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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
DYMAMICS OF FLUID FLOW
BASIC EQUATIONS (INTEGRAL FORM)

The study of fluid at rest is known as “Fluid Static”. When the fluids are at rest,
the only fluid property of significance is the specific weight of the fluids.

While in motion, various other fluid properties become significant. The science,
which deals with the geometry of the motion of the fluids without reference to
forces causing the motion, is known as “Fluid Kinematics”. The description of the
fluid motion is in terms of space-time relationship.

The science that deals with the action of the forces in producing or changing the
motion of the fluid is called “Fluid Kinematics”.

The dynamics of fluid flow is the study of fluid motion with forces causing the
fluid flow. The dynamic behavior of the fluid flow is analyzed by Newton’s
second law of motion.
Continuity Equation (Conservation of Mass)
The “control volume (CV)” is a finite region in space in which the attention is focused.
The boundary surface of this control volume is called the “control surface (CS)”. So,
conservation of mass for a control volume can be stated as,
Time rate of change of the mass of the system = 0, i.e.
DM sys
Dt
 0 where M sys 
  dV
(1)
sys
or,
D

 dV    dV    V.nˆ dA

Dt sys
t CV
CS

where,


  dV is the time rate of change of mass in the CV and
t CV
(2)
   V.nˆ  dA is the
CS
net mass flow through the CS and is given by,
   V.nˆ  dA   m
out
 min
(3)
CS
So, the general expression for continuity equation is,
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati

  dV  CS  V.nˆ dA  0
t CV
 
(4)
In some special cases,

When the flow is uniformly distributed over the opening of the control surface
(one-dimensional flow), the expression for mass flow rate is given by,
m Q   AV
(5)
where  is the fluid density, Q is the volume flow arte and V is the component of
fluid velocity perpendicular to area A . In case, the density changes (as in the case of
compressible flows), the average value of the component of velocity normal to the
area is considered and is defined as,
V

  V .nˆ  dA
A
A
(6)
When the flow is steady, the time rate of change of the mass of contents in the CV
is zero, so that
m
out

 min
(7)
For steady flow involving only one stream of specific fluid flowing through the
CV at section (1) and (2),
m  1 A1 V1  2 A2 V2

(8)
In case of incompressible flow,  is constant. So,
Q  A1 V1  A2 V2
(9)
Momentum Equation (Newton’s second law)
The Newton’s second law of motion for a system states that “the time rate of change of
the linear momentum of the system is equal to the sum of external forces acting on the
system”. Mathematically, it may be stated as,
D
 V dV   Fsys
Dt sys
(10)
Using “Reynolds Transport Theorem”, the left hand side of the above equation can be
written as,
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
D

V dV 

 V dV  CS V V.nˆ dA
Dt sys
t CV
(11)
or,
Time rate of change of linear momentum of the system = Time rate of change of linear
momentum of the contents of the control volume + Net rate of flow of linear momentum
through the control surface.
The right hand side of Eq. (10) i.e.
F
sys
is the vector sum of all the forces acting
on the control-volume. It includes surface forces on all fluids and solids intersected by
the control surface plus all body forces acting on the masses within the control volume.
For one dimensional momentum flux, a simplified relation is obtained from Eqs (10) and
(11) i.e.
F
sys


 V dV  CS VV.nˆ dA
t CV
(12)
The Eq. (10) is a vector relation and has the components in x, y and z direction. If
the flow is steady, then the time rate of change of linear momentum of the control volume
is zero i.e.

 V dV  0 . So, the Eq. (12) can be further simplified.
t CV
Moving control volumes
In most of the problems in fluid mechanics, the control volume is considered as a fixed
volume in space through which the fluid flows. There are certain situations for which the
analysis becomes simplified if the control volume is allowed to move or deform. The
main difference between the fixed and the moving control volumes is as follows;

 
It is the relative velocity W that carries fluid across the control surface of the
moving CV.

 
In case of moving CV, the absolute velocity V carries the fluid across the fixed
control surface.

