Question

3. (35 P) Diesel fuel (C,H,) at 25°C is burned in a steady flow combustion chamber with 40 percent excess air air that also enters at 25°C. The required mass flow rate of the diesel fuel is 49.5 g/s. a) Using conservation of atoms, find number of moles and a , b) Actual AF (A/F) c) Actual Air flow rate (g/s), d) Calculate the exit temperature of the product gases. (h - h.) kJ/kmole Mkg/kmolc, T=298 K kJ/kmole CH, 0 h.198 kJ/kmole -290870 CO, -393522 HO -241826 N, 0 0,00 44.01 56.21 18.015 43.87 28.013 33.71 31.999 0 35.59 CH, +(a/)(0,+3.76N)=nl CO,+ n2 H,0 +n3 N,+n4 O, Assume: h=h, + (h - h)= h, +C (T. - 298) kJ/kmole MO=16, MN=14, MC=12, MH=1,

Answer 1

#3] 92 H ₂6 + (as) (2 +3.76 N₂) = n, Co₂ + H₂O + n₂N₂ +nyo li 40% excess air means 1.4 x a th. g stiochiometric Condition :- G2 h 26 t ath (0₂ +3.76 N2) b co, & d hot eN from atom balance 3.76 ath = b = 12 e . 2d = 26 da 13 29th e = (3.76X18.5) - 69.56. 26 td 2 2*9th (2x12 +13) = 37 So, ath. = 18.5 Stichiometric Equation : So G2 H2 et 18.5 (0₂ +3.76N₂) 12 Co₂ + 13 Mo + 69.56 N with 40% excess air, Mole of air (1-4X 18:5)= 25.9 42 426 + 25.9 (Q2 + 3.76 N2) = nCO2 + 2H2O + N₂ N₂ + nyo from atom balance n ;/ 2₂ =26 => = 12 - 13 m - (25.93.76)=97-384 25.9x2 = 20,+ P₂ + 204 151.8= (2x12) + 13 + (214) = 7.4 Actual equation K2 h + 25.9 (02 + 3.76N2) = 12 Cą + 13 H₂o et 12 Cą + 13 H₂O + 97.384 N₂ et 7.40) Son 4 =

Equivalence ratio (0) (A/F Stio chiometric Alf) actual. 18-5* [32 +8.76* 28) 12 * 127 26*12 25.9 * (32 +6-76*28)] (12x12 + 26*1) 0.7143 as 25.9 = 18.5 sol $5.9 *0.7143) Ta as (Ansana) n = 97.384 -18.5 ma = 7.4 (Ansara) b). Actual Alf 25.9 [Mo₂ + 3.76*MA, MC12H26. - 25.9* [32 + (3.76*28)] (12 x 12 & 26x1) Go Actual Alf 20.915 (Ansarb c) we have, Mass flow rate of fuel (me) = 49.5g/s. 20.915 " mo 20.915*49.5 1035.293 g/s or m (Ans--0). This is the actual air flow rate.

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