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Additional Problems (1) A wet paper pulp is found to contain 71% water. After drying it is found that 60% of the original water has been removed. Calculate.

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Presentation on theme: "Additional Problems (1) A wet paper pulp is found to contain 71% water. After drying it is found that 60% of the original water has been removed. Calculate."— Presentation transcript:

1 Additional Problems (1) A wet paper pulp is found to contain 71% water. After drying it is found that 60% of the original water has been removed. Calculate the following: (a) The composition of the dried pulp. (b) The mass of water removed per kilogram of wet pulp Answer: (a) 50.5% dry pulp and 49.5% water (b) kg

2 Additional Problems (2) A cellulose solution contains 5.2% cellulose by weight in water. How many kilograms of 1.2% solution are required to dilute 100 kg of 5.2% solution to 4.2%? Answer: 33.3 kg

3 Additional Problems (3) A cereal product containing 55% water is made at the rate of 500 kg/hr. You need to dry the product so that it contains only 30% water. How much water has to be evaporated per hour? Answer: 178 kg/hr

4 Ch61: Industrial Chemistry The Material Balance
E. Abenojar 10 November 2010

5 Process! Process – a series of actions or operations that result in an end product. Examples: Chemical manufacture Fluid transport Distillation Batch vs Continuous Batch – specific time at specific conditions Continuous – precise control of process conditions Steady-state and non-steady state

6 Batch or Continuous Process
Batch and Continuous Processes. < Last accessed: 8 Nov 2010

7 Batch or Continuous Process?
Pharmaceutical Chemical Food Beverage Petrochemical Paint Fertilizer Cement Water purification Waste treatment Batch and Continuous Processes. < Last accessed: 8 Nov 2010

8 The Material Balance

9 The Material Balance System – arbitrary portion or whole of a process set out specifically for analysis (open or close). System boundary

10 The Material Balance Accounting for material flows and changes in inventory of material for a system. This can refer to a balance on the system for the: Total mass Total moles Mass of a chemical compound Mass of an atomic species Moles of a chemical compound Moles of an atomic species

11 Example Energy in Soybean Crushing and Transesterification. 15 March 2010. < Last accessed: 8 Nov 2010

12 The Material Balance: Example
1) A thickener in a waste disposal unit of a plant removes water from wet sewage sludge as shown in figure below. How many kilograms of water leave the thickener per 100 kg of wet sludge that enter the thickener? The process is in the steady state. Thickener 100 kg Wet Sludge 70 kg Dehydrated Sludge Water = ?

13 Example 2) Hydrogenation of coal to give hydrocarbon gases is one method of obtaining gaseous fuels with sufficient energy content for the future. The Figure shows how a free-fall fluidized-bed reactor can be set up to give a product gas of high methane content.

14 Example 2) Suppose, first, that the gasification unit is operated without steam at room temperature (25°C) to check the gas flow rate monitoring instruments. If 1200 kg of coal per hour (assume that the coal is 80% C, 10% H, and 10% inert material) is dropped through the top of the reactor without the air flowing, how many kg of coal leave the reactor per hour?

15 Example 2) Suppose, first, that the gasification unit is operated without steam at room temperature (25°C) to check the gas flow rate monitoring instruments. If, in addition to the coal supplied, 15,000 kg of air per hour is blown into the reactor, at 25°C, how many kg of air per hour leave the reactor?

16 Example 2) (c) Finally, suppose that the reactor operates at the temperature shown in the figure, and that 2000 kg of steam (H2O) per hour are blown into the reactor along with 15,000 kg/hr of air and the 1200 kg of coal. How many kg of gases exit the reactor per hour assuming complete combustion of the coal?

17 Example Reaction: C + O2 → CO2 3) Material Balances.
If 300 kg of air (29 g/mol) and 24.0 kg of carbon (12.0 g/mol) are fed to a reactor (see Figure) at 600°F and after complete combustion no material remains in the reactor, how many kg of carbon will have been removed? How many kg of oxygen? How many kg total? How many moles of carbon and oxygen enter? How many leave the reactor? How many total moles enter the reactor and how many leave the reactor? Reaction: C + O2 → CO2

18 Example (4) Continuous Distillation. A novice manufacturer of alcohol for gasohol (10% ethanol and 90% gasoline) is having a bit of difficulty with a distillation column shown in the Figure. Too much alcohol is lost in the bottoms (waste). Calculate the composition of the bottoms and the weight of alcohol lost in the bottoms.

19 Example (4) Continuous Distillation. A novice manufacturer of alcohol for gasohol (10% ethanol and 90% gasoline) is having a bit of difficulty with a distillation column shown in the Figure. Too much alcohol is lost in the bottoms (waste). Calculate the composition of the bottoms and the weight of alcohol lost in the bottoms.

20 Material Balance with Chemical Reactions

21 Example (1) Excess Air. Fuels for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does gasoline. Compressed propane has been suggested as a source of economic power for vehicles. Suppose that in a test 20 kg of propane (C3H8) is burned with 400 kg (29 g/mol) of air to produce 44 kg of CO2 and 12 kg of CO. What was the percent excess air. C3H8 + O2 → CO2 + H2O

22 Example (2) Preventing Corrosion. Corrosion of pipes in boilers by oxygen can be alleviated through the use of sodium sulfite. Sodium sulfite removes oxygen from boiler feedwater by the following reaction: Na2SO3 + O2 → Na2SO4 How many kg of sodium sulfite are theoretically required (for complete reaction) to remove the oxygen from 8,330,000 kg of water containing 10.0 parts per million (ppm) of dissolved oxygen and at the same time maintain a 35% excess of sodium sulfite.

23 Molecular Mass of Dry Air
Components in Dry Air Volume Ratio compared to Dry Air Molecular Mass - M (g/mol) Molecular Mass in Air Oxygen 0.2095 32.00 6.704 Nitrogen 0.7809 28.02 21.88 Carbon Dioxide 0.0003 44.01 0.013 Hydrogen 2.02 Argon 39.94 0.373 Neon 20.18 Helium 4.00 Krypton 83.8 Xenon 131.29 Total Molecular Mass of Air 28.97 < Last accessed: 10 Nov 2010

24 What to study? Material Balance Problems.
Relevant flow charts, diagrams, and important processes in the industry. Remember flow charts and appropriate labels. Questions will include problem solving, objective essay or process description. Chapters 3, 4, 19, and 37. All groups – please forward your presentations and papers to your classmates.

25 Reference Himmelblau, D. M. Basic Principles and Calculations in Chemical Engineering. 6th Ed. Prentice-Hall, 1996.


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