Question

Garbage is a potential source of energy. The average quantity of garbage produced each day per person in the US is approximately 2 kg. Assuming the energy content of the garbage to be approximately one-half of that of coal, determine the per-capita power, in watts, that could be obtained through its combustion. Based on a continuous per-capita consumption rate of 12 kW, determine the fraction of this quantity that could be supplied by garbage. Assume the heating value of one ton of coal to be 26 MBtu.

Answer 1

1 ton of coal BTU value =26 Mbtu= 26*10⁶btu ( 1 million =10,00,000)

1000 kg (1tons corrrespond to) =26*10⁶ Btu

1 kg of   cola prorduces =26*10⁶/1000 Btu=26*10³ Btu

1 kg of coal prduces the same energy as that of 2 kg of coal (since energy content of garbage is 1/2 that of coal)

2 kg of garbage prodces 26*10³ Btu/day =26*10³/(24*60*60) Btu/sec= 0.3008 Btu/sec

Consumpttion =12 Kw= 12 Kj/sec=12*0.9478 btu/sec (1 Kj= .9478 Btu) =11.3736 j/sec

Percentage of Garbage=100*0.3038/11.3736= 2.67%