β = 0.00650 K/m (6.5 ◦C/km or 3.6 ◦F/1000′). Hence in the troposphere we write

Θ = Θa − βz

where Θa is the temperature at sea level which for the standard atmosphere is taken to

be 15◦C. We can use the expression we found for compressible gasses (recall we allowed

for temperature variation!) and integrate to find

P = Pa

(

1 −βz

Θa

)

gRβ

CEE 3310 – Hydrostatics, Sept. 9, 2013 27

This is the expression for the pressure in the troposphere.

−60 −40 −20 0 200

5

10

15

20

25

Temperature (°C)

Alti

tude

(km

)

0 50 100 1500

5

10

15

20

25

Pressure (kPa)

Alti

tude

(km

)

Figure 3.1: U.S. standard atmsophere to 20.1 km

3.5 Measurement of Pressure

Absolute pressure is the pressure relative to a perfect vacuum, hence it is always a positive

(or zero) value.

Gage pressure is the pressure relative to the local atmospheric value

P > Pa ⇒ Pgage = P − Pa > 0

P < Pa ⇒ Pgage = P − Pa < 0 Pvacuum = Psuction = Pa − P > 0

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Units of pressure are force/area hence pounds per square inch (psi), pounds per square

foot (psf), newtons per square meter (N/m2=pascal (Pa) ).

We will take all pressures, unless otherwise noted, to be gage.

Example – Liquids with a Free Surface

3.5.1 Barometer

Patm − Pvapor = γh ⇒ Patm = γh + Pvapor

For mercury Pvapor ∼ 0 ⇒ Patm = γh.

Typical values of pressure (from the U.S. standard atmosphere at sea level) are: 14.7

psia, 101 kPa, or, 760. mm Hg, 29.9 in Hg, 33.9 ft water (recall γwater = 62.4 lbs/ft3)

and S.G.mercury = 13.6).