Answer to Question #216693 in Chemical Engineering for Waj

Question #216693
Blue water gas is obtained by passing steam over red hot carbon enriched
coal (coke). The gas is called "blue" water gas because of its blue flame, a characteristic of the combustion CO. A typical composition shows 4.7% CO2, 41% CO, 49% H2, 0.8% CH4, and 4.5% N2. If the blue water gas at 30 deg C, 700 torrs and 93% RH is burned in wet air, 38% in excess, supplied at the same condition as BWG with 85% RH, calculte
A.) Orsat analysis of the stack gas if the molar ratio of CO2 to CO is 1:8 and H2 to CO is 1:4
B.) GCV of BWG
C.) % Calorific value lost due to sensible heat
D.) Thermal Efficiency
Expert's answer

Part a

"T= 25^0 C = 298 K\\\\\nP_s= e^{73.649- \\frac{7258.2}{298}-7.3037 \\ln(298)+4.1653*10-6(298^2))(\\frac{760}{101325})}=23.57 torr\\\\\nN \\space \\space Dry \\space \\space BWD =100(\\frac{745-23.57*0.9}{760})(\\frac{273}{298})=87.25 kgmol \\space \\space BWG\\\\\n\\frac{nH_2O}{ n \\space dry \\space gas}=\\frac{ p H_2O}{ p \\space dry \\space gas}\\\\\nnH_2O = 87.24 (\\frac{ 23.57*0.9}{745-23.57*0.9})=2.556 kgmole\\\\\nO_2 \\space theo = 46.5 + \\frac{101.2}{4}-25.2 = 46.6 kgmole\\\\\nO_2 \\space supplied = 1.30*46.6=60.58 kgmole\\\\\nN_2 \\space air =60.58 + \\frac{79}{21} = 227.896 kgmole\\\\\nN_2 \\space total = N_2 \\space air + N_2 \\space fuel = 227.896 +4.5=232.4 kgmole\\\\\nCO_2 \\space formed = \\frac{8}{9}*46.6 = 41.33 kgmole\\\\\nCO \\space formed = \\frac{1}{9}*46.6 = 5.167 kgmole\\\\\nUnburned \\space H_2 = \\frac{1}{4}*5.167 = 1.2967 kgmole\\\\\nH_2 \\space combusted = 101.2( \\frac{1}{2})-1.29167 = 49.3083 kgmole=n H_2O \\space combustion\\\\\nNH_2O \\space from \\space air= (60.58+227.896)(\\frac{23.57*0.85}{745-23.57*0.85} = 7.9721 kgmole\\\\\nFree \\space O_2 from = 13.98+\\frac{5.167}{2}+\\frac{1.29167}{2} = 17.2092 kgmole\\\\"

Part b

"M^3 \\space air \/m^3 \\space BWG = 2.8848* \\frac{22.4}{1}* \\frac{760}{745-23.57*0.9}*\\frac{298}{273}*\\frac{1}{22.4}*\\frac{273}{298}* \\frac{745-23.57*0.9}{760}=2.88"

Part c

"\\frac{m^3 \\space stack \\space gas }{100 \\space moles \\space BWG} = 357.23 * \\frac{22.4}{1}*\\frac{101.325}{100}* \\frac{573}{273}=17017.84 \\frac{m^3}{kg mol}"

Part d


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be the first!

Leave a comment

New on Blog