Question

1 kmol of gas Ethane is burned with 175 percent theoretical air in a steady flow combustion chamber. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but 95 percent of the Carbon burns to CO2 with the remaining 5 percent forming CO. Treating the air as dry air, determine; (a)The balanced combustion equation. (b)The air-fuel ratio. (c)The amount of heat transfer (in kJ/kg of fuel) from the combustion process if the dry air and fuel enter the combustion chamber at reference state and the products leave the combustion chamber at 1540 K. (d)The dew point temperature for H2O in the products at a total pressure of 1 atm.

Answer 1

Solution writing stoichiometric Equation Ca Hà + 3s Out 27 a7 (Stoichiometric Reaction ) 2 3 7,0 +2 Co +35x 39 x NZ Now Excess air is 75%, ie 1078 8315= 6-125 Cathet 8-as ( cat ang mga 7300 + 1-960+ CaHg + 6-125 | 3420 + 10900gt Olco + 2.675 00 (Actual Reaction) | + 6 125 * 37 Na The above reaction ů Balanced Reaction HAFA TILL (b) m = 6-05 (82+ x 287 MA (2X12+681) na – 28.03 kg of Air sig of fuel

Calculating dew point. Terop. Nego = 3 (from produce side) Proo – Nemo - 3. Ntotal 30.2167 Products. Moles (N) H2O - 3 Cog – 1991 Table 1 On - 2.675 Ng - 23.041 3007167 = Ntotal 11111 101.325 30.7167 Priso = 9.896 kPa Tcorresponding to Puro = 45.6030 à WBT - 45.6030

substance Substance REYOR TO 1983 and mak KJ/kemel KJ/kmal KI)cmal. o o 8682 8669 9364 -393,520 CP ( Table 2) Note : assuming atm. cond" to be uc (298 k). Here data is not available, however if data is obtained, this can be easily Calculated as TQ= ENR (Reth - 5) – E Np Che th to Jp ® is taken from table and value is calculate. Please Substitute Ñ and Values if available to Get the Answer u . .