Solution 07

8/3/2019 Solution 07

http://slidepdf.com/reader/full/solution-07 1/19



=! " #

$% & '

HHV

ch

~

68,315 cal/mol

We can find the lower heating value by once again using the heat of vaporization, giving

=! " # $

% & '

LHV

ch

~

57,795 cal/mol

This gives a difference of 0.87% for LHV and 0.71% for HHV, showing that they are in

excellent agreement.

c) Find the corresponding values of !hc( )

HHV and !h

c( ) LHV

, namely per unit mass of H2.

To convert to mass-specific enthalpy, divide the molar-specific enthalpy by the molecular

weight of H2, namely 2.016 g/mol, or 2.016(10-3) kg/mol, which gives

( )

( ) kg cal mol kg

mol cal h

kg cal mol kg

mol cal h

LHV c

HHV c

/)10(89.2/)10(016.2

/300,58

/)10(41.3/)10(016.2

/800,68

7

3

7

3

==!

==!

"

"

d) Use the result from part (c) and the curve fit for !c p (T )!" #$ H 2Oin the combustion products to

obtain a value for the “adiabatic flame temperature” for stoichiometric combustion of

gaseous hydrogen in oxygen when both reactants are initially at 300 K.

The curve fit table gives, for water:

!c p (T ) = !a + !bT + !cT 2cal / (mol ! K )

where !a = 7.256, !b = 2.298(10-3), and !c = 0.283(10-6). Using the molecular weight of water,

18(10-3) kg/mol, we convert this to the mass-specific heat for water:

c p (T ) =!c p (T )

MW H 2O=

!a + !bT + !cT 2

18(10!3) kg / mol

= a + bT + cT 2cal / (kg ! K )

where a = 403.11, b = 0.128, and c = 1.57(10-5).

We can now set the enthalpy increase in the products as the temperature rises from T i = 300

K to the adiabatic flame temperature T f , to be equal to the enthalpy of combustion produced

by the reaction. However the lower heating value computed above is per kg H2, and the

c p(T ) above is per kg H2O. Since we know that 1 mole of H2 forms 1 mole of H2O, we can

say that 2.016 kg H2 will form 18 kg H2O, and multiply each side by the appropriate values



2.016 ! "hc( ) LHV

= 18 ! dh#

= 18 ! c p (T ) dT T i

T f

#

= 18 ! a + bT + cT 2 dT T i

T f

#

= 18 ! a(T f $ T i ) +b

2(T f

2 $ T i2) +

c

3(T f

3 $ T i3)

%

&'(

)*

which now allows us to solve for T f . We can use an iterative guess-and-check method, a

simple spreadsheet, or a tool like Matlab to do this, and we find that

= f T 4328 K

Notice that, strictly speaking, this temperature is out of the range of valid temperatures for

the curve fit that we used, which the table says is valid only for T < 1500 K. Thus we should

really do this with a curve fit that applies even at these high temperatures, though the

procedure is exactly the same as we did here. The difference is likely not to be tremendous,

and our main point in this problem is to see the method used to compute this adiabatic flame

temperature.

e) Use the results above to obtain a value for the “adiabatic flame temperature” for

stoichiometric combustion of gaseous hydrogen in air when both reactants are initially at

300K. Keep in mind that air is composed of 21% O2

and 79% N2

, and that N2

is essentially

inert due to the high bond energy in the triple bond required by its valence state.

This problem is very similar to part (d), but now our product composition has changed:

( )22222

2

76.376.3

2

1 N O H N O H +!++

Since we are burning the same fuel, and the nitrogen is inert, we have the same enthalpy of

combustion as in part (d). But we must now calculate the specific heat for the products.

