please show full explanation and steps for answer A coal-fired power plant bums
ID: 1068432 • Letter: P
Question
please show full explanation and steps for answer
A coal-fired power plant bums bituminous coal at a rate of 50.0 kg/s. Composition analysis shows that the coal contains 2.0 wt% of sulfur. Assume that the combustion efficiency of sulfur in the coal is 95% (i.e., 95% of the sulfur will convert to SO_2(g). and 5% of the sulfur will be present in the ash). What is the daily rate of emission of SO_2(g), without the installation of any Hue gas desulfurization (FGD) system? Give your answer in kg/d. A FGD system is installed to remove SO_2(g) by the reaction: SO_2(g) + CaCO_3(s) + 2H_2O + 0.5O_2(g) = CO_2(g) + CaSO_4 2H_2O(s) What mass of the product CaSO_4 2H_2O_(s) will be produced in one day? Assume a 100% yield o the above reaction (i.c.. all SO_2(g) can be converted to CaSO_4 middot 2H_2O). Give your answers in kg/d.Explanation / Answer
Given:
Bituminus coal burns at a rate of 50 kg/s.
Sulphur content in Coal is 2%
Combustion efficiency of sulphur = 95%
Hence 50 kg of coal contains 50 x 2/100 = 1 kg of Sulphur
Combustion efficiency of 1 kg is 950 g
therefore 950 g of S burns every sec
S + O2 SO2
32 g of S forms 64 g of SO2
Hence, 950 g of S forms 1900 g of SO2
1900 g of SO2 is formed per second.
Therefore the amount of SO2 formed per day = 1900x60x60x24 ( secminhrday) = 164160000 g /day
= 164160 kg /day
SO2 + CaCO3 + 2 H2O + 0.5 O2 CO2 + CaSO4 . 2 H2O
According to the above equation 64 g of SO2 forms 172 g of CaSO4.2 H2O.
Therefore, 164160 kg of SO2 will form (172/64)/ 164160 = 441180 kg / day.
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