During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).
(a) Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L.
(b) How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C3H8: density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO2(g) and H2O(l)] and the furnace used to burn the LPG has the same efficiency as the gas furnace.
(c) What mass of carbon dioxide is produced by combustion of the methane used to heat the house?
(d) What mass of water is produced by combustion of the methane used to heat the house?
(e) What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L.
(f) How many kilowatt–hours (1 kWh = 3.6 × 106 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house.
(g) Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?
(a)
Calor promedio producido
q = (3500 kWh) (3.6x10 ^ 6 J / 1kWh) (1 / .89) = 1.416x10 ^ 10 J = 1.416x10 ^ 7 kJ
entalpía de combustión del CH_4 =890.8 [kJ/mol]
𝑛_𝐶𝐻4 = 14157303.4/890.8 = 15892.7968 [𝑚𝑜𝑙]
Masa molar del CH_4 =16.04 [g/mol]
𝑚_𝐶𝐻4 = 16.04(15892.7968) = 254920.4607 [𝑔]
𝑣 = 𝑚/ 𝜌 = 254920.4607 /0.681 = 374332.5414 [𝐿]
Realizando la conversión tenemos:
𝑣 = 13217.6820 [𝑓𝑡^ 3 ]
(b)
Sabemos:
La entalpía de combustión del GLP es de 2219 kJ / mol
Calor promedio producido es de 1.4157x10 ^ 10 [J]
mol_LPG = (1.4157x10 ^ 7 [kJ]) / 2219 [kJ / mol] ≈6380mol
Masa molar del propano (C3H8) = 44 g / mol
Masa necesaria de LPG = (6380 [mol]) (44 g / mol) = 280 720 [g]
Volumen necesario = ((280720 [g])) / ((0.5318 [g / mL])) = 527 867.6 [mL] = 139.45 [galones]
(c)
Relación estequiométrica
CH_4 + 2O_2 + 3.76N_2 → CO_2 + 2H_2 O + 3.76N_2
mol CH_4 = mol de dióxido de carbono = 15892
Con masa molar del CO2 = 44 g / mol
Masa CO_2 = (15892 [mol]) (44 [g / mol) = 699248 [g] ≈ 700 [kg]
(d)
Relación estequiométrica
CH_4 + 2O_2 + 3.76N_2 → CO_2 + 2H_2 O + 3.76N_2
mol H_2 O = (2) (15892 [mol]) = 31784 [mol]
Utilizando la masa molar del H2O = 18
masa H_2 O = (31784 [mol]) (18 [g / mol) = 572 112 [g] ≈ 572 [kg]
(e)
Se necesitan 2 mol de oxigeno por cada uno de metano
Número de moles de oxígeno = 2 (15895.82) = 31791.64 mol
31791.64 mol (32 g / mol) = 1017332.48 g = 1017.33 kg
21% del aire es oxigeno
1017.33 kg / 0.21 = 4844.43 kg
Densidad del aire = 1.22 g / L
Vol = (4844.43 kg x1000) / 1.22 g / L
Vol = 3970843 [L]
(f)
Energía eléctrica = (1.4157x10 ^ 10 [J] (1 kWh) (0.89)) / ((3.6x10 ^ 6 J)) = 3500 [kWh]
(g)
Energía eléctrica de la central
q = 3500 kWh / 0.4 = 8750 kWh
Carbón requerido:
8750 kWh / 2.26 kWh / lb = 3871.68 lb
Conversión de libras a kg
3871.68 lb x 0.454 kg / lb = [1757.74 kg]
Problema bolsa de aire: https://sites.google.com/site/201908giselleruizlopez/201908_giseller_03