6HC Mass Transfer I - April'2000

Part A ( 20 x 2 = 40 Marks)

1. Explain the term "FLUX"
2. What is meant by molecular diffusion and eddy diffusion?
3. Define the term diffusivity
4. Give the Wilke-Chang equation and explain the terms involved in it
5. What is Lewis number?
6. Why various theories of mass transfer have been proposed?
7. Explain the term stage and stage efficiency
8. What is meant by absorption factor?
9. Explain the theory of humidification
10. What is meant by wet bulb temperature approach?
11. What is the principle of 'recirculating liquid gas humidifier'?
12. Define Dew point
13. Define the term "Bound moisture"
14. Explain the effect of temperature and mass flow rate of air on the constant rate of drying, NC.
15. What is meant by holdup in a rotary dryer?
16. Which type of drier is used in the manufacture of (a) tablets (b) Paraffin wax?
17. What is crystallization?
18. State McCabe's DL law of crystal growth
19. What is the purpose of agitator in the crystallizer?
20. Give examples of batch crystallizer and continuous crystallizer.
21. Part B (5 x 12 = 60 Marks)

22. (a) (i) Derive an expression for finding the mass flux of diffusion of A through non diffusing B, A and B are liquids. (8)
23. (ii) What is the equation of continuity and how Fick's second law of diffusion can be derived from it?

Or

(b) Through an accidental opening of a valve, water has been split on the floor of an industrial plant in a remote, difficult to reach area. Determine the time required to evaporate the water in the surrounding air. The water layer is 15 mm thick and may be assumed to remain at 25oC and 1 atm pressure with an absolute humidity of 0.002 kg water / kg dry air. Evaporation is assumed to take place by molecular diffusion through a gas film of 10 cm thick. The mass diffusivity of water vapor in air is 2.6 x 10-5 m2/sec. Vapor pressure of water at 25oC is 24 mm of Hg.

24. (a)Explain how the height of packing for an absorber can be calculated.
25. Or

(b) An absorption tower operating at 20oC and 1 atm pressure was used to absorb SO2 from an air mixture into water. At one point in the equipment, the partial pressure of SO2 was 30 mm Hg and concentration of the contacting liquid film was 1.2148 kmol/m3 of solution. The individual film mass transfer coefficients at 20oC and 1 atm were kC = 6.347 kmol / [hr.m2 (kmol/m3)] and kG = 1.44 kmol / hr.m2.atm.

The equilibrium data at 20oC are as follows:

 Partial pressure of SO2, mm Hg 0.5 3.2 8.5 26 59 Concentration of SO2 kmol/m3 0.306 1.459 2.784 6.2151 10.909

(i) Evaluate the interfacial composition CAi and PAi

(ii) Find KG and KL

(iii) What percentage of overall resistance to mass transfer lies in the gas phase?

26. (a) Explain the theory of adiabatic saturation curves and wet bulb temperature theory
27. Or

(b) Water is to be cooled in a cooling tower counter currently from 55oC to 30oC, using air counter currently. The entering air is at 20oC and an absolute humidity of 0.007 kg water / kg dry air. The value of Kya is 0.3 kg/m3.sec and the liquid flow rate is 1500 kg/hr.m2. Estimate the minimum air requirement and the height of the cooling tower, if 1.3 times the minimum air flow rate is used

 Temperature, oC 20 30 40 50 55 Enthalpy, H' kJ/kg 60 102 166 279 355

28. (a) Explain a typical drying rate curve and bring out its salient features.
29. Or

(b) A time of 5 hr was taken to dry a material from an initial moisture of 30% to a final moisture of 7%. Critical and equilibrium moisture are found to be 15% and 2% respectively. How much further time would be required to dry the material to a final moisture of 4%. All moisture contents are on wet basis.

30. (a) (i) Explain the crystallizers classification. (4)
31. (ii) Explain with neat sketch, working of a Swenson-Walker crystallizer. (8)

Or

(b) 5000 kg of KCl solution at a temperature of 80oC is cooled to 20oC in an open tank. The solubilities of KCl at 80oC and 20oC are 55 parts and 35 parts per 100 parts of water. Estimate the yield of KCl crystals by

(i) Assuming 5% water is lost by evaporation

(ii) Assuming no loss of water by evaporation.
{"ft_lang_label":"__label__en","ft_lang_prob":0.9005862,"math_prob":0.97536343,"size":2664,"snap":"2021-31-2021-39","text_gpt3_token_len":788,"char_repetition_ratio":0.10413534,"word_repetition_ratio":0.015296367,"special_character_ratio":0.3111862,"punctuation_ratio":0.091854416,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9775283,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T03:38:00Z\",\"WARC-Record-ID\":\"<urn:uuid:1c891f73-7d1f-4241-b273-f77f4709e779>\",\"Content-Length\":\"6085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4350cb03-a461-4f6f-9d6f-489eaa952ce1>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc0eb96e-458c-4d36-83f1-0d8bf6842316>\",\"WARC-IP-Address\":\"143.110.252.3\",\"WARC-Target-URI\":\"http://msubbu.in/ChemicalEngg_Resources/qb/mt/mt-I-2.htm\",\"WARC-Payload-Digest\":\"sha1:SD3QRTKQHD6RJOIMTXS774D2HLVRVRP2\",\"WARC-Block-Digest\":\"sha1:5N4SPIFSXPISIMVEINIYSXF7QMGBWSHY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154500.32_warc_CC-MAIN-20210804013942-20210804043942-00448.warc.gz\"}"}