Basics of Fluid Mechanics, 2014a

6.4. MORE EXAMPLES ON MOMENTUM CONSERVATION 193
rocket have no power for propulsion. Additionally, the initial take off is requires a larger
pressure.
The mass conservation is similar to the rocket hence it is
dm
dt = −U e A e
(6.VII.a)
The mass conservation on the gas zone is a byproduct of the mass conservation of the
liquid. Furthermore, it can be observed that the gas pressure is a direct function of the
mass flow out.
The gas pressure at the initial point is
P 0 = ρ 0 RT
(6.VII.b)
Per the assumption the gas mass remain constant and is denoted as m g . Using the
above definition, equation (6.VII.b) becomes
P 0 = m g RT
(6.VII.c)
V 0g
The relationship between the gas volume
V g = h g A
(6.VII.d)
The gas geometry is replaced by a virtual constant cross section which cross section of
the liquid (probably the same as the base of the gas phase). The change of the gas
volume is
dV g
= A dh g
= −A dh l
(6.VII.e)
dt dt dt
The last identify in the above equation is based on the idea what ever height concede
by the liquid is taken by the gas. The minus sign is to account for change of “direction”
of the liquid height. The total change of the gas volume can be obtained by integration
as
V g = A (h g0 − Δh l )
(6.VII.f)
It must be point out that integral is not function of time since the height as function
of time is known at this stage.
The initial pressure now can be expressed as
P 0 = m g RT
(6.VII.g)
h g0 A
The pressure at any time is
Thus the pressure ratio is
P
P 0
= h g0
h g
=
P = m g RT
h g A
h g0
1
= h g0
h g0 − Δh l
1 − Δh l
h g0
(6.VII.h)
(6.VII.i)