(a) A 70-kg person at rest has an oxygen consumption rate Qhum = 14.5 liter/h, 2

Question

(a) A 70-kg person at rest has an oxygen consumption rate Qhum = 14.5 liter/h, 2% of which is supplied by diffusion through the skin. Assuming that the skin surface area of this person is Ahum = 1.7 m2 , calculate the diffusion rate for oxygen through the skin in units of liters/(h cm2 ).

(b) What is the maximum diameter of a spherical animal, whose oxygen consumption at rest can be supplied entirely by diffusion through its skin? Make the following assumptions: i. The density of animal tissue is = 1 g/cm3 . ii. At rest, all animals require the same amount of oxygen per unit volume. iii. The diffusion rate for oxygen through the skin is the same for all animals.

Explanation / Answer

(a)

The required oxygen supply for the person through diffusion = 0.02 x 14.5 = 0.29 liters/h

Now

diffusion rate for oxygen x surface area = 0.29

diffusion rate for oxygen = 0.29/1.7 =0.17 liters/( h m2 ) = 1.7 x 10-5 liters/(h cm2 )

(b)

To make sense, the second assumption has to be " At rest, all animals require the same amount of oxygen per unit time per unit mass " instead of  "At rest, all animals require the same amount of oxygen per unit volume".

Let 'r' be the radius of the spherical animal.

Now, the amount of oxygen needed per unit mass per unit time from part (a)

= 14.5/70 = 0.21 liter/(h kg)

For the present animal density = 1 g/cm3 = 1 x 10-3 kg/cm3

Amount of oxygen needed by the spherical animal per unit time = 0.21 x Mass of the animal

= 0.21 x Density x Volume of the animal

= 0.21 x 10-3 x (4/3)r3 liters/h

Amount of oxygen supplied by the diffusion per unit time = Diffusion rate x surface area of the animal

= 1.7 x 10-5 x 4r2  liters/h

For the animal to survive

oxygen supply rate oxygen required rate

or, 1.7 x 10-5 x 4r2 0.21 x 10-3 x (4/3)r3

or, 0.24 r

Thus the maximum value of the diameter of the animal = 0.24 x 2 = 0.48 cm

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