1. Reynolds transport theorem relates the rate of change of a property within a control volume to the rate of change of the property convected with a moving fluid plus the net flux of the property entering and leaving the control volume. 2. The continuity equation states that for a fixed mass of fluid, the net mass flow entering and leaving a control volume is zero. For steady one-dimensional flow, the mass flow rate is constant. 3. The momentum equation equates the net external forces on a control volume to the rate of change of momentum entering and leaving the control volume. For steady one-dimensional flow, the momentum flow rate is constant.

1. 1
Nozzles and Diffusers

2. 2
Nozzle
Google Image

3. 3
Diffuser
Google Image

4. 4
Gas Dynamics

5. 5
REYNOLD’S TRANSPORT THEOREM
This relation, known as Reynolds’s transport theorem.
The rate of change of N for a given mass as it is moving around
is equal to the rate of change of N inside the control volume plus
the net efflux (flow out minus flow in) of N from the control
volume.
It can be interpreted in words as:
1
derivative
ˆ
( . )
material cv cs
dN
dv V n dA
dt t
 

 
 
 

 
 
…………………………… (1)

6. 6
CONSERVATION OF MASS
If we observe a given quantity of mass as it moves around, we
can say by definition that the mass will remain fixed. Another
way of stating this is that the material derivative of the mass is
zero:
d(mass)
=0
dt
This is the continuity equation for a control mass.
What corresponding expression can we write for a control volume?
If N represents the total mass, η is the mass per unit mass, or 1.
Substitution into equation (1) yields
…………………………… (2)

7. 7
cv cs
d(mass)
ˆ
ρdv+ ρ(V. n)dA
dt t

 

 

 
 
…………………………… (3)
But we know by equation (2) that this must be zero; thus the transformed
equation is
cv cs
ˆ
0 ρdv+ ρ(V. n)dA
t


  
This is the continuity equation for a control volume.
…………………………… (4)

8. 8
For steady flow, any partial derivative with respect to time is
zero and the equation becomes
cs
ˆ
0 ρ(V. n)dA
  …………………………… (5)
Let us now evaluate the remaining integral for the case of one-
dimensional flow. Figure below shows fluid crossing a portion
of the control surface. Recall that for one dimensional flow any
fluid property will be constant over an entire cross section.
Thus both the density and the velocity can be brought out from
under the integral sign. If the surface is always chosen
perpendicular to V , the integral is very simple to evaluate:
cs cs
ˆ ˆ
ρ(V. n)dA = ρ(V. n) dA = ρVA
 

9. 9
This integral must be evaluated over the entire control surface, which yields
ˆ
(V. n)
…………………………… (6)
This summation is taken over all sections where fluid crosses the control surface
and is positive where fluid leaves the control volume (since is positive
here) and negative where fluid enters the control volume. For steady, one-
dimensional flow, the continuity equation for a control volume becomes
cs
ˆ
ρ(V. n)dA= ρVA



10. 10
ρVA = 0
 …………………………… (7)
If there is only one section where fluid enters and one section where fluid leaves the
control volume, this becomes
  in
out
ρVA - (ρVA) =0
or
  in
out
ρVA = (ρVA) …………………………… (8)
We usually write this as
•
m = ρVA = constant
…………………………… (9)

11. 11
An alternative form of the continuity equation can be obtained
by differentiating equation (9). For steady one-dimensional flow
this means that
d(ρVA) = AV dρ + ρV dA+Aρ dV = 0 …………………………(10)
Dividing by ρAV yields
dρ dA dV
+ + = 0
ρ A V
…………………………(11)

12. 12
Equation of Mass Flow Rate through a Nozzle
m AV


The velocity of the flow is related to the static and stagnation enthalpies.
o P 0 P 0
o
T
V= 2(h -h)= 2C (T -T)= 2C T (1- )
T
Let's obtain an expression for the flow rate through a converging nozzle at any
location as a function of the pressure at that location. The mass flow rate is given by
and
k-1 k
P
0
P
V= 2C 1-
P
 
 
 
 
 
 
 
k-1 k
0 0
T P
=
T P
 
 
 
Hence,

13. 13
Write the mass flow rate as
o
o
ρ
m=AVρ
ρ
We note from the ideal-gas relations that
o
o
o
P
RT

1 k
0 0
ρ P
=
ρ P
 
 
 
