compute the simple operating margin Co2 factor for the year 2020, if the system

Question

compute the simple operating margin Co2 factor for the year 2020, if the system load of 6000MW is supplied by 10 coal genrator unit with a capacity of 100MW, 10 oil-firing generators with the capacity of 50MW and 10 gas fire generators with the capacity of 50MW

(a) the system load of 6000MW supplied from only coal unit.

(b) The system loads of 6000MW supplied equally from gas.

Question 2

If emission factor of producing electric power by photo voltaic cells is 100g of Co2 per KWh, by wind power is 15g of Co2 per KWh, and by coal is 1000g of Co2 per KWh, then find the ratio of Co2 emission when:

(a) 15% of power comes from wind farm

(b) 5% from photo voltaic source and

(c) the rest from coal as opposed to when all power is supplied by coal-run power stations

Explanation / Answer

Gievn the system load is 6000MW which is supplied by 10 coal units with the capacity of 100MW, 10 oil firing generators with the capacity of 50 MW and 10 gas firing generators with the capacity Of 50 MW.
The CO2 factor is nothing but the emission factor which is varying with the type of the fuel.
(a) If the system load of 6000MW supplied only from coal unit. For the coal generator let us consider the fuel as Lignite coal. It has the CO2 factor of 97.72kg CO2 per mm BTU = 97.72*103*103 CO2 per btu
we know 1MW = 34.12*105 btu/hr
hence the CO2 factor for 1MW load is 97.72*106/(34.12*105) = 2.86*10-5 *106 *60 = 17.16 *10
hence the CO2 factor for 6000MW load is 6000*171.6= 1029.6*103 = 1029.6 Kg per btu
(b) If the system loads of 6000MW from gas generator.
Let us consider the natural gas , the CO2 factor of natural gas in 2020 , is given by 53.06 kg CO2 per mm btu
hence the
It has the CO2 factor of 53.06 kg CO2 per mm BTU = 53.06*103*103 CO2 per btu
we know 1MW = 34.12*105 btu/hr
hence the CO2 factor for 1MW load is 53.06106/(34.12*105) = 1.55*10-5 *106 *60 = 93 *10
hence the CO2 factor for 6000MW load is 6000*930= 5580*103 = 5580 Kg per btu
Question 2 :

Given the emission factor of producing electric power by photo voltaic cells is 100g of CO2 per KWh
emission factor of producing electric power by wind is 15g of CO2 per KWh
emission factor of producing electric power by coal is 1000g of CO2 per KWh
We know the molecular weight of CO2 which consists of one mole of carbon and 2 moles of oxygen = (2*6)+(2*8*2)
= 12+32= 44grams.
Therefore emission factor of producing electric power by photo voltaic cells is 100/44 = 2.27per KWh
emission factor of producing electric power by wind is 15/44= 0.34per KWh
emission factor of producing electric power by coal is 1000/44= 22.7per KWh

Given 15% of power comes from wind= 0.15 * 0.34 per KWh = 0.051per KWh
5% from photo voltaic cell = 0.05 * 2.27 = 0.1135 per KWh
and remaining 80% from coal = 0.80 * 22.7 = 18.16 per KWh
Hence total Co2 emission is 0.051+0.1135+18.16 = 18.3245.
Hence the ratio 0.051 : 0.1135 : 18.16

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