Chapter 4
Fluid Statics
4.1 Variation of pressure with elevation
Fluid at rest cannot support shearing stress. It can only support normal stress or pressure that
can result from gravity or various other forces acting on the fluid. Pressure is an isotropic
stress since the force acts uniformly in all directions normal to any local surface at a given
point in the fluid. An isotropic stress is then a scalar since it has magnitude only and no
direction. By convention, pressure is considered a negative stress because it is compressive,
whereas tensile stresses are positive. The direction of pressure force is always pointing
inward the control volume. We now investigate how the pressure in a stationary fluid varies
with elevation z as shown in Figure 4.1-1.
P|z+z
A
z
z
P|z
Figure 4.1-1 Forces acting on control volume Az
Applying a momentum or force balance in the z-direction on the control volume Az we
obtain
Fz = maz = 0
AP|z  AP|z+z  g Az = 0
Dividing the equation by the control volume Az and letting z approach zero we obtain
lim P |z  z  P |z
dP
=
=  g
z  0
z
dz
(4.1-1)
Equation (4.1-1) is the basic equation of fluid statics. It can be integrated if the density and
the acceleration of gravity are known functions of elevation. We will assume g a constant
since the change in elevation is usually not significant enough for g to vary. The integration
will depend on the variation of density. If the density is not a constant, a relation between 
and z or P must be obtained. For constant density fluids, equation (4.1-1) can be easily
integrated
4-1

P2
P1
z2
dP =  g  dz  P2  P1 =  g(z2  z1)
z1
This equation can also be written as
P2 + gz2 = P1 + gz1 =  = constant
The sum of the local pressure P and static head gz is called the potential  that is constant
at all points within a given incompressible fluid.
Example 4.1-1. 1 ---------------------------------------------------------------------------------The manometer system shown in Figure 4.1-2 contains oil and water, between which there is
a long trapped air bubble. For the indicated heights of the liquids, find the specific gravity of
the oil. The two sides of the U-tube are open to the atmosphere.
h1 = 2.5 ft
h2 = 0.5 ft
P1
Oil
Air
h3 = 1.0 ft
P2
h4 = 3.0 ft
Water
Figure 4.1-2 A manometer system with oil, air, and water
Solution -----------------------------------------------------------------------------------------The pressure P1 at the water air interface on the left side of the U-tube has the same value as
the pressure P2 in the water at the same elevation on the right side of the U-tube.
P1 = Patm + ogh1 + agh2
P2 = Patm + wg (h4  h3)
If the pressure due to the weight of air is neglected compared to that of oil and water, we can
determine the specific weight so of oil as follow
ogh1 = wg (h4  h3)  so =
1
o
h  h3
3.0  1.0
= 4
=
= 0.80
2.5
w
h1
Wilkes, J., Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 28
4-2
If the fluid can be described by the ideal gas law then  =
PM
and equation (4.1-1)
RT
becomes
dP
PM
=
g
dz
RT
(4.1-2)
If the temperature is constant for all z, equation (4.1-2) can be integrated

P2
P1
ln
dP
Mg
=
P
RT

z2
z1
dz
P2
Mg
Mg
=
(z2  z1) = 
z
RT
RT
P1
 Mg 
z 
P2 = P1exp  
 RT

Hence
(4.1-3)
For an isentropic process (adiabatic and reversible), the temperature and pressure are not
constants. Therefore an expression for temperature as a function of pressure is required to
integrate equation (4.1-2). We can accomplish this by applying an energy balance (First Law
of Thermodynamics) to the system and then use ideal gas law to substitute volume V in terms
of pressure P and temperature T. For an adiabatic system, the change in internal energy is
equal to the work supplied to the system
dU = dW  CvdT =  PdV
We now use ideal gas law to obtain an expression for PdV in terms of P, dP, T, and dT
PV = RT  PdV + VdP = RdT   PdV = VdP  RdT
Therefore
CvdT = VdP  RdT =
RT
dP  RdT
P
We can separate the variables to obtain
(Cv + R)dT =
Since
C p  Cv
Cp
=
C p / Cv  1
C p / Cv
RT
dT
dP
dP  Cp
= (Cp  Cv)
P
T
P
=
Cp
k 1
, where k =
k
Cv
dT
k  1 dP
=
T
P
k
4-3
Integrating the equation we obtain
P
T
=  
T1  P1 
k 1
k
(4.1-4)
From equation (4.1-2)
dP
PM
=
g
dz
RT
(4.1-2)
dP
MgP  P1 
=
 
dz
RT1  P 
( k 1) / k
MgP (1k ) / k ( k 1) / k
Mg 1 / k ( k 1) / k
P1
=
P
P P1
RT1
RT1
=
Separating the variables yields

P2
P1
P 1 / k dP = 
Mg ( k 1) / k
P1
RT1

z2
z1
dz
Integrating over the limits we obtain


k
Mg ( k 1) / k
( k 1) / k
( k 1) / k
P2
 P1
P1
=
(z2  z1)
k 1
RT1
P2
P2
( k 1) / k
( k 1) / k
 P1
( k 1) / k
= P1
( k 1) / k
=
k  1 Mg ( k 1) / k
P1
z
k RT1
[1 
k  1 Mg
z]
k RT1
 k  1 Mgz 
P2 = P1 1 
k
RT1 

k /( k 1)
(4.1-5b)
From equation (4.1-4)
P 
T2
=  2 
T1
 P1 
(4.1-5a)
k 1
k
The pressures in equation (4.1-5) can be replaced with the temperatures to give
 k  1 Mgz 
T2 = T1 1 
k RT1 

4-4