Chemical Engineering Question Carbon dioxide (along with other gases) has been i

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Chemical Engineering Question

Carbon dioxide (along with other gases) has been implicated in global warming. A process is being developed which will separate CO2 from flu gas in a coal fired power plant and subsequently sequester the CO2 underground. A schematic of the proposed process is shown below.

A cleaned and dry flu stream (m1), comprised of 26 wt% CO2 and 74 wt% N2, enters into an absorber at a flow rate of 5.0 × 104 kg/hr where it contacts liquid-phase amine solvent (MEA) in an absorber. The amine solvent selectively absorbs CO2. The flu gas stream exiting the absorber (m2) is composed of 98 wt% N2 and 2.0 wt% CO2. The liquid stream exiting the absorber (m3) contains only amine and CO2. The liquid stream is subsequently fed to a stripper where it is heated and 95% of the available CO2 is desorbed from the amine. The CO2 exiting the stripper (m4) is subsequently pressurized and injected into subterranean rock formations. The MEA exiting the stripper (m5), comprised of 99 wt% amine and 1.0 wt% CO2, is recycled back into the absorber.

a.) What percent of the CO2 in stream m1 is removed?

b.) What is the concentration (in wt%) of CO2 in stream m3?

c.) What is the flow rate (in kg/hr) of the recycle stream (m5)?

Explanation / Answer

a)
Lets consider for 1 hour:
In m1:
CO2 = 26% of 5*10^4 Kg = 13000 Kg
N2 = 74% of 5*10^4 Kg = 37000 Kg

In m2:
Whole of N2 goes:
So N2 = 37000 Kg which is 98 %
0.98*m2 = 37000
m2=37755 Kg

CO2 = 2% of m2 =2% of 37755 Kg = 755.1 Kg

CO2 removed from m1 = 13000 - 755.1 =12244.9 Kg

Removed % = 12244.9*100 / (5*10^4) = 24.5 %

b)
In m5:
mass of CO2 = 1% of m5 =0.01 m5 Kg
mass of MEA = 99% of m5 =0.99 m5   Kg
Absorber adds 12244.9 Kg osf CO2

Before stripper:
Total CO2 = 0.01 m5 + 12244.9 Kg
Total MEA = 0.99m5 Kg

95 % of CO2 is removed by stripper
Co2 removed = 0.95* (0.01 m5 + 12244.9 )

After stripper:
CO2 = 5% of (0.01 m5 + 12244.9 ) = 0.05*(0.01 m5 + 12244.9 )
This account for 1 % of m5

so,
0.01*m5 = 0.05*(0.01 m5 + 12244.9 )
0.2 m5 = 0.01 m5 + 12244.9
m5=64446.8 Kg

In m3:
Total CO2 = 0.01 m5 + 12244.9 Kg = 0.01*64446.8 + 12244.9 = 12889.4 Kg
Total MEA = 0.99m5 Kg = 0.99*64446.8 =63802 Kg
Concentration of CO3 in m3 = 12889.4*100 /(12889.4 + 63802) = 16.8 %

c)
m5 as calculated above= 64446.8 Kg
Time assumed was 1 hour
So rate is 64446.8 Kg/hr

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