A 600,000 kilowatt natural gas power station burns its fuel with an average stoi

Question

A 600,000 kilowatt natural gas power station burns its fuel with an average stoichiometry of C3H7 and a LHV of 20,000 BTU/lbm. The plant is 42% efficient with energy recovery systems operation. CO2 is not captured, but emitted to the atmosphere.

a.) What is the rate of consumption of the fuel in kilograms/second?

b.) How many grams of CO2 are produced per gram of fuel burned?

c.) What is the rate of CO2 production in kilograms/second?

d.) What is the CO2 "footprint" in kilograms of CO2 per kilojoule of electrical output?

Explanation / Answer

Ans. #1. Electrical Energy production rate = 600000 kilowatt

                                                            = 6.0 x 105 kJ/s                      ; [1 W = J/s]

#2. Lower Heat Values (LHV) of fuel = 20000 BTU/ lbm

                                                = 20000 x 2.326 kJ/ kg                     ; [1 BTU/lbm = 1 kJ/kg]

                                                = 4.652 x 104 kJ/ kg

#3. Efficiency of the plant = 42%

Let the rate of fuel consumption = X kg/s

Total energy produced from fuel using LHV = Total mass of fuel x LHV of fuel

                                                          = (X kg/s) x (4.652 x 104 kJ/ kg)

                                                            = (4.652 x 104 kJ/ s) X

Thus, total rate of energy production = (4.652 x 104 kJ/ s) X

At 42% efficiency, electrical energy production rate =

= 42% of Total energy production rate

                                                = 42 % of (4.652 x 104 kJ/ s) X

                                                = (1.95384 x 104 kJ/ s) X

So, (1.95384 x 104 kJ/ s) X = Actual value of electrical energy production rate

            Or, (1.95384 x 104 kJ/ s) X = 6.0 x 105 kJ/s

            Or, X = 6.0 x 105 kJ/s / (1.95384 x 104 kJ/ s) = 3.071

Thus, rate of fuel consumption = X kg/s = 3.0741 kg/s

#4. Balanced reaction: 2 C3H7 + (13/2) O2 --------> 6CO2 + 7H2O

Molar mass of C3H7 = 43.09 g/ mol

Molar mass of CO2 = 44.00 g/ mol

According to the balanced stoichiometry, 2 moles C3H7 produces 6 mol CO2. Since mass if directly proportional to number of moles (no need of converting to moles), the mass of C3H7 consumed is directly proportional to mass of CO2 produced in same stoichiometry.

            So, 43.09 g C3H7 is equivalent to 44.00 g CO2

            Or, 1.0 g         -           -           -           -(44.0 /43.09) g

                                                                        = 1.02 g

Therefore, 1.0 g of fuel produces 1.02 g CO2.

Also, 1.0 kg of fuel produces 1.02 kg CO2.

#5. Rate of CO2 production = Total Rate of fuel consumption x CO2 equivalent of fuel

                                                = 3.0741 kg fuel/s x (1.02 kg CO2 / kg fuel)

                                                = 3.136 kg CO2 /s

#6. CO2 footprint = kg of CO2 / kJ of electrical output

Mass of fuel consumed to produce 1 kJ electricity=

                                    Fuel consumption rate / Electrical Energy production rate

                                    = (3.0741 kg fuel/s) / 60000 kJ/s

                                    = 5.1235 x 10-5 kg fuel / kJ electricity

CO2 footprint = (Mass of fuel consumed /kJ electricity) x CO2 equivalent of fuel

                        = (5.1235 x 10-5 kg fuel / kJ electricity) x (1.02 kg CO2 / kg fuel)

                        = 5.226 x 10-5 kg CO2 / kJ electricity

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