Basics of Fluid Mechanics, 2014a

194 CHAPTER 6. MOMENTUM CONSERVATION
Equation (6.VII.a) can be written as
m l (t) =m l0 −
∫ t
0
U e A e dt
(6.VII.j)
From equation (6.VII.a) it also can be written that
dh l
dt = U e A e
ρ e A
(6.VII.k)
According to the assumption the flow out is linear function of the pressure inside thus,
U e = f(P )+gh l rho ⋍ f(P )=ζP
(6.VII.l)
Where ζ here is a constant which the right units.
The liquid momentum balance is
−g (m R + m l ) − a (m R + m l )=
=0
{ }} {
d
dt (m R + m l ) U +bc +(U R + U l ) m l
(6.VII.m)
Where bc is the change of the liquid mass due the boundary movement.
End Solution
Example 6.8:
A rocket is filled with only compressed gas. At a specific moment the valve is opened
and the rocket is allowed to fly. What is the minimum pressure which make the rocket
fly. What are the parameters that effect the rocket velocity. Develop an expression for
the rocket velocity.
Example 6.9:
In Example 6.5 it was mentioned that there are only two velocity components. What
was the assumption that the third velocity component was neglected.
6.4.1 Qualitative Questions
Example 6.10:
For each following figures discuss and state force direction and the momentum that act
on the control volume due to .
Example 6.11: