Chemical process calculation in petroleum engineering

Chemical process calculation involves stochiometry solution to chemical Engineering design problems

1. Ideal gases, Real gases, Single component two phase systems, Multiple component phase systems, Phase rule, Phase equilbria , Combustion processes . UNIT III 3/9/2016 1 Dr.SS/Dr.DB

2. 3/9/2016 Dr.SS/Dr.DB 2 Gas Calculations

3. 3/9/2016 3 Dr.SS/Dr.DB Problem-1 What is the volume of 25kg of chlorine at standard condition? Required: Volume of Cl 2 =?m 3 MWT- Molecular weight

4. 3/9/2016 4 Dr.SS/Dr.DB Problem-2 What is the weight of one liter of methane (CH 4 ) at standard conditions? Required: Weight of 1 l CH 4

5. 3/9/2016 5 Dr.SS/Dr.DB Problem-3 A compound whose molecular weight is 103 analyses Carbon = 81.5%, Hydrogen = 4.9% and N 2 =13.6%, by weight. Find its formula? Element Atomic weight(g/ g.atom ) Weight, g Atom weight, g.atom Rounding of atoms Weight.g Carbon 12 81.5 81.5/12=6.8 7 84 Hydrogen 1 4.9 4.9/1=4.9 5 5 Nitrogen 14 13.6 13.6/14=0.9 1 14 Total 100g 103g Required: Chemical Formula Basis: 100 g of substance The chemical formula is C 7 H 5 N

6. 3/9/2016 6 Dr.SS/Dr.DB Problem-4 Chimney gas has the following composition:CO 2 – 9.5%:CO-0.2%: O 2 -9.6% and N 2 -80.7%. Using ideal gas law, calculate Its weight percentage Volume occupied by 0.5kg of gas at 30°C and 760 mmHg Density of the gas in kg/m 3 at condition of (ii) Specific gravity of the gas mixture (Density of air may be taken as 1.3 g/cc) Required: i . Its weight percentage Basis: 100 kmoles of gas mixture Component No.of moles MWT , kg/kmole Weight, kg Weight (%) CO 2 9.5 44 418 13.96% CO 0.2 28 5.6 0.19% O 2 9.6 32 307.2 10.26% N 2 80.7 28 2259.6 75.47% Total 100 2990.4 100

7. 3/9/2016 7 Dr.SS/Dr.DB ii. Volume occupied by 0.5kg of gas at 30°C and 760 mmHg: AVMWT- Average Molecular weight

8. 3/9/2016 8 Dr.SS/Dr.DB III.Density of the gas in kg/m 3 at condition of (ii) IV. Specific gravity of the gas mixture (Density of air may be taken as 1.3 g/cc)

9. 3/9/2016 9 Dr.SS/Dr.DB Problem-5 A gas mixture contains 0.274 kmol of HCl , 0.337 kmol of N 2 and 0.089 kmol of O 2 . Calculate i.Average molecular weight of gas and ii.Volume occupied by this mixture at 405.3kPa and 303K Component No.of Moles MWT(kg/kmole) Weight(kg) HCl 0.274 36.5 0.274x36.5= 10 N 2 0.337 28 0.337x28= 9.436 O 2 0.089 32 0.89x32 = 2.848 Total 0.7 22.875 Required: ( i )AVMWT

10. 3/9/2016 10 Dr.SS/Dr.DB II.Required:Volume

11. 3/9/2016 11 Dr.SS/Dr.DB Problem-6 A gas has the following composition by volume. CO 2 =9.5%, CO=0.2%,O 2 =9.6% and N 2 =80.7%. using ideal gas low, calculate Composition by wt % Volume occupied by 1 kg of the gas at 40 °C and 740 mm Hg Density of the gas in kg/m 3 at40 °C and 740 mm Hg . Component No.of Moles MWT(kg/kmole) Weight(kg) Weight(%) CO 2 9.5 44 418 14.09 CO 0.2 28 5.6 0.19 O 2 9.6 32 307.2 10.35 N 2 80 28 2240 75.40 Total 100 2971 100 Basis:100 kmole From ideal gas law Volume%= Mole%

12. 3/9/2016 12 Dr.SS/ Dr.DB II.Required:Volume III.Required :

13. 3/9/2016 13 Dr.SS/Dr.DB Problem-7 1000 litres of mixture of H 2 , N 2 and CO 2 at 150°C was found to have the following ratio for the partial pressure of the gases:PH 2 :PN 2 :PCO 2 is 1:4:3. If the total pressure is 2 atm absolute, calculate Mole fraction of each of these gases Weight per cent of each of these gases Average molecular weight Weight of CO 2 in kg ( i ) Required: Mole fraction According to Dalton ’ s law P-Total Pressure Pi- Partial pressure of i th component Take “ y ” is the partial pressure of H 2

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15. 3/9/2016 15 Dr.SS/Dr.DB Component Mole fraction No.of Moles(g .mole) MWT g/gmole) Weight(g) Weight% H 2 0.125 0.125x57.59=7.199 2 2x7.199=14.4 0.81 N 2 0.5 0.5x57.59=28.80 28 28x28.8=806.7 45.54 CO 2 0.375 21.591 44 950.4 53.65 Total 1 57.59 1771.4 100 ii. Weight % of each of these gases Weight of CO 2 in kg =0.9504kg

16. 3/9/2016 16 Dr.SS/Dr.DB Problem-8 Estimate the density of chlorine gas at temperature 503K(230°C) and 15.2MPa pressure using The ideal gas law and The Van der waals equation ( i ) Ideal gas law:

17. 3/9/2016 17 Dr.SS/Dr.DB Vander ’ s Waals equation: By solving the above equation

18. 3/9/2016 18 Dr.SS/Dr.DB Problem-9 The strength of phosphoric acid usually represented as weight % of P 2 O 5 . A sample of phosphoric acid analyzed 40% P 2 O 5 . What is the % by weight of H 3 PO 4 in the sample? Pure H 3 PO 4 MWT Pure% New H 3 PO 4 P 2 O 5 142 72.5 142 H 2 O 54 (54+x) 196 Units of the acid contains 142 units of P 2 O 5 “ x ” – the quantity of newly added water MWT MWT-molecular weight Material balancing the P 2 O 5

19. 3/9/2016 19 Dr.SS/Dr.DB Problem-10 An aqueous solution contains 40% of Na 2 CO 3 by weight. Express the composition in mole percent? Component Weight (g) MWT(g/gmole) No. of Moles (gmole) Mole% Na 2 CO 3 40 106 10.16 H 2 O 60 18 3.333 89.84 Total 100 3.710 100 Basis: 100 g of solution

20. 3/9/2016 20 Dr.SS/Dr.DB Fuels and Combustion

21. 3/9/2016 21 Dr.SS/Dr.DB Combustion is defined as the rapid chemical combination of oxygen with carbon, hydrogen and sulphur, accompanied by the diffusion of heat and light. That portion of the substance thus combined with the oxygen is called combustible. Combustion is perfect when the combustible unites with the greatest possible amount of oxygen, as when one atom of carbon unites with two atoms of oxygen to form carbon dioxide, CO 2 . The combustion is imperfect when complete oxidation of the combustible does not occur, or where the combustible does not unite with the maximum amount of oxygen, as when one atom of carbon unites with one atom of oxygen to form carbon monoxide, CO, which may be further burned to carbon dioxide.

22. 3/9/2016 22 Dr.SS/Dr.DB Analysis of Coal   Two methods: Ultimate analysis Proximate analysis. The ultimate analysis determines all coal component elements. Proximate analysis determines only the fixed carbon, volatile matter, moisture and ash percentages.

23. 3/9/2016 Dr.SS/Dr.DB 23 Proximate Analysis Proximate analysis indicates the percentage by weight of the Fixed Carbon, Volatiles, Ash, and Moisture Content in coal. The amounts of fixed carbon and volatile combustible matter directly contribute to the heating value of coal. Fixed carbon acts as a main heat generator during burning. High volatile matter content indicates easy ignition of fuel. The ash content is important in the design of the furnace grate, combustion volume, pollution control equipment and ash handling systems of a furnace.

