Dodecane gas (C12H26, M=170.328, Enthalpy of fonnation=-291,010 kJ/kmol) is burn

Question

Dodecane gas (C12H26, M=170.328, Enthalpy of fonnation=-291,010 kJ/kmol) is burned in a combustor with a given percentage of excess air. The pressure and temperature of both the air and fuel are 101 kPa and 298 K respectively. Assume that the mole fractions are 79% nitrogen and 21 % oxygen for air (use M=28.97 kg/kmol and R=0.287 kJ/kg-K) and that water is a vapor in the exhaust. Given the values below, determine the following: Given Values m_foel (kg) =102 T_exhaust (K) = 1270 Excess Air (%) = 30% Determine the air fuel ratio (kmol_air/kmol_fuel). Determine the amount (kmol) of fuel burned. Determine the amount (kmol) of air used. Determine the amount (kmol) of C02 produced. Determine the amount (kmol) of H20 produced. Determine the amount (kmol) of N2 produced. Determine the amount (kmol) of 02 produced. Determine the Enthalpy (kJ) of C02 in the products. Determine the Enthalpy (kJ) of H20 in the products. Determine the Enthalpy (kJ) of N2 in the products. Determine the Enthalpy (kJ) of 02 in the products. Determine the total heat transfer (kJ) into the combustion chamber (a negative value indicates heat transfer out of the chamber).

Explanation / Answer

Mass of fuel =102 kg

Molecular weight of C12H26= 12*12+26=170.328

Moles of fuel =100 kg/170.328 kg/kgmole=kg moles

2.The reaction of Dodecane with oxygen is

C12H26+ 18.5 O2- ---->12CO2+13H2O

1 Kmol of Dodecane requires 18.5 Kg mol of oxygen

0.998074 kmol of dodecane requires 0.998074*18.5/1= 18.4643 Kg moles of oxygen

But air contains 79% N2 and 21% oxygen. Hence moles of air required= 18.4643/0.79=87.925 Kg moles of air

But air is supplied 30% in excess, hence moles of air supplied =1.3*87.925= 114.3025 Kg moles

3. Moles of CO2 produces =12* Moles of dodecane= 12*0.998074=11.97689 kg moles

4. Moles of H2O produced= 13*0.998074= 12.97497 Kgmoles

5. Kg moles of Nitrogen used (not produced)= 0.79* 114.3025= 90.29898 kgmolres (since air contains 79% nitrogen by volume and air supplied is 114.3025 kg moles)

6. Oxygen supplied= Total moles of air- Moles of nitrogen= 114.3025-90.2989=24.0036 kg mole

Oxygen consumed= 18.4643 kg moles and Oxygen remaining =24.0036- 18.4643=5.5393 kg moles

7.Exhaust gas temperature = 1270 K

Enthalpy of CO2 in the product= moles of CO2* average specific heat* ( temperature difference)

=11.97689 kg moles*11.92 Kcal/kgmol.K *(1270-298)=138767.1 Kcal =138767.1*4.2 Kj=582822 KJ

8. Enthalpy of H2O produced= 12.97497*39.6 Kj/Kg mole*(1270-298)=499422.2 Kj

9. Enthalpy of N2 produced= = 90.29898*7.507*4.2*(1270-298)= 2767355 KJ

10. Enthalpy of O2 produced ( Remaining)= 5.5393*7.941*4.2*(1270-298)=179575 Kj

Enthalpy of formation of dodecane at 25deg.c= -291090 Kj/Kmol*0 .998074 Kmol =-290529.4Kj

Enthalpy of products = 582822+499422.2+2767355+179575=4029174 KJ

Heat transfer = enthalpy of products+ Heat of reaction- enthalpy of reactants

Since standard temperature is 298, Enthalpy of reactants= 0

Heat transfer= 4029174-290529.4=373865 Kj

Heat is transferred into the combustion chamber.

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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