2 k k+1 k
•
0
0 0 0
2k P P
m =AP -
(k-1)RT P P
 
   
 
   
 
   
 
Hence,
Where,

14. 14
MOMENTUM EQUATION
If we observe the motion of a given quantity of mass, Newton’s second law tells us
that its linear momentum will be changed in direct proportion to the applied forces.
This is expressed by the following equation:
c
1 d(momentum)
F =
g dt

Again, the question is: What corresponding expression can we write for a control
volume?
If we let N be the linear momentum of the system, η represents the momentum per
unit mass, which is V. Substitution into equation (1) yields
d(momentum)
ˆ
( . )
dt cv cs
V dv V V n dA
t
 

 
  
…………………………(12)
…………………………(13)

15. 15
and the transformed equation which is applicable to a control
volume is
c c
1 1
ˆ
F = ( . )
g g
cv cs
V dv V V n dA
t
 



   …………………………(14)
This equation is usually called the momentum or momentum flux
equation. The ∑ F represents the summation of all forces on the
fluid within the control volume
In the event that one-dimensional flow exists, the last integral in
equation (14) is easy to evaluate, as ρ and V are constant over any
given cross section. If we choose the surface A perpendicular to the
velocity, then
•
cs
ˆ
Vρ(V. n)dA= VρV dA = VρVA = mV
  
 
…………………………(15)

16. 16
The summation is taken over all sections where fluid crosses
the control surface and is positive where fluid leaves the control
volume and negative where fluid enters the control volume.
If we now consider steady flow, the term involving the partial
derivative with respect to time is zero. Thus for steady one-
dimensional flow, the momentum equation for a control volume
becomes •
c
1
F = mV
g
  …………………………(16)
If there is only one section where fluid enters and one section
where fluid leaves the control volume, we know (from
continuity) that
• • •
in out
m = m = m

17. 17
and the momentum equation becomes
•
out in
c
m
F = (V - V )
g
 …………………………(17)
This is the form of the equation for a finite control volume.
What assumptions have been fed into this equation? In using this
relation one must be sure to:
1. Identify the control volume.
2. Include all forces acting on the fluid inside the control volume.
3. Be extremely careful with the signs of all quantities.

18. 18
We want to show that the stagnation properties are related to
the Mach number M of the flow where
M
V
C


C is the speed of sound in the fluid.
VELOCITY OF SOUND & MACH NUMBER
A pressure disturbance propagates through a compressible fluid with a
velocity dependent upon the state of the fluid. The velocity with which this
pressure wave moves through the fluid is called the velocity of sound, or
the sonic velocity.
SPEED OF SOUND
If M <1 flow is subsonic , M =1 flow is sonic and for M >1 flow
is supersonic

19. 19
Consider a small pressure wave caused by a small piston displacement in a
tube filled with an ideal gas as shown below.

20. 20
It is easier to work with a control volume moving with the wave
front as shown below.

21. 21
Apply the conservation of energy for steady-flow with no heat transfer, no work, and
neglect the potential energies.
2 2
2 2 2
C (C-dV)
h+ = (h+dh) +
2 2
C (C -2CdV+dV )
h+ = (h+dh) +
2 2
Cancel terms and neglect ; we have
dV
2
dh - CdV=0
Now, apply the conservation of mass or continuity equation to the control
volume.

m AV
 

ρAC=(ρ+dρ)A(C-dV)
ρAC=A(ρC-ρdV+Cdρ-dρ dV)
Cancel terms and neglect the higher-order terms like . We have
d dV


C dρ-ρ dV=0

22. 22
Also, we consider the property relation
dh=T ds+v dP
1
dh=T ds+ dP
ρ
Let's assume the process to be isentropic; then ds = 0 and
dh dP

1

Using the results of the first law
dh dP C dV
 
1


From the continuity equation
C dρ
dV=
ρ
Now
1 Cdρ
dP=C
ρ ρ
 
 
 

23. 23
Thus
2
dP
=C
dρ
Since the process is assumed to be isentropic, the above becomes
For a general thermodynamic substance, the results of may be used to show that the
speed of sound is determined from
where k is the ratio of specific heats, k = CP/CV.
2
s
dP
=C
dρ
 
 
 
2
T
P
C k

 

  

 

24. 24
IDEAL GAS RESULT
Gas Equation:
P=ρRT
T
P
=RT
ρ
 

 

