The document discusses the application of the steady flow energy equation to various flow processes including turbines, nozzles, throttles, and pumps. It explains that for turbines, the steady flow energy equation relates the enthalpy drop of the fluid to the work produced. For nozzles, the equation shows the relationship between the enthalpy drop and the kinetic energy increase of the fluid. For throttles, the equation indicates that the enthalpy remains constant. And for pumps, the equation relates the enthalpy rise to the work input into the system. Examples are provided to demonstrate how to use the steady flow energy equation to analyze different flow processes.

1. FLOW PROCESS J2006/7/1
UNIT 7
FLOW PROCESS
OBJECTIVES
General Objective: To understand the application of steady flow energy equation.
Specific Objectives : At the end of the unit you will be able to:
 apply the steady-flow energy equation to :
• turbine
• nozzle
• throttle
• pump

2. FLOW PROCESS J2006/7/2
INPUT
7.0 APPLICATION OF STEADY FLOW EQUATION
Do you know the
application of steady
flow equation for
turbine?
The steady flow energy equation may be applied to any apparatus through which a
fluid is flowing, provided the conditions stated in Unit 6 are applicable. Some of the
most common cases found in engineering practise will now be dealt with in detail.

3. FLOW PROCESS J2006/7/3
7.0.1 Turbine
A turbine is a device which uses a pressure drop to produce work energy
which is used to drive an external load.
FLUID IN
1
Q
SYSTEM
W
BOUNDARY
2
FLUID OUT
Figure 7.0.1 Turbine
The steady flow energy equation gives
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  

Points to note :
i. The average velocity of flow of fluid through a turbine is normally
high, and the fluid passes quickly through the turbine. It may be
assumed that, because of this, heat energy does not have time to flow
into or out of the fluid during its passage through the turbine, and
hence Q = 0 .
ii. Although velocities are high the difference between them is not large,
and the term representing the change in kinetic energy may be
neglected.
iii. Potential energy is generally small enough to be neglected.
iv. W is the amount of external work energy produced per second.

4. FLOW PROCESS J2006/7/4
The steady flow energy equation becomes
- W = m( h2 − h1 )

or W = m( h1 − h2 )
 (7.1)
7.0.2 Nozzle
A nozzle utilises a pressure drop to produce an increase in the kinetic energy
of the fluid.
1
2
BOUNDARY
SYSTEM
FLUID FLUID
IN OUT
2
1
Figure 7.0.2 Nozzle
The steady flow energy equation gives
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  

Points to note :
i. The average velocity of flow through a nozzle is high, hence the fluid
spends only a short time in the nozzle. For this reason, it may be
assumed that there is insufficient time for heat energy to flow into or
out of the fluid during its passage through the nozzle, i.e. Q = 0.
ii. Since a nozzle has no moving parts, no work energy will be
transferred to or from the fluid as it passes through the nozzle,
i.e. W = 0.

5. FLOW PROCESS J2006/7/5
iii. Potential energy is generally small enough to be neglected.
Hence the equation becomes
  C 2 − C12 
0 = m ( h2 − h1 ) +  2
  


  2 
Often C1 is negligible compared with C2. In this case the equation becomes
 C2 
0 = m ( h2 − h1 ) +  2
  2 


  
2
C2
or = ( h1 − h2 )
2
or C 2 = 2( h1 − h2 ) (7.2)

6. FLOW PROCESS J2006/7/6
Example 7.1
A fluid flows through a turbine at the rate of 45 kg/min. Across the turbine the
specific enthalpy drop of the fluid is 580 kJ/kg and the turbine loss 2100 kJ/min
in the form of heat energy. Determine the power produced by the turbine,
assuming that changes in kinetic and potential energy may be neglected.
Solution to Example 7.1
The steady flow energy equation gives
  C 2 − C12
2
 
Q − W = m ( h2 − h1 ) + 
   + ( gZ 2 − gZ 1 ) 


