Chapter 4

59Chapter 5: Flow of Liquids

CHAPTER 4 FLOW OF LIQUIDS:

Flow of ideal liquids or liquids of small or negligible viscosity 59 The equation of continuity 59 Bernoulli’s equation 61 Flow meter 64 Flow of incompressible viscous liquid 67 Turbulent flow 68 Flow resistance 70 Blood flow in blood vessels 72 Questions 76

The flow of liquids in human, animal and plant circulatory systems is very important. Because of their flow, food could be transported inside these organisms and the metabolite wastes are rejected outside.

As a matter of fact the flow of liquids is affected by their physical properties such as the viscosity. Accordingly and based on the liquids viscosity behaviour the study of their flow is divided into two main categories :

1) Ideal liquids or incompressible liquids of very small viscosity. 2) Incompressible liquids of moderate viscosity.

Flow of ideal liquids or liquids of small or negligible viscosity: The flow of ideal liquids or liquids of small viscosity is usually

governed by what is called the equation of continuity and Bernoulli’s equation. The flow of liquids in this case is usually in the form of streamline or laminar layers having approximately the same velocity.

The equation of continuity: Consider the situation where an incompressible fluid flowing in a

certain vessel. Then if more fluid enters one end of this vessel, an

60 Chapter 5 : Flow of Liquids

equal amount must leave the other end. This principle is called the equation of continuity. That is when a liquid is entering a certain tube with a rate of flow Q1, then it should leave with the same flow rate Q2. Thus the equation of continuity is :

Q1 = Q2 (1)

Now let us apply the equation of continuity on the flow of liquid through two connecting tubes of different cross-sectional areas A1 and A2 as shown in Fig. (1).

Fig. (1). Flow of an ideal liquid through tubes of different radii Since the rate of flow of liquid Q is the volume of liquid flowing past a point in a tube or channel per unit time:

Q = tV

ΔΔ m3s−1 (2)

Then

Q1 = 111

11 vA

tLA

tV

=Δ

Δ=

ΔΔ

where ΔL is the change in the stream of the liquid moving with an average velocity v1 during a period of time Δt.

Similarly the rate of flow in the other tube is :

Q2 = 222

22 vA

tLA

tV

=Δ

Δ=

ΔΔ

Thus according to the equation of continuity.

Q1 =Q2 that is A1v1 = A2v2 (3)

This means that the product of the cross-sectional area and the velocity of the liquid streaming in a vessels of different cross-sections is constant.

A1 v1

Δ L1

A2 v2

Δ L2

Q1 Q2


Example (1): Water flows through a tube, of radius 1 cm ended with a hose of radius 0.5 cm, at rate of 3 liters per minute. Calculate the velocity in both the tube and hose.

Solution:

Q = tV

ΔΔ =

s60m10cm3000 63 −× = 5×10−5 m3s−1

Since Q = A1v1 = A2v2 and A1 = 21rπ and A2 = 2

2rπ Where r1 and r2 are the radii of the tube and hose respectively, then the velocity v1 in the tube is

v1 = =×

×=

π= −

−−

m10cm)1(14.3sm105

rQ

AQ

42

135

221

0.637 ms−1

Similarly the velocity v2 in the hose is

v2 = =×

×=

π= −

−−

m10cm)5.0(14.3sm105

rQ

AQ

42

135

222

0.159 ms−1

By comparing the values of v1 and v2, it is clear that the velocity v2 of the water in the hose of the smaller radius, r2, is greater than its velocity v1 in the tube of greater radius, r2. Accordingly we can rewrite equation (3) to be in the form:

v2 = v1 2

2

112

1

21

12

1 )rr( v

rr v

AA

=ππ

= (4)

That is the water flows faster in the narrow channel or tube.