From the curve fit table:

!c p (T )!" #$ H 2O= !a

1+ !b

1T + !c

1T

2

!c p (T )!" #$ N 2= !a

2+

!b2T + !c

2T

2

where !a1= 7.256, !b

1= 2.298(10 –3), !c

1= 0.283(10 –6), !a

2= 6.524, !b

2= 1.250(10 –3), and !c

2= –

0.001(10 –6). Using what we learned earlier for the thermodynamics of mixtures



!c p (T )!" #$ products = X H 2O !c p (T )!" #$ H 2O+ X N 2 !c p (T )!" #$ N 2

= X H 2O !a1+ X N 2 !a

2( )+ X H 2O!b1+ X N 2

!b2( )T + X H 2O !c

1+ X N 2 !c2( )T 2

% !a0 +

!b0T + !c

0T 2

Since X H 2O

= 1/(1+ 3.76/2) = 0.347 and X N 2

= 1 – X H 2O

= 0.653, this gives !a0

= 6.778, !b0

=

1.614(10-3), and !c0= 0.0975(10-6).

As in part (d) we convert to mass-specific heat by dividing by the molecular weight, namely

MW products = X H 2O MW H 2O + X N 2 MW N 2 = 24.53 g / mol

and then

c p (T )!" #$ products =!c p (T )!" #$ products

MW products

= a + bT + cT 2cal / (kg-K )

where a = 276.32, b = 0.0658 and c = 3.97(10-6).

As before, we now set the enthalpy increase in the products as the temperature rises (from T i

= 300 K to the adiabatic flame temperature T f ) to be equal to the enthalpy of combustion

produced by the reaction. In this case, one mole of H2 will form 2.88 moles of products,

therefore 2.016 kg of H2 will form 2.88* 24.53 = 70.646 kg of products, thus

2.016 ! "hc( ) LHV

= 70.646 ! dh#

= 70.646 ! c p (T ) dT T i

T f

#

= 70.646 ! a + bT + cT 2 dT T i

T f

#

= 70.646 ! a(T f $ T i ) +b

2(T f

2 $ T i2) +

c

3(T f

3 $ T i3)

%

&'(

)*

which allows us to solve for T f . Again using an iterative guess-and-check method, a simple

spreadsheet, or a tool like Matlab, we find that

= f T 2487 K

Notice that this adiabatic flame temperature for combustion in air is significantly lower than

the result we got in part (d) for combustion pure oxygen. This is due to the fact that, while

the amount of enthalpy released from burning the H2 is the same, the presence of the N2 in

the products means that the enthalpy is distributed over a far greater number of molecules.

Since the reaction has to heat up more products, the final temperature is lower.



f) Use the results above to obtain a value for the “adiabatic flame temperature” when gaseous

hydrogen is burned with three times the stoichiometric volume (hence moles) of air when

both reactants are initially at 300 K.

As in part (e), we must first find the product composition. Since we are burning with threetimes the number of moles of air needed to burn all of the H2, the overall reaction must be

( )222222

64.576.32

3 N OO H N O H ++!++

Since we now have excess oxygen remaining in the product mixture, we need to include its

contribution to the specific heat of the products. From the curve fit table, the specific heat

for oxygen:

!c p (T )!" #$O2

= !a3 +

!b3T + !c

3T

where !a3

= 6.148, !b3

= 3.102(10 –3) and !c3

= –0.923(10 –6). Using the same mixture formulasas in part (e), but now with X

H 2O

= 0.131, X O2

= 0.131 and X N 2

= 0.738 gives

!c p (T )!" #$ products =!a0+

!b0T + !c

0T 2

where !a0

= 6.571, !b0= 1.630(10 –3) and !c

0= –0.0846(10 –6).