  2  

Q = heat energy flow into system = -2100 kJ/min
W = work energy flow from system kJ/min
m = fluid flow rate = 45 kg/min
h2 - h1 = -580 kJ/kg
C1& C2 = neglected
Z1& Z2 = neglected
Therefore the steady flow energy equation becomes
-2100 kJ/min – W = 45 kg/min (-580 kJ/kg)
W = (26100 – 2100) kJ/min
= 24000 kJ/min
= 400 kJ/s
= 400 kW

7. FLOW PROCESS J2006/7/7
Example 7.2
Fluid with a specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with
negligible velocity at the rate of 14 kg/s. At the outlet from the nozzle the
specific enthalpy and specific volume of the fluid are 2250 kJ/kg and
1.25 m3/kg respectively. Assuming an adiabatic flow, determine the required
outlet area of the nozzle.
Solution to Example 7.2
The steady flow energy equation gives
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  

When applied to the nozzle, this becomes
  C 2 − C12 
0 = m ( h2 − h1 ) +  2
  


  2 
Since the inlet C1 is negligible, this may be written as
C 22 = 2( h1 − h2 )
= √ [2(2800 – 22500]
= 1050 m/s
Applying the equation of continuity at outlet gives
A2C2
m=

v2
A2 x 1050 m/s
14 kg/s =
1.25 m 3 /kg
A2 = 0.01668 m2

8. FLOW PROCESS J2006/7/8
Activity 7A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
7.1 Steam enters a turbine with a velocity of 16 m/s and specific enthalpy
2990 kJ/kg. The steam leaves the turbine with a velocity of 37 m/s and
specific enthalpy 2530 kJ/kg. The heat lost to the surroundings as the steam
passes through the turbine is 25 kJ/kg. The steam flow rate is 324000 kg/h.
Determine the work output from the turbine in kilowatts.
7.2 In a turbo jet engine the momentum of the gases leaving the nozzle produces
the propulsive force. The enthalpy and velocity of the gases at the nozzle
entrance are 1200 kJ/kg and 200 m/s respectively. The enthalpy of the gas at
exit is 900 kJ/kg. If the heat loss from the nozzle is negligible, determine the
velocity of the gas jet at exit from the nozzle.

9. FLOW PROCESS J2006/7/9
Feedback to Activity 7A
7.1 Neglecting the changes in potential energy, the steady flow energy equation
is
  C 2 − C12 
Q − W = ( h2 − h1 ) +  2
 


  2 
Q is negative since heat is lost from the steam to the surroundings
 C12
− C2
2

∴specific W = ( h1 − h2 ) + 
 -Q

 2 
(16 2 − 37 2 )
= (2999-2530) + − 25
2 x10 3
= 434.443 kJ/kg
The steam flow rate = 324000/3600 = 90 kg/s
∴ W = 434.443 x 90
= 39099.97 kJ/s or kW
≈ 39100 kW
≈ 39.1 MW

10. FLOW PROCESS J2006/7/10
7.2 The steady energy flow equation for nozzle gives
  C 2 − C12 
0 = m ( h2 − h1 ) +  2
  


  2 
On simplification,
C 2 = 2( h1 − h2 ) +C12
= √2(1200x 103- 900x 103) + 2002
= 800 m/s

11. FLOW PROCESS J2006/7/11
INPUT
7.0.3 THROTTLE
A throttling process is one in which the fluid is made to flow through a restriction ,
e.g. a partially opened valve or orifice, causing a considerable drop in the pressure
of the fluid.
2
1
Figure 7.0.3 Throttling process
The steady flow energy equation gives
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  

Points to note:
i. Since throttling takes place over a very small distance, the available area
through which heat energy can flow is very small, and it is normally
assumed that no energy is lost by heat transfer, i.e. Q = 0.
ii. Since there are no moving parts, no energy can be transferred in the form of
work energy, i.e. W = 0.
iii. The difference between C1 and C2 will not be great and consequently the
term representing the change in kinetic energy is normally neglected.
iv. The potential energy is generally small enough to be neglected.