 
2
C =kRT or C = kRT
Differentiate at constant temperature,
Hence,

25. 25
Consider the nozzle and control volume shown below.
The first law for the control volume is
dh VdV
 
 
0
The continuity equation for the control volume yields
d dA
A
dV
V


  

 0
Also, we consider the property relation for an isentropic process
Tds dh
dP
  

0
VARYING-AREA ADIABATIC FLOW

26. 26
and the Mach Number relation
dP
d
C
V
M

 
2
2
2

Putting these four relations together yields
dA
A
dP
V
M
 

2
2
1
( )
Let’s consider the implications of this equation for both nozzles and diffusers.
A nozzle is a device that increases fluid velocity while causing its pressure to drop;
thus, d > 0, dP < 0.

V
 Nozzle Results dA
A
dP
V
M
 

2
2
1
( )
2
2
2
Subsonic: M<1 dP(1-M )<0 dA<0
Sonic: M=1 dP(1-M )=0 dA=0
Supersonic: M>1 dP(1-M )>0 dA>0

27. 27
 To accelerate subsonic flow, the nozzle flow area must first decrease in the flow
direction. The flow area reaches a minimum at the point where the Mach number
is unity. To continue to accelerate the flow to supersonic conditions, the flow
area must increase.
 The minimum flow area is called the throat of the nozzle.
 We are most familiar with the shape of a subsonic nozzle. That is, the flow area
in a subsonic nozzle decreases in the flow direction.
 A diffuser is a device that decreases fluid velocity while causing its pressure to
rise; thus, d < 0, dP > 0.

V
 Diffuser Results
2
2
dA dP
= (1-M )
A ρV
2
2
2
Subsonic: M<1 dP(1-M )>0 dA>0
Sonic: M=1 dP(1-M )=0 dA=0
Supersonic: M>1 dP(1-M )<0 dA<0

28. 28
To diffuse supersonic flow, the diffuser flow area must first decrease in the flow
direction. The flow area reaches a minimum at the point where the Mach number is
unity. To continue to diffuse the flow to subsonic conditions, the flow area must
increase.
We are most familiar with the shape of a subsonic diffuser. That is the flow area in a
subsonic diffuser increases in the flow direction.

29. 29
GRAPHICAL REPRESENTATION

30. 30
 Consider a fluid flowing into a diffuser at a velocity , temperature T, pressure P,
and enthalpy h, etc. Here the ordinary properties T, P, h, etc. are called the static
properties; that is, they are measured relative to the flow at the flow velocity.
 The diffuser is sufficiently long and the exit area is sufficiently large that the
fluid is brought to rest (zero velocity) at the diffuser exit while no work or heat
transfer is done. The resulting state is called the stagnation state.
We apply the first law per unit mass for one entrance, one exit, and neglect the
potential energies.
V
STAGNATION PROPERTIES

31. 31
Since the exit velocity, work, and heat transfer are zero,
h h
V
o  
2
2
The term ho is called the stagnation enthalpy (some authors call this the total
enthalpy). It is the enthalpy the fluid attains when brought to rest adiabatically while
no work is done.
If, in addition, the process is also reversible, the process is isentropic, and the inlet
and exit entropies are equal.
s s
o 
q h
V
w h
V
net net o
o
    
 
2 2
2 2
Let the inlet state be unsubscripted and the exit or stagnation state have the subscript o.

32. 32
The actual stagnation pressure for irreversible flows will be somewhat less than the
isentropic stagnation pressure as shown below
The stagnation enthalpy and entropy define the stagnation state and the isentropic
stagnation pressure, Po.

33. 33
PROBLEM 1
Steam at 400o
C, 1.0 MPa, and 300 m/s flows through a pipe.
Find the properties of the steam at the stagnation state.

