Heat of Reaction Activity
(Methane, Propane, Methanol, Nitromethane)
- Write a balanced reaction for your assigned fuel with air under stoichiometric conditions. Verify the mass balance for each atom.
Methane
Propane
Methanol
Nitromethane
- Sketch and label isothermal and adiabatic reactions on an enthalpy-temperature diagram. Show reactant-side and product-side enthalpies.
- 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
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
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
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
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
- 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
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
- 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
Methane
a)
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)
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)
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)
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)
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)
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)
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)
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.
- 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.