Question

(1) Diatoms are common marine algae that account for most of the photosynthesis that occurs in the ocean. As Vogel (2004) discusses, although diatoms live in the top layers of the ocean, they are denser than the water around them, so they sink slowly in still water. Only the constan tmixing of water in the ocean prevents them from sinking into the deeper layers of the ocean. Diatoms are also apparently adapted to sink as slowly as possible, by evolving shapes that have very high fluid dynamic re- sistance. One such shape is a long filament. Why do they have this shape? In this problem you will compare sinking of filamentous diatoms and spherical diatoms. The sinking speed of a spherical diatom of radius r is given by the formula: U - 2 r2pg 9 where u is the viscosity (stickiness) of seawater, p is the buoy- ancy corrected density of the diatom (that is, the mass of 1 cm3 of diatom, minus the mass of 1 cm3 of seawater) and g is the gravitational acceleration constant. (a) We want to consider the effect of changing the volume of the diatom upon its sinking velocity. Supposing that the diatom is a sphere, its volume is given by . If V increases by 5%, predict the corresponding percent change in U. (b) In contrast a filamentous diatom with length L and radius a will sink at a speed: U = g (1(24) + ) If the volume of the diatom is V = maL and V is varied by varying L (i.e., changing the length of the filament but not its radius) predict the percentage change in U that would occur if V were changed by 5%. Assume that initially L/a = 20.

Answer 1

. Vi = y ñ v3 [initial volume] final volume Vi + 5%. Vi = YE LOIY чту, 10 хутка > (105) Ti => : 1:01Y Now, u drill Us arghfg Dividing both eg relation of and using = (1.01) Y Uf 0 Y 44 1.02.01 ui

= Us Uit dool ui 160 => 1 increase in U = 2.01% 6 awen given Vi - T a2 Li if is V = increused by St. na? Lq = 105 Vi I 105 100 TQ2 Lg - 105 tot 4 4 = 105 4 7 4 = - 100 46 10 105 given alven that i 20 - 100 x 105 4 – 20 Y L 20x105 100 ui pang / ln (24) + 7 ) you

? pang (34) + yn u in (42)+% 4.23 4.18 - ln (40) + 1 - Uf = 1001 ui ... % incruese in U = inquase in 1.19%. on 54. v T by si i I of v dureased Then Lf - 19 Us in 138) & Ya Ui = 4.13 Ui 4.16 4.18 7 Up = y Uf = 0.989 Ui Ui - 1.014 ui l-oly al in devuase 5 U = derrease 1.014% in v .

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