34. 34
From Steam Table, At T = 400oC and P = 1.0 MPa,
h = 3264.5 kJ/kg & s = 7.4670 kJ/kgK
SOLUTION

35. 35
2
2
o 2
2
kJ
m
300
V kJ kJ
kg
s
h = h+ = 3264.5 + = 3309.5
m
2 kg 2 kg
1000
s
 
 
 
Then
and
= = 7.4670
o
kJ
s s
kg K

h h P s
o o o
 ( , )
Since,

36. 36
We can find Po by trial and error . The resulting stagnation properties are
3
1.16
422.2
1
3.640
o
o
o
o
o
P MPa
T C
kg
v m



 

37. 37
 Ideal Gas Result
Rewrite the equation defining the stagnation enthalpy as
2
2
o
V
h h
 
For ideal gases with constant specific heats, the enthalpy difference becomes
C T T
V
P o
( )
 
2
2

38. 38
where To is defined as the stagnation temperature.
T T
V
C
o
P
 
2
2
For the isentropic process, the stagnation pressure can be determined from
or
Using variable specific heat data
o
R@T
o o ref
ref R@T
P
P P /P
= =
P P/P P
(k-1)/k
o o
T P
=
T P
 
 
 
k/(k-1)
o o
P T
=
P T
   
   
   

39. 39
PROBLEM 2
An aircraft flies in air at 5000 m with a velocity of 250 m/s. At
5000 m, air has a temperature of 255.7 K and a pressure of
54.05 kPa. Find To and Po.

40. 40
2
2
o 2
P
2
kJ
m
250
V kg
s
T =T+ =255.7 K+
kJ m
2C 2(1.005 ) 1000
kg×K s
=(255.7+31.1) K
=286.8K
 
 
 
k/(k-1)
o
o
1.4 (1.4-1)
T
P = P
T
286.8 K
=54.05
255.7 K
=80.77 kPa
 
 
 
 
 
 
SOLUTION

41. 41
EFFECT OF FRICTION
 The effects of nozzle friction are
• Reduction in enthalpy drop
• Reheating of steam i.e improving the quality of vapour at
the exit
• Reduction in exit velocity
• Increase in specific volume
• Decrease in mass flow rate

42. 42
NOZZLE PERFORMANCE
Nozzle efficiency is defined as:
“ The ratio of the actual enthalpy drop to the isentropic enthalpy drop
between the same pressures.”
 Due to friction between
• the fluid and the walls of the nozzle
• within the fluid itself
the expansion process is irreversible although still approximately adiabatic.
 In nozzle design:
• all the calculations are based on isentropic flow.
• and then we make friction allowance by using a co efficient or an
efficiency.
For steam flowing through a nozzle, its final velocity for a given pressure
drop is reduces due to:

43. 43
T-S DIAGRAM
Fig: for a vapour Fig: for a perfect gas
 The line 1-2s represents the ideal isentropic expansion.
 The line 1-2 represents the actual irreversible adiabatic expansion.
Expansion between P2 and P1 in a nozzle

44. 44
1 2
1 2
Nozzle efficiency =
s
h h
h h


For a perfect gas this equation is reduces to:
1 2 1 2
1 2 1 2
( )
Nozzle efficiency =
( )
p
p s s
c T T T T
c T T T T
 

 
If the actual velocity at exit from the nozzle is C2 and Velocity at exit
when the flow is isentropic is C2s . Then, using the steady–flow energy
equation in each case we have
2
2
2s
1
1 2s
2 2
2s 1
1 2s
C
C
h + =h +
2 2
or,
C C
h -h =
2

2 2
1 2
1 2
2 2
2 1
1 2
C C
h + =h +
2 2
or,
C C
h -h =
2

………………(1)

45. 45
Substituting these vales in equation (1):
2 2
2 1
2 2
2 1
Nozzle efficiency
s
C C
C C



When inlet velocity C1 is negligibly small then
2
2
2
2
Nozzle efficiency
s
C
C


46. 46
Choked Flow
 The pressure ratio that makes the mass flow rate a maximum is the
same pressure ratio at which the Mach number is unity at the flow
cross-sectional area.
 This value of the pressure ratio is called the critical pressure ratio
for nozzle flow. (For air it is 0.528)
 For pressure ratios less than the critical value, the nozzle is said to
be choked.
 When the nozzle is choked, the mass flow rate is the maximum
possible for the flow area, stagnation pressure, and stagnation
temperature.
 Reducing the pressure ratio below the critical value will not
increase the mass flow rate.

47. Assuming ideal gas behaviour, steady-state choked
flow occurs when the downstream pressure falls
below a critical value. That critical value can be
calculated from the dimensionless critical pressure
ratio equation
For air gamma = 1.4

48. 48
Maximum mass flow

49. 49
PROBLEM 3
Air at 800°R and 80 psia feeds a converging-only
nozzle having an efficiency of 96%. The receiver
pressure is 50 psia. What is the actual nozzle outlet
temperature?