12. FLOW PROCESS J2006/7/12
The steady flow energy equation becomes
0 = m (h2 - h1)

or h2 = h1 (7.3)
i.e. during a throttling process the enthalpy remains constant.
7.0.4 Pump
The action of a pump is the reverse of that of a turbine, i.e. it uses external
work energy to produce a pressure rise. In applying the steady flow energy
equation to a pump, exactly the same arguments are used as for turbine and
the equation becomes
- W = m( h2 − h1 )
 (7.4)
Since h2 > h1, W will be found to be negative.
OUTLET
2
Q
SYSTEM
W
BOUNDARY
1
INLET
Figure 7.0.4 Pump

13. FLOW PROCESS J2006/7/13
7.1 Equation of Continuity
This is an equation which is often used in conjunction with the steady flow
energy equation. It is based on the fact that if a system is in a steady state,
then the mass of fluid passing any section during a specified time must be
constant. Consider a mass of m kg/s flowing through a system in which all
conditions are steady as illustrated in Fig.7.5.
1
2
C1 C2
AREA A2
2
AREA A1
1
Figure 7.1 Mass flowing through a system
Let A1 and A2 represent the flow areas in m2 at the inlet and outlet
respectively.
Let v1 and v2 represent the specific volumes in m3/kg at the inlet and outlet
respectively.
Let C1 and C2 represent the velocities in m/s, at the inlet and outlet
respectively.
Then mass flowing per sec = volume flowing per sec m3/s
volume per kg m3/kg
A1C1 kg
= at inlet
v1 s
A2 C 2 kg
= at outlet
v2 s
A1C1 A2 C 2
i.e. m=
 = (7.5)
v1 v2

14. FLOW PROCESS J2006/7/14
Example 7.3
A fluid flowing along a pipeline undergoes a throttling process from 10 bar to 1
bar in passing through a partially open valve. Before throttling, the specific
volume of the fluid is 0.3 m3/kg and after throttling is 1.8 m3/kg. Determine the
change in specific internal energy during the throttling process.
Solution to Example 7.3
For a throttling process, the steady flow energy equation becomes
0 = m (h2 - h1)

or h2 = h1
But h2 = u2 + P2v2
and h1= u1 + P1v1
Therefore the change in specific internal energy
= u2 – u1
= ( h2 - P2v2 ) - ( h1 – P1v1)
= ( h2 - h1 ) – ( P2v2 - P1v1 )
= 0 – ( 1 x 1.8 – 10 x 0.3 ) bar m3/kg
= 120 x 10 Nm/kg
= 120 kJ/kg
Example 7.4
A pump delivers fluid at the rate of 45 kg/min. At the inlet to the pump the
specific enthalpy of the fluid is 46 kJ/kg, and at the outlet from the pump the
specific enthalpy of the fluid is 175 kJ/kg. If 105 kJ/min of heat energy are lost to
the surroundings by the pump, determine the power required to drive the pump if
the efficiency of the drive is 85 %.
Solution to Example 7.4
The flow rate of fluid = 45 kg/min
= 0.75 kg/s
The steady flow energy is
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  


15. FLOW PROCESS J2006/7/15
Q = - 105 kJ/min = - 1.75 kJ/s
W = work energy flow (kJ/s)
h1 = 46 kJ/kg
h2 = 1.27 kJ/kg
m = 0.75 kg/s
The kinetic and potential energy may be neglected
Substituting the data above with the steady flow energy equation gives
- 1.75 – W = 0.75 (175 – 46)
W = -1.75 – (0.75 x 129)
= - 98.5 kJ/s
= - 98.5 kW
(N.B. The negative sign indicates work energy required by the pump)
Since the efficiency of the drive is 85 %
100
Power required by the compressor = 98.5 x
85
= 114.8 kW
Turbine
T
&
Pump
-

