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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
DYMAMICS OF FLUID FLOW
Energy Equation (Conservation of Energy)
In words, the conservation of energy can be stated as,
Time rate of increase in stored energy of the system = Net time rate of energy
addition by heat transfer into the system + Net time rate of energy addition by work
transfer into the system.
In symbolic form, the statement is given by,
D
 e  dV 
Dt sys
Q  Q 
in
out sys
  Win  Wout 
(1)
sys
or,

D
 e.. dV  Qnet ,in  Wnet ,in
Dt sys

(2)
sys
where e is the total stored energy per unit mass of the system and is related to internal
V 2 
energy per unit mass  u  , kinetic energy per unit mass   and potential energy per
 2 
unit mass  g z  , i.e.
eu

V2
 gz
2
(3)

The net rate of heat transfer into the system Qnet , in and the net rate of work transfer into


the system Qnet , in is considered to be ‘+ ve’ quantity and the outward flow of heat and
work is taken as ‘- ve’.
Expanding the left hand side, the Eq. (2) can be written for a control volume (CV) as,

 e.. dV  CS e V. nˆ dA  Qnet ,in  Wnet ,in  CS p V. nˆ dA
t CV
 
 
(4)
The last term in the RHS of Eq. (4) is the “flow work” associated with the control
surface. Now, using Eq. (3) in Eq. (4),



 
V2
p V2
u


gz

dV

u



gz



  V. nˆ dA  Qnet , in  Wnet , in   p V. nˆ dA
 2

t CV

2


CS 
CS
(5)
 
 
1
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
Discussion:

The first term in the LHS of Eq. (5) represents the time rate of change of the total
stored energy  e  of the contents of CV. This term is zero for the steady flow i.e.

 
V2
u

 gz   dV  0


t CV 
2


The integrand
(6)


p V2
u

CS    2  gz   V. nˆ dA can be non-zero only when the
 
fluid crosses the control surface i.e.
 V. nˆ  0 . If the properties like internal
energy, pressure, kinetic and potential energies are uniformly distributed over the
cross-sectional area involved, then the equation becomes simple and can be
written as,






p V2
p V2
p V2
ˆ
u



gz

V.
n
dA

u



gz
m

u


 gz  m






CS   2
 2
 2
flow out 
flow in 



 
(7)

If there is only one stream entering and leaving the CV, then,






p V2
p V2
p V2
ˆ
u



gz

V.
n
dA

u



gz
m

u



gz





 min
out
CS   2

2

2


out

in
 
(8)

The simplified expression for one-dimensional steady flow energy equation is
then written as,

 p
m  uout  uin    
  out


 p   V 2  Vin2 
     out
  g  zout  zin    Qnet , in  Wnet , in
2
  in 


(9)
This equation is valid for both compressible and incompressible flows. Here, we
introduce one more fluid property called “enthalpy” and is defined by,
h u
p

(10)
In terms of enthalpy, Eq. (9) is written as,
2


 Vout
 Vin2 
m  hout  hin  
  g  zout  zin    Qnet , in  Wnet , in
2






(11)
2
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
BASIC EQUATIONS (DIFFERENTIAL FORM)
Conservation of Mass (Continuity Equation)
Consider an infinitesimal control volume in the shape of a rectangular parallelepiped
(Fig. 1) fixed in x-y-z coordinate system for a general flow. Let u, v, and w be the
velocities of the fluid in directions x, y, and z respectively and  be the density of the
fluid. Now consider the flow through the surfaces 1 and 2, which are parallel to y-z plane.
The efflux rate per unit area through section 1 is -  u and varies continuously in the x
direction. The net efflux rate through the section 2 can be given by  u     u  x  dx .
The net efflux rate through these surfaces is given by    u  x  dx dy dz . Performing
similar computations for other pairs of sides and adding the results,
   u     v     w 
Net efflux rate = 


 dx dy dz
y
z 
 x
Net rate of decrease in mass inside the control volume = 

dx dy dz
t
Equating the above two expressions,
  u 
x
u

(1)
  v 
y

   w
z


t
(12)
(2)
u + x(u)dx
y
x
z
Fig. 1: Infinitesimal fixed control volume.
3
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22

For steady flow, Eq. (12) becomes,
  u 
x

Dr. Niranjan Sahoo
IIT-Guwahati

  v 
y

   w
z
0
(13)
For incompressible flow, the fluid density remains constant. Hence, Eq. (13)
reduces to,
u v w
 
0
x y z
(14)
or, .V  0



where   iˆ  ˆj  kˆ
and V  uiˆ  vjˆ  wkˆ
x
y
z
Euler’s Equation of Motion
According to Newton’s law, the net force F acting on a fluid element of mass m and
acceleration a is given by,
F  m.a
(15)
In general, following forces may be present in for a fluid in motion.

