Problems
1. Gaseous Propane at 25ºC is burned with moist air at 400 K in a steady state,
steady flow process. The combustion process is adiabatic, and the exiting
temperature is measured to 1200 K. A sample of the products is tested and
found to have a dewpoint temperature of 70ºC. Determine the percentage of
theoretical air used and the relative humidity of this air. Assume the combustion
is complete and that the pressure is 100 KPa throughout the process.
2. A mixture of 80% ethane and 20% methane on a mole basis is throttled
from10MPa, 65ºC, to 100 KPa and is fed to a combustion chamber where it
undergoes complete combustion with air, which enters at 100 KPa , 600 K. The
amount of air is such that the products of combustion exit at 100 KPa , 1200 K.
Assuming that the combustion process is adiabatic and that all components
behave as ideal gases except the fuel mixture, which behaves according to the
generalized tables or charts, with pseudocritical constants, determine :
a) the percentage of theoretical air used in the process
b) the dew-point temperature of the products
Solution:
1)
Notations:
Nx = number of kmols of component (x)
Mx = molar mass of component (x) (kg/kmol)
rx = the molar (volumetric) percentage of component “x” in a mixture
Q = molar heat exchange (kJ/kmol)
 = universal constant of gases,  = 8.314 kJ/kmol.K
Cp = molar specific heat at constant pressure (kJ/kmol.K)
cp = mass specific heat at constant pressure (kJ/kg.K)
 = air excess coefficient
 = volumetric ratio nitrogen/oxygen of air composition,  = 3.762
d = moisture content in the air, d = kg vapors / kg dry air
 = relative humidity of air (mixture)
The stoichiometric oxidation reaction of propane with dry air is:
C3H8  5O2  N2   3CO2  4H2O  5N2
( 1)
while the actual reaction with dry air in excess is
C3H8  5O2  N2   3CO2  4H2O  5  1O2  5N2
( 2)
Now, we consider a humid air coming into reaction. If (d) is the moisture content,
we can compute the molar fraction of water content in the air:
M
N M
Nw
M
29
d w  w w

d a d
 1.61d
Ma
N aM a
Na
Mw
18
Thus, for every kmole of dry air, we have 1.61d kmoles of water (gaseous).
Therefore, our equation becomes:
C3H8  5O2  N2   8.051   dH2O 
 3CO2  4  8.051   dH2O  5  1O2  5N2
( 3)
where
8.05 = 5 x 1.61
We have 2 parameters which need to be determined:
() = excess of air
d = the degree of moisture
The conditions which allow to compute these parameters are:
the dew point of products
the temperature of products
If we know the dew point of products, we can find out by means of Steam Tables
the partial pressure of water vapors in the exhaust mixture.
Thus, at a saturation temperature of 70 deg. Celsius, we find that the
corresponding saturation pressure is
ps = 0.312 bar
( 4)
According to Dalton’s law, we have
pi 
Ni
p
N
( 5)
for every component (i) in a mixture.
Since the exhaust gases pressure is 100 kPa = 1 bar, that means
N H 2O  0.312  N 
( 6)
where (N) = total number of kmoles of exhaust products:
N   3  4  8.051  d  5  1  5  2  8.051  d  51  
( 7)
But
N H 2O  4  8.051   d
( 8)
so that
4  8.051  d  0.3122  8.051  d  51  


( 9)
4  8.051  3.762d  0.624  2.511  3.762d  1.561  3.762
4  38.33d  0.624  11.95d  7.43

26.38d  7.43  3.376  d  0.282 
0.128

Now, we need to apply the energy equation which allows to determine the
exhaust temperature (that we know)
Since the burning process takes place at constant pressure, we can write:
( 10)




 Qreaction   N i Cp i 
 Tf
 N i Cp i Ti 
 i
 reac tan ts
 i
 products
( 11)
where (Qreaction) will be expressed in kJ/kmole fuel, since we have 1 kmole
propane in equation, and Tf = final temperature.
From databases, we find the following:
H2Ovap
cp = 2.84 kJ/kg.K

Cp  2.84  18  51.12 kJ/kmol.K
CO2
cp = 1.37 kJ/kg.K

Cp  1.37  44  60.28 kJ/kmol.K
N2
cp = 1.28 kJ/kg.K

Cp  1.28  28  35.84 kJ/kmol.K
O2
cp = 0.906 kJ/kg.K

Cp  0.906  32  29 kJ/kmol.K
Air
cp = 1.005 kJ/kg.K

Cp  1.005  29  29.15 kJ/kmol.K
C3H8
cp = 1.68 kJ/kg.K

Cp  1.68  44  73.92 kJ/kmol.K
Qreaction = 2210000 kJ/kmolfuel
By replacing in (1), we get


 Qreaction
 N i Cp i Ti 
 i
 reac tan ts
Tf 



 N i Cp i 
 i
 products
73.92  25  273)  5  (1  3.762)  29.15  400  8.05d  (1  3.762)  51.12  400   2210000
Tf 
3  60.28  4  8.05  1  3.762d   51.12  5  1  29  5  3.762  35.84

