Answer / kangchuentat

MICROBIOLOGICAL ENGINEERING - ANSWER 28.2 : (a) Number of death cells = Total cells x % of death cells / 100 = 57 x 5.3 / 100 = 3.021 rounded to 3 cells. Number of viable cells = Total cells - number of death cells = 57 - 3 = 54 cells. (b) Average number of viable cells / square = 54 / 5 = 10.8 cells / square. (c) Dilution = final volume / initial volume = (100 + 50) microlitres / 100 microlitres = 1.5. (d) Concentration = (Average number of viable cells / square) x dilution x (square / volume) = 10.8 cells x 1.5 x (1 / 0.0001 mL) = 162000 cells / mL. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

i m a frsh chemical engineer now i m confuse to attend the interview so anybody pls send model question papers to this christ_vivek33@yahoo.com

0 Answers   HPCL,

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0 Answers   IFFCO,

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BIOPROCESS ENGINEERING - EXAMPLE 14.1 : In differential centrifugation of cells with diameter D in centimeter, the square of D is given by D x D = [18n ln (RF / RI) ] / [ (RP - RFF) Wt ] where n is the fluid viscosity (poise), RF is the final radius of rotation (cm), RI is the initial radius of rotation (cm), RP is cell density (g / ml), RFF is the fluid density (g/ml), W the square for the rotational velocity in (radians / s) (radians / s), t is the time required to sediment from RI to RF (s). Derive an equation for W as a function for D, n, RF, RI, RP, RFF and t, with the stated units above, in radian & degree.

1 Answers  

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pls send me HPCL & reliance test paper for chemical engineering.....

4 Answers   BCPL, HPCL, NOCL, Reliance,

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