A coal that has unusually large hydrogen, methane, and water cornet has a LHV wi

Question

A coal that has unusually large hydrogen, methane, and water cornet has a LHV with magnitude of 8000 Htu/lbm. A chemical analysis suggests that the carbon and the volatile gases in the coal can be effectively modeled as 2 carbon and 3 hydrogen atoms; i.c., the combustible fuel can be modeled as "C_2H_3". Further testing reveals that the coal mixture is about 30% water and 70% "C_2H_3" (mole tractions). Determine a balanced reaction equation assuming that 20% excess air is used. Before proceeding, please CHECK THAT YOUR REACTION EQUATION COEFFICIENTS ARE CORRECT ! Find the heat liberated per lbm of coal if the reactants are at 77 degree F and the products an at 2000 degree R. If a power plant has an overall efficiency of 37% and is to provide a net power output of 200 MW. then determine the required coal flow rate (in tons/hr). Would the adiabatic flame temperature of this coal be more or less than 3000 degree R when burned with 20% EA ? (Guesses are not acceptable - this requires a calculation, but the calculation is fairly simple if you are clever.)

Explanation / Answer

2*C2H3 + (7/2)*O2 = 4*CO2 + 3*H2O

Mass of air = 7/(4*0.21) =8.33

Mass of carbon dioxide = (4/2)*0.7 = 1.4

Total mass of water = (3/2)*0.7 + 0.3 = 1.35

Let,

latent heat of vapourisation of water = 2258

specific heat of water vapour = 1.996

specific heat of water = 4.186

specific heat of carbon dioxide = 0.844

specific heat of air = 0.72

heat release per lbs of coal = heat gain by water + heat gain by carbon dioxide + heat gain by air

= [((1.75*4.186*(100-25))+(1.75*2258)+(1.75*1.996*(1093.33-100)))+(1.4*0.844*(1093.33-25))+(8.333*0.72*(1093.33-25))]/2.204

= 7094.186 KJ/lbs

power output of plant = 200 MW

Efficiency of plant = 0.37

heat required by plant = 200*10^6/0.37

                                           = 540540540.54 W

                                             = 540540540.54/3.412

                                             = 158423370.62 BTU/hr

Mass of coal required = 158423370.6/8000

                                        = 19802.92 lbs/hr

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