Gravity force, Fg

Pressure force, Fp

Viscous force, Fv

Force due to turbulence, Ft

Force due to compressibility, Fc
The net force becomes,
F  Fg  Fp  Fv  Ft  Fc
(16)
The Euler’s equation of motion represents a special case of a flow field in which the
viscous forces and force due to turbulence is absent.
The linear momentum of a fluid element of mass dm , moving at velocity V is given
as  dm V  . The fundamental statement of Newton’s law for an inertial reference is given
in terms of linear momentum as,
dF 
D
 dm V 
Dt
(17)
The above equation may be written as,
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NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
dF 
Dr. Niranjan Sahoo
IIT-Guwahati
D
V
V
V V
 dm V    dx dy dz  u  v  w  
Dt
x
x t 
 x
(18)
The Euler’s equation is restricted to the case of no shear stress with only gravity as a
body force. The negative z direction corresponds to the direction of gravity. So,
Gravity force    g  dx dy dz  kˆ   g  z   dx dy dz 
The surface force on the fluid element is only due to pressure p (Fig. 2). Hence,
 p
p ˆ p ˆ 
Pressure force    iˆ 
j  k   dx dy dz     p  dx dy dz 
y
z 
 x
where iˆ, ˆj and kˆ are the unit vectors in x, y and z directions respectively.
p +(p z)dz
p
p +(p y)dy
p
y
p +(p y)dy p
x
z
Fig. 2: Pressure variation in the x-y-z directions.
Incorporating the gravity and pressure forces in Eq. (18),

V
V V
 V
p  g  z   u
v
w


x
x t
 x
1
 DV

 Dt
(19)
The above equation can be expanded in rectangular coordinates as,
x direction, 
 u
1 p
u
u  u
 Bx   u
v w 
 x
y
z  t
 x
(20a)
y direction, 
 v
1 p
v
v  v
 By   u  v  w  
 y
y
z  t
 x
(20b)
 w
1 p
w
w  w
 Bz   u
v
w 
 z
y
z  t
 x
(20c)
z direction, 
In streamline coordinate system, Eq. (19) is expressed as,
5
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Along the direction of streamline (s), 
Dr. Niranjan Sahoo
IIT-Guwahati
1 p
V V
 Bs  V

 s
s t
(21a)
1 p
V 2 Vn
 Bn 

 n
R
t
(21b)
Along the direction perpenticular to ' s ', 
where B stands for gravity force in the respective direction. Eqs. (20a-c and 21a-b) are
known as Euler’s equation of motion.
Integration of Steady-State Euler Equation: Bernoulli’s Equation
Consider a general steady-state flow of an inviscid, incompressible, irrotational fluid. The
Euler’s equation of motion can be expressed in streamline coordinate system as,

1

p  gz  V
V
s
(22)
where s is the coordinate along a stream line. Now, take the dot product of each term
given for streamline coordinates. Thus,

1

p.ds  gz.ds  V
V
.ds
s
The term p.ds becomes dp , the differential change in pressure along a streamline;
z.ds becomes dz , the differential change in elevation along the streamline. The righthand side of the equation simplifies to V .dV , where dV is the velocity change along a
streamline. Hence,

V 2 
 g dz  V dV  d  

 2 
dp
Taking g as constant and integrating along a streamline, the above equation can be
written as,
p

0
V2
gz
 constant

2
dp
(23)
This equation is often called the compressible form of Bernoulli’s equation. If  is
expressed as a function of p only, i.e.     p  , the first expression is integrable. Flows
having this characteristic are called barotropic flows. On the other hand, if the flow is
incompressible, i.e.   constant , the Eq. (23) becomes,
6
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
p V2

 z  constant
 g 2g
(24)
Multiplying Eq. (24) by 1 g  and replacing   g  by  (i.e. specific weight or weight
density of the fluid),
p V2

 z  constant
 2g
(25)
Discussions

The terms in Eq. (25) are units of length and are frequently designated as
pressure, velocity and elevation “heads” respectively.

