Nozzles - Lecture B

Thermodynamics – 2
Fall Semester, 2014-2015
Lecture – B
Nozzles
Dr. Ahmed Rezk
1

Critical velocity
• The critical velocity is the velocity at the throat of a correctly designed
convergent-divergent nozzle, or the velocity at the exit of a
convergent nozzle when the pressure ratio across the nozzle is the
critical pressure ratio or less.
2
1
1 1
2
RatioPressureCritical











P
P
z c
c
1
2
RatioeTemperaturCritical
1 

T
Tc
cc TRV  

Maximum flow rate through the nozzle
• When the back pressure Pb, is equal to the inlet pressure P1, then
there no fluid flow.
• As Pb reduces the mass flow rate through the nozzle increases.
• When the back pressure reaches the critical value pc, then the
velocity at exit of the convergent nozzle is sonic and the mass flow
through the nozzle is maximum and the nozzle is said to be chocked.
3

Maximum flow rate through the nozzle
• Further reduction of the back pressure below the critical value dose
not affect the mass flow which remains at its maximum value. The
exit pressure remains at pc, and expansion to the back pressure
occurs outside the nozzle.
• For a convergent nozzle, the maximum mass flow is obtained when the
pressure ratio across the nozzle is the critical pressure ratio
• For convergent-divergent nozzle, the flow is maximum and fixed when the
pressure at the throat is critical pressure.
4

Convergent nozzle
5
Pb>Pc Over-expansion m<mmax
Pb=Pc Chocked Nozzle m=mmax
Pb<Pc Under-expansion m=mmax

Example
A fluid at 6.9 bar and 93°C enters a convergent nozzle with negligible velocity, and
expands isentropically into a space at 3.6 bar. Calculate the mass flow per square
meter of exit area.
a. When the fluid is helium (Cp = 5.23 kJ/kg K), (ɣ = 1.66).
b. When the fluid is ethane (Cp = 1.66 kJ/kg K), (ɣ= 1.2).
6

Example
• For helium gas, check the chocked nozzle conditions.
• The actual back pressure is 3.6 bar, which is higher than the critical pressure. At
this condition the nozzle is not chocked.
• Apply the isentropic relation between the actual inlet and outlet conditions.
7
488.0
66.2
2
1
2 66.0
66.1
1
1
















P
Pc
 barPc 37.3488.09.6 

Example
8
772.0
9.6
6.3 66.1
66.01
1
2
1
2
















P
P
T
T
   KT 5.282772.0273932 
 kgm
P
RT
v 3
2
2
2 63.1
1006.3
5.282079.2




       smTTChhV p 63.15.28236623.572.4472.4472.44 21212 
 smkg
v
V
A
m 2
2
2
2
/573
63.1
935



Example
• For ethane gas, check the chocked nozzle conditions.
• The actual back pressure is 3.6 bar, which is lower than the critical pressure. At
this condition the nozzle is chocked.
• Apply the isentropic relation between the actual inlet and outlet conditions.
9
566.0
2.2
2
1
2 2.0
2.1
1
1
















P
Pc
 barPc 91.3566.09.6 

Example
10
91.0
2.2
2
1
2
1



T
Tc
   KTc 7.33291.027393 
 kgm
P
RT
v
c
c 3
2 23.0
10091.3
7.332277.0




 smRTVV c /7.3327.33210277.02.1 3
2  
 smkg
v
V
A
m 2
2
2
2
/1412
23.0
7.332



Convergent-divergent nozzle
11
Chocked nozzle
Normal shock
wave
Oblique shock wave
Oblique expansion wave

12
• When the discharge pressure is slightly lower than the inlet pressure (case-a), the
flow is subsonic throughout the nozzle. In this case, the lowest pressure and
highest velocity occur at the nozzle throat where the cross-sectional area reaches
a minimum.
• As the discharge pressure is decreased (critical back pressure), the velocity at the
throat continues to increase until eventually it reaches the local speed of sound
at the throat (case-b). At this conditions the nozzle is chocked and the mass flow
rate reaches its maximum value.
• As the discharge pressure is reduced still further (case-c), the mass flow rate
through the nozzle remains constant but the flow accelerates from the speed of
sound at the throat to supersonic velocities in the diverging section of the nozzle.
At some point downstream of the throat, the flow suddenly changes from being
supersonic to being subsonic (shock wave).