Bernoulli’s Equation: It was found by Bernoulli in the 1930s that the work done on a fluid as it flows from one place to another is equal to the change in its mechanical energy. This means that the energy per unit volume at any point in the flowing fluid should be constant. Bernoulli’s equation can be used under the following conditions:

1. The fluid is incompressible; its density remains constant. 2. The fluid does not have appreciable frictional effect, it is non-

viscous. Consequently, no mechanical energy is lost due to friction.

3. The flow is streamline, not turbulent.


4. The velocity of the fluid at any point does not change during the period of observation.

Using the law of conservation of energy, it is easy to get the sum of energies per unit volume at the inlet and outlet of the tubes shown in Fig. (2). By equating these sum we get Bernoulli’s equation in the form

P1 + ½ ρ 21v + ρgh1 = P2 + ½ ρ 2

2v + ρgh2 (5)

Fig. (2). Flow of liquid through a tube of different diameters and heights from a

certain base line.

where P1 & P2 and v1 & v2 are the pressures and velocities of the liquid of density ρ, at the inlet and outlet of tubes of cross-sectional areas A1 and A2 respectively. h1 and h2 are the tube heights above the base line.

If the tubes are horizontally located, that is at the same heights, h1=h2, then equation (5) becomes :

P1 + ½ ρ 21v = P2 + ½ρ 2

2v (6) While if the liquid in the tubes of Fig (2) is at rest, that is v1 = v2, then according to equation (5) the difference in pressures at the inlet and outlet of the tube system is :

P1 + ρgh1 = P2 + ρgh2 that is

P1 – P2 = ρg(h2 – h1) (7)

If P2 is at atmospheric pressure, Patom, and h = h2 – h1, then P1 = Patm + ρgh

This equation is the same as that given before in chapter (4), Eq. (2).

v1

A1

V A2

h2

h1

Base line


Example (2): The pressure 1 m above a floor is measured to be normal atmospheric pressure, 1.03×105 Pa. How much greater is the pressure at the floor if the temperature is 0oC?

[The density of air at atmospheric pressure and 0oC is 1.29 Kgm-3].

Solution:

The pressure at the floor PF = Patm + ρ g h, then

PF = 1.013×105 Pa + (1.29 Kgm−3)(9.8 ms−2) (1 m)

= (1.013×105 + 12.6) Pa

Example (3): A liquid of density 850 Kgm−3 flows through a horizontal tube of radius 0.03 m at an absolute pressure of 1.6×105 Pa. At a certain section the tube is constricted so that its radius becomes a 0.02 m, consequently the pressure is changed to 1.1×105 Pa. Calculate the velocity and the rate of flow of liquid in the constricted tube section.

Solution: Since the tube is in the horizontal position, then

P1+½ρ 21v = P2 + ½ ρ 2

2v that is

2-255

21

222

121 sm 850

105.0850

10)1.16.1()vv(PP ×=

×−=−=

ρ− (i)

From the equation of continuity

A1v1 = A2v2, then

(0.03)2v1 = (0.02)2v2 , v2 = 12

2v

)02.0()03.0(

v2 = 2.25 v1 (ii) Apply from (ii) to (i) then

v1 = 5.9 ms−1 , v2 = 13.275 ms−1 The rate of flow Q = A1v1 or A2v2

= π 21r v1 = 3.14×(0.03 m)2 (5.9 ms-1) = 16.6×10−3m3s−1


Example (4): Calculate the percentage ratio of decrease in the blood pressure of the aorta due to atheroscleratic plaque and the consequences decrease in its cross-sectional area of 1/5 from the original (neglect the viscosity of the blood, at this small section, and the effect of height). Note that the average blood pressure and the density are 100 mmHg and 1000 Kgm−3 respectively and its velocity before blockage is 0.12 ms−1.

