GSEM5-LM1-Self Study 1-Solutions PDF

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Gas Separation Using Membranes 5 Learning Module 1 – Introduction and classification Self-study sheet 1 with solutions 1.1. To transport natural gas from a field in the North Sea to shore using a pipeline, high purity (>98 mol% CH4) is required. Therefore 400 kmol/h of natural gas containing 10 mol% of CO2 are to be treated using a membrane separation process. The gas is available at 60 °C and 54 bar. To reach the CO2 purity needed for reinjection in the field for Enhanced Oil Recovery (EOR) and reduce methane losses, the multistage configuration in figure is proposed using the same material for both separation stages M1 and M2. The final purity in stream 5 must be at least 86 mol% CO2. Assume that natural gas is a binary mixture. Calculate the methane lost in stream 5. Do you have enough uinformation to answer?

M1

M2

Yes, doing a mass balance around the total system (red dashed line), you do not need to know anything about the membranes to answer the question. Data: 𝑄1 = 400 𝑘𝑚𝑜𝑙/ℎ 𝑥𝐶𝑂2,1 = 0.1 𝑥𝐶𝐻4,1 = 0.9 𝑥𝐶𝑂2,3 = 0.02 𝑥𝐶𝑂2,5 = 0.86 𝑄𝐶𝐻4,5 =?

Total mass balance: 𝑄3 = 𝑄1 − 𝑄5 Methane mass balance: 𝑄𝐶𝐻4,5 = 𝑄5 ∙ 𝑥𝐶𝐻4,5 = 𝑄1 ∙ 𝑥𝐶𝐻4,1 − 𝑄3 ∙ 𝑥𝐶𝐻4,3 Substituting the second equation in the first: 𝑄1 ∙ (𝑥𝐶𝐻4,1 − 𝑥𝐶𝐻4,3 ) 𝑄5 = (𝑥𝐶𝐻4,5 − 𝑥𝐶𝐻4,3 )

1.2.

𝑄𝐶𝐻4,5 = 5.33 𝑘𝑚𝑜𝑙/ℎ

Convert 10 cm 3 (STP) of oxygen to SI units. STP refers to Standard Temperature and Pressure that are defined by IUPAC as 273.15 K and 1 bar. You can convert the known volume at STP to an amount of substance using the ideal gas law. Ideal Gas Law: 𝑃 ∙ 𝑉 = 𝑛 ∙ 𝑅 ∙ 𝑇 Data: 𝑉𝑆𝑇𝑃 = 10 𝑐𝑚 3 = 10−5 𝑚3 𝑇𝑆𝑇𝑃 = 273.15 𝐾 𝑃𝑆𝑇𝑃 = 100000 𝑃𝑎 𝐽 𝑅 = 8.314 𝑚𝑜𝑙 ∙ 𝐾 𝑃𝑆𝑇𝑃 ∙ 𝑉𝑆𝑇𝑃 𝑛= = 4.5 × 10−4 𝑚𝑜𝑙 𝑅 ∙ 𝑇𝑆𝑇𝑃

1.3.

Convert 10 Barrer in SI units. The definition of Barrer is: 1 𝐵𝑎𝑟𝑟𝑒𝑟 = 10−10

𝑐𝑚 3 (𝑆𝑇𝑃)∙𝑐𝑚 𝑐𝑚2 ∙𝑠∙𝑐𝑚𝐻𝑔

You can look for the conversion factors in the tables. From Baker (2012): 1 𝐵𝑎𝑟𝑟𝑒𝑟 = 3.347 ∙ 10−16 And therefore:

