Pressure variation with altitude in a static compressible fluid (e.g. air) with constant
temperature gradient:
________________________________________________________________________
Basic equation:
dp H    H gdH
(1)
This equation cannot be integrated since variation of  with height is not known.
Assume air is ideal:
pH
 RTH
(2)
dp H
dH
 g
pH
RTH
(3)
TH  T0   H
(4)
H
From (1) & (2):
In the troposphere:
where  = temperature lapse rate = 6.5 x 10-3 K/m = 6.5 K/km
from (3) & (4):
dp H
g
dH

pH
R  T0   H 
(5)
d T0   H    dH  dH 

pH
p0
d T0   H 

dpH
g H d  T0   H 
g
p

 ln pH  pH 
ln  T0   H 

0
pH
R 0  T0   H 
R


H
0
g
p H  T0   H  R


p0 
T0 
g
 T   H  R  T0   H 
H
p
RT
p
T0
 H  0  H
 0
 

 0 RTH p 0
p 0  T0   H   T0 
 T0 
 g

 1


 H  T0   H   R

 0  T0 
- 1-
1
Variation of pressure with altitude in a static compressible fluid (e.g. air) at constant
temperature
Lower stratosphere ( 11km  H  20.1km ):
TH  56.5o C  const  T *
Note: air is still treated as ideal.
dpH
g
g
p
H

dH  ln pH  pH  
HH

*
*
i
i
pH
RT
RT

p
 H e
pi
g H Hi 
RT *
Temperature lapse rate in the atmosphere under polytropic conditions
Static fluid:
dp    gdz
(1)
Polytropic relationship:
(n= polytropic index)
p
n
 const
(2)
Assume air is ideal:
p

 RT
(3)
From (2):
p  const   n  dp  const n n 1 d
dp 
From (3):
p

n
n n 1 d  dp  np
d

 nRTd (4)
dp  R  dT  RTd (5)
From (4) & (5):
dp  R  dT 
d
 1
 n 
 dp 1    R  dT  dp  
 R  dT
 n
 n  1
n
From (1) & (6):
- 2-
(6)
 n 
dp    gdz  
 R  dT
 n  1
dT
 n  1 g
 

 n R
dz
Note: (i) for adiabatic conditions:
n = k; k: adiabatic index (k = Cp / Cv)
(ii) n = 1 in the lower stratosphere
dT
0
dz

 T  const
Variation of pressure with altitude in the atmosphere under polytropic conditions
Basic pressure-height relationship:
Assume air to be ideal:
dp H    H gdz
(1)
p H   H RTH
(2)
Temperature variation with altitude:
 n  1 g
TH  T0  
 z
 n R
(3)
(1) divided by (2):
dp H
g

dz
pH
RTH
(4)
with (3), (4) becomes
dp H
g
dz

pH
R 
 n  1 g 
 T0   n  R z 



pH
p0
(5)
dp H
g Z
dz
 
pH
R 0 
 n  1 g 
T0   n  R z 


- 3-
ln
pH g  n  R Z
 

p 0 R  n  1 g 0


 n  1 g 
 n  1 g 
d T0  
T0  
 z
 z

 n R   n  
 n  R 


 ln
T0

 n  1 g   n  1
T0   n  R z 


 n 



 n  1 g    n 1
 n 
T

 z


  0 
 n  1

n
R


pH  
 n  1 g


 1  
z


p0 
T0
  n  RT0 




Note: the above is valid only in the troposphere. In the lower stratosphere, n = 1;
the above will “blow up”
Forces due to liquid pressure on plane submerged surfaces:

FR : resultant force due to liquid pressure
C: centriod of area
z
Atmospheric pressure
Free surface of liquid
O

Liquid density = 
dA h
h
dF
FR
y
edge view
of plane
y
x
dy
C(x,y)
y
P(x’,y’)
P: centre of pressure (i.e. point of application of FR)
- 4-
dA
A
A
A
A
A
Note: P is always below C unless the surface is horizontal in which case P = C
Need: magnitude of FR
direction of FR
line of action of FR
No shearing stresses in a static fluid; forces will be normal to surface independent of the
orientation of the surface


dF   pdA

(the negative on right-hand side of equation indicates that the direction of dF is opposite

to dA )