Solution:

Since the effect of height is neglected, then

P1 – P2 = )vv(2

21

22 −

ρ (i)

Since, A1v1 = A2v2 then v1 = 1/5 v2 or v2 = 5v1

Applying in (i) then

[ ] Pa8.172)ms12.0(242

1000)vv25(2

PPP 2121

2121 =×=−

ρ=−=Δ −

To get the blood percentage decrease, we have to convert the normal blood pressure, 100 mmHg, to Pascal unit

Pblood = hρg = 100 mmHg× 10−3 m × 13600 Kgm−3×9.8 ms−2

≅ 1.36×104 Pa

The percentage decrease in blood pressure

PPΔ

×100 = 13600

8.172×100 ≅ 1.27%

This means that the decrease in the aorta blood pressure at a certain section, due to partial block, leads to a corresponding increase of its velocity in that section.

Flow meter: The concepts of Bernoulli equation is applied to measure the flow in blood vessels, the airspeed of planes, and many other flow rates.


The horizontal constricted pipe illustrated in Fig. (3), known as a venture tube, can be used to measure flow velocities in an incompressible fluid. If the pressure difference, P1-P2, between the tube ends is measured then the flow velocity, v2, could be calculated according to the following treatment.

Since the pipe is horizontal, then Bernoulli’s equation is:

P1 + ½ ρ 21v = P2 + ½ ρ 2

2v

From the equation of continuity we get A1v1 = A2v2 or v1= 1

2AA v2

where A1 and A2 are the cross-sectional areas of the tube parts as shown in Fig. (3).

Fig. (3): Schematic diagram

of a venture tube. The pressure P1 is greater than P2, since v1 < v2.

Substituting the value of v1 in the previous Bernoulli Eq. gives

P1 + ½ ρ 22

2

1

2 vAA

⎟⎟⎠

⎞⎜⎜⎝

⎛ = P2 + ½ ρ 2

2v

v2 = A1( )

)AA(PP2

22

21

21

−ρ− (8)

Note that since A2 < A1, it follows that P1 is greater than P2. In other words, the pressure is reduced in the constricted part of the pipe.

Bernoulli’s equation has many applications in our life as an example in the operation of airplane. The shapes of airplane wings are designed such that the upper surface has a smaller radius of curvature than the lower surface, Fig. (4). Air flowing over the upper surface follows more of a curved path than the air flowing over the lower

P1 P2

v2 v1

A A 12

1


surface. Such difference in air flow over the upper and lower surfaces creates a difference in pressure and consequently a net upward force F on the wing, called the dynamic lift, depends on several factors such as the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal.

Fig. (4). Streamline flow around an airplane

wing. The pressure above is less than that below, and hence there is a dynamic upward lift force, F.

Also a number of devices operates in a manner described in Fig. (5). A stream of air passing over an open tube reduces the pressure above the tube. This reduction in pressure causes the liquid to rise into the air stream. The liquid is then dispersed into a fine spray of droplets. This is the atomizer action which is the same as the action of car carburetor.

Fig. (5). Schematic representation of the atomizer.

If a person has advanced arteriosclerosis, the Bernoulli principle produces a sign called vascular flutter. In this situation, the artery is constricted as a result of an accumulation of plaque on its inner walls. In order to maintain a constant flow rate through such a constricted artery, the driving pressure must increase. Such an increase requires a greater demand on the heart muscles. If the blood velocity is sufficiently high in the constricted region, the artery may collapse under external pressure, causing a momentary interruption in blood flow. At this point, there is no Bernoulli principle and the vessel reopens under arterial pressure. As the blood rushes through the constricted artery, the internal pressure drops and again the artery closes. Such variations in blood flow can be heard with a stethoscope.

F


If the plaque becomes dislodged and ends up in a smaller vessel that delivers blood to the heart, the person can suffer a heart attack.

Flow of incompressible viscous liquid: Since the viscosity of a certain liquid causes a loss in energy during its flow. Accordingly the total energy per unit volume, Bernoulli’s principle, is not constant along the path of such viscous fluid. This means that we can not apply Bernoulli’s equation to describe the flow of a viscous fluid.

