Question

Problem #3 (50 pts) Liquid ethanol burns steadily with air supplied at 33 phase) exit at 1000 K. 1 atm. A dry molar analysis CO, 16.6%; N, 80.24%. Atomic weights Ignore kinetic and potential energy effects. (1) (12 pts) Write down the chemical reac uid ethanol (CH.OH) at 298K. 1 atm enters a combustion chamber and Pplied at 350K. 1 atm. The combustion products (with water in the gas molar analysis of the combustion products gives CO., 3.16%; omic weights of C, H, O, N are 12, 1, 16, 14 g/mol, respectively. mass. e chemical reaction equation and determine the air-fuel ratio by Calz Hal ? a (CH₃OH + 3(0e + 37602) b. 22612) CO2+ 3 H2O + 11.2882 O = 16 N=14 + c# Dega (4) (18 pts) If there is no work involved, determine the rate of heat transfer from the burner to the environment, in kJ per kmol of fuel. T=2984

Answer 1

Selection, 2 2 Liquid ethanal - CG H₂ OH = C 460 XC H0+ Y0, +2 No 73.16C02 +16,600+80124 Ng taH, O Nitrogen Balence - 2=80.24 since all nitrogen comes from are, Ž -29 - ys 24 - 21132 Carbon Balance 22-3:16 +16,6 -) X-9.88 Hydrogen Balance a=3x3 3x9.88 229164 Oxygen Balance + Y=311641616+ 29,64 Y=2132 Substiteting these values fos X, Y,24 a we have 918862 H704211320,+80.24 N, --- 3.16C0g #1616 CO +80 24 N, +27. 6 4 11, 0 Now dividing both sides by 9.88 C2H60+2.16% +8.12N, 70.3200, +1682048:12N+3440 (combustion ean) (12.1/f Ratio - 2.16781/2)X28.97 " "1282 +6x7+16 M/F 76147 aislig full (Aus) (2) Hp = (C2H60 läh298 +(2:16 the t8.l264) Esok 5-276000 kr/moet (2116 X 1531) +(8•12x152) HR3-260424 kt (kmad.

Hp at 10ook -32XC-3935220* 33405) 41068 610529621686 +(8112821406)+3x6241827725978) HP 3-115237-1492577173817 -647547 #p= -738223 kJ/Kmala Heat transfers Hp Hp & 477799 kJ kmol of fuel.