Page 50 - Handbook of Energy Engineering Calculations

(1.2 × 3.78) N .
2

3. Compute the relative weights of the reactants and products of the
combustion process
Relative weight = moles × molecular weight. Coefficients of the chemical

equation in step 2 represent the number of moles of each component. Hence,
for the reactants, the relative weights are: for C = 1 × MC = 1 × 12 = 12; O =
2
1.2 × MO = 1.2 × 32 = 38.4; N = (1.2 × 3.78)MN = (1.2 × 3.78 × 28) =
2
2
2
127. For the products, relative weights are: for CO = 1 × MCO = 1 × 44 =
2
2
44; O , = 0.2 × MO = 0.2 × 32 = 6.4; N = 127, unchanged. It should be
2
2
2
noted that the total relative weight of the reactants equal that of the products
at 177.4.
4. Compute the relative weights of the products of combustion on the basis
of a per unit relative weight of carbon
Since the relative weight of carbon, C = 12 in step 3; hence, on the basis of a

per unit relative weight of carbon, the corresponding relative weights of the
products are: for carbon dioxide, wCO = MCO /12 = 44/12 = 3.667; oxygen,
2
2
wO = MO /12 = 6.4/12 = 0.533; nitrogen, wN = MN /12 = 127/12 = 10.58.
2
2
2
2
5. Compute the final product’s temperature
Since the combustion chamber is insulated, the combustion process is
considered adiabatic. Hence, on the basis of a per unit mass of carbon, the

heating value (HV) of the carbon = the corresponding heat content of the
products. Thus, relative to a temperature base of 77°F (25°C), 1 × HVC =
[(wCO × c CO ) + (wO × c O ) + (wH + c N )](t − 77), where the
2
2
p 2
p 2
2
2
2
p
3
heating value of carbon, HVC = 14,087 Btu/lb (32.74 × 10 kJ/kg); the
m
constant-pressure specific heat of carbon dioxide, oxygen, and nitrogen are
c CO = 0.300 Btu/lb (0.697 kJ/kg), cpO = 0.240 Btu/lb (0.558 kJ/kg), and
2
p
2
c N = 0.285 Btu/lb (0.662 kJ/kg), respectively; final product temperature is
p 2
t ; other values as before. Then, 1 × 14,087 = [(3.667 × 0.30) + (0.533 ×
2
0.24) + (10.58 × 0.285)(t − 77)]. Solving, t = 3320 + 77 = 3397°F
2
2
(1869°C).