Between any two points along a streamline, Eqs. (24) and (25) can be written as,
p1 V12
p2 V22

z 

z
 g 2g 1  g 2g 2
p1



V12
p V2
 z1  2  2  z2
2g
 2g
(26)
(27)
For most of the fluid flow, there are some energy losses (expressed in terms of
heads, hL ), which should be taken into account for considering Bernoulli’s
equation between two state points of a streamline.

One of the important features of Bernoulli’s equation is to define two grade lines
of a flow. One of the grade lines is the Bernoulli’s constant in Eq. (25) and named
as “energy grade line (EGL)”. The other grade line is the “hydraulic grade line
(HGL)” which shows the height corresponding to elevation and pressure head i.e.
EGL minus Velocity Head.
Comparison of Bernoulli and Steady Flow Energy Equation
Bernoulli’s equation is based on certain assumptions;
1. Steady uniform flow – a common assumption applicable for many kinds of
practical flows.
2. Incompressible flow – is acceptable for both the equations (i.e. Euler’s and
Bernoulli’s equation) if the Mach number is less than 0.3.
7
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
3. Frictionless flow – very restrictive because friction can be introduced from the
solid walls on to the flow.
4. Flow along a streamline – i.e. different streamlines will have different Bernoulli’s
constants (Eq. 25).
5. No shaft work – no pumps or turbines are involved between two sections.
6. No heat transfer is involved between two sections.
Example-1
Air enters in a compressor with negligible velocity with a flow rate of 10kg/s and is
discharged through a pipe of cross-sectional area of 0.1m2 with same flow rate. The
pressure and temperature at the inlet to compressor is atmospheric where as the
corresponding values in the discharge section is 3.5 bar and 400C. If the compressor takes
600hp power input, determine the rate of heat rejection.
Solution:
The control volume showing the inlet and exit sections are given in the Ex. Fig. 1.
The steady flow energy equation for the compressor can be written as given by Eq. (11)
i.e.


 V 2  V12 
m  h2  h1   2
  g  z2  z1    Q  W
2 





Since, air behaves as an ideal gas with constant specific heat, so h2  h1  c p T2  T1  .
Also, air enters with negligible velocity i.e. V22  V12  ~ 0 and also, z2  z1  0 . Hence
the above equation becomes,
V22
Q  mc p T2  T1   m
W
2
Control volume
(2)
p2 = 3.5 bar
p1 = 1 bar
T1 = 298 K
V1 = 0
(1)
T2 = 313 K
A2 = 0.1m2
m = 10 kg/s
Compressor power input = 600hp
Ex. Fig. 1: Control Volume for compressor.
8
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
Using continuity equation and equation of state for air, we can write,
m  1 AV
1 1   2 A2V2 i.e. V2 
m
m R.T
. 2  . 2
A2
A2 p2
For air, R  0.287 kJ kg.K , c p  1.005 kJ kg.K and atmospheric conditions at inlet to
compressor can be taken as, p1  1bar ; T1  250 C
so V2 
Q
10 0.287   40  273

 25.7 m s
0.1
350
10 1005  40  25 
 25.7 
 10 
2
2
  600  745 
1000
 292.95kW
Example-2
A tank of volume 0.1m3 is connected to a high-pressure air reservoir at 2MPa through a
valve. Initial pressure in the tank is 200kPa (absolute). The line connecting the reservoir
and the tank is sufficiently large so that the temperature may be assumed to be uniform at
250C. When the valve is opened, the tank temperature rises at the rate of 0.080C/s.
Determine the instantaneous flow rate of air into the tank by neglecting the heat transfer.
Solution:
The CV is chosen with reference to the following schematic diagram.
Control volume
Air reservoir
Tank
p = 2 MPa
T = 298 K
V = 0.1m3
p = 200 kPa
T = 298 K
Ex. Fig. 2: Control Volume for the tank.
Following assumptions are made;

Since, no heat transfer and shafts work are involved, W  Q  0 .

Velocities in the line and tank are small and can be neglected.
9
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22

Dr. Niranjan Sahoo
IIT-Guwahati
Since, there is no elevation difference, the potential energy changes can be
neglected.

Flow is uniform at the inlet of the tank m   V A

Properties are uniform in the tank.

