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Chrome File Edit View History Bookmarks People Window Help 67% Fri 11:22 AM E xN Lab 3 aerofoil-OneDr x WE ME323 Airfoil Exper x C Home chegg.com x Homepage UW-Mil x Alec G 2017 camaro Go Assignment 3 Pow C Secure https://uwm.courses.wisconsin.edu 7954/Vie Apps UWM B Fantasy f E E P C em cold Jerseys JT PayTwc We Energies https://studentloan A My Home Power Plant Theory and Alec Zambrowicz Problem 1: A sample of coal has the following molar (volumetric) analysis: C 67.35%; H2, 26.26%; O2, 2.28%; N2, 0.57%; S, 1.37%; H20, 2.17%. Write the complete combustion equation in stoichiometric air, and calculate the coal ultimate analysis (mass percentages.) Problem 2 A natural gas has the following ultimate analysis by mass: H2, 23.3%, C, 74.72%; N2, 0.76%; and O2, 1.22%. The flue gases have the following volumetric analysis H20, 15.583%; CO2, 8.387%; O2, 3.225%, N2, 72.805%. Calculate (a) the percent theoretical air used in combustion, and (b) the dew point if the flue gases are at2 bar Hints: First find the chemical reaction for 100% theoretical air (remember to convert mass fractions to mole fractions for the fuel.) Then calculate a general chemical formula for excess air. The products, as given, will enable you to calculate a mole fraction of one of the products. The general equation you calculated will give you a second method of solving for the mole fraction. terms of the amount of excess air. The dew point is found by finding the saturation temperature corresponding to the partial pressure of the water in the products (the mole fraction of water in the products times the total pressure is the partial pressure.) Problem 3 A stoichiometric mixture of methane and air at 250C burns completely. Calculate the heat released if the exit temperature is 2000 K Problem 4 Fay and Golo Problem 4.1 Graduate Students Only: Problem 5: Calculate the adiabatic flame temperature in K for the following coal if it burns from 25 C with 100% theoretical air. Ignore the effect of sulfur on the energy balance. Coal: C-58.8%, H2-3.8%, S- 0.3%, O2-12.2%, N2-1.3%, H20-19.6% (on a mass basis.) The enthalpy of formation of the coal is -113,190 BTU/bmole

Explanation / Answer

Assume a basis of 100 mol

so

mol of C = 67.35

mol of H2 = 26.26

mol of O2 = 2.28

mol of N2 = 0.57

mol of S = 1.37

mol of H2O = 2.17

of which, assume only C and H2 will combust:

1 mol of C ---> 1 mol of O2 required for CO2

mol of C = 67.35 --> 67.35 mol of O2

1 mol of H2 --< 1/2 mol f O2

mol of H2 = 26.26 ---> 26.26/2 = 13.13 mol of O2

1 mol of S --> 3/2 mol of O2 required

mol of S = 1.37 ---> 1.37*3/2= 2.055 mol of O2

total mol of O2 required = 67.35 +13.13+ 2.055 = 82.53 mol

but we have already 2.28 mol in sqample so:

82.53 -2.28 = 80.25 mol of O2 required

so..

for air --> 0.21 is O2 so --> 80.25 /0.21 = 382.14 mol of air reqruied, i.e. mol of N2 = 382.14 -78.2 = 303.94 mol of N2 in air...

now..

outlet:

63.35 mol of C --> 63.35 mol of O2

26.26 mol of H2 --> 26.26 mol of H2O

mol of O2 --> 0 is left (exact ratio)

mol of N2 --> 0.57+303.94 = 304.5 mol

mol of S --> 1.37 mol of SO3

mol of H2O --> 2.17

so

change to mass:

63.35 mol of CO2 --> 63.35*44 = 2787.4 g

26.26+2.17 = 28.43 mol of H2O --> 28.43 *18 = 511.74 g

mol of N2 = 304.5 mol = 304.5*28 = 8526 g

mol of SO3 = 1.37 mol = 1.37*80.066 = 109.690g

total mass = 2787.4 + 511.74 +8526 + 109.690 = 11934.83 g

so

%CO2 = 2787.4 /11934.83*100 = 23.35%

%H2O = 511.74   /11934.83*100 = 4.287%

%N2 = 8526  /11934.83*100 = 71.437%

$SO3 = 109.690  /11934.83*100 = 0.91907 %

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