A 250. MW coal-fired power plant burns fuel with an energy density of 35.0 MJ kg-1. Water enters the temp reduction tower at 293 K and leaves at 350. K at a rate of 4200. kg s-1.
4. Calculate the energy removed by the water each second.
5. Calculate the energy produced by the combustion of coal each second.
6. Calculate the overall efficiency of the power plant.
7. Calculate the mass of coal burned each second.
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1000 MW
To answer these questions, we will use the given information and apply the appropriate formulas. Let's break it down step by step:
4. To calculate the energy removed by the water each second, we can use the formula:
Energy removed by water = mass of water per second × specific heat capacity of water × (final temperature - initial temperature)
Given:
mass of water per second = 4200 kg/s
specific heat capacity of water = 4.18 kJ/kg·K (specific heat capacity of water is usually given in kJ/kg·K)
The initial temperature is 293 K, and the final temperature is 350 K.
Substituting the values into the formula:
Energy removed by water = 4200 kg/s × 4.18 kJ/kg·K × (350 K - 293 K)
5. To calculate the energy produced by the combustion of coal each second, we can use the formula:
Energy produced by coal per second = power output of the plant (250 MW) × 10^6 (to convert from MW to W)
6. To calculate the overall efficiency of the power plant, you need to find the ratio of the energy output to the energy input. The energy output is the power generated by the power plant (250 MW). The energy input can be calculated by multiplying the mass of coal burned each second by the energy density of the coal.
Overall efficiency = (energy output / energy input) × 100%
7. To calculate the mass of coal burned each second, we need to divide the power output of the plant by the energy density of the coal:
mass of coal burned per second = power output of the plant (250 MW) × 10^6 (to convert from MW to W) / energy density of the coal (35.0 MJ/kg)