Note: positive direction of dA is the outward drawn normal to the area
Resultant force:

FR  
A


dF   p dA
A
Pressure-height relationship in a static fluid:
dp
 g
dh
(h is positive downward from the free surface)
At the free surface i.e. at h = 0, p = patm
also,  = const (liquid)

p
patm
z
h
dp   g  dh
0
h
dF

h=ysin
p  patm   gh  p  patm   g y sin 



FR     p atm   g y sin   dA    p atm dA   g sin 
 A
A
FR
 k   

A


p atm dA k   g sin 
 FR   patm dA   g sin 
A

A



y
dA
k
A

y dA
- 5-


y
d
A
A

Recall

A
y dA : first moment of area of surface about the x-axis
  y dA  y A
A
FR  patm  A   g sin  y A  patm A   gh A
using gauge pressures: FR   gh A  pA
(*)
NOTE: (*) is based upon the assumption that pressure at the free surface of the liquid is
atmospheric. More on this soon -- imaginary free surface
 = const means that (*) is valid only for a single homogeneous liquid
Point of application of resultant force (centre of pressure)
Resultant force is equivalent to the individual forces if the moment of the resultant about any
axis is the same as the sum of the moments of the individual forces about the same axis

 
 

r ' FR   r dF    r  pdA
A
A
 




 
 
r '  i x' j y ' ,r  i x  j y, dA  dA k ,FR   FR k


i x' j y'   F k    i x  j y  pdAk



R

A
k




 j FR x'i FR y'   j  x pdA  i  y pdA
A
A
 y' FR   y pdA, x' FR   x pdA
A
A
j
y-location of centre of pressure
y' 

A
y pdA
FR

y' 
A

 y p
A
atm
  g y sin   dA
 patm   g y sin   A
p atm y dA    g sin  y 2 dA
A
p atm A   g y sin  A

p atm y A   g sin  I x
p atm A   g y sin  A
- 6-
i
NOTE:

y 2 dA
is the second moment of area about the x-axis
Parallel-Axis Theorem: I x  I x  y 2 A
dA
y’
B’
B
C
y
C: Centroid
d
A’
A
Moment of inertia with respect to the AA’ axis
2
I   y 2 dA, y  y ' d , I    y ' d  dA
A
A

I
A
y' 2 dA
2d  y' dA

moment of inertia
with respect to
centroidal axis
A

d2
 dA
A
first moment of area
with respect to
BB’ = 0 since

BB’ = I
A
y'dA  y A and y  0
 I  I d2 A
y' 

  Ay p
patm Ay   g sin  I x  y 2 A
patm A   g sin  yA
y'  y   g
atm
  g sin  y 
A patm   g sin  y 
Ix
h
y A  patm   g sin  y 
- 7-

 g sin  I x
A patm   g sin  y 
y'  y 
 p  patm  
p
Ix
Ay
where p  patm   gh
using gauge pressures, y'  y 
Similarly: x '  x 
Ix
Ay
 p  patm 
I xy
p
Ay
Using gauge pressures, x '  x 
where I xy : product of inertia
I xy
Ay
NOTE: 1) y and x are referred to the origin at the free surface
2) product of inertia may be positive or negative unlike Ix or Iy and Ixy = 0
when one or both axes are axes of symmetry
Example
An elliptical gate covers the end of a pipe 4m indiameter.
If the gate is hinged at the top, what normal force, F, is
required to open the gate when the water free surface is
8m above the top of the pipe and the pipe is open to
atmosphere on the other side? Neglect the weight of the
gate.
water
8m
Resultant hydrostatic force on gate
z
FR  p A
5m
Gauge pressures:
Patm
F
FR   gh A
FR  1000  9.81  10  
Hinge
4  5
 1541
. MN
 2  2
y
m
n n
m
- 8-
4m
Location of centre of pressure along the y-axis
b
3
1
 a3b
Ix
a2
4
y'  y 
 12.5 
 12.5
y A
12.5  a b
50
8
10
12
5
y ' 12.625 m
Taking moments about the hinge
M
0
H
FR  y H  F  5  0  1541
.
 12.625  10  F  5  0
F
1541
.
 2.625
 0.809 MN
5
APPROACH #1
Determination of the hydrostatic force on a curved submerged body
Consider the following case:
Fig. 3.6 (Fox & McDonald)
Patm
z
dA
dAx
dAy
y
dAz
x


