Fluid Mechanics II

Steady flow momentum equation

Newton’s second law applied to a control volume

Fluids, either in a static or dynamic motion state, impose forces on immersed
bodies and confining boundaries. In the engineering context these forces are
clearly of major relevance and importance for rational design requirements.
It is frequently possible to estimate the steady overall forces applied to either
immersed bodies or contained boundaries using basic momentum consider-
ations expressed in Newton’s second law. The present section of the
course develops the relevant background theory for this simple but powerful
and effective methodology with examples of its practical application.

Newton’s second law states that:

The net force acting on a body or system of bodies in a particular

direction is equal to the rate of change of momentum of the body or
system of bodies in that direction.

The mathematical formulation of Newton’s second law is:

→
−
→
− d I
F = , (1)
dt
→
− →
−
where F is the net force acting on the system and I is the total linear
momentum all particles of the system:
→
− X → −
I = mi V i .
i

→
− →
−
Note that both F and I are vector quantities and can be represented via the
components in a suitable coordinate system. For example, in two dimensions
for the x and y directions we have
d Ix d Iy
Fx = ; Fy = .
dt dt

1
 2

P (pressure)

Control volume

D (drag force)

R (reaction on fluid)
τ (shear stresses)

Fig. 1: Fig. 2:

The net force acting on the system is equal to the sum of all external forces
acting on the individual particles of the system. Internal forces, that is the
forces acting between different particles of the system, do not contribute to
the change of the momentum of the system because by Newton’s third
law for each such a force there is an equal opposite reaction (figure 1). For a
fluid it is convenient to apply physical principles to a system of fluid particles
occupying a certain suitable volume called a control volume. The force on
fluid particles within the control volume include the net force acting on the
bounding surface of the control volume, forces on solid surfaces of immersed
bodies and gravity or inertial forces.
Main applications of the momentum principle include evaluation of forces
on immersed objects produced by a flowing stream. By Newton’s third
law there must be an equal and opposite reaction on the fluid. We wish to
apply the law to a flowing fluid in order to deduce the magnitude of the force
→
−
F which must have acted on the fluid within the control volume in order to
→
−
produce the observed rate of change of momentum. The force F will then be
the total force on the fluid inside the control volume arising from all sources
→
−
which will include not only the reaction from the body drag D , but also due
to any pressure or shear forces over the bounding surface of of the control
volume (figure 2) plus any weight components.
Let us consider a stream tube of a small cross section such that the fluid
velocity can be assumed constant across the tube. It is convenient to chose
a part of the stream tube between two cross section 1 and 2 as a control
volume (region abcd on figure 3). The areas of the cross section 1 and 2 are
→
−
A1,2 and the corresponding fluid velocity vectors are V 1,2 . During a small
time interval ∆t the fluid in the region a 0 add 0 enters the control volume, and
the corresponding mass gain is

∆m1 = ρ A1 V1 ∆t.
 3

b V2
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d’ d
V1 ∆ t

Fig. 3:

Similarly, the fluid in the region bcc 0 b 0 exits the control volume, and the mass
loss is
∆m2 = ρ A2 V2 ∆t.
According to the continuity principle, the mass flow rates across the surfaces
1 and 2 are equal to each other, and we can write

∆m1 = ∆m2 = ρ Q ∆t,

where Q is the volumetric flow rate through the stream tube. Note that V1,2
are the velocity components normal to the surfaces 1, 2, and for the case when
the velocity vectors are perpendicular to that surfaces (as on figure 3) V1,2
→
−
are the absolute values of the corresponding velocity vectors: V1,2 = | V 1,2 |.
The momenta associated with masses ∆ m1,2 are
→
− →
−
∆ I 1,2 = ρ Q V 1,2 ∆ t.

Note the vector values in both sides of the equation. Therefore, during the
→
− →
−
time period ∆ t the control volume gains the momentum ∆ I 1 = ρ Q V 1 ∆ t
due to the fluid flow into the control volume trough the cross section 1, and
the corresponding loss of the momentum due to the flow out of the control
→
− →
−
volume trough the cross section 2 is ∆ I 2 = ρ Q V 2 ∆ t. For a steady flow the
→
−
momentum of the fluid inside the control volume does not change: ∆ I = 0.
This implies that certain force should act on the fluid inside the control
volume to compensate the change of momentum due to the fluid flow in and
out of the control volume. According to Newton’s second law (1) the change
of momentum caused by an action of such a force during the time period ∆t
 4

→
− →
−
is: ∆ I F = F ∆ t, and the overall balance of the change of momentum in
the control volume is
→
− →
− →
− →
−
∆ I = ρ Q ∆ V 1 ∆ t − ρ Q ∆ V 2 ∆ t + F ∆ t = 0.