50. 50
Table

51. 51
Note that since prec/pinlet = 50/80 = 0.625 > 0.528, the nozzle
will not be choked, flow will be subsonic at the exit, and p2 =
prec
From table,
M2s ≈ 0.85 and
2
t2s
= 0.8737
s
T
T
 
2s 2s t1
t2s t1 t2s
p p p 50
= = 1 =0.625
p p p 80
 
 
 

52. 52
   o
2s t2s
2s
t2s t1
T T
T = = 0.8737 1 (800) = 699 R
T T
1 2
1 2s
2
T -T
Nozzle efficiency =
T -T
800-T
0.96 =
800-669
Hence,
T2 = 703°R
Can you find the actual outlet velocity?
Another method of expressing nozzle performance is with a velocity coefficient,
which is defined as
v
actual outlet velocity
C =
ideal outlet velocity

53. 53
DIFFUSER PERFORMANCE
 Although the common use of nozzle efficiency makes this parameter well
understood by all engineers, there is no single parameter that is
universally employed for diffusers.
 For a definition of diffuser efficiency analogous to that of a nozzle, we
recall that the function of a diffuser is to convert kinetic energy into
pressure energy; thus it is logical to compare the ideal and actual
processes between the same two enthalpy levels that represent the
same kinetic energy change. Therefore, a suitable definition of diffuser
efficiency is
d
actual pressure rise
=
ideal pressure rise

or from Figure below
2 1
d
2 1
=
s
p p
p p




54. 54

55. 55
PROBLEM 4
Air enters a nozzle at a temperature of 420 K, pressure 0.6 MPa, and a
velocity of 150 m/s. Determine, the exit pressure, exit temperature,
and exit-to inlet area ratio for M = 1 at the exit.
Assumptions
1. Air is an ideal gas with constant specific heats at room temperature.
2. Flow through the nozzle is steady, one-dimensional, and isentropic.
Properties The properties of air are k = 1.4 and cp = 1.005 kJ/kg·K.
Analysis
The properties of the fluid at the location where Ma = 1 are the critical
properties, denoted by superscript *. We first determine the stagnation
temperature and pressure, which remain constant throughout the nozzle
since the flow is isentropic.

56. 56
Table A-32

57. 57
2 2
i
0 i 2 2
p
V (150 m/s) 1 kJ/kg
T = T + = 420K + 431.94 K
2c 2×1.005 kJ/kg. K 1000 m /s
 

 
 
and
k k-1 1.4/(1.4-1)
0
0 i
i
T 431.194 K
P = P = (0.6 MPa) =0.65786 MPa
T 420 K
   
   
 
 
From Table A-32 at Ma = 1, we read T/T0 = 0.8333, P/P0= 0.5283.
Thus,
T = 0.8333T0 = 0.8333(431.194 K) = 359.31 K ≈ 359 K
and
P = 0.5283P0 = 0.5283(0.65786 MPa) = 0.34754 MPa ≈ 0.348 MPa = 348
kPa
Also, 2 2
i i
1000 m /s
c = kRT = (1.4)(0.2587kJ/kg.K)(420K) =410.799 m/s
1 kJ/kg

58. 58
and
i
i
V 150 m/s
M= = = 0.3651
c 410.799 m/s
From Table A-32 at this Mach number we read Ai /A* = 1.7452. Thus the ratio of the
throat area to the nozzle inlet area is
i
1
= = 0.573
1.7452
A
A


59. 59
Air enters a converging nozzle at a specified temperature of 400K and
a pressure of 900 kPa with low velocity. Calculate and plot the exit
pressure, exit velocity, and the mass flow rate versus the back
pressure.
PROBLEM 5
Assumptions
1 Air is an ideal gas with constant specific heats at room temperature.
2 Flow through the nozzle is steady, one-dimensional, and isentropic.
Properties The properties of air are k = 1.4, R = 0.287 kJ/kg·K, and
cp = 1.005 kJ/kg·K.
Analysis The stagnation properties in this case are identical to the inlet properties
since the inlet velocity is negligible. They remain constant throughout the nozzle
since the flow is isentropic,
P0 = Pi = 900 kPa
T0 = Ti = 400 K

