This is given from my notes: Power Generation Oil - 40% efficient; CO2 Emissions

Question

This is given from my notes: Power Generation Oil - 40% efficient; CO2 Emissions (g/mol) 70 g/mol Natural gas - 23% efficient; CO2 Emissions (g/mol) 50g/mol Coal - 23% efficient; CO2 Emissions (g/mol) 90 g/mol Energy Density of various sources (In heat of combustion -MJ/kg) Coal - 30 MJ/kg Diesel - 45 MJ/kg Gasoline - 47 MJ/kg Keronsene - 46 MJ/kg Natural Gas - 39 MJ/kg @ STP Water - 10,000 J/m3 Please Show all work and give as much details aspossible. Questions: A coal power plant is 23% efficient. If it produces 50MW of power, how much energy does it actually consume persecond? Per hour? Per Day? How much coal does it burn to do this (I think you need thetable I wrote at the top) How much CO2 is produced per second? Per hour? Per day? Please help me with this - I can use this help for a labI have to do later. Thanks for your time. This is given from my notes: Power Generation Oil - 40% efficient; CO2 Emissions (g/mol) 70 g/mol Natural gas - 23% efficient; CO2 Emissions (g/mol) 50g/mol Coal - 23% efficient; CO2 Emissions (g/mol) 90 g/mol Energy Density of various sources (In heat of combustion -MJ/kg) Coal - 30 MJ/kg Diesel - 45 MJ/kg Gasoline - 47 MJ/kg Keronsene - 46 MJ/kg Natural Gas - 39 MJ/kg @ STP Water - 10,000 J/m3 Please Show all work and give as much details aspossible. Questions: A coal power plant is 23% efficient. If it produces 50MW of power, how much energy does it actually consume persecond? Per hour? Per Day? How much coal does it burn to do this (I think you need thetable I wrote at the top) How much CO2 is produced per second? Per hour? Per day? Please help me with this - I can use this help for a labI have to do later. Thanks for your time.

Explanation / Answer

Plant is 23% efficient. Therefore consumed energy = producedenergy divided by 23% Likewise, consumed power = power produced divided by 23% Therefore, the plant consumes 50 MW/23% = 217.4 MW = 217.4MegaJoules per second 217.4*3600 = 782,609 MJ/hour 782609*24 = 18,782,609 MJ/day 217.4 MJ/s divided by 30 MJ/kg = 7.25 kg/s of coal per second CO2 produced = 7246 g/s 7246 g/s divided by 90 g/mol = 80.5 mol/s of CO2 80.5*3600 = 289,855 mol/hr 289855*24 = 6,956,522 mol/day Source: cramster.com/physics-answers-5-738386-cpi0.aspx

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