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  • Calculate flow rate.
  • Define units of volume.
  • Describe incompressible fluids.
  • Explain the consequences of the equation of continuity.

Flow rate Q size 12{Q} {} is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in [link] . In symbols, this can be written as

Q = V t , size 12{Q= { {V} over {t} } } {}

where V size 12{V} {} is the volume and t size 12{t} {} is the elapsed time.

The SI unit for flow rate is m 3 /s size 12{m rSup { size 8{3} } "/s"} {} , but a number of other units for Q size 12{Q} {} are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter    (L) is 1/1000 of a cubic meter or 1000 cubic centimeters ( 10 3 m 3 size 12{"10" rSup { size 8{ - 3} } `m rSup { size 8{3} } } {} or 10 3 cm 3 size 12{"10" rSup { size 8{3} } `"cm" rSup { size 8{3} } } {} ). In this text we shall use whatever metric units are most convenient for a given situation.

The figure shows a fluid flowing through a cylindrical pipe open at both ends. A portion of the cylindrical pipe with the fluid is shaded for a length d. The velocity of the fluid in the shaded region is shown by v toward the right. The cross sections of the shaded cylinder are marked as A. This cylinder of fluid flows past a point P on the cylindrical pipe. The velocity v is equal to d over t.
Flow rate is the volume of fluid per unit time flowing past a point through the area A size 12{A} {} . Here the shaded cylinder of fluid flows past point P size 12{P} {} in a uniform pipe in time t size 12{t} {} . The volume of the cylinder is Ad size 12{ ital "Ad"} {} and the average velocity is v ¯ = d / t size 12{ {overline {v}} =d/t} {} so that the flow rate is Q = Ad / t = A v ¯ size 12{Q= ital "Ad"/t=A {overline {v}} } {} .

Calculating volume from flow rate: the heart pumps a lot of blood in a lifetime

How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min?


Time and flow rate Q size 12{Q} {} are given, and so the volume V size 12{V} {} can be calculated from the definition of flow rate.


Solving Q = V / t size 12{Q=V/t} {} for volume gives

V = Qt . size 12{V= ital "Qt"} {}

Substituting known values yields

V = 5 . 00 L 1 min ( 75 y ) 1 m 3 10 3 L 5 . 26 × 10 5 min y = 2 . 0 × 10 5 m 3 . alignl { stack { size 12{V= left ( { {5 "." "00"" L"} over {"1 min"} } right ) \( "75"" y" \) left ( { {1" m" rSup { size 8{3} } } over {"10" rSup { size 8{3} } " L"} } right ) left (5 "." "26" times "10" rSup { size 8{5} } { {"min"} over {y} } right )} {} #" "=2 "." 0 times "10" rSup { size 8{5} } " m" rSup { size 8{3} } {} } } {}


This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of water contained in a 6-lane 50-m lap pool.

Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. The precise relationship between flow rate Q size 12{Q} {} and velocity v ¯ size 12{ {overline {v}} } {} is

Q = A v ¯ , size 12{Q=A {overline {v}} } {}

where A size 12{A} {} is the cross-sectional area and v ¯ size 12{ {overline {v}} } {} is the average velocity. This equation seems logical enough. The relationship tells us that flow rate is directly proportional to both the magnitude of the average velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. [link] illustrates how this relationship is obtained. The shaded cylinder has a volume

V = Ad , size 12{V= ital "Ad"} {}

which flows past the point P size 12{P} {} in a time t size 12{t} {} . Dividing both sides of this relationship by t size 12{t} {} gives

V t = Ad t . size 12{ { {V} over {t} } = { { ital "Ad"} over {t} } } {}

We note that Q = V / t size 12{Q=V/t} {} and the average speed is v ¯ = d / t size 12{ {overline {v}} =d/t} {} . Thus the equation becomes Q = A v ¯ size 12{Q=A {overline {v}} } {} .

[link] shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2,

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Source:  OpenStax, College physics (engineering physics 2, tuas). OpenStax CNX. May 08, 2014 Download for free at http://legacy.cnx.org/content/col11649/1.2
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