Heat of Reaction Activity
(Methane, Propane, Methanol, Nitromethane)
1. Write a balanced reaction for your assigned fuel with air under stoichiometric
conditions. Verify the mass balance for each atom.
C x H y Oz  ( x 
y z
y
y z
 ) * (O2  3.76 N 2 )  x * CO2  * H 2O  ( x   ) * 3.76 N 2
4 2
2
4 2
Methane
CH 4  2 * (O2  3.76N 2 )  1* CO2  2 * H 2O  2 * 3.76N 2
Propane
C3 H 8  5 * (O2  3.76 N 2 )  3 * CO2  4 * H 2 O  5 * 3.76 N 2
Methanol
CH 4O  1.5 * (O2  3.76N 2 )  1* CO2  2 * H 2O  1.5 * 3.76N 2
Nitromethane
CH 3 NO2  0.75 * (O2  3.76 N 2 )  1* CO2  1.5 * H 2 O  (.5  .75 * 3.76) N 2
2. Sketch and label isothermal and adiabatic reactions on an enthalpy-temperature
diagram. Show reactant-side and product-side enthalpies.
H - reactants
Enthalpy
adiabatic
isothermal
H - products
T = 298K
Temperature
3. Using separate enthalpies for each species calculate the lower heating value on a
per mole of fuel and per kilogram of fuel. Assume standard temperature and
pressure conditions as well as an equivalence ratio of unity. Compare this with the
calorific value given in Table A.3. How does the lower heating value change as the
mixture is made leaner? How does the heat of reaction change if the water vapor
condenses?
LHV = h(products) – h(reactants)
H(products) =  (chemical coefficient * species enthalpy) for products
H(reactants) =  (chemical coefficient * species enthalpy) for reactants
Methane
CH 4  2 * (O2  3.76N 2 )  1* CO2  2 * H 2O  2 * 3.76N 2
h(products) = 1*(-393.5) + 2*(-241.8) + 2*3.76*(0) [MJ/kmol]
Enthalpy values from table A.4 in Stone Text
h(reactants) = 1*(-74.52) + 2*(0) + 2*3.76*(0) [MJ/kmol]
Enthalpy value for fuel is the Enthalpy of Formation (Hf) from table A.3 Stone text
LHV = -877.1 + 74.5 = -802.6 [MJ/kmol]
**agrees with calorific value from table A.3**
**Heating Value is the same as calorific value, but with different signs**
Note: This the energy release for one kmol of fuel
To normalize the answer per kg fuel you must divide by the molecular weight of the fuel.
LHV = -802.6 [MJ/kmol] * [1kmol/(12+4*1)kg] = -50.2 MJ/kgfuel
Propane
C3 H 8  5 * (O2  3.76 N 2 )  3 * CO2  4 * H 2 O  5 * 3.76 N 2
h(products) = 3*(-393.5) + 4*(-241.8) [MJ/kmol]
h(reactants) = 1*(-104.7) + 2*(0) + 5*3.76*(0) [MJ/kmol]
LHV = -2147 + 104.7 = -2042 [MJ/kmol]
LHV = -2042 [MJ/kmol] * [1kmol/(3*12+8*1)kg] = -46.4 MJ/kgfuel
Methanol
CH 4O  1.5 * (O2  3.76N 2 )  1* CO2  2 * H 2O  1.5 * 3.76N 2
h(products) = 1*(-393.5) + 2*(-241.8) [MJ/kmol]
h(reactants) = 1*(-201) + 2*(0) + 2*3.76*(0) [MJ/kmol]
LHV = -877.1 + 201 = -676.1 [MJ/kmol]
LHV = -676.1 [MJ/kmol] * [1kmol/(1*12+4*1+1*16)kg] = -21.1 MJ/kgfuel
Nitromethane
CH 3 NO2  0.75 * (O2  3.76 N 2 )  1* CO2  1.5 * H 2 O  (.5  .75 * 3.76) N 2
h(products) = 1*(-393.5) + 1.5*(-241.8) [MJ/kmol]
h(reactants) = 1*(-74.7) + 1.5*(0) + 1.5*3.76*(0) [MJ/kmol]
LHV = -756.2 + 74.7 = -681.5 [MJ/kmol]
LHV = -681.5 [MJ/kmol] * [1kmol/(1*12+3*1+1*14+2*16)kg] = -11.2 MJ/kgfuel
4. How does the lower heating value change as the mixture is made leaner?
How does the heat of reaction change if the water vapor condenses?
When displaying results on a “per fuel” basis, having excess air would add enthalpy of air
to both the products and reactants side, but since the enthalpy of air at STP is zero (0), it
has no effect on the heating value if being reported on a “per fuel” basis.
If water vapor condenses, it changes the enthalpy of water on the product side. You must
add the enthalpy of condensation for water to the tabulated value, effectively reducing the
enthalpy of the products. This is known as the ‘Higher Heating Value” or HHV.
For example: Find the HHV of Propane
Propane
C3 H 8  5 * (O2  3.76 N 2 )  3 * CO2  4 * H 2 O  5 * 3.76 N 2
h(products) = 3*(-393.5) + 4*(-241.8-43.99) [MJ/kmol]
** The enthalpy of H2O is reduced by 43.99 MJ/kmol**
h(reactants) = 1*(-104.7) + 2*(0) + 2*3.76*(0) [MJ/kmol]
HHV = -2324 + 104.7 = -2219 [MJ/kmol]
HHV = -2219 [MJ/kmol] * [1kmol/(3*12+8*1)kg] = -50.4 MJ/kgfuel
Alternatively, we could reduce the LHV on a kmol basis by 4*-43.99, or
HHV = -2042 + (4*-43.99) = -2218 MJ/kmol
5. Calculate the heat of an isobaric, isothermal reaction on a per kilogram mixture
basis for equivalence ratios of (a) unity and (b) one-half.
For an equivalence ratio of 0.5, you can use the following equation
1
y z
C x H y Oz  ( x   ) * (O2  3.76 N 2 ) 

