Question

Limestone (CaCO3(s)) is fed into a kiln and converted into quicklime (CaO(s)) via the following reaction:
CaCO3(s) CaO(s) + CO2 (g) The heat used to drive this reaction comes from the combustion of methane in the kiln using 50% excess dry air. The limestone and air enter at 25°C, and the quicklime exits at 900°C. The combustion gases exit at 200°C.
DHof, CH4, gas = -74.84 kJ/mol Cp (for all gases and vapors): 30 J/mol.K DHof, CaCO3, solid = -1206.9 kJ/mol Cp,CaCO3(s) = 234.5 J/mol.K DHof, CaO, solid = -635.6 kJ/mol Cp,CaO(s) = 111.8 J/mol.K DHof, CO2, gas = -393.51 kJ/mol    DHof, H2O, gas = -241.83 kJ/mol a) How many kg of CaCO3 can be processed per 1000 m3 of CH4 measured at standard conditions (i.e., 0 °C and 1 atm for gases, assuming CH4 is an ideal gas).
b) What is the dew point of the exit gases if the total pressure of the combustion gas stream is 760 mmHg?
Antoine Constants for H2O (for p* in mmHg, Temp. in K): A = 18.3036, B = 3816.44, C = -46.13
c) Comment on the phase of the water in the exit stream. What would happen if you cooled it down to the dew point temperature? What happens when you cool it to room temperature? What would happen if we raise the pressure of the exit stream at constant temperature?

Answer 1

The limestone reaction is

The heat required for this reaction is given by combustion of

Methane

Given data

∆H_f(CaO) = -635. 6 KJ/mol

∆H_f(CO₂) = -393. 51 KJ/mol

∆H_f(CaCO₃) = -1206.9 KJ/mol

∆H_r = ∆H_products - ∆H_reactants

∆H_r = -635. 6 -393. 51 - (-1206.9)

∆H_r (CaCO₃) = +177.79 KJ/mol

The combustion reaction of methane

∆H_f(CO₂) = -393. 51 KJ/mol

∆H_f(CH₄) = -74. 84 KJ/mol

∆H_f(H₂O) = -241. 83 KJ/mol

∆H_f(O₂) = 0

∆H_r = -393. 5-(2×241.83) -(-74.84)

∆H_r = -802. 32 KJ/mol of methane reacted

Methane present

V = 1000 m³

T = 0°C = 273 K

P = 1 atm = 1.013×10⁵ Pa

n = PV/(RT)

n = (1.013×10⁵) (1000) /(8.314×273) = 44631.016 moles

n = 44.631 kmol of methane

Heat needed to heat the quick lime to 900°C

C_p(CaO) = 111.8 J/mol K

Q = 111.8(900-25) =+ 97825 J/mol

The gases exit at 200°C

Heat needed to heat the CO₂  from 25 to 200°C

Q = 30(200-25) = + 5250 J/mol

Heat released by methane combustion

= -802. 32 KJ/mol

Methane present = 44.631 kmol

Total heat released = -802. 32(44.631) = - 35808.343 MJ

= - 3.5808×10¹⁰ J

Heat required to form quick lime

= +177.79 KJ/mol

Moles of H₂O formed per mole CH₄ = 2

Moles of CO₂ formed per mole CH₄ and CaO = 1+1 = 2

Total heat required to process 1 mole of product =

177.79(1000) +(2×30(200-25)) +(2×30(200-25)) = 198790 J

Heat needed = 1.9879×10⁵ J /mol product

Moles of CaCO₃ that can be processed

= (3.5808×10¹⁰) /(1.9879×10⁵) = 180.129 Kmol

CaCO₃ Processed = 180.129 kmol per 1000 m³ methane

B)

Moles of methane = 44.631 kmol

Moles of oxygen required according to stiochiometry = 44.631(2) = 89.262 kmol

But oxygen is 50% in excess

Oxygen supplied = 89.262(1.5) = 133.893 kmol

Air supplied = 133.893/0.21 = 637.585 kmol

Exit gases analysis

Component	kmol	mol%
N2	637.585(0.79) = 503.692	58.40
O2	133.893-89.262 = 44.631	5.175
CO2	1(44.631) +180.129 = 224.76	26.06
H2O	2(44.631) = 89.262	10.351
Total	862.345	100

Mole fraction of water vapor = 0.10351

P = 760 mmHg

Partial pressure of water = 760(0.10351) =78.667 mmHg

P* = e^{(18.3036-(3816.44/(T-46.13) ))}

P is in mmHg , T in K

The dew point occurs when partial pressure of water vapor is equal to vapor pressure of water vapor

Substituting P* = 78.667 mmHg

ln(78.667) = 18.3036-(3816.44/(T-46.13) )

T = 319.93 K

T = 46.93°C

The dew point of exit gas = 46.93°C

C)

The water in the exit stream will be in the vapor state

If gas stream is cooled down to dew point temperature the water vapor starts condensing thereby changing its phase from vapor to liquid state

If we raise the pressure of the exit gas stream then the partial pressure of water vapor increases thereby increasing the dew point.

If pressure increases at constant temperature the water vapor condenses at much higher temperature than 46.93°C

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