A glass melting furnace is burning ethene in pure oxygen (not air). The furnace operates at an equivalence ratio of 0.9 and consumes 30 kmol/hr of ethene. Determine the energy input rate based on the LHV of the fuel in terms of kW.
•Determine O2 consumption rate in kmol/hr.
Ethene, also known as ethylene, is C2H4. The lower heating value (LHV) is 11,270 kcal/kg.
You don't need that number to answer the question.
The balanced reaction is
C2H4 + 3 O2 -> 2 CO2 + 2 H2O
30 kmol/hr of ethene requires 90 kmol/hr of O2 for complete combustion. But with an equivalence ratio of 0.9, you use excess O2 by a ratio 1/0.9 (lean mixture). That means you use 100 kmol/hr of O2.
To determine the O2 consumption rate in kmol/hr, we first need to calculate the stoichiometric air-to-fuel ratio (AFR) for ethene and oxygen. The stoichiometric AFR is the ratio of the actual amount of air used to the theoretical amount of air required to burn the fuel completely.
The balanced chemical equation for the combustion of ethene (C2H4) with oxygen (O2) is as follows:
C2H4 + (3.5 * O2) -> 2 * CO2 + 2 * H2O
From the balanced equation, we can see that it requires 3.5 moles of oxygen (O2) to completely burn 1 mole of ethene (C2H4).
Given that the furnace operates at an equivalence ratio (ER) of 0.9, we can calculate the AFR as follows:
AFR = ER / (ER - 1)
AFR = 0.9 / (0.9 - 1)
AFR = 0.9 / (-0.1)
AFR = -9
The negative sign in AFR indicates that we are using an excess of fuel (ethene) rather than air/oxygen.
Now, to determine the O2 consumption rate in kmol/hr, we multiply the ethene consumption rate by the moles of oxygen required per mole of ethene (3.5 moles O2 per mole of ethene):
O2 consumption rate = 30 kmol/hr * 3.5 kmol O2/kmol ethene
O2 consumption rate = 105 kmol/hr
So, the O2 consumption rate in kmol/hr is 105 kmol/hr.
Next, to calculate the energy input rate based on the Lower Heating Value (LHV) of the fuel in terms of kW, we need to know the LHV of ethene. The LHV represents the amount of heat released when the fuel is burned without considering the latent heat in the water vapor formed.
The LHV of ethene is approximately 47.7 MJ/kg or 47.7 kJ/mol.
Now, we can calculate the energy input rate as follows:
Energy input rate = Ethene consumption rate * LHV
Energy input rate = 30 kmol/hr * 47.7 kJ/mol
Energy input rate = 1431 kJ/hr
Finally, to convert the energy input rate from kJ/hr to kW, we divide it by 3600 (since there are 3600 seconds in an hour):
Energy input rate in kW = 1431 kJ/hr / 3600 s/hr
Energy input rate in kW ≈ 0.397 kW
Therefore, the energy input rate based on the LHV of the fuel is approximately 0.397 kW.
To determine the O2 consumption rate, we can start by determining the stoichiometric ratio of ethene to oxygen in the reaction. The balanced chemical equation for the combustion of ethene (C2H4) can be written as:
C2H4 + O2 -> CO2 + H2O
From the balanced equation, we can see that 1 mole of ethene reacts with 3 moles of oxygen.
Given that the equivalence ratio (ER) is 0.9, we can calculate the amount of oxygen needed for the combustion:
Oxygen needed = ER * Stoichiometric Oxygen
= 0.9 * 3 kmol/hr
= 2.7 kmol/hr
Therefore, the O2 consumption rate is 2.7 kmol/hr.