22028.16  277624.6  783855.7d  2210000
 1200
240.32  819.15  1959.6d
( 12)

22028.16  277624.6  783855.7d  2210000  288384  98298  2351520d
1567664.3d  1943644.16  179326.6

( 13)
By replacing (d) with (10), one yields
0.128 

1567664.3 0.282 
  1943644.16  179326.6
 


442081.3  200661  1943644.16  179326.6
262754.7  2144305.16

d  0.282 

  8.161
0.128
 0.266
8.161
( 14)
( 15)
The formula which relates the relative humidity () to the moisture content (d) is:
d  0,622 
  ps
p    ps
( 16)
where
ps = saturation pressure at given temperature (in our case is 0.312 bar)
p = actual pressure of the humid air (or mixture)

d ( p    ps )  0.622  ps 



d
p

d  0.622 ps
( 17)
0.266
1

 0.96  96%
0.266  0.622 0.312
2)
a)
Throttling real gases is a complex process, which requires additional
information, like what type of transformation the gas is submitted to. It could be
adiabatic, isenthalpic or isentropic, depending of the throttle.
For us, it is important to know the temperature after throttling, in order to compute
the enthalpy addiction in the burning process.
However, since the temperature after throttle is much less than 65 deg Celsius,
we will simplify the calculations and will neglect the enthalpy of the fuel mixture.
We will consider only the enthalpy of the air and the combustion heat which,
together, will determine the exhaust temperature. The error will be less than 2%.
Since the molar ratio ethane/methane is 80% / 20% = 4 : 1, the chemical reaction
(stoichiometric) will be described by the equation
4C2H6  CH4  16O2  N2   9CO2  14H2O  16N2
( 18)
If we have an air excess defined by factor (), then he above equation becomes
4C2H6  CH4  16O2  N2   9CO2  14H2O  16  1O2  16N2
( 19)
The energy equation which allows to determine the exhaust temperature (that we
know) is the following, since the burning process takes place at constant
pressure:




i
  N i Qreact
  N i Cp i 
 Tf
 N i Cp i Ti 
i
 i
 reac tan ts
 i
 products
( 20)
where (Qireact) is the molar heat of reaction for the fuel “i”, which is expressed in
kJ/kmole fuel and Tf = final temperature.
By developing the above equation according to (19) and neglecting the enthalpy
of fuel, we can write:


16 Cp O 2  Cp N 2  Tair  4Q C 2H 6  Q CH 4 


 9Cp CO 2  14Cp H 2O  16  1Cp O 2  16Cp N 2  Tf
We need to solve this equation in ():
( 21)


16 Cp O 2  Cp N 2 Tf  Tair  


 4Q C 2H 6  Q CH 4  9Cp CO 2  14Cp H 2O  16Cp O 2 Tf



( 22)

4Q C 2H 6  Q CH 4  9Cp CO 2  14Cp H 2O  16Cp O 2 Tf


16 Cp O 2  Cp N 2 Tf  Tair 
( 23)
From Thermodynamic databases, we find:
(Q)C2H6 = 1560000 kJ/kmol
(Q)CH4 = 890000 kJ/kmol
H2Ovap
cp = 2.84 kJ/kg.K

Cp  2.84  18  51.12 kJ/kmol.K
CO2
cp = 1.37 kJ/kg.K

Cp  1.37  44  60.28 kJ/kmol.K
N2
cp = 1.28 kJ/kg.K

Cp  1.28  28  35.84 kJ/kmol.K
O2
cp = 0.906 kJ/kg.K

Cp  0.906  32  29 kJ/kmol.K
By replacing these values in (23), we get
4  1560000  890000  9  60.28  14  51.12  16  29   1200

 3.927
16  29  3.762  35.84   1200  600 
Remark:
In general, when burning fuels with excess of air, one neglect the enthalpy of the
fuels at inlet, since this is much less than the other relevant amounts of energy
involved in the energetic balance.
For example, I was dealing with burning process in the combustion chambers of
jet engines which use kerosene as fuel ( a mixture of liquid hydrocarbons) and I
never considered the enthalpy of the fuel at inlet. The results that I’ve got were in
good accordance with experimental data.
b)
In order to find the dew point of exhaust products, we need to know the
molar fraction of water vapors in the exhaust mixture, which allow to compute the
partial pressure of them.
According to (19), we have
rH 2O 
N H 2O
14

N
9  14  16  1  16
( 24)
By replacing with numercal values, we get
rH 2O 
14
 0.0457
9  14  16  3.927  1  16  3.762  3.927
( 25)
Therefore, according to Dalton’s law, the partial pressure of water vapors in the
exhaust mixture will be:
pg  100 kPa  1 bar

pH 2O  0.0457 bar
Considering this as a saturation pressure, from Steam Tables we can find the
corresponding saturation temperature:
TS = 304.4 K = 31.4C
This is the dew point of exhaust mixture that we were looking for.