Question

The inner diameter of the human aorta is about 2.50 cm, while that of a typical capillary is about 6 ????m = 6.00 × 10−6 m. In a person at rest, the average flow speed of blood is about 20.0 cm/s in the aorta and about 1.00 mm/s in the capillary. Calculate;

(a) The volume flow rate of blood in the aorta
(b) The volume flow rate in a single capillary
(c) The total number of open capillaries into which the blood from the aorta is distributed at any one time?

Answer 1

(a) 98.2 cm^3/s

(b) 2.826 x 10^-8 cm^3/s

(c) 3.5 x 10^9

Step-by-step explanation:

diameter of arota, d = 2.5 cm

diameter of capillary, D = 6 x 10^-6 m

velocity in arota, v = 20 cm/s

velocity of blood in capillary, V = 1 mm/s

Area of arota, a = πd²/4 = 4.91 cm^2

Area of capillary, A = πD²/4 = 2.826 x 10^-7 cm^2

(a) Volume flow in arota = velocity in arota x area of arota

= 20 x 4.91 = 98.2 cm^3/s

(b) Volume flow in capillary = velocity in capillary x area of capillary

= 0.1 x 2.826 x 10^-7 = 2.826 x 10^-8 cm^3/s

(c) Number of cappilaries = Volume flow in arota / volume flow in a capillary

= 98.2 / (2.826 x 10^-8) = 3.5 x 10^9

Thus, the number of capillaries required are 3.5 x 10^9

Answer 2

a) Flow rate of aorta = 9.818 x
`10^(-5)`
`m^(3)`/s

b) Flow rate of capillariy = 2.827 x
`10^(-14)`
`m^(3)`/s

c) Number of capillaries = 3.473 x
`10^(9)`

Step-by-step explanation:

This question focuses on the flow rate of a fluid. So let us look what flow rate is.

Flow rate is the volume of fluid that flows through the tube per unit time.

i.e. Flow rate =
`(dV)/(dt)`

where,

V = Volume of the fluid

t = time

We know that Volume = Area x distance

therefore,

Flow rate =
`(d(Ax))/(dt)`

where,

A = cross sectional area of the tube

x = distance

Since in this problem area is a constant, we can take A out of differentiation.i.e.

Flow rate =
`(Adx)/(dt)`

Now, we got
`(dx)/(dt)` which is equal to velocity of the fluid.

Thus,

Flow rate =
`(Adx)/(dt)` = Av

where, v = velocity

a) diameter of aorta = 2.50cm = 2.5x
`10^(-2)`

Area, A = π
`(d^(2) )/(4)` = π
`((2.5X10^(-2))^(2) )/(4)`

= 4.909 x
`10^(-4)`
`m^(2)`

v = 20.0 cm/s = 0.2m/s

We know that Flow rate = Av = 4.909 x
`10^(-4)` x 0.2 = 9.818 x
`10^(-5)`
`m^(3)`/s

b) diameter of capillary = 6.00 × 10−6 m

Area, A = π
`(d^(2) )/(4)` = π
`((6.00X10^(-6))^(2) )/(4)` = 2.827x
`10^(-11)`
`m^(2)`

v = 1.00 mm/s =
`10^(-3)`m/s

We know that Flow rate = Av = 2.827x
`10^(-11)` x
`10^(-3)` = 2.827 x
`10^(-14)`
`m^(3)`/s

c)Since the fluid is incompressible, the flow rate before and after should be same. Let there be n capillaries.

Then total flow rate in the capillaries = n x 2.827 x
`10^(-14)`
`m^(3)`/s

Flow rate in aorta = 9.818 x
`10^(-5)`
`m^(3)`/s

These two should be equal. i.e.

n x 2.827 x
`10^(-14)` = 9.818 x
`10^(-5)`

n =
`(9.818 X 10^(-5))/(2.827 X 10^(-14))`

n = 3.473 x
`10^(9)`