Unit 41: Fluid Mechanics

Unit code: T/601/1445

QCF Level: 4
Credit value: 15

OUTCOME 4

TUTORIAL 6 – MOMENTUM AND PRESSURE FORCES

4 Understand the operating principles of hydraulic machines

Impact of a jet: power of a jet; normal thrust on a moving flat vane; thrust on a moving
hemispherical cup; velocity diagrams to determine thrust on moving curved vanes; fluid
friction losses; system efficiency
Operating principles of turbines: operating principles, applications and typical system
efficiencies of common turbo-machines including the Pelton wheel, Francis turbine and Kaplan
turbine
Operating principles of pumps: operating principles and applications of reciprocating and
centrifugal pumps; head losses; pumping power; power transmitted; system efficiency

This is another major outcome requiring a lot of study time and the tutorial probably contains more
than required.

CONTENTS

1. PRESSURE FORCES

2. MOMENTUM FORCES
2.1. Newton’s Second Law
2.2. Application to Pipes Bends

3. COMBINED PRESSURE AND MOMENTUM FORCES

4. APPLICATION TO STATIONARY VANES

4.1. Flat Plate Normal to Jet
4.2. Flat Plate at Angle to Jet
4.3. Curved Vanes

5. MOVING VANES
5.1. Moving Flat Plane
5.2. Moving Curved Vane

Let’s start with forces due to changes in the pressure of the fluid.
1. PRESSURE FORCES

Consider a duct as shown in fig.1. First identify the control volume on which to conduct a force
balance. The inner passage is filled with fluid with pressure p1 at inlet and p2 at outlet. There will
be forces on the outer surface of
the volume due to atmospheric
pressure. If the pressures of the
fluid are measured relative to
atmosphere (i.e. use gauge
pressures) then these forces
need not be calculated and the
resultant force on the volume is
due to that of the fluid only. The
approach to be used here is to
find the forces in both the x and
y directions and then combine
them to find the resultant force.

Fig.1
The force normal to the plane of the bore is pA.
At the inlet (1) the force is Fp1= p1A1
At the outlet (2) the force is Fp2 = p2A2 These forces must be resolved vertically and horizontally to
give the following. Fpx1 = Fp1 cos θ1 (to the right) Fpx2 = Fp1 cos θ2 (to the left) The total
horizontal force FH = Fpx1 -Fpx2
Fpy1 = Fp1 sin θ1 (up) Fpy2 = Fp2 sin θ2 (down) The total vertical force FV = Fpy1 -Fpy2

WORKED EXAMPLE No. 1

2
A nozzle has an inlet area of 0.005 m and it discharges into the atmosphere. The inlet gauge
pressure is 3 bar. Calculate the resultant force on the nozzle.

Fig.2
SOLUTION

Since the areas are only in the vertical plane, there is no vertical force. FV = 0

Using gauge pressures, the pressure force at exit is zero. Fpx2 = 0

Fpx1 = 3 x 105 x 0.005 = 1500 N
FH = 1500 – 0 = 1500 N to the right.
 WORKED EXAMPLE No. 2

The nozzle shown has an inlet area of 0.002 m2 and an outlet area of 0.0005 m2. The inlet
gauge pressure is 300 kPa and the outlet gauge pressure is 200 kPa. Calculate the horizontal
and vertical forces on the nozzle.

Fig.3
SOLUTION
Fp1= 300 x 103 x 0.002 = 600 N
Fpx1= 600 N
Fpy1=0 N since the plane is vertical.
Fp2 = 200 x 103 x 0.0005 = 100 N
Fpx2 = 100 x cos 60o = 50 N
Fpy2 = 100 x sin 60o = 86.67 N

Total Horizontal force FH = 600 - 50 = 550 N

Total vertical force FV = 0 - 86.67 N = - 86.67 N

2. MOMENTUM FORCES

When a fluid speeds up or slows down, inertial forces come into play. Such forces may be produced
by either a change in the magnitude or the direction of the velocity since either change in this vector
quantity produces acceleration.
For this section, we will ignore pressure forces and just study the forces due to velocity changes.