From this equation the total force on the fluid in the control volume is given
by
→
− →
− →
− →
− →
−
F = ρ Q V 2 − ρ Q V 1 = ρ Q ( V 2 − V 1) . (2)
Verbally, equation (2) says:

Momentum principle: For a steady flow the total force on the fluid in the
control volume is equal to the momentum flow rate out of the control
volume minus the momentum flow rate into the control volume.

Equation (2) is written in a vector form. Equation for any specific direction
can be obtained from (2) by taking the projection on this direction. For
example, in the case of 3 dimensions we have 3 coordinate axis (x, y, z). The
corresponding components of force and velocity vectors are (Fx , Fy , Fz ) and
(u, v, w) respectively. The vector equation (2) is then equivalent to three
equations in each of three coordinate directions:

Fx = ρ Q (u2 − u1 ); Fy = ρ Q (v2 − v1 ); Fz = ρ Q (w2 − w1 ).

For a collection of stream tubes forming a larger control volume with perhaps
non-uniform velocity distributions across the inlet and outlet boundaries we
may integrate equation (2) to obtain the general steady flow momentum
equation
→
− →
− →
−
Z Z
F = ρ V Vn dA − ρ V Vn dA,
S2 S1

where the integrals are taken over the inlet and outlet surfaces S1 and S2 ,
and Vn is the component of velocity normal to an area element dA of the
corresponding integration surface.

Applications of steady flow momentum equation

General approach
1. Draw the flow system.

2. Make and justify necessary assumptions.

 5

3. Chose a coordinate system and a suitable control volume.

4. Define velocities on the boundaries of the control volume and their

components in the chosen system of coordinates.

5. Define and label forces on bodies, ducts, etc. and their reactions to the
fluid.

6. Assess total force on fluid in control volume including any drag of

internal immersed bodies, boundary pressure and shear forces, gravity
forces. Consider components of the total force in the coordinate system.

7. Assess momentum flows into and out control volume and apply the
momentum principle in chosen directions.

Assumptions
The following assumptions are used in this section, if not stated otherwise:

1. Flow is two-dimensional.

2. Flow is frictionless.

3. Gravity effects are neglected.

4. Flow in ducts and jets are assumed parallel and uniform with a constant
velocity over the cross section.

Nomenclature
Selected notations used in this section:

x, y Coordinates.
→
−
V Velocity vector.
V Absolute value of velocity.
u, v Velocity components along axes x and y respectively.
Fx , F y Components of the net force acting along axes x and y on
fluid inside a control volume.
X, Y Force components acting along axes x and y on solid surfaces
(bodies, ducts, etc.)
 6

y
V2

11
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00
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00
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00
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00
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00
11 U
00
11 x
00
11
00
11
A 00
11
00
11
00
11
00
11
V1 00
11
00
11 Fx
00
11
00
11
00
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00
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00
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11

V3

Fig. 4:

Impact of jets on surfaces

Normal plate
Formulation:
A jet of water of cross section A and velocity V1 strikes a flat plate perpendic-
ular to the direction of the jet and moving with a constant velocity U < V1
in the jet direction (figure 4). Find the force applied by the jet to the plate.
Note: Steady flow momentum principle can be applied not only to a stationary
control volume but also to a control volume moving with a constant
velocity. In this case fluid motion is considered in the coordinate system
moving together with the control volume.
Solution:
We chose a control volume and coordinate axes (x, y) moving together with
the plate (figure 4). The pressure in the jet is equal to an atmospheric
pressure. Therefore, the pressure on the boundary of the control volume is
constant and does not make a contribution into the total force acting on the
control volume, which means that in this particular case the net force on the
fluid is equal to the reaction from the wall (Fx = −X; Fy = −Y ). It is useful
to note that the flow has a horizontal symmetry axis.
 7

y
V2

U
11
00
00
11
x
Fy 11
00
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11 Fx
00
11
00
11
00
11
00
11
A 00
11
00
11
θ V1 00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11

