Chemical Tutorial Solution P1 PDF

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Chemical Tutorial Solution P1...

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Part 1 – Things you should know Find out the following things and note them – these are all things you will need to know in the future. 1. What is atmospheric pressure (in bar Pa and in mmHg)? 1.01325 bar, 101.325 kPa, (1 bar ≡ 100,000 Pa) 760mmHg

2. What is the composition of air (by volume and on a molar basis)? (78.1% N2, 20.9% O 2, plus minor components. Unless there is a good reason to do otherwise we would normally use 79%N2 / 21% O2 for convenience (or even 80%/20%).) 3. What is the molecular mass of water? 18 g/mol 4. What is the molecular mass of air? You could work this out from a weighted average of oxygen and nitrogen, or look it up and find it to be 28.97g/mol we normally use 29g/mol) 5. What is the density of water? At 4°C water has a density of 1000kg/m3. Water’s specific gravity is 1 since water is the reference for specific gravity) 6. What is the acceleration due to gravity, ? 9.81 m/s2 7. What is the kilowatt-hour (kWh) a measure of? Energy

Part 2 – Things to look up Some conversion factors to find – google for unit conversions or dimensions and units or check a textbook1 (you may already know some of them) 1. Btu to joules. 1 btu = 1 055.05585 joules 2. Tonnes to kg 1 tonne = 1000kg 3. Centipoise to Pa s (and then work out what Pa s are in basic SI units) 1cP = 10-3 Pa s

4. lb-mol to kmol 1lb-mol = 0.45359 kmol

Part 3 – Conversions You may need to find some data for these questions. Don’t just look for conversion factors between complex units, work them out from first principles! 1. In the USA the unit the acre-foot is used to measure volumes in water storage. What would 12 acre-feet be expressed in m3? The acre foot is the volume occupied by an area of 1 acre and a depth of 1 foot 1acre

4 046.85642 m2 and 1ft = 0.3046 m

12

2. The most unusual unit I have seen is the circular inch, the area of a circle one inch in diameter. It was intended, apparently, as a convenient measure of the cross-

1

Elementary Principles of Chemical Processes, Richard M. Felder and Ronald W. Rouseau, Wiley, 1986 or Chemical and Energy Process Engineering, Sigurd Skogestad, CRC Press, 2009 – a book which is too new to have made it onto the recommended book list but which I rate very highly.

sectional area of pipes. What is the area of a pipe with a diameter of 40cm, in circular inches?

1in = 0.0254m D2 4

r2

3.14 0.4 4

  

2

4

0.1256   

2

5.065 10

4 1 5.065 10

2

4

248.0

4

Alternatively The area of a circle with diameter D inches is D2 circular inches. So the area of a 2 circle 40cm in diameter is (0.40 /0.0254) circular inches = 248.0 circular inches.

3. The molecular mass of hexane is 86.18 g/mol. If there is a flow of 2.5 te/h going through a pipe what is that flow expressed as kmol/s? 2.5

1000

1 86.18

1000

1 3600

1 1000

8.06 10

3

4. The mole fraction of a substance in a mixture is equal to the ratio of the number of moles of that substance present to the total number of moles present. What is the mole fraction of salt (NaCl) in a mixture of 40g of salt in 500ml of water? The molecular mass of NaCl is 54.443g/mol so 40g is 0.735 moles of NaCl The molecular mass of water is 18g/mol. The mass of water is equal to the volume times density. The density of water is 1000kg/m3 or 1g/ml (1000kg = 106 g and 1m3=106ml. the density could also be expresses as 1kg/l) So the mass of water is 500g, which is equivalent to 27.78moles So the mole fractionof NaCl is

0.735 0. 735 27. 78

0.0258 =2.58%

The weight fraction would have been 0.074=7.4%

5. Hydrogen and oxygen, both in the gas phase are (rather unwisely) mixed together in equal volumes. What is the mole fraction of hydrogen?