The difference between the absolute and relative velocities is the velocity of CV
 V  i.e.
CV
V = W + VCV
(13)
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Eq. (11) can thus be written as,
D

V dV 

 V dV  CS V W.nˆ dA
Dt sys
t CV
(14)
Using Eq. (13) and (14), Eq. (10) can be expressed as,

 W+VCV  dV  CS W  VCV  W.nˆ dA   Fsys
t CV
(15)

 W+VCV  dV  CS W  W.nˆ dA  VCV CS  W.nˆ dA   Fsys
t CV
(16)




or,


In case of steady flow, the first and third term on the left hand side of the above equation
becomes zero. So, linear momentum equation for a moving, non-deforming CV involving
steady flow becomes,
 W W.nˆ dA   F
sys
(17)
CS
The linear momentum equation is very useful in engineering applications. However,
some specific applications related to vanes and pipe bends are discussed in subsequent
examples.
Example-1
Air flows steadily in a long cylindrical pipe of 15cm diameter. The pressure and
temperature are measured between two sections 1 and 2 of the pipe; at section 1, the
pressure and temperatures are 7 bar and 300 K respectively. The corresponding values at
section 2 are 1.2 bar and 250 K respectively. If the average air velocity at section 2 is
300m/s, find the velocity of air at section 1.
Solution:
The continuity equation is given by,

  dV  CS  V. n dA  0
t CV
Since the flow is steady, the first term of the above equation is zero. Hence,
  V. n dA  m
2
 m1  0
CS
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
i.e. m2  m1 or 1 A1 V1   2 A2 V2
V1 
so that,
2
p T
V2  2 1 V2
1
p1 T2
Here, p1  7bar; T1  300K; p2  1.2bar; T2  250K; V2  300m/s
So, V1 
1.2  300
 300  61.7m/s
7  250
Example 2 (Stationary vanes)
A horizontal jet of water strikes a vane and is turned at an angle  as shown in the Ex.
Fig. 1. The cross-sectional area and velocity at the inlet of the vane are 60cm 2 and 5m/s
respectively. Neglecting the gravity and viscous effects, determine the anchoring force
required to hold the vane stationary.
A1 = 60cm2
V1 = 5m/s
Vane

(2)
y
(1)
x
Ex. Fig. 1: Forces due to water jet.
Solution:
The control volume (CV) is selected that includes the vane and a portion of the water as
shown in Ex. Fig. 2. Since water enters and leaves CV as a free jet at atmospheric
pressure, so the speed of the jet remains constant as 5m/s (Bernoulli’s equation; to be
discussed later).
Hence, A1  A2 (By, continuity equation)
At section 1, u  V1 , v  0 (-vesign is because of water entering theCV) and at section 2,
u  V1 cos  , v  V1 sin 
Now, apply the linear momentum equation to this fixed CV. Then, horizontal and vertical
component of anchoring force can be written as,
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Control volume
(2)
V2 = 5m/s

(1)
y
x
Fx
Fy
Ex. Fig. 2: Forces due to water jet.
Fx  V1. .  V1  . A1  V1 cos  . . V1  . A2
Fy   0  . .  V1  . A1  V1 sin  . . V1  . A2
The above equations can be simplified as,
Fx    . A1.V12 . 1  cos  
Fy   . A1.V12 .sin 
With the given data and taking the density of water as 1000kg/m3, anchoring forces can
be expressed in terms of  as,
Fx  150 1  cos   N
Fy  150sin  N
In some extreme cases,

If   00 ; Fx  Fy  0 , i.e. water does not turn in the vane and the anchoring force is
zero. The fluid only slides without applying any force on it.

If   900 ; Fx  150N and Fy  150N . These forces are necessary to push the vane
to the left and up in order to change the direction of water flow from horizontal to
vertical.

If   1800 ; Fx  300N and Fy  0 , the water jet turns back on itself
Example 3 (Moving vanes)
The same vane of length 1m as shown in Ex. Fig. 3 moves at a velocity of 30cm/s where
the jet of water enters the vane at 150m/s. If the water jet turns by 450, determine the
magnitude and direction of the force exerted by the stream of water on the vane surface.
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Vane
A1 = 60cm2
V1

(2)
y
(1)
V0
x
Ex. Fig. 3: Forces on a moving vane.
Solution:
The magnitude and direction of the force  F  , exerted by the water on the vane can be
estimated by using Eq. (17).
(2)
V2
Control volume

