3.
Deen 11.3
(neglect gravity)
Solution
a)
In problems with one inlet and one outlet, the macroscopic momentum balance in the form of Eq.
(11.3-9) can be used.
If the velocity fields are flat (i.e., replacing the velocity with the average
velocity), then
a
1
=
a
2
=
1
.
Neglect the effect of gravity for this rapid flow through a nozzle, so the force F
* that the enclosed fluid
exerts on the solid part of the control volume (the nozzle, in this case) is
F
*
=
−
p
1
+
ρ
v
1
2
(
)
n
1
A
1
−
p
2
+
ρ
v
2
2
(
)
n
2
A
2
.
But
n
1
=
−
e
x
, n
2
=
e
x
, A
1
=
π
D
1
2
4
,A
2
=
π
D
2
2
4
,
v
1
=
U, p
2
=
p
0
and
v
2
A
2
=
v
1
A
1
so the net force on the solid nozzle from the fluid inside it is
F
x
*
=
p
1
A
1
−
p
0
A
2
+
ρ
U
2
A
1
1
−
A
1
A
2
⎛
⎝
⎜
⎞
⎠
⎟
.
The net force on the nozzle, including the surrounding air at pressure p
0
, is obtained by replacing
absolute pressures by gauge pressures (see Example 11.3-1 in Deen).
Replacing p
1
by p
1
-p
0
and p
2
by 0
gives
F
x
0
=
p
1
−
p
0
(
)
A
1
+
ρ
U
2
A
1
1
−
A
1
A
2
⎛
⎝
⎜
⎞
⎠
⎟
.
If gravity is negligible, then the y-component force is zero.
b)
The analysis is the same as part (a), except that now

n
2
=
e
x
cos
θ −
e
y
sin
θ
.
Consequently there are both x- and y- components to the force.
For the x-direction these turn out to be
F
x
*
=
p
1
A
1
−
p
0
A
2
cos
θ
+
ρ
U
2
A
1
1
−
A
1
A
2
cos
θ
⎛
⎝
⎜
⎞
⎠
⎟
and
F
x
0
=
p
1
−
p
0
(
)
A
1
+
ρ
U
2
A
1
1
−
A
1
A
2
cos
θ
⎛
⎝
⎜
⎞
⎠
⎟
.
For the y-direction they are
F
y
*
=
p
0
+
ρ
U
2
A
1
A
2
⎛
⎝
⎜
⎞
⎠
⎟
2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
A
2
sin
θ
and
F
y
0
=
ρ
U
2
A
1
2
A
2
⎛
⎝
⎜
⎞
⎠
⎟
sin
θ
.
4.
Deen 11.6
(a,b)
a)
The total flow rate across the plane at the exit of the jet must equal the total flow rate out, so that
with plug flow
v
an
π
4
D
2
− λ
2
D
2
(
)
+
v
jet
π
4
λ
2
D
2
=
v
out
π
4
D
2
or
v
an
1
− λ
2
(
)
+
v
jet
λ
2
=
v
out
.
In terms of flow rates, an equivalent relation is
Q
jet
+
Q
an
=
Q
out
.
b) The momentum transfer between the jet and the liquid in the annulus occurs via viscous friction, so it
is unlikely that the mechanical energy balance will yield a useful relation between the pressures.
However, if the shear force on the wall is negligible, the momentum balance can be used,
d
dt
ρ
v
dV
V
a
(t)
∫
+
ρ
v
v
−
w
(
)
⋅
n
dA
A
a
(t)
∫
=
ρ
g
dV
V
a
(t)
∫
+
n
⋅
T
dA
A
a
(t)
∫
.

We use a control volume bounded by planes “1” and “2” at the annulus/jet exit and the tube outlet,
respectively, in Fig. P11.6, and following the wall of the pipe.
The problem is steady and the control
volume is fixed.
Take the z component of the momentum balance, where e
z
is the direction of flow, and
neglect viscous stresses.
Then we find that
−
ρ
v
z
2
dA
A
an
∫
−
ρ
v
z
2
dA
A
jet
∫
+
ρ
v
z
2
dA
A
out
∫
=
p
jet
dA
A
jet
+
A
an
∫
−
p
0
dA
A
out
∫
.
Here A
out
=A
jet
+A
an
.
We assume the velocity profiles are flat, and replace the velocities with their
average values (flow rate divided by area).
Then we have
−ρ
Q
jet
2
A
jet
− ρ
Q
an
2
A
an
+
ρ
Q
jet
+
Q
an
(
)
2
A
out
=
p
jet
−
p
0
(
)
A
out
.
The pressure rise between the jet and the outlet is therefore
p
0
−
p
jet
=
ρ
Q
jet
2
A
jet
A
out
+
ρ
Q
an
2
A
an
A
out
− ρ
Q
jet
+
Q
an
(
)
2
A
out
2
.
Then
p
0
−
p
jet
=
ρ
A
jet
A
out
v
jet
2
+
ρ
A
an
A
out
v
an
2
− ρ λ
2
v
jet
+
1
− λ
2
(
)
v
an
(
)
2
and
p
0
−
p
jet
=
ρλ
2
v
jet
2
+
ρ
1
− λ
2
(
)
v
an
2
− ρ λ
2
v
jet
+
1
− λ
2
(
)
v
an
(
)
2
.
Multiplying out the squared term,
p
0
−
p
jet
ρ
=
λ
2