dF   pdA
- 9-
a
“Hydrostatic force acts, as in plane submerged surface, normal to surface. However,
differentiation due to pressure on each element of the surface acts in a different direction
because of the surface curvature”


Differential force: dFR   pdA


Resultant force: FR    pdA
A




FR  i FRx  j FRy  k FRz
 
 
FRx   dFx  FR  i    p dA  i    p dA cos x    x p dAx
A
Similarly
A
A
Ax
FRy   p dA cos y    p dAy
A
Ay
FRz   p dA cos z    p dAz
A
Az

: angle between A and respective unit vectors

A
: angle between
and respective unit vectors


It is convenient to obtain the components of FR first and then FR can be expressed as the
vector sum of the components

Line of action of each component of FR

 

r ' x  i FRx   r  dFx i

 

r ' y  j FRy   r  dFy j

 

r ' z k FRz   r  dFz k
e.g. 2-D curved submerged surface
- 10-
z
z
dA
FRz
dAy
FRy
z’
y
z’
dAz
z’
y’
 
dA y  dA  j  dA  1  cos  2
 
dAz  dA  k  dA  1  cos 3
 
dA  k  dA1 cos 3  dAcos 3
 
dA  j  dA1 cos 2  dAcos 2
3
k
2
j


dA
dAcos
=-dAcos2=dAy
dAsin
=dAcos3
=dAz
ALTERNATIVE APPROACH
dA=wds
dAv=dAsin
y
p = y
dF
A
dFy
dy
dFx
dA
ds

x
B
dx
dAh=dAcos
y
Enlarged view of dA
- 11-
Consider a 2-D curved surface AB as shown above. It is convenient to determine the
horizontal and vertical components of the hydrostatic force and to obtain the resultant
vectorially from the components.
dF   ywds
dFx   ywds sin    ydAv
-subscript v means projection of area to a vertical plane
Fx    ydAv  

ydAv   y Av
Note:

ydA  yA
 Fx  p Av (magnitude of Fx)
y'  y 
Ix
y Av
w
Fx
dFy   yw ds cos    y dAh
--subscript h means projection of area to a horizontal plane
dFy   dV
 Fy   V
- 12-
x
dAx
dA
y
dAy
Ah  Ay
Line of action

n
x'Fy   Wi xi'

i 1
Wi: contribution to the weight of liquid by part i
y = axn
y
x’
Fx
y’
x
FR
Fy
y dF
h
b
dA
x
a
METHOD #1


dF   p dA



FR   dF    pdA
A
A
- 13-
 
 
Fy  FR  j    p dA  j    pb dAy
h=(b-y)
j
Fx     b  ydyw
y
0
dAx
 
 
Fx  FR  i    pdAb  i    pdAx
i
x
dAy
ph
Fy     b  y dx w
b
0
Fy     h dAy ;
Fx     h dAx
Line of action
y
x’
x
Fy
dA
x






x' i  Fy j   xi  dF j x' Fy k  k  x p dAy
A
A
x' F y   x b  ywdx
b
0
x' 

b
0
x b  y  wdx
Fy




y ' j  Fx i   y j  dFx i
- 14-
Fy  pbottom  A   b a w   V
n
x' Fy   Wi xi
i 1
y
h
b
Fy
w
x
a
a
METHOD #2
Projection of surface onto plane made up of
y & w results in a plane area; this area is
acted upon by the x-component of the
resultant force
Fx  p A
Fx   h A
A  bw
w
y’
y'  y 
Ix
yA
b
Fx
- 15-
h