Ry
(1)
Rx
y
m
W
V0
x
Vcv = V0
Ex. Fig. 4: Forces on a moving vane.
Referring to Ex. Fig. 4, if R is the reaction force, then Eq. (17), can be rewritten as,
W
x
 W .nˆ dA   Rx
CS
W
y
 W .nˆ dA  Ry  ww
CS
where ww is the weight of the water in the moving CV. The above equations are
simplified as,
 W1  m1    W2 cos 450   m2    Rx
 W2 sin 450   m2   Ry  ww
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Since, the ambient pressure is atmospheric, so, the speed of the water relative to moving
CV is constant i.e. W1  W2 . Density of water at inlet and exit can be taken as,
1   2  1000 kg m3 and ww  .g. A1.l . By continuity equation,
m1  m2  1.W1. A1  2 .W2 . A2
So, the reaction forces can be written as,

Rx   .W12 . A1. 1  cos 450

Ry   .W12 . A1.sin 450  .g. A1.l
Relative speed of water entering the CV, W1  = 150 - 30 = 120 cm/s =1.2 m/s.
Hence,
Rx  1000. 1.2  .  60 104 1  cos 450   2.53N
2
Ry  1000. 1.2  .  60 104  .sin 450  1000  9.81  60 104  .(1m)  64.96N
2
Combining the components, the resultant force is given by,
R  Rx2  Ry2  65.05N
 Ry 
0
  88
 Rx 
  tan 1 
The angle of R from x-direction is,
Example 4 (Pipe bends)
Water flows horizontally through a constant area pipe with cross-sectional area of
0.001m2 and with 1800 bend as shown in Ex. Fig. 5. The flow is axial and its velocity is
5m/s throughout. The absolute pressures at the entrance and exit of the bend are 2.2 bar
and 1.6 bar respectively. Calculate the anchoring force required to hold the bend in place.
Assume the atmospheric pressure as 1 bar.
Cross-sectional area = 0.01m2
(1)
V1 = 15m/s
(2)
y
V2 = 15m/s
v
u
x
Ex. Fig. 5: Water flow in a 1800 pipe-bend.
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Solution:
In order to find the anchoring force to hold the pipe bend in place, it is required to choose
the control volume (Ex. Fig. 6).
Cross-sectional area = 0.01m2
(1)
p1, A1, V1
Fy
(2)
Fx
p 2, A 2, V 2
Control volume y
v
W
u
x
Ex. Fig. 6: Water flow in a 1800 pipe-bend.
In this figure, all the forces exerted on the fluid and pipe bend are resolved and combined
into two resultant components, Fx and Fy . At, section (1) and (2), the flow is in xdirection, so there will be no y-direction momentum flow into or out of the CV, rather the
weight of the water will be balanced by the force in y-direction. Now, the CV reduces to
one-dimensional flow for which the momentum equation can be written as,
 v1  m1    v2  m2   Fx  p1.A1  p2 .A2
By continuity equation,
m1  m2  m  1.A1.v1  1000 0.001 5  5kg s
Fx  m  v1  v2   p1.A1  p2 .A2
Solving for Fx ,
Fx   5  5  5  0.001 p1  patm    p2  patm 
  5  5  5  0.001 2.2  1.6  2  105 
 230N
Now, the anchoring force is the force that the bend puts on the fluid inside of it. It can be
found by choosing the CV as shown in Ex. Fig. 7 and is given by,
Rx  Fx  patm  A1  A2   230 1105 (0.001  0.001)  430N
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
Cross-sectional area = 0.01m2
(1)
p1, A1, V1
patm (A1+A2)
(2)
p2, A2, V2
Rx
Fx
y
Control volume
v
u
x
Ex. Fig. 7: Water flow in a 1800 pipe-bend.
EXERCISES
1. Water is pumped from a reservoir and flows steadily through a nozzle. The exit
diameter of the nozzle is 5cm. Determine the pumping capacity so that the velocity at
the nozzle exit does not fall below 25m/s.
2. A rectangular tank is filled with water at steady rate of 30litres/min from a tap. The
size of the tank is 12cm5cm4cm. Estimate the time rate of change of depth of
water at any instant.
3. A tank of volume 0.06m3 contains air at 700 kPa and 250C. The air begins to escape
through a circular valve of 10mm diameter. The speed of the air through the valve is
300m/s and the density at that instant is 6 kg/m3. Determine the instantaneous rate of
change of density in the tank.
4. Consider the incompressible, laminar water flow in a straight pipe of radius R . At
certain section of the pipe, the flow velocity is U and the profile is parallel to the
