Limestone (CaCO3) decomposition in a kiln (40 points) A vertical kiln is charged

Question

Limestone (CaCO3) decomposition in a kiln (40 points) A vertical kiln is charged with pure limestone (CaCO3) and pure coke (carbon), both at 25°C. Air (21% O2, 79 N2) is blown at the bottom. The reaction between O2 in air and coke provides the heat for decomposition of the CaCO3.

Combustion reaction: C(s) + O2(g) > CO2(g)

Limestone decomposition reaction: CaCO3(s) > CaO(s) + CO2(g)

Coke is burned with 20% excess air. The flue gases exit the kiln at 320°C. The solid mixture leaves the bottom at 950°C and consists of 94% CaO and 6% CaCO3 by mass. Assume that the kiln is perfectly insulated.

QUESTION: How many kg/hour of coke is needed if CaCO3 enters the kiln at 1000 kg/hour?

Molecular weight (g/mol): H = 1, C = 12, N = 14, O = 16, S = 32, Ca = 40

Hf, CaCO3, solid = -1206.9 kJ/mol Cp, CaCO3(s) = 117.1 J/mol.K

Hf, CaO, solid = -635.6 kJ/mol Cp, CaO(s) = 57.3 J/mol.K

Hf, CO2, gas = -393.51 kJ/mol

Cp for all gases and vapors: 30 J/mol.K

Explanation / Answer

taking the data on the basis of one hour,

let, x kg of CaCO3(s) decomposes

molar mass of CaCO3 is = 100 gram/mole = 0.1 kg/mole

so number of moles decomposes per hour = x/0.1 = 10x mole

Limestone decomposition reaction: CaCO3(s) > CaO(s) + CO2(g)

according to the chemical reaction, 1 mole of CaCO3(s) gives 1 mole of CaO(s)

so, 10x mole of CaCO3(s) will give 10x mole of CaO(s)

molar mass of CaO = 56 gram/mole = 0.056 kg/mole

So, mass of CaO(s) = 10x*0.056 = 0.56x kg CaO

mass of unreacted CaCO3(s) = (1000-x) kg

So, 0.56x/(1000-x) = 94/6

by solving the above equation, it,s found that x's value is 965.5 kg which means 965.5 kg of CaCO3 decomposes per hour

965.5 kg of CaCO3 = 965.5*1000/100 = 9655 moles of CaCO3

from the above data, enthalpy of CaCO3(s) formation is -1206.9 kJ/mol

So, the enthalpy of CaCO3(s) decomposition is  1206.9 kJ/mol

So. heat required for decomposition of 9655moles of CaCO3(s) is = 1206.9*9655 = 11652619.5 kJ

unreacted CaCO3(s) = 1000-965.5 = 34.5 kg = 345 moles

heat required to increase the temperature of unreacted CaCO3(s) from 25C to 950C = 345*117.1*(950-25) /1000

= 37369.5375 kJ

by the decomposition of CaCO3(s), CaO(s) formed = 9655 mole

To increase the temperature of CaO(s) from 25C to 950C, heat required =9655*57.3*(950-25)/1000 =511739.1375 kJ

Let, y kg/hour coke is needed which means y*1000/12 = 83.33y moles of coke

So, according to the formation reaction of CO2, 83.33y moles coke will produce 83.33moles of CO2 and for that reaction, heat will produced = 83.33y*393.51 =32791.1883y kJ

For CO2 formation, 83.33y moles of O2 is required, which is equal to 83.33y*32 = 2666.56y kg of mass

So, air required, 2666.56y*100/21 =12697.9y kg

As 20% excess air is used, so total air in the kiln is = 12697.9y*120/100 =15237.48571y kg air

Average molar mass of air = (0.21*32+0.79*28) = 28.84 grams per mole

So, 15237.48571y kg air = 15237.4857y*1000/28.84 =528345.5513y moles

Total CO2 produced in the process = 9655+83.33y moles

Total flue gas = (528345.5513+9655+83.33y) moles=(538000.5513+83.33y) moles

Amount of heat required to increase the temperature of flue gas from 25C to 320C =(538000.5513+83.33y)*30*(320-25)/1000 = (4761304.879+737.47y) kJ

By the energy balance of the kiln,

32791.1883y kJ = 11652619.5 kJ +37369.5375 kJ+511739.1375 kJ+ (4761304.879+737.47y)

By solving this equation, y = 529.206

So, the 529.206 kg of coke is needed per hour

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.