As in part (d) we convert to mass-specific heat by dividing by the molecular weight:

MW products

= X H 2O

MW H 2O

+ X N 2 MW

N 2+ X

O2

MW O2

= 27.21g/mol

c p (T )!" #$ products =!c p (T )!" #$ products

MW products= a + bT + cT 2 cal/(kg-K)

where a = 241.49, b = 0.060 and c = –3.109(10 –6). As before, we set the enthalpy of

combustion equal to the enthalpy change of the products rising from T i = 300 K to the

adiabatic flame temperature T f , as

2.016 ! "hc( ) LHV

= 207.21 ! dh#

= 207.21 ! c p (T ) dT T i

T f

#

= 207.21 ! a + bT + cT 2 dT T i

T f

#

= 207.21 ! a(T f $ T i ) +b

2(T f

2 $ T i2) +

c

3(T f

3 $ T i3)

%

&'(

)*



which allows us to solve for T f using an iterative “guess-and-check” method, or a simple

spreadsheet, or a tool like Matlab. We find that

= f T 1279 K

Note that since we are burning fuel lean (i.e., with significantly more air needed to burn all of

the fuel), we are heating up excess products and thus the adiabatic flame temperature has

decreased substantially. Gas turbine combustors routinely operate with substantial amounts

of such excess air to keep the combustion product temperature (T 3 in our Brayton cycle

analysis) at values that the turbine blades can survive.

2. Using CEA for Constant-Pressure Combustion Processes

In this problem, you will use CEA to examine constant-pressure combustion processes usingfour different fuels. Consider the following four fuels: hydrogen (H2), methane (CH4),

propane (C3H8) and n-butane (C4H10), each in the gaseous state.

a) Begin with gaseous hydrogen (H2) fuel. Let’s first examine a simple case, where the pressurein 1 atm and both the fuel and air are initially at 300 k. We will use CEA to determine the

temperature that results when the fuel is burned with air at constant pressure and without heatloss.

• Hydrogen

Overall methane-air combustion balance is:

H 2+1

2O

2+ 3.76 N

2( )! H 2O+

3.76

2 N

2

The molecular weights of the reactants are

MWH2 = 2, MWO2 = 32, and MW N2 = 28

So the fuel mass is 2 and the air mass is 0.5(32 + 3.76 x 28) = 68.64, giving

%28.2028.0

2

46.681

1

1

1==

+

=

+

=

fuel

air

mm

fractionmass fuel

Thus the stoichiometric fuel mass fraction is 2.8%, but the maximum will actually vary from

this. In order to find where the maximum temperature occurs, use the following 16 valuesfor fuel mass fraction.

% f =1.0, 1.5, 2.0, 2.5, 2.8, 2.9, 3.0, 3.1, 3.2, 3.5, 4.0, 4.5, 5.0, 6.0, 8.0, 10.0



The plot and the data below indicate that the maximum temperature occurs at a value slightly

higher than stoichiometric. The maximum temperature can be seen to occur at a fuel massfraction of approximately 3.0 with a temperature of 2395 K.

Problem 1 Part A

0

500

1000

1500

2000

2500

3000

0 2 4 6 8 10

Fuel Mass Fraction

T e m p e

r a t u r e ( k )

% Fuel Temperature (k)

1.00E+00 1.30E+03

1.50E+00 1.69E+03

2.00E+00 2.01E+03

2.50E+00 2.27E+03

2.80E+00 2.37E+03

2.90E+00 2.39E+03

3.00E+00 2.40E+03

3.10E+00 2.39E+03

3.20E+00 2.39E+03

3.50E+00 2.35E+03

4.00E+00 2.27E+03

4.50E+00 2.20E+03

5.00E+00 2.13E+03

6.00E+00 2.00E+03

8.00E+00 1.78E+03

1.00E+01 1.60E+03

Table 1: Data for Problem 1 part (a)



b) Now you will repeat this calculation for conditions that are more typical of a gas turbinecombustor. Thus take the air entering the combustor to be at a pressure of 40 atm and a

temperature of 700 K. Take the fuel entering the combustor to be at the same pressure but ata temperature 400 K.

The plot given below shows that increasing the pressure/temperature of the fuel/air cause the

maximum temperature to also increase. Notice that the fuel mass ratio that the maximumtemperature occurs at approximately 3.0%, but the temperature has increased to 2676 K.