pipe axis. At some other section, the velocity profile is axi-symmetric and parabolic
i.e. zero velocity at the pipe wall and maximum at the centerline  umax  . Find
(i) the relationship between U and umax .
(ii) average velocity at section 2.
5. An aircraft is cruising at 900 km/hr in an altitude where the density of air is 0.75
kg/m3. The engine intake area is 0.75 m2 and the exhaust area is 0.6 m2. The exhaust
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
gas comes out of the engine at a velocity of 1000 km/hr and density of 0.6 kg/m3.
Estimate the mass flow rate of the fuel into the engine.
6. The water is to be spread in a garden through a lawn sprinkler having two nozzles as
shown in the following figure. The exit area of each nozzle is 40mm 2. The water
enters at the base of the sprinkler at a rate 1.5 liters/s. Determine the average speed of
water in each of the nozzle, if
(i)
the sprinkler head is stationary
(ii)
the sprinkler head rotates at 500rpm.
(iii)
the sprinkler head accelerates from 0 to 500 rpm.
Sprinkler head
Nozzle
(section 2)
section 1
Nozzle
(section 2)
Q
Q
7. A syringe consisting of a plunger and cylinder is used to inject medicines into animals
as shown in the following figure. The plunger has a face area of 400mm2. The liquid
in the syringe is to be injected at the rate 200 cm3/min. The leakage rate is 8% of the
volume flow arte out of the needle. Determine the speed advancement of the plunger.
L
Qleak
Area = 400 mm2
VP
Q
Plunger motion
Qleak
8. Air flows steadily in a straight, long pipe with inside diameter of 8cm. The pressure
temperature and velocity are measured between two sections (1) and (2) of the pipe
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 21
Dr. Niranjan Sahoo
IIT-Guwahati
as, p1  7bar , T1  300K , V1  70 m s; p2  1.2bar , T2  250K , V2  300 m s .
Assuming uniform velocity distributions between two sections, determine the
frictional force exerted by the pipe wall on the airflow.
9. Air enters to the engine of an aircraft at a velocity of 220m/s and static pressure of
0.7bar (absolute). The static temperature and cross-sectional area at intake is 0.8m2.
The exhaust gases leave the engine at 1bar and with a velocity 550m/s. Determine
the thrust developed by the aircraft.
10. A fixed vane with a uniform cross-sectional area A turns a jet of fluid through an
angle  without changing the magnitude of the velocity. The flow is steady,
atmospheric and friction on the vane is negligible. Find the magnitude and direction
of the resultant force acting on the vane.
11. Water jet impinges on a flat plate moving at a velocity of 12m/s and splits into an
equal upward and downward jet. The area of the jet and velocity are 5cm2 and 25m/s
respectively. Assuming steady flow and neglecting the weight of water and the jet,
determine the force required to keep the plate moving at same uniform velocity.
12. A vane with a turning angle of 700 moves at a constant velocity of 12m/s. It receives a
jet of water with a velocity of 40m/s from a stationary nozzle. If the nozzle has an exit
area of 50cm2, determine the forces acting on the vane.
13. A steam power plant uses a conveyor belt to supply coal to the boiler from a hopper.
Initially, the belt is empty and begins to fill the coal from the hopper while moving at
a velocity of 1.2m/s. The coal falls vertically from the hopper with a velocity of
1.8m/s and with a flow rate 800tonnes/hr. If the friction in the drive system and roller
is negligible, find the force required to pull the belt while filling the conveyor belt.
Take the density of coal as 1380kg/m3.
14. Water is to be turned by an angle 900 by an elbow. The cross-sectional area at inlet
and exit are 0.008m2 and 0.001m2 respectively. The absolute pressure at inlet is 2.5
bar while the velocity at the exit is 20m/s. If the elbow discharges water to
atmosphere, then find the anchoring force required to hold the vane.
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