FUEL fuel to the combustion chamber include hot gas mixture and pure hydrogen n
ID: 954520 • Letter: F
Question
FUEL fuel to the combustion chamber include hot gas mixture and pure hydrogen n shown below Two streams of the hot gas mixture from the fuel side and the oxidizer side enter the combustion chamber The hot gas mixture is comprised of hydrogen and water vapor The mass flow rate, molar composition, temperature and pressure for each stream is provided m the schematic Two streams of pure hydrogen enter the combustion chamber for each stream, mass flow rate, temperature and pressure is provided m the schematic below. OXIDIZER The oxidizer used m the combustion chamber is oxygen which enters the combustion chamber at 104 K and 22 MPa. The equivalence ratio (pi) is 0.6 What is the balanced chemical reaction? What is the flow rate of oxygen to the combustion chamber in kg/s and kmol/s? What is the adiabatic flame temperature in K? Please provide detailed solutions. I have already "solved" these problems but my answers are not within an acceptable tolerance of the given values. I really need to find out what I am doing wrong. Thanks!Explanation / Answer
1. Balanced Chemical reaction
2H2 + O2 à 2H2O
Even though water is used with H2. Its main purpose is to act as a humidifying agent as dry H2 is explosive in nature.
2. Flow rate of oxygen to the combustion chamber in kg/s and kmol/s.
As we have seen from the chemical reaction 2 moles of H2 requires 1 mol of oxygen to produce water.
We need to calculate accurate amount of H2 entering into chamber.
We will be listing all entry points and then calculating flow rate of Hydrogen. And then using that we will be calculating flow rate of Oxygen.
1) H2 gas – 7.7 kg/s + 5 kg/s = 12.7 kg/s
2) Fuel side 89% of 67.6 kg/s = 60.167 kg/s
3) Oxidizer side 92.5% of 30.84 kg/s = 28.527 kg/s
Total mass flow rate is = 12.7 + 60.167+28.527 = 101.394 kg/s of hydrogen.
As per the reaction we need half the oxygen. But equivalence ratio is 0.6. It means combustion is lean with excess of air.
Mass flow rate of oxygen required is 101.394/2 = 50.697
Actual Mass flow rate as per equivalence ratio = 50.697/0.6 = 84.495 kg/s
Kmol/s = (kg/s) / molar mass of oxygen
= 84.495/15.994 = 5.2811 kmol/sec
3. Adiabatic flame temperature in K (Tad)
We will require specific heat of reactants and products
2H2 + O2 à 2H2O
Standard enthalpy of gases is zero.
We will take all the values at 104K
Enthalpy of reactants and enthalpy of products are same because reaction is adiabatic.
Hreactants = Hproducts
Cp*H reactants = Cp*H products
H(H2) + H(O2) = H(H2O)
0 + 0 = [-244500 + 42.44(Tad-104)]
= -244500 + 42.44*Tad – 4413.760
= -248913.760 * 42.44Tad
Tad = 946.200K
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