Problem 1 Part B

0

500

1000

1500

2000

2500

3000

0 2 4 6 8 10 12

Fuel Mass Fraction

T e m p e r a t u r e ( k )


1.00E+00 1.62E+03

1.50E+00 1.97E+03

2.00E+00 2.28E+03

2.50E+00 2.54E+03

2.80E+00 2.66E+03

2.90E+00 2.67E+03

3.00E+00 2.68E+03

3.10E+00 2.67E+03

3.20E+00 2.66E+03

3.50E+00 2.61E+03

4.00E+00 2.52E+03

4.50E+00 2.44E+03

5.00E+00 2.36E+03

6.00E+00 2.22E+03



1.50E+00 1.97E+03 2.00E+00 1.50E+03 2.00E+00 1.46E+03 4.00E+00 2.07E+03

2.00E+00 2.28E+03 4.00E+00 2.15E+03 4.00E+00 2.08E+03 5.00E+00 2.34E+03

2.50E+00 2.54E+03 4.50E+00 2.29E+03 5.00E+00 2.35E+03 5.50E+00 2.46E+03

2.80E+00 2.66E+03 5.00E+00 2.41E+03 5.50E+00 2.47E+03 5.70E+00 2.50E+03

2.90E+00 2.67E+03 5.20E+00 2.46E+03 5.70E+00 2.51E+03 5.90E+00 2.53E+03

3.00E+00 2.68E+03 5.30E+00 2.47E+03 5.80E+00 2.52E+03 6.00E+00 2.54E+03

3.10E+00 2.67E+03 5.40E+00 2.49E+03 5.90E+00 2.54E+03 6.10E+00 2.55E+03

3.20E+00 2.66E+03 5.50E+00 2.50E+03 6.00E+00 2.55E+03 6.20E+00 2.56E+03

3.50E+00 2.61E+03 5.60E+00 2.51E+03 6.10E+00 2.56E+03 6.30E+00 2.56E+03

4.00E+00 2.52E+03 5.70E+00 2.51E+03 6.20E+00 2.56E+03 6.40E+00 2.56E+03

4.50E+00 2.44E+03 5.80E+00 2.50E+03 6.30E+00 2.56E+03 6.50E+00 2.56E+03

5.00E+00 2.36E+03 6.00E+00 2.48E+03 6.50E+00 2.54E+03 6.80E+00 2.53E+03

6.00E+00 2.22E+03 6.50E+00 2.40E+03 7.00E+00 2.48E+03 7.00E+00 2.50E+03

8.00E+00 1.98E+03 8.00E+00 2.16E+03 8.00E+00 2.33E+03 8.00E+00 2.36E+03

1.00E+01 1.79E+03 1.00E+01 1.86E+03 1.00E+01 2.05E+03 1.00E+01 2.07E+03

Table 3: Data for Problem 1 part (c)

d) Now let’s consider a more realistic model for Jet-A fuel, composed of equal molar parts of

iso-octane (C8H18), naphthalene (C10H8), and biphenylene (C12H10). Take the air enteringthe combustor to be at 40 atm and 700 K, and the fuel to be at 40 atm and 400 K. First let’s

look at the stoichiometric fuel mass fraction for Jet-A fuel.

For the Jet-A fuel blend problem, find the stoichiometric fuel mass fraction as:

1C 8 H

18+1C

10 H

8+1C

12 H

10! 1C

30 H

36

The general equation for stoichiometry is

1C n H

m+ n +

m

4

! " #

$ % & O

2+ 3.76 N

2( )' nCO2+m

2 H

2O + 3.76 n +

m

4

! " #

$ % & N

2

Thus the stoichiometric air-to-fuel mass ratio is

ma

m f

=

n +m

4

! " #

$ % &

32 + 3.76 '28( )

n '12 + m '1= 137.28

n +m

4

! " #

$ % &

12n + m( )

Thus the stoichiometric fuel mass fraction is

m f

m f + m

a

=1

1+m

a

m f

=12n + m

149.28n + 35.32m

For our Jet-A surrogate blend specified above, n = 30 and m = 36, and thus the stoichiometric

fuel mass fraction is 0.0689 = 6.89%. This number compares well to what was found above.



