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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.

Making connections: syringes

A horizontally oriented hypodermic syringe has a barrel diameter of 1.2 cm and a needle diameter of 2.4 mm. A plunger pushes liquid in the barrel at a rate of 4.0 mm/s. Calculate the flow rate of liquid in both parts of the syringe (in mL/s) and the velocity of the liquid emerging from the needle.

Solution:

First, calculate the area of both parts of the syringe:

A 1 =   π ( d 1 2 ) 2 =   π ( 0.006 ) 2 =   1.13   ×   10 4 m 2
A 2 =   π ( d 2 2 ) 2 =   π ( 0.0012 ) 2 =   4.52   ×   10 6 m 2

Next, we can use the continuity equation to find the velocity of the liquid in the smaller part of the barrel ( v 2 ):

A 1 v 1 =   A 2 v 2
v 2 =   ( A 1 A 2 ) v 1
v 2 =   ( 1.13 × 10 4 4.52 × 10 6 ) ( 0.004 ) =   0.10   m / s

Double-check the numbers to be sure that the flow rate in both parts of the syringe is the same:

Q 1 =   A 1 v 1 = ( 1.13 × 10 4 ) ( 0.004 ) =   4.52 × 10 7   m 3 / s
Q 2 =   A 2 v 2 = ( 4.52 × 10 6 ) ( 0.10 ) =   4.52 × 10 7   m 3 / s

Finally, by converting to mL/s:

( 4.52 × 10 7   m 3 1   s ) ( 10 6   mL 1   m 3 ) = 0.452   mL / s

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 "." } {}

Questions & Answers

Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
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for the answer to complete, the units need specified why
muqaddas Reply
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
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how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
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1.75cm
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the same behavior thru the prism out or in water bud abbot
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If this will experimented with a hollow(vaccum) prism in water then what will be result ?
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Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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