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n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 2 , size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}

where n 1 size 12{n rSub { size 8{1} } } {} and n 2 size 12{n rSub { size 8{2} } } {} are the number of branches in each of the sections along the tube.

Calculating flow speed and vessel diameter: branching in the cardiovascular system

The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is 8.0 μ m , calculate the number of capillaries in the blood circulatory system.

Strategy

We can use Q = A v ¯ size 12{Q=A {overline {v}} } {} to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.

Solution for (a)

The flow rate is given by Q = A v ¯ size 12{Q=A {overline {v}} } {} or v ¯ = Q πr 2 size 12{ {overline {v}} = { {Q} over {πr rSup { size 8{2} } } } } {} for a cylindrical vessel.

Substituting the known values (converted to units of meters and seconds) gives

v ¯ = 5.0 L/min 10 3 m 3 /L 1 min/ 60 s π 0 . 010 m 2 = 0 . 27 m/s . size 12{ { bar {v}}= { { left (5 "." 0`"L/min" right ) left ("10" rSup { size 8{ - 3} } `m rSup { size 8{3} } "/L" right ) left (1`"min/""60"`s right )} over {π left (0 "." "010 m" right ) rSup { size 8{2} } } } =0 "." "27"`"m/s"} {}

Solution for (b)

Using n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 1 size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {} , assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for n 2 size 12{n rSub { size 8{2} } } {} (the number of capillaries) gives n 2 = n 1 A 1 v ¯ 1 A 2 v ¯ 2 . Converting all quantities to units of meters and seconds and substituting into the equation above gives

n 2 = 1 π 10 × 10 3 m 2 0.27 m/s π 4.0 × 10 6 m 2 0.33 × 10 3 m/s = 5.0 × 10 9 capillaries . size 12{n rSub { size 8{2} } = { { left (1 right ) left (π right ) left ("10" times "10" rSup { size 8{ - 3} } " m" right ) rSup { size 8{2} } left (0 "." "27"" m/s" right )} over { left (π right ) left (4 "." 0 times "10" rSup { size 8{ - 6} } " m" right ) rSup { size 8{2} } left (0 "." "33" times "10" rSup { size 8{ - 3} } " m/s" right )} } =5 "." 0 times "10" rSup { size 8{9} } " capillaries"} {}

Discussion

Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per mm 3 size 12{"mm" rSup { size 8{3} } } {} , or about 200 × 10 6 size 12{"200" times "10" rSup { size 8{6} } } {} per 1 kg of muscle. For 20 kg of muscle, this amounts to about 4 × 10 9 size 12{4 times "10" rSup { size 8{9} } } {} capillaries.

Section summary

  • Flow rate Q size 12{Q} {} is defined to be the volume V size 12{V} {} flowing past a point in time t size 12{t} {} , or Q = V t size 12{Q= { {V} over {t} } } {} where V size 12{V} {} is volume and t size 12{t} {} is time.
  • The SI unit of volume is m 3 size 12{m rSup { size 8{3} } } {} .
  • Another common unit is the liter (L), which is 10 3 m 3 size 12{"10" rSup { size 8{ - 3} } `m rSup { size 8{3} } } {} .
  • Flow rate and velocity are related by Q = A v ¯ size 12{Q=A {overline {v}} } {} where A size 12{A} {} is the cross-sectional area of the flow and v ¯ size 12{ {overline {v}} } {} is its average velocity.
  • For incompressible fluids, flow rate at various points is constant. That is,
    Q 1 = Q 2 A 1 v ¯ 1 = A 2 v ¯ 2 n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 2 . size 12{ left none matrix { Q rSub { size 8{1} } =Q rSub { size 8{2} } {} ##A rSub { size 8{1} } {overline {v}} rSub { size 8{1} } =A rSub { size 8{2} } {overline {v}} rSub { size 8{2} } {} ## n rSub { size 8{1} } A rSub { size 8{1} } {overline {v}} rSub { size 8{1} } =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v}} rSub { size 8{2} }} right rbrace "." } {}

Conceptual questions

What is the difference between flow rate and fluid velocity? How are they related?

Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which it flows.)

Identify some substances that are incompressible and some that are not.

Problems&Exercises

What is the average flow rate in cm 3 /s size 12{"cm" rSup { size 8{3} } "/s"} {} of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L?

2.78 cm 3 /s size 12{"cm" rSup { size 8{3} } "/s"} {}

Questions & Answers

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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Source:  OpenStax, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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