Question

Two problems needed to be worked out below:

1. If natural gas (98 percent CH4 by weight) is used to fuel a power plant, compute (a) the amount of oxygen required (in ton/hr) to produce 100 MW of power and (b) the amount of CO2 emission. Assume the calorific value of gas is 50 MJ/kg and all the heating value of gas is converted to electricity power.

2. Determine the saturation concentration of O2 (in mg/L) in water at 10 oC at 1 atm. Will the saturation concentration higher or lower at 20 oC? Use the following assumptions:
O2 is 21% of air (v/v)
Henry’s constant of oxygen in water is 3.27 ´ 104atm.

Answer 1

1) Let x grams of natural gas is used for generation of 100 MW power

The amount of CH4 will be = 0.98x grams

The comubstion equation is

          CH4 + 2O₂ ---> CO2(g) + 2H2O

the mass of O₂ needed for 16 grams of CH4 = 2 X 32 = 64grams

The mass of O2 needed for 0.98x gram of CH4 = 64 x 0.98x / 16 = 3.92 x grams

50MJ of energy is released by 1Kg therefore 50MW of power will be produced from 1Kg of gas

the power produced by 1 grams of gas = 50MW / 1000

The power produced by x grams of gas = 50xMW / 1000 = 100MW

x = 2000

So oxygen required = 3.92 x grams / second = 7840 grams / second = 7840 X 60 X 60 grams / hour

oxygen required = 28224 Kg / hour

oxygen required = 31.11 tons / hour of oxygen

1 mole of CO2 produced from 2 moles of O2

Hence 64 grams of O2 will give 44 grams of CO2

1 gram of O2 will give 44/64 grams of CO2

31.11 tons / hour of oxygen will give = 44 X 31.11 / 64 tons of CO2 / hour = 21.39 tons / hour

2) The solubility of gas = p / K_H

where p = partial pressure of gas K_H = Henry.s constsnt

Partial pressure of oxgen = 0.21 X 1 atmosphere

concentration = 0.21 / 3.27 X 10^4 = 6.422 X 10^-6 moles / L = 6.422 X 32 X 10^-6 grams / L = 205.504 X 10^-3 mg / L

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