Using CEA the max temperature is found to be 2632 K and occurs at a fuel mass fraction of 7.20%.


7.16E+00 2631.6

7.17E+00 2631.77.18E+00 2631.8

7.19E+00 2631.9

7.20E+00 2632

7.21E+00 2632

7.22E+00 2632

7.23E+00 2632

7.24E+00 2632

7.25E+00 2631.9

7.24E+00 2632

7.26E+00 2631.8

7.27E+00 2631.8

7.28E+00 2631.67.29E+00 2631.5

7.30E+00 2631.3

Table 4: Data for Problem 1 part (d)

e) Now find the combustion product composition for a range of fuel mass fractions around the

value that corresponds to the peak flame temperature in part (d). To do this keep the fuel and

air composition, pressure, and temperature the same as in part (d). Use the following fuelmass fractions: % f = 1.0, 3.0, 5.0, 6.0, 6.5, 6.75, 7.0, 7.1, 7.2, 7.3, 7.4, 7.5, 7.75, 10.0, 50.0,

90.0. The graph below shows the resulting mole fractions of the major species CH4, O2, N2,

CO2, H2O, CO over this range of fuel mass fractions.Problem 1 Part E

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100

Fuel Mass Fraction

M o l e F r a c t i o n s

CH4

O2

N2

CO2

H2O

CO



% Fuel Temperature (k) CH4 O2 N2 CO2 H2O CO

1.00E+00 1.06E+03 0.00E+00 1.79E-01 7.76E-01 2.23E-02 1.32E-02 3.23E-12

3.00E+00 1.69E+03 1.84E-27 1.18E-01 7.64E-01 6.67E-02 3.98E-02 1.95E-06

5.00E+00 2.25E+03 6.03E-20 5.39E-02 7.51E-01 1.11E-01 6.60E-02 6.28E-04

6.00E+00 2.48E+03 3.06E-17 2.43E-02 7.43E-01 1.30E-01 7.82E-02 4.42E-03

6.50E+00 2.57E+03 5.63E-16 1.24E-02 7.39E-01 1.35E-01 8.39E-02 1.03E-02

6.75E+00 2.61E+03 2.40E-15 7.86E-03 7.36E-01 1.35E-01 8.66E-02 1.51E-02

7.00E+00 2.63E+03 1.02E-14 4.49E-03 7.32E-01 1.34E-01 8.90E-02 2.18E-02

7.10E+00 2.63E+03 1.82E-14 3.48E-03 7.31E-01 1.32E-01 8.99E-02 2.50E-02

7.20E+00 2.63E+03 3.21E-14 2.66E-03 7.29E-01 1.31E-01 9.07E-02 2.85E-02

7.30E+00 2.63E+03 5.62E-14 2.00E-03 7.27E-01 1.29E-01 9.15E-02 3.22E-02

7.40E+00 2.63E+03 9.71E-14 1.49E-03 7.25E-01 1.27E-01 9.22E-02 3.62E-02

7.50E+00 2.62E+03 1.65E-13 1.10E-03 7.23E-01 1.25E-01 9.28E-02 4.05E-02

7.75E+00 2.61E+03 5.76E-13 5.12E-04 7.17E-01 1.18E-01 9.40E-02 5.16E-02

1.00E+01 2.33E+03 1.03E-09 8.69E-07 6.62E-01 6.15E-02 8.73E-02 1.45E-01

5.00E+01 1.23E+03 4.48E-02 7.35E-20 1.98E-01 5.00E-03 2.21E-02 7.44E-02

9.00E+01 1.06E+03 1.52E-01 1.09E-23 3.12E-02 3.57E-04 1.20E-02 4.22E-03

Table 5 : Data for Problem 1 part (e)

f) Now you will make a similar plot for the same conditions as in part (e), but showing the

minor species concentrations varies with the fuel mass fraction: OH, H2, H, O, NO.

