Question

nominal temperature of 660K and heptane 3. Assume that air enters a combustor at a (C-H16) enters at the reference temperature (298.15 K) creating products of combustion leaving at a nominal temperature of 1540 K. Values given in the table below are in units of kJ/kmol. Values are also given for the heat of formation. In the case of the H2O this would be for a gaseous product. (a) Write a reaction equation for heptane burning with air with c*100% theoretical air. (b). Write an energy balance across the combustor based on one mole of fuel assuming Qdot,cv 0 and Wdot,cv 0. (c). Find the percent theoretical air needed to produce the exit temperature of 1540 K. (d). Determine the fuel/air ratio assuming the molecular weight of C,Hi6-100.21 kg/kmol. Const MW h Tref 298 KTair-660 K Tp-1540 K 73417 59888 19870 50756 CO2 44.01 -393520 9364 HOI 18.021-24 18201 99041 32 N2 28.01 866919380 48470 CHi6 100.21 -224400

Answer 1

a) Heptane C_7 H_14 burning with air (with c* 100% theoretical air) C_7 H_14 + 2C(O_2 + 3.76N_2) rightarrow CO_2 + 2H_2O + 2(C - 1)O_2 + 7.52CN_2 (b) h_react = h_products hf C_7 H_ice + 2C(ho_2 Vertical-bar 660k + 3.76hN_2 Vertical-bar 660k) implies hfco_2 + (h/1540k - hf co_2) + 2(h + H_2o + h/1540 - hfn_2 o)) + 2(c - 1) (ho_2/1540) + 7.5c h/1540 N_2. (c) - 74850 + 2c(14770 + 54824.56) = -393520 + 57244 + 2(- 241820 + 47022) + 2(c - 1)40594 + 7.52c times 38807 implies - 74850 + 139189.12c = -807060 + 373016.64c c = 3.1314 % theoretical air Needed = 313.14% (d) F/A = 100.21/4.76 times 2 times 3.13 times 28.8 F/A = 0.1667 = 0.1167