2.1 NEWTON'S 2nd LAW OF MOTION

This states that the change in momentum of a mass is equal to the impulse given to it. Impulse =
Force x time Momentum = mass x velocity Change in momentum = ∆mv
Newton’s second law may be written as ∆mv = Ft Rearrange to make F the subject. ∆mv/t = F
Since ∆v/t = acceleration ‘a’ we get the usual form of the law F = ma The mass flow rate is m/t and
at any given moment this is dm/dt or m' and for a constant flow rate,
only the velocity changes.
In fluids we usually express the second law in the following form. F = (m/t) ∆v = m'∆v
m'∆v is the rate of change of momentum so the second law may be restated as

F = Rate of change of momentum

F is the impulsive force resulting from the change. ∆v is a vector quantity.
2.2 APPLICATION TO PIPE BENDS

Consider a pipe bend as before and use the idea of a control volume.

Fig.4

First find the vector change in velocity using trigonometry. tanΦ= v2sinθ/(v2cosθ-v1)
2 2 ½
∆v= {(v2 sinθ) + (v2 cosθ - v1) } Alternatively ∆v could be found by drawing the diagram to
scale and measuring it. If we had no change in magnitude then v1 = v2 = v then
½
∆v= v {2(1 -cosθ)}

The momentum force acting on the fluid is Fm = m'∆v The force is a vector quantity which must be
in the direction of ∆v. Every force has an equal and opposite reaction so there must be a force on
the bend equal and opposite to the force on the fluid. This force could be resolved vertically and
horizontally such that FH = FmcosΦ and FV = FmsinΦ

This theory may be applied to turbines and pump blade theory as well as to pipe bends.

SELF ASSESSMENT EXERCISE No. 1

1. A pipe bends through an angle of 90o in the vertical plane. At the inlet it has a cross sectional
area of 0.003 m2 and a gauge pressure of 500 kPa. At exit it has an area of 0.001 m 2 and a
gauge pressure of 200 kPa.
Calculate the vertical and horizontal forces due to the pressure only. (200 N and 1500 N).

2. A pipe bends through an angle of 45o in the vertical plane. At the inlet it has a cross sectional
area of 0.002 m2 and a gauge pressure of 800 kPa. At exit it has an area of 0.0008 m 2 and a
gauge pressure of 300 kPa.
Calculate the vertical and horizontal forces due to the pressure only. (169.7 N and 1430 N).

3. Calculate the momentum force acting on a bend of 130o which carries 2 kg/s of water at 16 m/s
velocity.
Determine the vertical and horizontal components. (Ans.24.5 N and 52.6 N)

4. Calculate the momentum force on a 180o bend that carries 5 kg/s of water. The pipe is 50 mm
bore diameter throughout. The density is 1000 kg/m3.
(Ans. 25.46 N )

5. A horizontal pipe bend reduces from 300 mm bore diameter at inlet to 150 mm diameter at
outlet. The bend is swept through 50o from its initial direction. The flow rate is 0.05 m3/s and
the density is 1000 kg/m3. Calculate the momentum force on the bend and resolve it into two
perpendicular directions relative to the initial direction. (Ans.108.1 N and 55.46 N).
3. COMBINED PRESSURE AND MOMENTUM FORCES

Now we will look at problems involving forces due to pressure changes and momentum changes at
the same time. This is best done with a worked example since we have covered the theory already.

WORKED EXAMPLE No.3

A pipe bend has a cross sectional area of 0.01 m2 at

inlet and 0.0025 m2 at outlet. It bends 90o from its
initial direction.
The velocity is 4 m/s at inlet with a pressure of
100 kPa gauge. The density is

Fig.5
SOLUTION

v1 = 4m/s Since A1v1 = A2v2 then v2 = 16 m/s

We need the pressure at exit. This is done by applying Bernoulli between (1) and (2) as
follows.
p1 + ½ v12 = p2 + ½ v22
100 x 103 + ½ 1000 x 42 = p2 + 1000 x ½ 162
p2 = 0 kPa gauge

Now find the pressure forces.

