08/03/2017
Lecture 22
Linear Momentum Principle Through RTT
Yesterday, we discussed about the conservation of linear momentum principle through RTT.
DB
d
   dU    (vr .nˆ )dA
Dt system dt CV
CS
For linear momentum,
D(mv )
d
  F   v  dU   v  (vr .nˆ )dA
Dt system
dt CV
CS
 We worked on a couple of examples.
Today, we will see one more example on the topic.
Example:(Adopted from F.M. White’s Fluid Mechanics-Text Book)
A sluice gate controls the flow in open channels. The flow is uniform at sections 1 and 2.
Pressure of liquid is assumed as hydrostatic. Neglect bottom friction and atmospheric pressure.
Derive a formula for the horizontal force F required to hold the gate.
Soln: Given the figure of sluice gate problem:
Fig.1: Problem Figure
(Source: Fluid Mechanics by Frank White)
 We need to appropriately choose the control volume.
 Sections 1 and 2 have uniform flow. So we include those sections as part of control
surfaces.
 As the flow is uniform, it automatically implies the flow upstream and downstream are
steady.
 Assuming the coordinate system and v  ueˆ1  veˆ2
Conservation of mass equation suggests :
  (v.nˆ ) dA  0 for steady state flow
CS
i.e., ρu2bh2 – ρu1bh1 = 0 (b is width of section into the paper)
m2  m1  0
u2  u1
h1
h2
 Pressure is considered hydrostatic at section 1 and 2.
 P1b=patm+ρgh1
Neglecting atmospheric pressure, p1b= ρgh1
Similarly, at section 2, p2b= ρgh2
 The force on the sluice gate wall should be horizontal, so as to balance the hydrostatic
and the other forces:
d
 Fx   u  dU   u  (vr .nˆ )dA
dt CV
CS
Since,
d
u  dU  0
dt 
CV
 Fx  u2  (v2 .nˆ )bh2  u1  (v1.nˆ )bh1
Recall,
v2  u2eˆ1  0eˆ2
v1  u1eˆ1  0eˆ2
 Fx   u22bh2   u12bh1
h
h
 Fx   g 1 bh1   g 2 bh2  Fgate
2
2
2
2
h h
i.e. gb( 1  2 )  Fgate   b(u22 h2  u12 h1 )
2 2
 gb
(h1  h 2 )(h1  h 2 )  Fgate   b(h2u22  h1u12 )
2
Fgate 
Substitute the relation u2  u1
h1
to get the final expression for Fgate .
h2
Example:
A liquid jet of velocity Vj and area Aj strikes a single 1800 bucket on a turbine wheel
rotating at angular velocity ω . Derive an expression for the power P delivered to this
wheel at this instant as a function of the system parameters. At what angular velocity is
the maximum power obtained?
Soln: As like previous analyses, we need to choose appropriate control volume as shown
below.
Fig. 2: A 1800 jet bucket. The control volume is shown dotted.
(Image Source: http://www.mne.psu.edu/cimbala/Learning/Fluid/CV_Momentum/home.htm)
You know velocity v = Radius x Angular velocity
=Rxω
Also, inlet : v j  v j eˆ1  0eˆ2
Outlet: v j  v j eˆ1  0eˆ2
Liquid jet enters control volume at speed vj
The bucket moves right at a speed v = R ω
 The relative speed of jet entering the control volume = vj – v = vj - R ω
Apply mass conservation first, we get, 𝑚̇1 = 𝑚̇2 = 𝑚̇ = ρAj(vj-Rω)
Applying momentum principle in steady state conditions:
 Fx   Fbucket  mu out mu in
 m[(v j  R )]  m[v j  R ]
 2 m[v j  R ]  2  Aj [v j  R ]2
P  Fbucket *Velocity
 2  Aj [v j  R ]2 * R
Maximum power will be felt for the angular velocity ω, as such
dP
 2  Aj [ R(v j  R )2  R.2(v j  R )( R)]
=
0
which
gives
finally
d
𝑑𝜔
0
R(v j  R )(v j  R )  2 R 2  (v j  R )
𝑑𝑃
2 R  v j  R
dP
For
 0 3R  v j
d
vj
R 
3
Conservation of Angular Momentum
If in the Reynolds Transport Theorem, the extensive property B is taken as angular momentum,
⃗ , the angular momentum.
then B = 𝐻
Recall, angular momentum at any point O can be written as :
H0 
 (r  v )dm
system
Where r is the position vector from O to elemental mass dm, v is the velocity of that element.
 The angular momentum per unit mass that is β is given by:

dH 0
 r v
dm
 The RTT will be :
DH 0
d
ˆ dA
 [  (r  v)  dU]   (r  v)  (vr .n)
Dt system dt CV
CS
 The rate of change of angular momentum of a system should be equal to the sum of all
moments about the point O in that system.
DH 0
d
ˆ dA
  M 0  [  (r  v)  dU]   (r  v)  (vr .n)
Dt
dt CV
CS