Problem 1 Part F

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 20 40 60 80 100

Fuel Mass Fraction


H2

OH

NO



Problem 1 Part F

0

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

0 20 40 60 80 100

Fuel Mass Fractions


H

O

% Fuel Temperature (k) OH H2 NO H O

1.00E+00 1.06E+03 5.87E-08 2.18E-12 5.24E-05 2.28E-15 5.53E-113.00E+00 1.69E+03 8.73E-05 3.42E-07 2.09E-03 1.28E-08 2.25E-06

5.00E+00 2.25E+03 1.55E-03 6.86E-05 6.96E-03 9.63E-06 1.34E-04

6.00E+00 2.48E+03 3.11E-03 4.34E-04 7.38E-03 7.69E-05 3.28E-04

6.50E+00 2.57E+03 3.57E-03 9.98E-04 6.14E-03 1.72E-04 3.62E-04

6.75E+00 2.61E+03 3.55E-03 1.50E-03 5.14E-03 2.40E-04 3.34E-04

7.00E+00 2.63E+03 3.31E-03 2.22E-03 4.00E-03 3.16E-04 2.76E-04

7.10E+00 2.63E+03 3.15E-03 2.59E-03 3.54E-03 3.47E-04 2.47E-04

7.20E+00 2.63E+03 2.98E-03 3.01E-03 3.10E-03 3.77E-04 2.18E-04

7.30E+00 2.63E+03 2.78E-03 3.49E-03 2.69E-03 4.05E-04 1.88E-04

7.40E+00 2.63E+03 2.57E-03 4.02E-03 2.31E-03 4.30E-04 1.61E-04

7.50E+00 2.62E+03 2.36E-03 4.62E-03 1.97E-03 4.52E-04 1.35E-04

7.75E+00 2.61E+03 1.87E-03 6.34E-03 1.30E-03 4.93E-04 8.50E-051.00E+01 2.33E+03 1.52E-04 3.62E-02 3.11E-05 3.38E-04 8.65E-07

5.00E+01 1.23E+03 2.12E-11 2.21E-01 7.20E-14 1.87E-08 1.47E-18

9.00E+01 1.06E+03 1.25E-13 1.61E-01 8.36E-17 3.89E-10 2.73E-22

Table 6: Data for Problem 1 part (f)



3. Comparing Constant-Pressure and Constant-Volume Combustion Processes

In this problem, you will use CEA to compare the adiabatic flame temperatures that result

from constant-pressure and constant-volume combustion processes to gain insights into the

differences between these two types of processes.

Consider 1 mole of gaseous iso-octane (C8H18) fuel being burned with 21.4 moles of air.

Over a range of fuel-air equivalence ratios:

Φ = 0.825, 0.850, 0.875, 0.900, 0.925, 0.950, 0.975, 1.000, 1.025, 1.050, 1.100, 1.125, 1.150,

1.175, 1.200

For the constant-pressure case, take the pressure to be 8 atm and the initial temperature to be

540 K for both the fuel and the air. For the constant-volume case, set the density to be 4.4

kg/m3

and again take the initial temperatures to be 540 K.

Notice in the results below that the two curves look very similar and follow the same trends,

but that constant-volume combustion produces higher adiabatic flame temperatures than

constant-pressure combustion.

