Basic Equation of Fluid Statics – Non Vertical Direction
dP/dz= 
g
(barometric equation)
dP/da (?)
dP/dz= (dP/da).(da/dz)= 
g
cos (
) =
= dz/da
dP/da=

g .(dz/da)
= 
g cos(
)
x
y
z
x
y
z
Bottom
Top
z=0
(x=0, y=0, z=0)
z=
z

+
a
z/
a
Lecture 3
Jan 16 2015
1
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Force= Mass . Acceleration
lbm
lbf
lb
kgm
kgf
kg
Gravitational acceleration= 9.8 m/s
2
or 980 cm/s
2
or
32.2 ft/s
2
1 lbf = 1 lbm . 32.2 ft/s
2
(1)
1 kgf=
1 kgm. 9.8 m/s
2
(2)
Kgf ≠
kgm
why?
If we divide kgf to both sides in eq (2)
lbf ≠
lbm
1 kgf/kgf=
1 kgm/kgf 9.8 m/s
2
and assume
kgm=kgf
1=9.8
m/s
2
(absurd)
Jan 16 2015
2
PRESSURE DEPTH RELATIONSHIP
P/
z= 
g
(P
2
– P
1
) = 
g (z
2
z
1
)
How can we use the above eq to find
pressure at some depth?
Classroom problem
An oil storage tank is 60 ft deep and contains an oil of density
55 lbm/ft3. If top is open to atmosphere. What is the gaugepressure
at the bottom of the tank?
http://www.engineeringtoolbox.com/pressured_587.html?
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 Fall '17

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