60. 60
The critical pressure is determined to be
 
k/k-1 1.4/0.4
*
0
2 2
P = P = 900 kPa = 475.5 kPa
k+1 1.4+1
   
   
   
Then the pressure at the exit plane (throat) will
be
Pe = Pb for Pb ≥ 475.5 kPa
Pe = P* = 475.5 kPa for Pb < 475.5 kPa (choked flow)
Thus the back pressure will not affect the flow when 100 < Pb < 475.5 kPa. For a
specified exit pressure Pe, the temperature, the velocity and the mass flow rate can
be determined from
k-1 k 0.4/(1.4)
e e
e o
o
P P
Temperature T = T = (400 K)
P 900
   
   
 
 

61. 61
2 2
p o e e
1000 m /s
Velocity V = 2c (T -T ) = 2(1.005kJ/kg.K)(400-T )
1 kJ/kg
 
 
 
e e
e 3
e e
P P
Density ρ = =
RT (0.287 kPa.m / kg. K)T
•
2
e e e e e
Mass flow rate m = ρ V A = ρ V (0.001 m )
The results of the calculations are tabulated below

62. 62
Pb ,kPa Pe ,kPa Te ,K Ve ,m/s ρe, kg/m3
m
•
,kg/sec
900 900 400 0 7.840 0
800 800 386.6 162.9 7.206 1.174
700 700 372.3 236.0 6.551 1.546
600 600 356.2 296.7 5.869 1.741
500 500 338.2 352.4 5.151 1.815
475.5 475.5 333.3 366.2 4.971 1.820
400 475.5 333.3 366.2 4.971 1.820
300 475.5 333.3 366.2 4.971 1.820
200 475.5 333.3 366.2 4.971 1.820
100 475.5 333.3 366.2 4.971 1.820

63. 63
SUPERSATURATED FLOW
 When steam flows/expands through a nozzle, it would be normally
expected that the discharge of steam through the nozzle would be
slightly less than the theoretical value.
 But experiments on flow of wet steam show that discharge is slightly
greater than calculated by the formulae. This can be explained as
follows:
 The converging part of nozzle is so short and the steam velocity is so
high that the molecules of steam have insufficient time to collect and form
droplets.
 So, normal combustion does not take place.
 There is a rapid expansion phase, which is said to be metastable and it
produces supersaturated state.
 In supersaturated state, steam is undercooled to a temperature less
than that corresponding to its pressure.
 So, density of the steam increases and hence the weight of
discharge.

64. 64
WILSON LINE
Prof. Wilson showed it experimentally that:
“When dry saturated steam is suddenly expanded in absence of dust it
does not condense until its about Eight times that of the saturated vapour
at the same pressure.”
The limiting condition of under-cooling at which condensation commences
and is assumed to restore conditions of normal thermal equilibrium is called
WILSON LINE.

65. 65
SUPERSATURATED FLOW
Figure: Super Saturated Expansion of Steam in a Nozzle
 The process 1-2 is the isentropic expansion. The change of phase will begin to
occur at point 2
 Consider the h-s diagram as shown in figure.
 But, Vapour continues to expand in a dry state.

66. 66
 Steam remains in this unnatural superheated state until its density is
about eight times that of the saturated vapour density at the same
pressure.
 When this limit is reached, the steam will suddenly condense.
 Point 3 is achieved by extension of the curvature of constant pressure
line p3 from the superheated region which strikes the vertical expansion
line at 3 and through which Wilson line also passes. The point 3
corresponds to a metastable equilibrium state of the vapour.
 The process 2-3 shows expansion under super-saturation condition
which is not in thermal equilibrium. It is also called under cooling
 At any pressure between p2 and p3 i.e., within the superheated zone, the
temperature of the vapour is lower than the saturation temperature
corresponding to that pressure.

67. 67
 Since at 3, the limit of super saturation is reached, the steam will now
condense instantaneously to its normal state at the constant pressure,
and constant enthalpy which is shown by the horizontal line 3-3’ where is
on normal wet area pressure line of the same pressure p3.
 To be noted that 4 and 4’are on the same pressure line.
 3’-4’ is again isentropic, expansion in thermal equilibrium.
Thus the effect of super saturation is to reduce the enthalpy drop slightly
during the expansion and consequently a corresponding reduction in final
velocity. The final dryness fraction and entropy are also increased and the
measured discharge is greater than that theoretically calculated.

68. 68
THANK YOU