Fpx1 = p1A1 = 1200 N

Fpy2 = p2A2 = 0 N Next solve the momentum forces.
m' = Av = 40 kg/s
v = (42 + 162)½ = 16.49 m/s
Fm = m'v = 659.7 N
 = tan-1(16/4) = 75.96o

RESOLVE Fig.6

Fmy = 659.7 sin 75.96 = 640 N Fmx = 659.7 cos 75.96 = 160 N

Total forces in x direction = 1200 + 160 = 1360 N

Total forces in y direction = 0 + 640 = 640 N

ALTERNATIVE SOLUTION

Many people prefer to solve the complete problems by solving pressure and momentum forces
in the x or y directions as follows.

x direction m'v1 + p1A1 = FX = 1200 N

y direction m'v2 + p2A2 = FY = 640 N

When the bend is other than 90o this has to be used more carefully because there is an x
component at exit also.
4. APPLICATIONS TO STATIONARY VANES

When a jet of fluid strikes a stationary vane , the vane decelerates the fluid in a given direction.
Even if the speed of the fluid is unchanged, a change in direction produces changes in the velocity
vectors and hence momentum forces are produced. The resulting force on the vane being struck by
the fluid is an impulsive force. Since the fluid is at atmospheric pressure at all times after leaving
the nozzle, there are no forces due to pressure change.

4.1 FLAT PLATE NORMAL TO JET

Fig.7

The velocity of the jet leaving the nozzle is v1. The jet is decelerated to zero velocity in the original
direction. Usually the liquid flows off sideways with equal velocity in all radial directions with no
splashing occurring. The fluid is accelerated from zero in the radial directions but since the flow is
equally divided no resultant force is produced in the radial directions. This means the only force on
the plate is the one produced normal to the plate. This is found as follows.

m' = mass flow rate. Initial velocity = v1.

Final velocity in the original direction = v2 = 0.
Change in velocity =v = v2 – v1= - v1
Force = m'v = -mv1

This is the force required to produce the momentum changes in the fluid. The force on the plate
must be equal and opposite so

F = m'v1 = A v1

WORKED EXAMPLE No.4

A nozzle has an exit diameter of 15 mm and discharges water into the atmosphere. The gauge
pressure behind the nozzle is 400 kPa. The coefficient of velocity is 0.98 and there is no
contraction of the jet. The jet hits a stationary flat plate normal to its direction. Determine the
force on the plate. The density of the water is 1000 kg/m 3. Assume the velocity of approach
into the nozzle is negligible.

SOLUTION

The velocity of the jet is v1 = Cv(2p/)½

½
v1 = 0.98 (2 = 27.72 m/s
The nozzle exit area A =  x 0.0152/4 = 176.7 x 10-6 m2.
The mass flow rate is Av1 = 1000 x 176.7 x 10-6 x 27.72 = 4.898 kg/s.
The force on the vane = 4.898 x 27.72 = 135.8 N
4.2 FLAT PLATE AT ANGLE TO JET

If the plate is at an angle as shown in fig. 4.9 then

the fluid is not completely decelerated in the
original direction but the radial flow is still equal
in all radial directions. All the momentum normal
to the plate is destroyed. It is easier to consider the
momentum changes normal to the plate rather than
normal to the jet.

Fig. 8
Initial velocity normal to plate = v1 cos.
Final velocity normal to plate = 0.
Force normal to plate = m'v =0 - A v1 cos.
This is the force acting on the fluid so the force on the plate is

m' v1 cos or A v12 cos.

If the horizontal and vertical components of this force are required then the force must be resolved.

WORKED EXAMPLE No. 5

A jet of water has a velocity of 20 m/s and flows at 2 kg/s. The jet strikes a stationary flat plate.
The normal direction to the plate is inclined at 30 o to the jet. Determine the force on the plate
in the direction of the jet.

SOLUTION

Fig. 9

The force normal to the plate is mv1 cos o = 34.64 N.

The force in the direction of the jet is found by resolving.

FH = F/cos30o = 34.64/cos 30o = 40 N

4.3 CURVED VANES

When a jet hits a curved vane, it is

usual to arrange for it to arrive on
the vane at the same angle as the
vane. The jet is then diverted from
with no splashing by the curve of
the vane. If there is no friction
present, then only the direction of
the jet is changed, not its speed.

