You are working as a summer intern at a company and a fellow intern (obviously a

Question

You are working as a summer intern at a company and a fellow intern (obviously an U of A engineering student) accidently opened the wrong valve and dumped some unknown volume of oil ( specific gravity 0.918) into a vat of sea water ( specific gravity of water 1.028) forming a stratified (layered) fluid column. The vat is known to have a diameter of 3 m. The engineers at the firm need to know the volume of the oil that had been dumped into the vat and you are tasked with giving them a value. Earlier in the summer you had been playing (I mean working) with a 20 cm diameter, 15 cm tall piece of wood which you found floats with 3.75 cm above the surface in pure water. a) Develop an equation for the volume of the oil as a function of the height from the surface of the oil to the top of the wooden cylinder when it is placed in this stratified fluid column and b) the volume of the oil when the height above the oil surface is 3.5 cm.

Explanation / Answer

. by the law of floatation weight of floating body = weight of fluid displaced;

density of wood *   (pi)/4 r2*h= density of water *(pi)

/4 r2*h' ;

density of wood = 1000*11.25/15 (height of wood below the surface);

density of wood = 750 kg/m3;

let the level of wood above surface be x

immersed= 15-x;

again by the law of floation ;

750 *(pi)

/4 r2*15= volume of oil* density + vol of sea water dispalecd * density;

vol of sea water dispalced h above water surface = 11cm

let bet the change in water level when oil is immersed

height above suface for sea water = 15 -4-t

for oil t

volume of oil = 750 *15- (11-t)*1028/d oil

d. volume of oil t=.5 ;

volume of oil = 1.61 m3

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