## Question

###### Thermodynamics Your assigned value of y is _2___ (You will lose 1 pt if this is...

thermodynamics

Your assigned value of y is _2___ (You will lose 1 pt if this is blank). Methane (CH) is burned with (110 + 0.3 y)% of theoretical air. (Hence, if y = 10 the methane is burned with 113% theoretical air). The mixture enters the combustion chamber at 25°C. a) Find the Air/Fuel ratio for this mixture: AF = (+ 0.03) b) Find the adiabatic flame temperature for the mixture. Tadiabitic flame = OC (+ 20°C) c) Find the mole fraction of the N2 in the exhaust. YN2 =

## Answers

buned with theoretical CHA is 110.67 (110+ 0.322 14. air so considu reaction as + bco₂+ 1.106 a (02 + 3-76N2) CHqt CH2O + d N2 + 0.10690₂ Balance C b=1 1x1 = b H-Balance 4x I- 2X C ) С 2. O Balance +0.106X2x a (1.106x2) x2 2x btc 20 2 btc -N-Balance 1-106x9 x 3.76x2 d= 8.31712 - co₂+ 2H₂O +8.317 N₂ + 0.21202 Chat 2.212 (02+3.76 N2) atase Combo Product Tadiabatic 2.212 x (1 +3.71) x 29 19.084 kg-au mig mit 2 A IX 16 keg-fuel exhaust mole fraction of Nz on Y No NN2 nTot 8.317 Li + 2+ 8-317+0-212) - 0.7214snehe Qc.v.= Enpho р P- Produch 97 Meactant at 298< or fer a5c adiabatic net Qc-v=0 Enphp = Ene he Enphp = (histh(t)) + a(rift + 8.3174 het hl())420 i theoret +0.212/rtthit), cay (hºt)oz=2 formation enthalpy of is zero. Cal. lenown value from Table + 8-314 entT)Nz +0.218 h(t)g H2O :(-3 •393522 h th(t)) cox + 2(-244826 +24t) 817174 .212-415), + 8-314 h(T) Ny to ACT) CO2 + 2h(T) H2O (rittent S -74873 kJ Temol Enne = 122 802301 h(T) + 0.212 thLT), = N2 +8.314 an(T) H2O HIT) CO2 + hit e trical. for Now By Tadê Let Tadi= 2000K 726 28 3.33 kJ 0.212 X 59176 = amol 8.314 561377 91439 + 2X 72788 + + 802 301 kg het Tadi: 2200k 810814.908 K Temol 103572 + 2x83153 7 63362x8.314 + 0.212 x66770 = 7 802.301 less kamol so 2000 < Tadia 22ook it lies blu there.using interpolation al I + 83153-72758) +78788 Tadi-2000 9200-200 (Tadi –2000) x (10356 2 - 91439] + 91439 ( 2 902 – 2000) и, 56137) +56137 Tadi-2000 2200-2000 (63362 + 8.314 +51137) N Tadi-2000 to.21 = 802 301 213 [ x( 59170] 66770.-59176) +59176 2200 zoow simplenty Tadi-2000) * 600615 + 103.65( Tadi-2000) + 301.2823 ( Tade-2000) + 8-0496 ( Tadi -2000) 802301 - 726 283.33 760 1767 Tadi- 2000) = Tadi - 2160:5113°C (60.615 + 103.65+ 301-2825 +8.0496) ALS.

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