Fig. 10
This is the same problem as a pipe bend with uniform
size. v1 is numerically equal to v2.

If the deflection angle is  as shown in figs. 10 and 11

then the impulsive force is

F = m'v = m' v1{2(1 - cos)}1/2 Fig. 11

The direction of the force on the fluid is in the direction of v and the direction of the force on the
vane is opposite. The force may be resolved to find the forces acting horizontally and/or vertically.

It is often necessary to solve the horizontal force and this is done as follows.

Fig.12

Initial horizontal velocity = vH1 = v1

Final horizontal velocity = vH2 = -v2 cos (180 - ) = v2 cos 
Change in horizontal velocity = vH1
Since v2 = v1 this becomes vh= {v2 cos - v1 } = v1{cos - 1}
Horizontal force on fluid = m'v1{cos - 1}
The horizontal force on the vane is opposite so

Horizontal force = m'vH = m'v1{1 - cos}

 WORKED EXAMPLE No. 6

A jet of water travels horizontally at 16 m/s with a flow rate of 2 kg/s. It is deflected 130 o by a
curved vane. Calculate resulting force on the vane in the horizontal direction.

SOLUTION
½
The resulting force on the vane is F = m' v1{2(1 - cos)
½
F = 2 x 16 {2(1 -cos 130o)} = 58 N

The horizontal force is

FH = m' v1{cos - 1}
FH = 2 x 16 x (1 - cos130)
FH = 52.6 N

SELF ASSESSMENT EXERCISE No. 2

Assume the density of water is 1000 kg/m 3 throughout.

1. A pipe bends through 90o from its initial direction as shown in fig. 4.7. The pipe reduces in
diameter such that the velocity at point (2) is 1.5 times the velocity at point (1). The pipe is 200
mm diameter at point (1) and the static pressure is 100 kPa. The volume flow rate is 0.2 m 3/s.
Assuming that there is no friction calculate the following.

a) The static pressure at (2).

b) The velocity at (2).

c) The horizontal and vertical forces on the bend FH and FV.

d) The total resultant force on the bend.

Fig. 13
2. A nozzle produces a jet of water. The gauge pressure behind the nozzle is 2 MPa. The exit
diameter is 100 mm. The coefficient of velocity is 0.97 and there is no contraction of the jet.
The approach velocity is negligible. The jet of water is deflected 165 o from its initial direction
by a stationary vane. Calculate the resultant force on the nozzle and on the vane due to
momentum changes only. (29.5 kN and 58.5 kN).

3. A stationary vane deflects 5 kg/s of water 50 o from its initial direction. The jet velocity is 13
m/s. Draw the vector diagram to scale showing the velocity change. Deduce by either scaling or
calculation the change in velocity and go on to calculate the force on the vane in the original
direction of the jet. (49.8 N).

4. A jet of water travelling with a velocity of 25 m/s and flow rate 0.4 kg/s is deflected 150 o from
its initial direction by a stationary vane. Calculate the force on the vane acting parallel to and
perpendicular to the initial direction. (Ans.18.66 N and 5 N)

5. A jet of water discharges from a nozzle 30 mm diameter with a flow rate of 15 dm 3/s into the
atmosphere. The inlet to the nozzle is 100 mm diameter. There is no friction nor contraction of
the jet. Calculate the following.

a) The jet velocity.

b) The gauge pressure at inlet.
c) The force on the nozzle.

The jet strikes a flat stationary plate normal to it. Determine the force on the plate.

5. MOVING VANES

When a vane moves away from the jet as shown on fig.4.14, the mass flow arriving on the vane is
reduced because some of the mass leaving the nozzle is producing a growing column of fluid
between the jet and the nozzle. This is what happens in turbines where the vanes are part of a
revolving wheel. We need only consider the simplest case of movement in a straight line in the
direction of the jet.