𝑚𝑜𝑙∙𝑚 𝑚2 ∙𝑠∙𝑃𝑎

10 𝐵𝑎𝑟𝑟𝑒𝑟 = 3.347 ∙ 10−15

𝑚𝑜𝑙 ∙ 𝑚 𝑚2 ∙ 𝑠 ∙ 𝑃𝑎

To avoid remembering the conversion factor we can use the definition of Barrer and of STP and the Ideal Gas Law to do the calculation. Ideal Gas Law: 𝑃 ∙ 𝑉 = 𝑛 ∙ 𝑅 ∙ 𝑇 𝑐𝑚 3 (𝑆𝑇𝑃) ∙ 𝑐𝑚 10 𝐵𝑎𝑟𝑟𝑒𝑟 = 10 ∙ 10−10 𝑐𝑚 2 ∙ 𝑠 ∙ 𝑐𝑚𝐻𝑔 3 −6 3 𝑐𝑚 (𝑆𝑇𝑃) = 10 𝑚 (𝑆𝑇𝑃) 𝑚3 (𝑆𝑇𝑃) ∙ 𝑐𝑚 𝑚3 (𝑆𝑇𝑃) ∙ 𝑚 −14 10 𝐵𝑎𝑟𝑟𝑒𝑟 = 10 ∙ 10−16 = 10 ∙ 10 𝑐𝑚 2 ∙ 𝑠 ∙ 𝑐𝑚𝐻𝑔 𝑚2 ∙ 𝑠 ∙ 𝑐𝑚𝐻𝑔 1 𝑎𝑡𝑚 𝑐𝑚𝐻𝑔 = 76 𝑚3 (𝑆𝑇𝑃) ∙ 𝑚 10 ∙ 10−14 ∙ 76 𝑚3 (𝑆𝑇𝑃) ∙ 𝑚 10 𝐵𝑎𝑟𝑟𝑒𝑟 = 10 ∙ 10−14 2 = 𝑚 ∙ 𝑠 ∙ 𝑐𝑚𝐻𝑔 101325 𝑚2 ∙ 𝑠 ∙ 𝑃𝑎 𝑇𝑆𝑇𝑃 = 273.15 𝐾 𝑃𝑆𝑇𝑃 = 100000 𝑃𝑎

𝐽 𝑚𝑜𝑙 ∙ 𝐾 10 ∙ 10−14 ∙ 76 𝑚3 (𝑆𝑇𝑃) ∙ 𝑚 10 ∙ 10−14 ∙ 76 ∙ 100000 𝑚𝑜𝑙 ∙ 𝑚 10 𝐵𝑎𝑟𝑟𝑒𝑟 = = 𝑚2 ∙ 𝑠 ∙ 𝑃𝑎 101325 101325 ∙ 8.314 ∙ 273.15 𝑚2 ∙ 𝑠 ∙ 𝑃𝑎 𝑚𝑜𝑙 ∙ 𝑚 10 𝐵𝑎𝑟𝑟𝑒𝑟 = 3.303 ∙ 10−15 2 𝑚 ∙ 𝑠 ∙ 𝑃𝑎

𝑅 = 8.314

The results is not exactly the same and depends on the definition of STP. Baker used 273.15 K and 1 atm (not 1 bar as in the IUPAC definition that changed in 1982).

1.4. Calculate the change in chemical potential of a perfect gas that is isothermally compressed from 2 to 30 bar at 40 °C. As we have a single gas that is compressed isothermally the change in the chemical potential can be calculated as: 𝑑𝜇 = 𝑉 𝑑𝑝 where 𝑉 is the molar volume of the gas. Since the gas is ideal we can use the ideal gas law: 𝑝𝑉 = 𝑅𝑇 And the expression becomes: 𝑑𝜇 = Data: 𝑝2 = 30 𝑏𝑎𝑟 𝑝1 = 2 𝑏𝑎𝑟 𝑇 = 313.15 𝐾 𝑅 = 8.314

𝑝2 𝑅𝑇 𝑑𝑝 = 𝑅𝑇𝑙𝑛 𝑝 𝑝1

𝐽 𝑚𝑜𝑙 ∙ 𝐾 𝑑𝜇 = 7050

𝑘𝐽 𝑚𝑜𝑙...