V3

Fig. 5:

Flow rate through the moving boundary: Q = A (V1 − U )

→
− →
−
Flow symmetry: V 2 = − V 3 ; Q2 = Q3
Velocity components in the moving coordinate system:

x-direction: u1 = V1 − U ; u2 = u3 = 0
y-direction: v1 = 0; v2 = −v3

Momentum principle:

x-direction: Fx = ρ Q2 u2 + ρ Q3 u3 − ρ Q u1 = −ρ Q (V1 − U )
y-direction: Fy = ρ Q2 v2 + ρ Q3 v3 − ρ Q v1 = 0

Force components acting on the plate:

X = −Fx = ρ A (V1 − U )2 ; Y = −Fy = 0.

The walue of X is positive, which means that the force acts in the positive
x-direction (from left to right).
 8

y
Fy
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Fx
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θ
V2
Fig. 6:

Inclined plate
Formulation:
The jet has an angle θ to the plate normal. The plane is moving with velocity
U < V1 in the jet direction (figure 5). Find the force applied by the jet to
the plate.
Solution:
It is convenient to consider axes oriented as shown. Assuming no friction, no
forces can be applied to the fluid by the plane in the y-direction: Fy = 0.
Flow rate: Q = A (V1 − U )
Velocity components in the x-direction: u1 = (V1 − U ) cos(θ); u2 = u3 = 0
Momentum principle in the x-direction:

Fx = ρ Q2 u2 + ρ Q3 u3 − ρ Q u1 = −ρ Q (V1 − U ) cos(θ)

Force components acting on the plate:

X = −Fx = ρ A (V1 − U )2 cos(θ); Y = −Fy = 0.

Curved vane
Formulation:
A curved vane deflects a jet of water of diameter A through angle θ. (figure 6).
The mean velocity of water across the jet is V1 . The vane is moving in the
jet direction Assuming that on inlet the jet is tangential to the vane surface
 9

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x

Fig. 7:

find the force exerted by the jet on the vane.

Solution:
Flow rate: Q = A (V1 − U )
For frictionless flow the magnitudes of jet velocities relative to the vane are
the same at inlet and outlet. This gives the following velocity components:

x-direction: u1 = V1 − U ; u2 = (V1 − U ) cos(θ)

y-direction: v1 = 0; v2 = −(V1 − U ) sin(θ)

Momentum principle:

x-direction: Fx = ρ Q (u2 − u1 ) = ρ Q (V1 − U )( cos(θ) − 1)

y-direction: Fy = ρ Q (v2 − v1 ) = −ρ Q (V1 − U ) sin(θ)

Force components acting on the plate:

X = −Fx = ρ A (1 − cos(θ))(V1 − U )2 ;
Y = −Fy = ρ A sin(θ)(V1 − U )2 .

Flows in contractions and nozzles

Straight contraction
Formulation:
Water flows through a pipe with a straight contraction which cross section
gradually redices from A1 to A2 . The mean flow velocity at the inlet of the
contraction is V1 and pressure here is P1 . Determine the force exerted by the
 10

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Fig. 8:

water on the contraction

Solution:
Continuity: Q = A1 V1 = A2 V2 ⇒ V2 = A1 V1 /A2 .
Assuming frictionless uniform flow we can relate pressures at inlet and outlet
by Bernoulli equation:
1 1
P2 = P1 + ρ V12 − ρ V22 .
2 2
Pressures at inlet and outlet are different, which means the there is pressure
contribution into a net force on the control volume. Therefore the total force
on the fluid in the control volume includes the pressure component and the
reaction from the contraction.
Momentum principle:

Fx = −X + (P1 A1 − P2 A2 ) = ρ Q (V2 − V1 ).

Force on the contraction:

X = −ρ Q (V2 − V1 ) + (P1 A1 − P2 A2 ).

Nozzle
Formulation:
A pipe with cross section area A1 terminates in a conical nozzle from which
a jet of water is discharged to atmosphere. The cross section of the jet is A2
and the gauge pressure before the nozzle is P1 . Find the total force acting
 11

on the nozzle.
Solution:
Pressure across the jet is constant and equal to the atmospheric pressure
(gauge pressure is 0). Therefore, the gauge pressure on the right boundary
of the control volume is constant and equal to the atmospheric pressure:
P2 = 0.
Continuity:
A1 V1 = A2 V2 = Q (3)
Bernoulli:
1 1
ρ V12 = P2 + ρ V22 .
P1 + (4)
2 2
With known P1 and P2 = 0 simultaneous solution of (3) and (4) gives values
of V1 and V2 .
The total force on the control volume includes the pressure component and
the reaction from the nozzle.
Momentum principle:

Fx = −X + P1 A1 = ρ Q (V2 − V1 ).