5.1 MOVING FLAT PLATE

The velocity of the jet is v and the velocity of the

vane is u. If you were on the plate, the velocity of
the fluid arriving would be v - u. This is the
relative velocity, that is, relative to the plate. The
mass flow rate arriving on the plate is then

m' = A(v-u)

Fig. 14

The initial direction of the fluid is the direction of the jet. However, due to movement of the plate,
the velocity of the fluid as it leaves the edge is not at 90 o to the initial direction. In order to
understand this we must consider the fluid as it flows off the plate. Just before it leaves the plate it
is still travelling forward with the plate at velocity u. When it leaves the plate it will have a true
velocity that is a combination of its radial velocity and u. The result is that it appears to come off
the plate at a forward angle as shown.
We are only likely to be interested in the force in the direction of movement so we only require the
change in velocity of the fluid in this direction.

The initial forward velocity of the fluid = v

The final forward velocity of the fluid = u
The change in forward velocity = v-u
The force on the plate = m'v = m' (v-u)
Since m' = A(v-u) then the force on the plate is
F = A(v-u)2

5.2 MOVING CURVED VANE

Turbine vanes are normally curved and the fluid

joins it at the same angle as the vane as shown in
the diagram.

The velocity of the fluid leaving the nozzle is v1.

This is a true or absolute velocity as observed by
anyone standing still on the ground.
Fig.15

The fluid arrives on the vane with relative velocity v1-u as

before. This is a relative velocity as observed by someone
moving with the vane. If there is no friction then the
velocity of the fluid over the surface of the vane will be v 1-
u at all points. At the tip where the fluid leaves the vane, it
will have two velocities. The fluid will be flowing at v1-u
over the vane but also at velocity u in the forward direction.
The true velocity v2 at exit must be the vector sum of these
two.
Fig. 16

If we only require the force acting on the vane in the direction of movement then we must find the
horizontal component of v2. Because this direction is the direction in which the vane is whirling
about the centre of the wheel, it is called the velocity of whirl v w2. The velocity v1 is also in the
direction of whirling so it follows that v1 = vw1.

Vw2 may be found by drawing the vector diagram (fig.4.16) to scale or by using trigonometry. In
this case you may care to show for yourself that vw2 = u + (v1-u)(cos)

The horizontal force on the vane becomes FH = m' (vw1-vw2) = m' (v1-vw2)

You may care to show for yourself that this simplifies down to Fh = m'(v1-u)(1-cos)
This force moves at the same velocity as the vane. The power developed by a force is the product of
force and velocity. This is called the Diagram Power (D.P.) and the diagram power developed by a
simple turbine blade is
D.P. = m'u(v1-u)(1-cos)

This work involving the force on a moving vane is the basis of turbine problems and the geometry
of the case considered is that of a simple water turbine known as a Pelton Wheel. You are not
required to do this in the exam. It is unlikely that the examination will require you to calculate the
force on the moving plate but the question in self assessment exercise 5 does require you to
calculate the exit velocity v2.
 WORKED EXAMPLE No. 7

A simple turbine vane as shown in fig.15 moves at 40 m/s and has a deflection angle of 150 o.
The jet velocity from the nozzle is 70 m/s and flows at 1.7 kg/s.

Calculate the absolute velocity of the water leaving the vane and the diagram power.

SOLUTION

Drawing the vector diagram (fig.4.15) to scale, you may show that v2 = 20.5 m/s. This may also
be deduced by trigonometry. The angle at which the water leaves the vane may be measured
from the diagram or deduced by trigonometry and is 46.9o to the original jet direction.

D.P. = m'u(v1-u)(1+cos) = 1.7 x 40(70-40)(1 - cos 150) = 3807 Watts

SELF ASSESSMENT EXERCISE No. 3

1. A vane moving at 30 m/s has a deflection angle of 90 o. The water jet moves at 50 m/s with a
flow of 2.5 kg/s. Calculate the diagram power assuming that all the mass strikes the vane.
(1.5 kW).

2. Figure 10 shows a jet of water 40 mm diameter flowing at 45 m/s onto a curved fixed vane. The
deflection angle is 150o. There is no friction. Determine the magnitude and direction of the
resultant force on the vane.

The vane is allowed to move away from the nozzle in the same direction as the jet at a velocity
of 18 m/s. Draw the vector diagram for the velocity at exit from the vane and determine the
magnitude and direction of the velocity at exit from the vane.