Equilibrium of Fluids
A fluid can support no shearing stress when in equilibrium. The stress for compression is
simply the pressure. If there are no external forces (such as gravity), the pressure is the
same everywhere in the fluid when the fluid is in equilibrium.
1. Gravity on fluids
When dealing with a fluid, we no longer have point masses. Forces such as gravity act
on fluid elements which are the same as volume elements. This leads to the idea of a
body force (force per volume). For gravity, this becomes:
f = ρg = -ρg y , where ρ is the mass per volume, dm/dV.
Newton’s 2nd law becomes (with P being the pressure):
Fx = +Pleft dAleft - Pright dAright = 0
Fz = +Pin dAin – Pout dAout = 0
Fy = +Pbottom dAbottom – Ptop dAtop -ρg dV = 0 .
The volume element can be written as:
dV = dx dy dz = dAxz dy , so that, with dAtop = dAbottom = dAxz, we have
Pbottom – Ptop = ρg Δy .
This can be generalized for a body force to be:
P2 – P1 = ΔP =
r2
r1
f • dr ,
or
f = P
.
Note that the constant pressure lines (actually surfaces) are perpendicular to the body
force, f. Also, since   P = 0, we have   f = 0, which allows us to have:
P(r) – Po =
r
ro
f  dr
which is similar to a potential energy.
2. Bulk Modulus
The Bulk Modulus, B, is defined as: B = stress/ strain, where stress is simply the
pressure, P; and strain is ΔV/V : B = PV/ΔV, so we have -dV/V = dP/B. The minus
sign comes from converting the ΔV into a dV, but noting that the ΔV is a decrease in the
volume, not an increase.
Let’s now consider density, ρ = m/V. The quantity dρ is
dρ = d(m/V) = m d(1/V) = m d(1/V)/dV dV = m (-1/V2) dV, so that the
quantity dρ/ρ becomes:
dρ/ρ = m (-1/V2) dV / (m/V) = -dV/V .
Note that whenever dV increases, dρ decreases, and when dV decreases, dρ increases.
Hence the difference in the sign. Putting -dV/V = dP/B together with dρ/ρ = -dV/V
gives:
dρ/ρ = dP/B ,
and integrating both sides gives:
ln [ρ/ρo] = PoP (1/B) dP
, or ρ = ρo exp [PoP (1/B) dP ] .
3. Three special cases
Case 1: incompressible fluid
If we have a fluid that is incompressible (B ≈ ∞), then ρ ≈ constant, so using Pbottom – Ptop
= ρg Δy we have: Pbottom = Ptop + ρgh , where h = height (or depth) of the fluid.
Numerical example
If we assume that water is incompressible with a density of 1.0 gr/cc = 1,000 kg/m3, then
the pressure at a depth of 100 meters would be
ΔP = Pbottom - Ptop = ρgh
= (1 x 103 kg/m3) (9.8 m/s2) (100 m) = 9.8 x 105 Nt/m2 .
At this depth with this pressure, with the actual finite value of the bulk modulus of water
being 2.1 x 109 Nt/m2, water would be compressed by:
B = PV/ΔV,
or
ΔV/V = P/B = 9.8 x 105 Nt/m2 / 2.1 x 109 Nt/m2 = .000467
which says that the water would be compressed to only about 0.05% at that depth. In
this case, our approximation of constant density holds up well.
If we change the depth to 8 kilometers (about the depth of the deepest ocean trench), then
ΔP = (1 x 103 kg/m3) (9.8 m/s2) (8,000 m) = 7.84 x 107 Nt/m2 , and
ΔV/V = P/B = 7.84 x 107 Nt/m2 / 2.1 x 109 Nt/m2 = .037 (about 4%).
Case 2: Bulk modulus is constant
Here we consider the case where the bulk modulus is a constant, independent of pressure
and density. In this case, the density equation gives:
ρ = ρo exp [PoP 1/B) dP ] = ρo exp [P/B – Po/B] .