16. FLOW PROCESS J2006/7/16
Activity 7B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
7.3 A rotary air pump is required to deliver 900 kg of air per hour. The enthalpy
at the inlet and exit of the pump are 300 kJ/kg and 500 kJ/kg respectively.
The air velocity at the entrance and exit are 10 m/s and 15 m/s respectively.
The rate of heat loss from the pump is 2500 W. Determine the power
required to drive the pump.
7.4 In activity 7A, for question No. 7.2, if the diameter of the nozzle at exit is
500 mm, find the mass flow rate of gas. The gas density at the nozzle inlet
and exit are 0.81 kg/m3 and 0.39 kg/m3 respectively. Also determine the
diameter of the nozzle at the inlet.

17. FLOW PROCESS J2006/7/17
Feedback to Activity 7B
900
7.3 Data : m =
 = 0.25 kg/s
3600
h1 = 300 kJ/kg
h2 = 500 kJ/kg
C1= 10 m/s
C2= 15 m/s
Q = 2500 W = 2.5 kW
W=?
The steady flow energy equation gives
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   + ( gZ 2 − gZ 1 ) 


  2  

Neglecting the change in Potential energy since it is negligible
  C 2 − C12  
Q − W = m ( h2 − h1 ) +  2
   +


  2  
15 2 − 10 2
-W = 0.25 [( 500- 300) + ( )] + 2.5
2 x10 3
-W = 52.5 kW

18. FLOW PROCESS J2006/7/18
0.5 2
7.4 Data : A2 = π = 0.196 m2
4
ρ1 = 0.81 kg/m3
ρ2 = 0.39 kg/m3
m =?

d =?
Mass flow rate at exit,
m = A 2 C2 ρ 2

= 61.2 kg/s
From the mass balance,
Mass entering the nozzle = mass leaving the nozzle = m
m = A 1 C1 ρ 1 = A 2 C2 ρ 2

On substitution
A1 x 200 x 0.81 = 61.2
On simplification
A1 = 0.378 m2
or
d1 = 0.694 m
= 694 mm

19. FLOW PROCESS J2006/7/19
SELF-ASSESSMENT
You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback to Self-Assessment on the
next page. If you face any problem, discuss it with your lecturer. Good luck.
1 Steam flows through a turbine stage at the rate of 4500 kJ/h. The steam
velocities at inlet and outlet are 15 m/s and 180 m/s respectively. The rate of
heat energy flow from the turbine casing to the surroundings is 23 kJ/kg of
steam flowing. If the specific enthalpy of the steam decreases by 420 kJ/kg in
passing through the turbine stage, calculate the power developed.
2 A rotary pump draws 600 kg/hour of atmospheric air and delivers it at a
higher pressure. The specific enthalpy of air at the pump inlet is 300 kJ/kg
and that at the exit is 509 kJ/kg. The heat lost from the pump casing is
5000 W. Neglecting the changes in kinetic and potential energy, determine
the power required to drive the pump.
3 A nozzle is supplied with steam having a specific enthalpy of 2780 kJ/kg at
the rate of 9.1 kg/min. At outlet from the nozzle the velocity of the steam is
1070 m/s. Assuming that the inlet velocity of the steam is negligible and that
the process is adiabatic, determine:
a) the specific enthalpy of the steam at the nozzle exit
b) the outlet area required if the final specific volume of the steam is
18.75 m3/kg.
4 Fluid at 10.35 bar having a specific volume of 0.18 m 3/kg is throttled to a
pressure of 1 bar. If the specific volume of the fluid after throttling is
0.107 m3/kg, calculate the change in specific internal energy during the
process.

20. FLOW PROCESS J2006/7/20
Feedback to Self-Assessment
Have you tried the questions????? If “YES”, check your answers now.
1. 476 kW
2. 353 kW
3. 2208 kJ/kg; 2660 mm2
4. 175.7 kJ/kg
CONGRATULATIONS!!!!
…..
You can continue with the
next unit…