4 2
y
1
y z
1
y z
x * CO2  * H 2 O  (  1) * ( x   ) * O2  * ( x   ) * 3.76 N 2
2

4 2

4 2
Methane
a) CH 4  2 * (O2  3.76N 2 )  1* CO2  2 * H 2O  2 * 3.76N 2
LHV (from prob 3) = -802.6 [MJ/kmol]
Per kgmix = -802.6 [MJ/kmol] * kmol/((1*12)+(4*1)+(2*32)+(2*3.76*28))kgmix
LHV= -2.76 MJ/kgmix
b) CH 4  4 * (O2  3.76N 2 )  1* CO2  2 * H 2O  2 * O2  4 * 3.76N 2
LHV (from prob 3) = -802.6 [MJ/kmol]
Per kgmix = -802.6 [MJ/kmol] * kmol/((1*12)+(4*1)+(4*32)+(4*3.76*28))kgmix
LHV= -1.42 MJ/kgmix
Propane
a) C3 H 8  5 * (O2  3.76 N 2 )  3 * CO2  4 * H 2 O  5 * 3.76 N 2
LHV (from prob 3) = -2042 [MJ/kmol]
Per kgmix = -2042 [MJ/kmol] * kmol/((3*12)+(8*1)+(5*32)+(5*3.76*28))kgmix
LHV= -2.79 MJ/kgmix
b) C3 H 8  10 * (O2  3.76 N 2 )  3 * CO2  4 * H 2 O  5 * O2  10 * 3.76 N 2
LHV (from prob 3) = -2042 [MJ/kmol]
LHV = -2042 [MJ/kmol] * kmol/((3*12)+(8*1)+(10*32)+(10*3.76*28))kgmix
LHV= -1.44 MJ/kgmix
Methanol
a) CH 4O  1.5 * (O2  3.76N 2 )  1* CO2  2 * H 2O  1.5 * 3.76N 2
LHV (from prob 3) = -676.1 [MJ/kmol]
LHV = -676.1 [MJ/kmol] *kmol/((1*12)+(4*1)+(1*16)+(1.5*32)+(1.5*3.76*28))kgmix
LHV= -2.84 MJ/kgmix
b) CH 4O  3 * (O2  3.76N 2 )  1* CO2  2 * H 2O  1.5 * O2  3 * 3.76N 2
LHV (from prob 3) = -676.1 [MJ/kmol]
LHV = -676.1 [MJ/kmol] * kmol/((1*12)+(4*1)+(1*16)+(3*32)+(3*3.76*28))kgmix
LHV= -1.52 MJ/kgmix
Nitromethane
a) CH 3 NO2  0.75 * (O2  3.76 N 2 )  1* CO2  1.5 * H 2 O  (.5  .75 * 3.76) N 2
h(products) = 1*(-393.5) + 1.5*(-241.8) [MJ/kmol]
h(reactants) = 1*(-74.7) + 1.5*(0) + 1.5*3.76*(0) [MJ/kmol]
LHV = -756.2 + 74.7 = -681.5 [MJ/kmol]
LHV = -681.5 [MJ/kmol]*
kmol/((1*12)+(3*1)+(1*14)+(2*16)+(.75*32)+(.75*3.76*28))kgmix
LHV= -4.15 MJ/kgmix
b) CH 3 NO2  1.5 * (O2  3.76 N 2 )  1* CO2  1.5 * H 2 O  0.75 * O2  (.5  1.5 * 3.76) N 2
h(products) = 1*(-393.5) + 1.5*(-241.8) [MJ/kmol]
h(reactants) = 1*(-74.7) + 1.5*(0) + 1.5*3.76*(0) [MJ/kmol]
LHV = -756.2 + 74.7 = -681.5 [MJ/kmol]
LHV = -681.5 [MJ/kmol]*
kmol/((1*12)+(3*1)+(1*14)+(2*16)+(1.5*32)+(1.5*3.76*28))kgmix
LHV= -2.55 MJ/kgmix
Reflection:
We can see that equivalence ratio has no bearing on the heating value of fuel when
reported on a ‘per kg of fuel’ basis. However, when displaying results on a ‘per kg of
mix’ it’s easy to see that excess air (with no heating value of its own) effectively reduces
the energy of the mixture.
Another point of interest is comparing the heating value of the various fuels. The energy
“per kg fuel” is very different for the various fuels. On the surface, Methanol has about ½
the energy of gasoline, and nitromethane has about ¼ the energy. However, when looking
at the energy of the mixture nearly all the fuels are the same – about 2.75 MJ/kgmix.
However, nitromethane is nearly double that value. That is one reason, despite the lower
energy on a ‘per kg of fuel’ basis, that nitromethane is such a popular fuel for racing
applications where extreme power is desired.
6. Explain how to estimate the adiabatic isobaric combustion temperature for a
stoichiometric mixture. How does this vary with equivalence ratio? Why?
Using tables, this would be done by iteration. You would have to guess a final
temperature for the products, and then calculate the product enthalpy at that temperature.
The goal would be to keep iterating on your temperature guess until the enthalpy of the
products was equal to the enthalpy of the reactants (unchanged from previous
calculations). In actual practice, we will use software to do this for us.
On either side of stoichiometric there are excess molecules present. Because of the
change in temperature between the reactants and products, the excess air molecules now
detract from the product enthalpy. Energy used in heating these extra molecules detracts
from the max flame temperature.