Force on the nozzle:

X = −ρ Q (V2 − V1 ) + P1 A1 .

Pipe bend
Formulation:
A pipe with a flowing water bends through an angle θ and reduces cross
section from A1 to A2 . The pressure and velocity before the bend are P1 and
A1 respectively. Find the magnitude and direction of the force exerted on
the bend by the water.
Solution:
Continuity: Q = A1 V1 = A2 V2 ⇒ V2 = A1 V1 /A2 .
1 1
Bernoulli: P2 = P1 + ρ V12 − ρ V22 .
2 2
Velocity components:

x-direction: u1 = V1 ; u2 = V2 cos(θ)
y-direction: v1 = 0; v2 = V2 sin(θ)
 12

A2
V2
θ

Fy
P2
Fx
A1 V1
y P1

Fig. 9:

Total force components on the control volume:

x-direction: Fx = −X + P1 A1 − P2 A2 cos(θ)
y-direction: Fy = −Y − P2 A2 sin(θ)
Momentum principle:
x-direction: Fx = ρ Q (u2 − u1 )
y-direction: Fy = ρ Q (v2 − v1 )
Force components on the bend:
x-direction: X = P1 A1 − P2 A2 cos(θ) − ρ Q (V2 cos(θ) − V1 )
y-direction: Y = −P2 A2 sin(θ) − ρ Q V2 sin(θ)
√
Force magnitude: R = X2 + Y 2
Force direction (angle from x-axis): α = sin−1 (Y /R)

Body drag
Formulation:
A cylindrical model is placed across a working section of a wind tunnel of
height 4d (figure 10). Air flow at the inlet of the working section can be
assumed parallel and uniform with velocity V1 . At the outlet of the working
section, where the flow is also parallel, the velocity profile is measured. It
consists of a viscous wake of thickness 2d with profile
Vw = V2 ( 1 + 2 (y/d)2 − (y/d)4 )/2,
 13

y=2d
y
V1 V2
y=d
P1
D x Vwake =V(y)

y=−d
1
2
y=−2d

Fig. 10:

and a uniform flow outside of the wake with velocity V2 . Assuming the flow
to be two-dimensional, find the drag per one meter of span of the model.
Solution:
Continuity:
Flow rates through inlet and outlet are the same:

Zd
Q = 4 d V1 = 2 d V2 + V2 ( 1 + 2 (y/d)2 − (y/d)4 )/2 dy.
−d

15
After taking the integral we find the value of V2 : V2 = 13
V1 .

Bernoulli:
We can apply Bernoulli equation in the inviscid region outside of the viscous
wake to specify the pressures at the outlet. The flow here is parallel, and the
pressure will be the same across the whole section. For the stream line 1–2
we have:
ρ V12 ρ V22
P1 + = P2 + ,
2 2
and substituting the known values of velocities we have the following pressure
28
difference: P1 − P2 = 169 ρ V12 .
Momentum:
We neglect friction on the walls of the wind tunnel. Then the total force on
the fluid inside the control volume includes the pressure forces at the inlet
and outlet of the control volume and the reaction due to drag of the model
R = −D. The flow at the outlet is not uniform, therefore, we have to apply
 14

the momentum principle in the general integral form. Momentum flow rate
at inlet:
Z2d
I1 = ρ V12 dy = 4d ρ V12 .
−2d

At the outlet the integral for momentum flow rate consists of tree parts:

Z−d Zd Z2d
I2 = ρ V22 dy + ρ V22 ( 1 + 2 (y/d)2 − (y/d)4 )2 /4 dy + ρ V22 dy,
−2d −d d

After taking the integrals and substituting the value for V2 this gives gives:
380
I2 = d ρ V12 .
91
Now we can apply the momentum principle:

Fx = 4 d (P1 − P2 ) − D = I2 − I1 ,

and substituting all values we obtain:

576
D= d ρ V12
1183