24. 3/9/2016 Dr.SS/Dr.DB 24 Significance of Various Parameters in Proximate Analysis   Fixed carbon : Fixed carbon is the solid fuel left in the furnace after volatile matter is distilled off. It consists mostly of carbon but also contains some hydrogen, oxygen, sulphur and nitrogen not driven off with the gases. Fixed carbon gives a rough estimate of heating value of coal   Volatile Matter: Volatile matters are the methane, hydrocarbons, hydrogen and carbon monoxide, and incombustible gases like carbon dioxide and nitrogen found in coal. Thus the volatile matter is an index of the gaseous fuels present. Typical range of volatile matter is 20 to 35%. Volatile Matter Proportionately increases flame length, and helps in easier ignition of coal. Sets minimum limit on the furnace height and volume. Influences secondary air requirement and distribution aspects. Influences secondary oil support

25. 3/9/2016 Dr.SS/Dr.DB 25 Significance of Various Parameters in Proximate Analysis   Ash Content : Ash is an impurity that will not burn. Typical range is 5 to 40% Ash Reduces handling and burning capacity. Increases handling costs. Affects combustion efficiency and boiler efficiency Causes clinkering and slagging.   Moisture Content: Moisture in coal must be transported, handled and stored. Since it replaces combustible matter, it decreases the heat content per kg of coal. Typical range is 0.5 to 10% Moisture Increases heat loss, due to evaporation and superheating of vapour Helps, to a limit, in binding fines. Aids radiation heat transfer.

26. Ultimate Analysis: The ultimate analysis indicates the various elemental chemical constituents such as Carbon, Hydrogen, Oxygen, Sulphur, etc. It is useful in determining the quantity of air required for combustion and the volume and composition of the combustion gases. This information is required for the calculation of flame temperature and the flue duct design etc. 3/9/2016 Dr.SS/Dr.DB 26

27. 3/10/2016 27 Dr.SS/Dr.DB Problem-1 What is the approximate molar composition of air? What is the approximate molar ratio N 2 to O 2 in air? Solution : Basis: 100 moles of air Therefore Air composition: O 2 – 21moles N 2 -79moles

28. 3/10/2016 28 Dr.SS/Dr.DB Problem-2 A gas contains 1 mol H 2 , 1 mol O 2 , and 2 mol H 2 O. What is the molar composition of this gas on a wet basis? On a dry basis? Element Mole Wet basis Dry basis H 2 1 25 50 O 2 1 25 50 H 2 O 2 50   Total 4 100 100

29. 3/10/2016 29 Dr.SS/Dr.DB Problem-3 A flue gas contains 5 mol% H 2 O . Calculate the ratios kmole flue gas / kmole H 2 O kmole dry flue gas / kmole dry flue gas kmole H 2 O/ kmole dry flue gas. Compound kmole Dry borne flue gas 95 H2O 5 Total 100 Solution: Basis: 100kmole of flue gas

30. 3/10/2016 30 Dr.SS/Dr.DB Problem-4 One hundred mol/h of butane (C 4 H 10 ) and 5000mol/h of air are fed into a combustion reactor. Calculate the % excess air? Solution: th .-Theoretical Chemical Reaction: From stiochiometric equation,

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32. 3/10/2016 32 Dr.SS/Dr.DB Solution: Proble m -5 Methane burns in the reactions One hundred mol/h of methane is fed to a reactor. What is the theoretical O 2 flow rate if complete combustion occurs in the reactor? What is the theoretical air flow rate If 100% excess air is supplied, what is the flow rate of air entering the reactor? If the actual flow rate of air is such that 300mol O 2 /h enters the reactor, what is the percent excess air? Complete combustion Reaction:

33. 3/10/2016 33 Dr.SS/Dr.DB