please box answers A fuel composed, by weight, of 70% carbon, 28% hydrogen, and
ID: 532274 • Letter: P
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please box answers
A fuel composed, by weight, of 70% carbon, 28% hydrogen, and 2% sulfur is burned with 10% excess air For an 800 MWe plant with a 40% thermal efficiency: What is the input fuel requirement [lbm/hr]? What are the emissions [in lbm/hr]? In the combustion Example of Lecture 7, a "divide and conquer" approach was taken to solve the problem. Repeat that exercise except (1) perform all the calculations (for example, the heating value and the fuel input mass flow rate) exclusively using SI units (i.e., kJ/kg and k and (2) do not use 3 separate reaction equations, rather use the single long stoichiometric balance equation and explicitly calculate the number of moles of the reactants and products to determine the flue gas emissions in kg/hr. (Do not compute the scrubber system input/output masses.)Explanation / Answer
a)
E = 800 MW 40% efficiency
Etotal required = Eoutput / efficient = 800/0.4 = 2000 MW required
C + O2 = CO2 H = -393.5
S + O2 = SO2 H = -296.8
H2 + 1/2O2 = H2O H = -241.8
Assume a basis of : 100 g of fuel
70 g is C, 28 g is H and 2 g is S
mol of C = mass/MW = 70/12 = 5.833 mol of C --> 5.833 mol of CO2
mol of H = masS/MW = 28/1 = 28 mol of H = 14 mol of H2 = 14 mol of H2O
mol of S = mass/MW = 2/32 = 0.0625 mol of S = 0.0625 mol of SO2
Total Energy = n1*Hf1 + n2*Hf2 + n3*Hf3 = 5.833 *-393.5 + 14*-241.8 + 0.0625*-296.8 = -5699.0355 kJ
this is per 100 g
Energy density = 5699.0355 kJ / 100 g
E = 2000 MW = 2000*10^3 kW = 2*10^6 kW = 2*10^6 kJ/s --> (2*10^6)(3600) = 7200000000 kJ/h
Total flow =(7200000000 kJ /h) / ( 5699.0355 kJ / 100 g ) = 7200000000 /5699.0355 *100 = 126337167.0 g/h
454 g = 1 lbm
126337167.0 g = x
x = 126337167.0 g / 454 g / lb = 278275.698238 lbm/h
b)
find emissions:
278275.698238 lbm/h --> 278275.698238 *0.70 * /12*44 = 714240.95 lbm of CO2 / h
278275.698238 lbm/h --> 278275.698238 *0.02/32*64.066= 11142.50 lbm of SO2 / h
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