Basics of Fluid Mechanics, 2014a

7.5. EXAMPLES OF INTEGRAL ENERGY CONSERVATION 223
U p + 4 ρLπr2
K p
dU p
dt
= 2(P 1 − P 2 )
K p
(7.II.k)
where K p is the resistance in the pipe and U p is the (averaged) velocity in the pipe.
Using equation (7.II.c) eliminates the U p as
dh
dt + 4 ρLπr2 d 2 ( ) 2
h R
K dt 2 = 2(P 1 − P 2 )
r K p
Equation (7.II.l) can be rearranged as
K
(
p r
) ( 2 dh
2 ρ R dt + 4 ρLπr2 d 2 )
h
K dt 2 = (P 1 − P 2 )
ρ
(7.II.l)
(7.II.m)
Solution
The equations (7.II.m) and (7.II.i) provide the frame in which the liquid velocity in tank
and pipe have to be solved. In fact, it can be noticed that the liquid velocity in the
tank is related to the height and the liquid velocity in the pipe. Thus, there is only
one equation with one unknown. The relationship between the height was obtained by
substituting equation (7.II.c) in equation (7.II.m). The equations (7.II.m) and (7.II.i)
have two unknowns (dh/dt and P 1 ) which are sufficient to solve the problem. It can
be noticed that two initial conditions are required to solve the problem.
The governing equation obtained by from adding equation (7.II.m) and (7.II.i) as
] ⎞
2
d
([U t
dt 2 + gh
V
{}}{
hA⎠ − 1 ( ) 2 ( ) 2 dh A3
U 1 A 1 + K ( ) 2
t dh
2
2 dt A 1 2 ρ dt
+ K (
p r
) ( 2 dh
2 ρ R dt + 4 ρLπr2 d 2 )
(7.II.n)
h
K dt 2 = (P 3 − P 2 )
ρ
The initial conditions are that zero initial velocity in the tank and pipe. Additionally,
the height of liquid is at prescript point as
h(0) = h 0
dh
dt (0) = 0 (7.II.o)
The solution of equation can be obtained using several different numerical techniques.
The dimensional analysis method can be used to obtain solution various situations which
will be presented later on.
Qualitative Questions
End Solution
ˆ A liquid flows in and out from a long pipe with uniform cross section as single
phase. Assume that the liquid is slightly compressible. That is the liquid has a