Basics of Fluid Mechanics, 2014a

6.2. MOMENTUM EQUATION APPLICATION 189
Noticing that the derivative with time of control volume mass is the flow out in equation
(6.V.n) becomes
mass
rate
dm c.v.
U x + d U out
x
dt dt m {}}{
c.v. = − ṁ 0
Combining all the terms results in
U x = −m 0
LA 0 B
2 A
(6.V.o)
LA 0 B
−F x + a (m f + m t )=−m 0
2 A − U 0 m 0 (6.V.p)
Rearranging and noticing that a = dU T /dt transformed equation (6.V.p) into
(
F x
LA0 B +2AU 0 (m f + m t )
a = − m 0 (6.V.q)
m f + m t 2 A (m f + m t )
(
If the F x ≥ m LA0 B
0 2 A + U 0)
the toy will not move. However, if it is the opposite the
toy start to move. From equation (6.V.d) the mass flow out is
The mass in the control volume is
U 0
{ }} {
h
{ }} {
m 0 (t) = B h 0 e −tA 0 B
A A 0 ρ
m f = ρ
V
{ }} {
Ah 0 e −tA 0 B
A
)
(6.V.r)
(6.V.s)
The initial condition is that U T (t =0)=0. Substituting equations (6.V.r) and (6.V.s)
into equation (6.V.q) transforms it to a differential equation which is integrated if R x
is constant.
For the second case where R x is a function of the R y as
R x = μR y (6.33)
The y component of the average velocity is function of the time. The change in the
accumulative momentum is
d [ ] dm f
(mf ) U y =
dt
dt U y + dU y
dt m f
(6.V.t)
The reason that m f is used because the solid parts do not have velocity in the y
direction. Rearranging the momentum equation in the y direction transformed
⎛
⎞
m f
{ }} {
F y =
⎜
m t + ρAh 0 e −tA 0 B
(
A
ρh0 A 2 0 B 2
g +2
⎟
A
⎝
⎠
) 2
e
− tA 0 B
A
(6.V.u)