The 5th lecture on Aerodynamics for 1st year Aeronautical Engieering students at Imperial College London covering the foundations of the subjects in depth and in detail
A101 Introduction to Aerodynamics Section 1-5
__________________________________________________________________________
Outline
• We have previously considered a control volume when considering the conservation of
mass.
• We have also introduced/reviewed partial differentiation.
• Today we will use both these ideas to form the basic equation governing hydrostatics
• Whilst you may be familiar with the results from your previous studies try and follow the
form of the formulation since we will re-use these ideas in deriving the governing
equations of conservation of mass and momentum.
Hydrostatics
There is no motion of the fluid to consider.
Z
Although we typically use ‘y’ as the vertical
coordinate we shall use adopt the convention of
using ‘z’ in our discussion of hydrostatics since
this is the normal convention
yl
I
Therefore consider the forces on a control volume in a fluid in which the
pressure only varies with ‘z’, i.e. P=P(z).
P2 Ga)
' 1111
-
baby
wt b-
! Eddy
(
weight) f ← son
G)
P1
We want to determine the force balance between the pressure acting on the control volume
and the weight of the mass in the control volume.
1
, A101 Introduction to Aerodynamics Section 1-5
__________________________________________________________________________
Vertical force acting on control volume due to pressure.
= Area Lover Area
Upper pressure x -
pressure x
= - P2 ΔxΔy + P1 ΔxΔy
= -
( Pa + Pi ) Doody
=
D
Doody
-
p
This force must be balanced by the weight of the fluid in the control volume, i.e. W = mg
where, g, is the acceleration due to gravity. Therefore the force due to gravity (which acts
downwards)
pbxbybzg
-W= -mg = -
Net force = 0 = Force on CV due to pressure + Force on CV due to weight
Dp doc by Doody bag
=
p
-
-
O =
Dada
-
pg
-
Taking the limit as Δz →0 we obtain
dp
= − ρg (1)
dz
If we know how ρ varies with ‘z’ and g is a constant then we can integrate this equation.
Z constant
pg
For example if ρ = constant then P = -
t
Or
P constant
pg
z
-
t -
Pg 6)
z
Pat
h -
p a
pgh =
P
Pa
pgh
- -
-
0 P=Pa
If P = Pa at z = 0 then at z = h we have
2
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