(the reason why it is rather unwise to mix hydrogen and oxygen is that virtually all mixtures of H2 and O2 are explosive) A mole of any gas occupies 22.4l under standard conditions so if we have equal volumes of the two gases the mole fraction of H2 is 0.5

6. The following equation is used in modelling liquid flow through packed beds of solids Cf

2 P L

3

cos 3 u2 a

Deduce the units of . Note:  is a pressure is a length ε is the ratio of two volumes ρ is a density is a velocity and a has units m-1 Setting out all the units gives: 3 2 m 1 m s m kg m 2 1

N m2

Ns 2 kg m

Now we need to get rid of the derived unit N Kgm

s2

s2

m kg

Everything cancels so Cf is dimensionless. 7. Convert 12kWh to SI units. 1kWh is the energy corresponding to 1kW over 1 k = 1000J/s for 3600 s ≡ 3.6 MJ So 12 kWh = 12 x 2.6MJ = 43.2 MJ

Part 4 – Types of Process; basic process considerations . State whether the following are processes:

processes,

processes or

a) Hot water and tea bags are placed in a teapot to brew tea. b) Greenhouse plants are watered via a continuous trickle feed of water.

c) Crude oil is fed to a fractionating column at a steady rate and several fractions, corresponding to different boiling rates are withdrawn at various points on the column at constant rates.

4.1 a) This is a batch process, all the feeds are added to the process at the beginning and then poured out at the end. b) This is semi-batch, the plants and pot etc are placed in situ at the beginning and the water flows continuously through the system. c) This is continuous; all streams are constant over time. (There will be variations of flow rates if the composition of the crude oil changes with time, but the process is still continuous)

Preliminary question: of a gas are sent through a 1 m diameter column. What is the average gas velocity in the column (this is called the superficial velocity)? 50 of CO 2 are to be extracted from various gas streams, using a solvent in an absorption tower. To allow effective contact of the liquid and gas phase, the average velocity of the gas mixture inside the tower must be 0.5 m/s at most. Two processes are considered: : Ambient air at 0.0387 % by volume CO2 is used. What is the required flow rate of air? From this result, calculate the diameter of the column that is needed. : This time, flue gas at 12 % by volume CO2 from a coal-fired power station is used instead of ambient air. Answer the same questions as for Process 1. Compare your results for the two processes. Which one is likely to be the cheapest and why?

4.2 Preliminary question: (please note: this question was edited on 16/10/2012) If uniformly distributed over a surface area inside a circular section of diameter D = 1 m, a flow rate F = 10 m3 /s of gas will have a superficial velocity 2 v = F / { D /4 } v = (10 / 3600) / { 1 2/4} v = 3.5 m/s, Process 1 uses a much more dilute CO2 source than Process 2, and so will require a much greater flow rate of gas. Assuming that all the CO2 could be extracted, Process 1: 50 m3 /hr CO2 (presumably at atmospheric pressure) from a 0.0387 vol.% feed gives a total flowrate for the feed 3 3 3 F = 50 / (0.0387/100) = 12910 m /hr = 35.9 m /s In figure 4.1, the cylinder represents the absorption tower. Its cross section area is A = D2 /4. The gas is going at velocity v = 0.5 m/s, which means that a volume v ·A per unit time can be processed through the column. Hence the allowable flow rate of gas, F = v ·A.

A Cross section of column

Flow rate

= 

v = 0.5 m/s Figure 4.1: the absorption tower

Hence, D √ F v D = 9.56 m Likewise with process 2, you will find F = 0.116 m3/s and D = 0.54 m. This means that the equipment required in Process 1 is going to be much larger than for process 2.This will result in capital costs for process 1 being much higher than for process 2 2 . Handling costs (e.g. energy for compression, pumping of absorbent etc.) will also be much higher for Process 1 due to the much large flowrates involved.

2

and that is without even taking into account other aggravating factors (rate of absorption of very dilute gases is bound to be slower than concentrated gases due to lower gradients of CO 2 concentration between solvent and gas. Slower rates often translates into larger size of equipment to allow more contact time between streams)...