Problem 2

1500

1700

1900

2100

2300

2500

2700

2900

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

Phi

T e m p e r a t u r e ( k )

HP

UV



HP UV

Phi Temperature (k) Phi Temperature (k)

8.25E-01 2.26E+03 8.25E-01 2.62E+03

8.50E-01 2.29E+03 8.50E-01 2.66E+03

8.75E-01 2.33E+03 8.75E-01 2.69E+03

9.00E-01 2.36E+03 9.00E-01 2.71E+03

9.25E-01 2.39E+03 9.25E-01 2.74E+03

9.50E-01 2.41E+03 9.50E-01 2.76E+03

9.75E-01 2.43E+03 9.75E-01 2.78E+03

1.00E+00 2.45E+03 1.00E+00 2.79E+03

1.03E+00 2.46E+03 1.03E+00 2.80E+03

1.05E+00 2.46E+03 1.05E+00 2.81E+03

1.08E+00 2.46E+03 1.08E+00 2.82E+03

1.10E+00 2.45E+03 1.10E+00 2.82E+03

1.13E+00 2.44E+03 1.13E+00 2.82E+03

1.15E+00 2.42E+03 1.15E+00 2.81E+03

1.18E+00 2.40E+03 1.18E+00 2.80E+031.20E+00 2.39E+03 1.20E+00 2.79E+03

Table 7: Data for Problem 2

4. Using CEA for Chapman-Jouguet Detonation Processes

In this problem, you will use CAE to determine Chapman-Jouguet detonation wave speeds.

We will consider two different fuel types, methane (CH4) and hydrogen (H2), mixed with

varying amounts of air, and find the C-J wave speed for each as a function of fuel-air

equivalence ratio.

a) For the first C-J detonation take pure gaseous hydrogen fuel at 300 k and the oxidizer to

be air at 300 k. Set the initial “state 1” temperature and pressure of the un-reacted fuel-air mixture to be 300 k and 1 atm. Run CEA with the following 16 equivalence ratios:

Φ = 0.825, 0.850, 0.875, 0.900, 0.925, 0.950, 0.975, 1.000, 1.025, 1.050, 1.100, 1.125,1.150, 1.175, 1.200



Problem 3 Part A

1860

1880

1900

1920

1940

1960

1980

2000

2020

2040

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Phi

D e t e n a t i o n V e l o c i t y ( m / s e c )

Phi Velocity (m/sec)

8.25E-01 1.88E+03

8.50E-01 1.89E+038.75E-01 1.90E+03

9.00E-01 1.92E+03

9.25E-01 1.93E+03

9.50E-01 1.94E+03

9.75E-01 1.95E+03

1.00E+00 1.96E+03

1.03E+00 1.97E+03

1.05E+00 1.98E+03

1.08E+00 1.99E+03

1.10E+00 2.00E+03

1.13E+00 2.01E+03

1.15E+00 2.01E+031.18E+00 2.02E+03

1.20E+00 2.03E+03

Table 8: Data for Problem 3 part (a)



b) Now repeat the same procedure, but with methane as the fuel.

Problem 3 Part B

1720

1740

1760

1780

1800

1820

1840

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Phi

D e t e n a t i o n V e l o c i t y ( m / s e c )

Phi Velocity (m/sec)

8.25E-01 1.73E+03

8.50E-01 1.74E+03

8.75E-01 1.75E+03

9.00E-01 1.76E+03

9.25E-01 1.77E+03

9.50E-01 1.78E+03

9.75E-01 1.79E+03

1.00E+00 1.80E+03

1.03E+00 1.81E+03

1.05E+00 1.81E+03

1.08E+00 1.82E+03

1.10E+00 1.82E+03

1.13E+00 1.83E+03

1.15E+00 1.83E+03

1.18E+00 1.83E+03

1.20E+00 1.83E+03

Table 9: Data for Problem 3 part (b)



c) Now make a single plot that compares the detonation wave speed versus equivalence

ratio for these two fuels. Notice the hydrogen fuel has a higher detonation velocity, andalso does not drop off as much as methane does at higher equivalence ratios.

Problem 3 Part C

1700

1750

1800

1850

1900

1950

2000

2050

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Phi

D e t e n a t i o n V

e l o c i t y ( m / s e c )

H2

CH4

Solution 07

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