13
• The shock wave occurring inside the nozzle is called a normal shock because it is
perpendicular to the flow direction. Note that shock waves can only occur at
locations where the upstream flow is supersonic and the downstream of the
shock wave, the flow must be subsonic.
• The location of the normal shock moves towards the nozzle exit as the discharge
pressure is further reduced. (Case-d) illustrates the behavior for the back
pressure value that causes a shock wave exactly at the nozzle exit.
• If the back pressure is decreased still further, an oblique shock wave occurs
outside of the nozzle as the pressure adjusts from the supersonic condition at the
nozzle exit to the imposed back pressure (case-e). The oblique shock wave will, in
general, not be perpendicular to the flow. This condition of a discharge pressure
that is lower than the imposed back pressure is referred to as overexpansion.

14
• When the back pressure is reduced to the design pressure (case-f), supersonic
flow leaves the nozzle with an exit pressure that is exactly equal to the applied
back pressure and no shock occurs. If the back pressure is reduced to a value that
is below the design pressure (case-g), an oblique expansion wave results. This
behavior is not a pressure discontinuity but rather a continuous wave that is a
result of the flow exiting the nozzle coming to equilibrium with the low external
pressure. This situation is called under-expansion.

Nozzle efficiency
15ss hh
hh
VV
VV
21
21
2
1
2
2
2
1
2
2







Nozzle efficiency
16
• The ideal process for the expansion in the nozzle is the reversible adiabatic
(isentropic) process. Due to frication either between the fluid and the walls of the
nozzle or within the fluid itself, the expansion process is irreversible but still
adiabatic.
• In nozzle design it is common to build all calculations on isentropic flow and then
taking the frication into consideration through an efficiency called the adiabatic
or the isentropic efficiency.

Calculation coefficients.
17
• Nozzle velocity coefficient CV.
• The nozzle velocity coefficient is defined as the ratio of the actual velocity at exit
from the nozzle to the velocity at exit when the flow is isentropic (ideal).
• Nozzle coefficient of discharge Cd.
• The coefficient of discharge of a nozzle is defined as the ratio of the actual mass
flow through the nozzle (m), to the mass flow through the nozzle if the flow is
isentropic (mS).
sV
V
CV
2
2

sm
m
Cd 

Nozzle length restrictions
18
• To void the separation of the fluid from the nozzle wall the divergent portion of
the nozzle is usually made very long compared with the convergent portion with
a divergent angle less than or equal 20°.
• Because the divergent portion is made very long and velocities in this portion are
higher, most of the friction losses take place in this portion. In fact it is reasonable
to assume that all friction losses occur after the throat of the nozzle. This
assumption implies that friction does not affect the mass flow through the nozzle
is choked.

Example
19
• Gases expand in a propulsion nozzle from 3.5 bar and 425°C down to a back
pressure of 0.97 bar at the rate of 18 kg/s. Taking a coefficient of discharge of
0.99 and a nozzle efficiency of 0.94, calculate the required throat and exit areas
of the nozzle. For the gases take Cp = 1.11 KJ/Kg and ɣ = 1.333. Assume that the
inlet velocity is negligible.
• Solution: The critical pressure ratio and the corresponding critical pressure are:
• Since Pb < Pc, then the nozzle is chocked.
541.0
333.2
2
1
2 333.0
333.1
1
1
















P
Pc
 barPc 894.1541.05.3 

Example
20
• All these calculations based on the isentropic assumption.
857.0
333.2
2
1
2
1



T
Tc
   KTc 598857.0273425 
 smRTVc /471598102775.011.1 3
 
 kgm
P
RT
v
c
c
c
3
875.0
100984.1
598277.0





Example
21
• Based on the coefficient of discharge we can find the isentropic mass flow rate as:
sm
m
Cd 
sm
18
999.0   skgms 18.18
c
cc
s
v
VA
m 
 2
0338.0
471
875.018.18
mAc 



Example
22
• Calculate the area at the exit section:
• Apply isentropic efficiency:

 1
1
2
1
2








P
P
T
T s
 KT s 506
5.3
97.0
698
333.1
333.0
2 






506698
698
94.0 2
21
21






T
TT
TT
s

 KT 5.5172 

Example
23
 kgm
P
RT
v 3
2
2
2 48.1
10097.0
5.5172775.0




     smTTCV p 6335.5176981011.172.4472.44 3
212 
2
22
v
VA
m 
 2
2 0422.0
633
48.118
mA 



Nozzles - Lecture B

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