For a different power plant the power input (not output) is 3570 MW. The coal has an energy content of 28,000 kJ per kg. How much coal must be put into the plant each day

31475404.8 kg/day

Explanation:

From the information given:

The power plant capacity W = 3570 MW

Energy content of the coal = 28000 kJ/kg

let assume that the thermal efficiency = 35%

Recall that:

1 kw = 3600 kJ/hr provided that the energy conversion is 100% efficient

But assuming the thermal efficiency = 35%.

Then:

Heat input = 3600/0.35 = 10286 kJ/kw.hr

Now, for producing 1 kw.hr, the quantity of the required coal = 10286/28000

= 0.36736 kg

For 3570 MW, the amount of coal that must be input is:

= 0.36736 kg × 3570000

= 1311475.2 kg/hr

= 1311475.2 × 24 kg/day

= 31475404.8 kg/day

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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