The pressure equation gives us:
Pbottom – Ptop = ρg Δy , or dP = ρg dy = ρo exp{P/B – Po/B} g dy
= ρo eP/B e-Po/B g dy.
We can separate this equation to get:
e-P/B dP = ρo e-Po/B g dy
, or
P
Po
e-P/B dP =
yo
y ρo e-Po/B g dy
and with yo = 0 (measured from the surface) and Po = atmospheric pressure (at the
surface), we have upon integration:
(-B) e-P/B – (-B)e-Po./B = ρo e-Po/B g y ,
or
e-P/B =
e-Po./B [1 - ρo g y/B]
,
or
P = -B ln{ e-Po/B [1 - ρogy/B] }
.
Check: if y = 0, then P = -B ln{ e-Po./B } = Po as required.
Check: if P/B is small, then the equation above, e-P/B = e-Po./B [1 - ρo g y/B]
simplified using the approximation that e-x ≈ (1 – x) to get:
1 – P/B ≈ (1 - Po/B) [1 - ρo g y/B]
can be
, or
P = Po + ρo g y + terms of order (P/B) .
Numerical example
Let’s consider the pressure at the deepest part of the ocean, y = 8,000 meters. With Po =
Patmospher = 1.01 x 105 Nt/m2 and B = 2.1 x 109 Nt/m2, so that Po/B = 4.81 x 10-5, we get
P = -B ln{ e-Po./B [1 - ρogy/B] }
= - (2.1 x 109 Nt/m2) ln{ e-4.81 x 10^-5 [1 – (1 x 103 kg)(9.8 m/s2)(8,000 m)/(2.1 x 109) }
= 8.00 x 107 Nt./m2 .
Note that this is about 2% larger than the value we obtained by considering the water to
be incompressible (7.84 x 107 Nt/m2 ).
Case 3: Ideal gas at constant Temperature
For an ideal gas, we have the ideal gas law:
PV = nRT
where n is the number of moles, R is the gas constant (R = 8.3 J/mole-K), and T is the
temperature (in an absolute scale such as Kelvin). This relates the volume to the
pressure, and hence the density to the pressure:
ρ = m/V = m / {nRT/P} = Pm/nRT .
The m is the mass, and n is the number of moles. We can combine these two to get the
mass per mole, M = m/n, so the
ρ = PM/RT .
We can now use this in the relation for pressure:
dP = ρg dy = (PM/RT)g dy .
This can be separated and integrated to give:
P
Po
(1/P)dP =
yo
y (Mg/RT) dy
, or
ln {P/Po} = Mgy/RT , or P = Po e(Mg/RT)y .
The above has y as the “depth” where P increases. To make it into a height (where P
decreases), we must put in a negative sign for the y:
P = Po e-(Mg/RT)y .
Numerical Example
Let’s consider the air pressure above the earth’s surface. If we assume that air is
incompressible, how high up would the atmosphere extend? From measurements, we
have: Pat surface = Patmospheric = 1.01 x 105 Nt/m2, ρo = 1.2 kg/m3, we are assuming Ptop = 0.
ΔP = Psurface – Ptop = Patmosphere – 0 = 1.01 x 105 Nt/m2
= ρgh = (1.2 kg/m3)(9.8 m/s2)h , or h = 8,588 meters.
Let’s now see what the ideal gas model at constant temperature will predict for the
atmosphere: We will start by assuming that air consists of ¾ N2 and ¼ O2 for an average
mass per mole, M, of 29 grams/mole, or .029 kg/mole. We will also assume a constant
temperature of 32oF (or 273 Kelvin). Since the prediction for an ideal gas is for an
exponential decrease in pressure as we increase height, there will be no finite distance at
which the pressure will reach zero. Instead, let’s see what the pressure would be when
the height is 8,588 meters (the cut-off for the atmosphere in the incompressible
prediction):
P = Po e-(Mg/RT)y
= 1.01 x 105 Nt/m2 * exp{-(.029 kg/mole * 9.8 m/s2 )/(8.3 J/mole-K * 273 K) 8,588 m}
= 3.44 x 104 Mt/m2 = .341 Patmosphere .