A typical coal-fired plant burns 300 metric tons of coal per hour to generate 75

Question

A typical coal-fired plant burns 300 metric tons of coal per hour to generate 750MW of electricity. A metric ton = 1000kg. The density of coal is 1500kg per cubic meter, and its heat of combustion is 28MJ per kg. Assuming ideal efficiency, in that we can ignore friction in moving parts, inefficient heat transfer, etc., find the
(a) height to which one day's worth of coal must be piled in a square room 10m on a side
(b) the thermal efficiency of the power plant

Please help!! You'll be a lifesaver!!

Explanation / Answer

A typical coal fired plant burns 300 metric tons density of coal is 1500 kg per cubic meter            = 300 tons/1500 Kg           = 300000 kg/1,500kg/m^3             = 200 m^3 per hour
     200 * 24 = 4,800 cubic meters/day
    Area of the room is A = 100 *100                                     = 1000 m^2
                                    = 4800m^3  / 1000m^2
                       height of pile is 4.8 m b )    Q=300 tons * 1000 kg/ton * 28 MJ/kg                = 8.4 * 10^6 J of energy. However, the plant can generate only          750 MW of electricity (in one hour)          efficiency = 750*10^6 / 8.4 * 10^6 J                              = 89.2%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---