Air can be taken as ideal gas with state equation, p  .R.T and du  cv dT
With these assumptions, Eq.(5) reduces to,

 u.. dV  u  pv  VA  0
t CV
Since the properties are uniform, so

d
can be replaced by , and then
t
dt
d
u.M    u  pv  m  0
dt
dM
du
or, u.
 M .  u.m  pv.m  0
dt
dt
From continuity equation,
dM
m
dt
Hence the simplified expression becomes,
m

m
M .cv  dT dt 
p.v

.V .cv  dT dt 
R.T
ptank 200 103

 2.35kg m3
RT
287  293
2.35  0.1 718  0.08
 1.6 104 kg s
287  293
10
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
EXERCISES
1. The front of a jet engine acts as a diffuser receiving air at 900km/h, -50C, 50kPa
bringing it to 80m/s before entering the compressor. If the flow area is reduced to
80% of the inlet area, find the temperature and pressure in the compressor inlet.
2. Air at a temperature of 150C passes through a heat exchanger at a velocity of 30 m/s
where its temperature is raised to 8000C. It then enters a turbine with the same
velocity of 30 m/s and expands until the temperature falls to 6500C. On leaving the
turbine, the air is taken at a velocity of 60 m/s to a nozzle where it expands until the
temperature falls to 5000C. If the air flow rate is 2 kg/s, calculate (a) the rate of heat
transfer to the air in the heat exchanger, (b) the power output from the turbine
assuming no heat loss, and (c) the velocity at the exit of the nozzle, assuming no heat
loss. Take the enthalpy of air as h  c p T , where c p is the specific heat at constant
pressure equal to 1.005 KJ/kg.K and T is the temperature.
3. Air flows steadily through a turbine that produces a power output of 750hp. The
diameter of inlet and exit port is 15cm. The static pressure and temperature at the inlet
section are 1MPa and 1500C respectively. The corresponding values in the exit
section are 0.3MPa and 20C and respectively. If the inlet velocity is 30cm/s,
determine, (i) exit velocity; (ii) rate of heat transfer.
4. A balloon is to be partially filled at atmospheric pressure with helium from a wellinsulated storage tank that has a volume of 6.2m3. At the time of filling the balloon,
atmospheric pressure is 95kPa. The helium in the tank is initially at gage pressure of
500kPa and 200C before the valve on the tank is opened. Assuming the conditions to
be uniform throughout the tank at any instant and heat transfer is negligible;
determine the mass of the balloon that has been put into the balloon when a pressure
gage in the storage tank reads 300kPa. For helium under these conditions,
h  1.667u  5.193T and R  2.007 kJ kg.K with h and u in kJ kg and T in K .
5. The steam supply to an engine comprises two streams, which mix before entering the
engine. One stream is supplied at the rate of 0.01 kg/s with an enthalpy of 2952 kJ/kg
and a velocity of 20 m/s. The other stream is supplied at the rate of 0.1 kg/s with an
enthalpy of 2569 kJ/kg and a velocity of 120 m/s. At the exit from the engine, the fluid
leaves as two streams, one of water at the rate 0.001 kg/s with enthalpy of 420 kJ/kg
11
NPTEL Course Developer for Fluid Mechanics
Module 04; Lecture 22
Dr. Niranjan Sahoo
IIT-Guwahati
and the other stream; the fluid velocities at the exit are negligible. The engine develops
a shaft power of 25 kW and the heat transfer is negligible. Evaluate the enthalpy of the
second stream.
6. A nozzle is a device for increasing the velocity of a steadily flowing stream. At the inlet
to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the velocity is
60 m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The nozzle is horizontal and
there is negligible heat loss from it. (i) Find the velocity at the exit of the nozzle; (ii) If
the inlet area is 0.1 m2 and specific volume at the inlet is 0.187 m3/kg, find the mass
flow rate; (iii) If the specific volume at the nozzle exit is 0.498 m3/kg, find the exit
area of the nozzle.
7. A turbine operates under steady flow conditions, receiving stream at the following state:
pressure 1.2 MPa, temperature 1880C, enthalpy 2785 kJ/kg, velocity 33.3 m/s and
elevation 3 m. The steam leaves the turbine at the following state: pressure 20 kPa,
enthalpy 2512 kJ/kg, velocity 100 m/s and elevation 0 m. Heat is lost to the
surroundings at the rate of 0.29 kJ/s. If the rate of steam flow through the turbine is
0.42 kg/s, what is the power output in kW?
8. Steam enters to a turbine with a velocity 40m/s and enthalpy of 3560 kJ/kg and leaves
as liquid-vapour mixture with a velocity of 75m/s and enthalpy of 2478 kJ/kg. If the
flow through the turbine is adiabatic and elevation changes are negligible, determine
the specific work output for the turbine.
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