Converging-diverging nozzle
Solution:
First off all, I have to specify that I prefer to work in SI units. I don’t think that it is
a problem to convert the SI units in English-American ones.
C
A
B
VA
VB
VC
B
A
C
a) The inlet conditions are the standard atmospheric conditions, which means
pA  1.013 bar  1.013  105 N/m 2
TA  288 K (  15C)
(1)
From here, we can get the density at inlet station, by applying the ideal gas law:
A 
pA
R TA
(2)
where R = the gas constant (in our case, this is the air thermodynamic constant).
The general formula for computing the gas constant is the following:
R

M
(3)
where  = the universal gas constant, which is  = 8.314 kJ/kmol.K
M = the molecular weight of the gas, which is M = 29 kg/kmol (for air)
As a result, the air constant will be:
8.314
R
 0.2867 kJ/kg.K  286.7 J/kg.K
29
Introducing the numerical values in (2), we get the following:
A 
1.013  105
 1.227 kg/m 3
286.7  288
(4)
The speed at station (A) is VA = 380 ft/s = 115.8 m/s and the cross area of
section A-A is
AA 
 2
DA
4
(5)
where DA = 5” = 5x25.4 = 127 mm = 0.127 m.
By replacing in equation (5), we will have the following:
AA 

 0.1272  0.0127 m2 (  19.63 in2 )
4
(6)
Now, we are able to compute the air mass flow through the nozzle, by using the
formula:
M a  A AAVA

M a  1.227  0.0127  115.8  1.804 kg/s (  3.977 lbm/s)
(7)
(8)
b)
In order to compute the velocities at stations (B) and (C), we will apply the
mass flow conservation law:
M A  M B  M C

 A AAVA  B ABVB  C ACVC
We can compute easily the cross section areas at stations (B) and (C):
DB  2"  2  25.4  50.8 mm  0.0508 m
DC  6"  6  25.4  152.4 mm  0.1524 m
 2 
DB   0.05082  0.002 m2 (  3.14 in2 )
4
4


AC  DC2   0.15242  0.0182 m2 (  28.27 in2 )
4
4
AB 
The densities at stations (B) and (C) are:
0.4536
B  0.060 lbm/ft 3  0.060 
kg/m 3  0.961kg/m 3
3
0.3048
0.4536
C  0.050 lbm/ft 3  0.050 
kg/m 3  0.801kg/m 3
3
0.3048
By using (8) and (10), one yields:
M a
1.804
VB 

 938.6 m/s (  3079.4 ft/s)
B AB 0.961 0.002
M a
1.804
VC 

 123.7 m/s (  405.8 ft/s)
C AC 0.801 0.0182
(9)
(10)
c)
The pressure at station (B) will be computed, using the ideal gas law:
pB  BR TB
(11)
where
TB  30.5F 

5
 30.5  32  0.83C  - 0.83  273  272.17 K
9
pB  0.961 286.7  272.17  0.75  105 N/m 2  0.75 bar
(12)
(13)
In order to compute the pressure at station (C), we will apply the isentropic law:
pB
Bk

pC
 
pC  pB  C 
 B 

Ck
k
(14)
where k = adiabatic exponent:
k
cp
cV
= 1.4 for air
(15)
1.4
 0.801

pC  0.75  
  0.581bar
 0.961
d)
The potential energy is a result of gravity or other fields, which are not
present in our problem. Therefore, the potential energy will be neglected in the
energetic balance.
If we denote
h*  h 
V2
p V2
u 
2

2
(16)
we obtain the total enthalpy, thus the energy conservation law leads to:
h*  const.
(17)
So, if we compute the total enthalpy, we can apply the above equation (16) and
find out all kinds of energies involved in the process.
h*  h 
V2
V2
k
V2
 c pT 

RT 
2
2
k 1
2
(18)
where cp = specific heat at constant pressure
and
h  h0  c pT  h0 
kR
T
k 1
h0 = a reference value, which usually is taken zero.
(19)
For station (A) we will have the following:
h* 
1 .4
115.82
 286.7  288 
 295698 J/kg  295.7 kJ/kg
1 .4  1
2
VA2
115.82
 295.7 
 289 kJ/kg
2
2  103
VB2
938.62
hB  h * 
 295.7 
 144.8 kJ/kg
2
2  103
(The enthalpy can be negative, since it is a relative feature, depending on a
reference level of the total enthalpy, which in our case, was taken as zero)
V2
123.72
hC  h *  C  295.7 
 288 kJ/kg
2
2  103
The kinetic energies (for one mass flow unit) are:
VA2 115.82
ek , A 

 6704.8 J/kg  6.7 kJ/kg
2
2
V 2 938.62
ek ,B  B 
 440485 J/kg  440.5 kJ/kg
2
2
V 2 123.72
ek ,C  C 
 7650.8 J/kg  7.7 kJ/kg
2
2
From (16), the specific internal energy values will be:
p
1.013  105 1
u A  hA  A  289 
 3  206.4 kJ/kg
A
1.227
10

hA  h * 
uB  hB 
pB
0.75  105 1
 144.8 

 222.8 kJ/kg
B
0.961 103
uC  hC 
pC
0.581 105 1
 288 
 3  215.5 kJ/kg
C
0.801
10
Final remark: The above results are specific features; if we want to calculate
absolute features, we have to multiply with the actual mass flow rate.
I hope this is the answer that you are looking for.
(20)