Combustion Efficiency Calculation after Preheating

The Python code for the effect of preheating temperature on Adiabatic flame Temperature for stoichiometric combustion of one mole of methane

#The effect of preheating on the adiabatic flame temperature

import cantera as ct 
import matplotlib.pyplot as plt 
gas = ct.Solution(\'gri30.cti\')
#specifying the temp range
for Temp in range (298,600,5):
	Air = ct.Quantity(gas)
	Air.TPX = Temp,ct.one_atm,{\'O2\':0.21, \'N2\':0.79}
	Air.moles = 2/0.21

	Fuel = ct.Quantity(gas)
	Fuel.TPX = 298.15,ct.one_atm,\'CH4:1546\'
	Fuel.moles = 1
	products = ct.Quantity(gas)
	products.TPX = 1700,ct.one_atm,{\'CO2\':0.095, \'H2O\':0.1901, \'N2\':0.7148}
	products.moles = 10.52

	Combustion = Air+Fuel
	Combustion.equilibrate(\'HP\')
	print 
	plt.plot(Temp,Combustion.T,\'+\', color=\'red\')
plt.show()

In the above plot, we can observe that as the air inlet temperature increase it proportionally increases the Adiabatic flame Temperature it\'s because when we increase the air inlet temperature it leads to increasing the enthalpy of the air, in other words, the increases temperature of the air will proportionally increase the providing energy to the furnace. We are dealing with adiabatic conditions so the energy cant goes through boundaries so the supplied energy will ultimately lead to rising in the Adiabatic temperature of the air.

2) The recuperator is used to preheat the inlet air with the help of exhaust gases and it works like a heat exchanger. This process is generally used to increase the efficiency of the model, with this process the same amount of energy is produced by burning less amount of fuel it ultimately saves the fuel also. The recuperator uses the wastage energy that is the heat remained in the flue gases so the maximum amount of energy can be utilized from the furnace this leads to an increase in combustion efficiency also.

percentage of fuel-saving = (mass of fuel with recuperator - mass of fuel without recuperator)/(mass of fuel without recuperator)

Combustion Efficiency: It is the ratio of the sum of heat gains and losses to the total energy provided

`eta_(combustion) = (Q_(reactants)-Q_(exhaust))/(m_f*LHV)`

The python code for the calculation of combustion efficiency and percentage of saving fuel for stoichiometric combustion of one mole of methane

import cantera as ct 
import matplotlib.pyplot as plt 

gas = ct.Solution(\'gri30.cti\')

fuel = ct.Quantity(gas)
fuel.TPX = 298.15,ct.one_atm,\'CH4:1\'
fuel.moles = 1
mass_fuel = fuel.mass 
H_fuel = fuel.h

products = ct.Quantity(gas)
products.TPX = 1700,ct.one_atm,{\'CO2\':0.095,\'H2O\':0.1901, \'N2\':0.7148}
products.moles = 10.52
mass_products = products.mass
H_products = products.h 

air = ct.Quantity(gas)
air.TPX = 298.15,ct.one_atm,{\'O2\':0.21,\'N2\':0.79}
air.moles = 9.52
mass_air = air.mass 
H_air = air.h

Q = (mass_air*H_air) + (mass_fuel*H_fuel) -(mass_products*H_products)
LHV = 50e6
LHV = 50e6
# calculating efficiency without preheat condition
efficiency_STP = (Q/((mass_fuel)*(LHV)))*100

for T in range (298,600,5):
	air.TPX = T,ct.one_atm,{\'O2\':0.21,\'N2\':0.79}
	air.moles = 9.52
	mass_air = air.mass 
	H_air = air.h

	Q = (mass_air*H_air) + (mass_fuel*H_fuel) -(mass_products*H_products)
	LHV = 50e6
	efficiency = (Q/((mass_fuel)*(LHV)))*100

	saving = ((efficiency/efficiency_STP)-1)*100

	plt.plot(T,efficiency,\'+\',color = \'red\')
	plt.plot(T,saving,\'o\',color = \'blue\')

plt.xlabel(\'preheat_temp in (K)\')
plt.ylabel(\'combustion_efficiency\')
plt.legend([\'combustion_efficiency\',\'percentage_of_saving_the_fuel\'])
plt.show()

The change in combustion efficiency for the change in preheat temperature

The plot showing both combustion efficiency and the percentage of change in fuel for changing in preheat temperature.

It is better to maintain the standard output temperature this will indirectly keep the constraint for the maximum requirement of energy for that particular model because above the production of that energy will do no good for that model in the worst case it may fail also. If the output energy is maintained it also increases the life span of the recuperator so the maintaining of standard output temperature is good