A power plant burns coal and generates an average of 600.0 Megawatts (MW) of electrical power while discharging 900.00 MW as waste heat. Find the total electrical energy generated by the plant in a 30-day period.
Coals vary a lot in their energy content. The energy released by the burning of coal is typically between 23 and 33 Mega Joules per kg of coal. The power plant above uses coal rated at 26.7 MJ/kg to heat the boilers. Some of that heat is converted to electrical energy, but most is waste heat discharged into the environment. Assume that it takes about 2.20 kg of oxygen to burn 1 kg of coal (The actual amount varies with the type coal). Find the total mass of the reagents used in operating for 30 days.
physics - Damon, Monday, January 6, 2014 at 4:25pm
Well, now you are getting somewhere.
Go back to the original solution I gave you.
630 MW for 30 days ---> 1.63*10^15 Joules
so 600 MW for 30 days ---> 1.55*10^15 Joules
and 900 MW waste heat ---> 2.33*10^15 Joules
all of which means we must produce
(1.55+2.33)10^15 = 3.88*10^15 Joules with our coal and O2 in 30 days
coal ---> 26.7 *10^6 Joules/Kg
so we need
3.88*10^15/(2.67*10^7) = 1.45*10^8 Kg of coal for 30 days
then we will need 2.2*1.45*10^8 = 3.2*10^8 kg of O2
(1.45+3.2)10^8 = 4.64*10^8 Kg reagents
= 464 million Kg of coal and oxygen for 30 days
physics - Nika, Monday, January 6, 2014 at 4:52pm
something is still incorect at the first part..
physics - Damon, Monday, January 6, 2014 at 5:21pm
600*10^6 watts * 30 days * 24 hr/day *3600 seconds/hr = 1.55 * 10^15 Joules
seems to be what I got before
physics - Damon, Monday, January 6, 2014 at 5:24pm
You realize that originally in the earlier problem you gave me a 630 MW plant and later changed it to 600MW ?
physics - Nika, Monday, January 6, 2014 at 5:27pm
yeah, thank's I got it..
sorry for bothering you, you gave me a great help.