Answer to Question #216693 in Chemical Engineering for Waj

Question

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

Answer 1

Part a

$T= 25^0 C = 298 K\\ P_s= e^{73.649- \frac{7258.2}{298}-7.3037 \ln(298)+4.1653*10-6(298^2))(\frac{760}{101325})}=23.57 torr\\ N \space \space Dry \space \space BWD =100(\frac{745-23.57*0.9}{760})(\frac{273}{298})=87.25 kgmol \space \space BWG\\ \frac{nH_2O}{ n \space dry \space gas}=\frac{ p H_2O}{ p \space dry \space gas}\\ nH_2O = 87.24 (\frac{ 23.57*0.9}{745-23.57*0.9})=2.556 kgmole\\ O_2 \space theo = 46.5 + \frac{101.2}{4}-25.2 = 46.6 kgmole\\ O_2 \space supplied = 1.30*46.6=60.58 kgmole\\ N_2 \space air =60.58 + \frac{79}{21} = 227.896 kgmole\\ N_2 \space total = N_2 \space air + N_2 \space fuel = 227.896 +4.5=232.4 kgmole\\ CO_2 \space formed = \frac{8}{9}*46.6 = 41.33 kgmole\\ CO \space formed = \frac{1}{9}*46.6 = 5.167 kgmole\\ Unburned \space H_2 = \frac{1}{4}*5.167 = 1.2967 kgmole\\ H_2 \space combusted = 101.2( \frac{1}{2})-1.29167 = 49.3083 kgmole=n H_2O \space combustion\\ NH_2O \space from \space air= (60.58+227.896)(\frac{23.57*0.85}{745-23.57*0.85} = 7.9721 kgmole\\ Free \space O_2 from = 13.98+\frac{5.167}{2}+\frac{1.29167}{2} = 17.2092 kgmole\\$

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

$\frac{T_C}{T_H}*100\\ \frac{25}{30}*100\\ 83.33\%$