In the nineteen century the French scientist Poiseulle noticed that the rate of laminar flow, Q, of a viscous liquid through a cylindrical tube, Fig. (6), is directly proportional to the difference in pressure, ΔP, along the length of the tube L, that is the pressure gradient ΔP/L) and also to the fourth power of the tube radius, r. Then the rate of flow Q, Poiseulle’s law, can be written as

L8PrQ

4

ηΔπ

= (9)

where η is the viscosity of the liquid.

Poiseuille’s law indicates that high viscosity leads to low flow rates, which is reasonable. It also shows that the flow rate is proportional to r4. That is the rate is extremely dependent on the radius of the tube. This implies, for example, that in blood vessels moderate adjustments in the radius can produce large changes in the flow rates. This gives the body a very effective way of adjusting the blood flow and responding to changing requirements.

Example (5): A large artery in a dog has an inner radius of 4×10-3m. If the blood flows through this artery at the rate of 10-6m3s-1.

Find: (a) The average and maximum velocities of the blood. (b) the pressure drop in a 0.1 m long segment of the artery. Given that the viscosity of the blood is 2.084×10−3 Pas.

Solution: The average velocity is :


1223

136

2 m1099.1)m104(

sm10rQ

AQv −−

−

−−

×=×π

=π

==

vmax = 2 v = 2(1.99×10-2ms-1) = 3.98×10-2ms-1

ΔP = 2rvL8η

= 2.07 Pa Fig. (6). Laminar flow of a viscous

liquid. The maximum velocity vmax occurs at the center.

Turbulent flow: If a gradual increase in the velocity of a viscous fluid flowing in a tube, due to the gradual reduction in its radius, the velocity of this fluid will reach a critical velocity, cv , above which the laminar flow changes into turbulent flow as shown in Fig. (7).

Fig. (7). The conversion of liquid flow from laminar to turbulent due to the increase in the liquid velocity.

In order to determine whether the flow is laminar and thus whether Poissouille’s law can be applied, we can make use of one of an empirical rule. It states that the value of a dimensionless quantity called Reynolds number NR determines whether the flow is turbulent or laminar.

Consider a fluid of viscosity η and density ρ. If it is flowing in a tube of radius, r, and has an average velocity, v , then the Reynolds number is defined by

ηρ

=rv2NR (10)

In tubes, it is found experimentally that if,

NR < 2000 flow is laminar.

Vmax


NR > 3000 flow is turbulent. 2000 < NR < 3000 flow is unstable (may change from laminar to turbulent or vice versa).

Laminar flow is more efficient than turbulent flow. This is clearly illustrated in Fig. (8a) in which the slope of the curve in the laminar flow region is greater than that in the turbulent flow region. That is, a given increase in pressure causes a greater increase in the laminar flow rate than in the turbulent flow rate. The reduction in efficiency is apparent in the blood flow through an artery with an obstruction as shown in Fig. (8b).

(a) (b) Fig (8). (a) The relation between the rate of flow of blood and the pressure.

The increase in the blood pressure and consequently the velocity to a value greater than the critical value Pc leads to change the rate of flow from laminar to turbulent. (b) If an artery is obstructed then it needs more pressure ΔP2 to give the same rate of flow (QB-QA) as in case of normal, ΔP1 (ΔP2 > ΔP1).

It is clear that if both arteries are required to deliver the same flow rate, Q = QB-QA, the change in pressure ΔP2 in the obstructed artery will be much greater than that of normal artery, ΔP1, because the flow in the obstructed case is turbulent.

Example (6): Calculate the Reynolds number and the type of blood flow streaming through an artery, of radius 4×10−3 m, with average velocity 1.99×10−2 ms−1.Given that the viscosity and density of blood are 2.084×10−3 Pas and 1.0595×103 Kgm−3 respectively.

Solution:

Flowrate

Obstructedartery

Pc

ΔP1

ΔP2

QB

QA

Normalartery

Turbulent

Laminar

Q

Flowrate

PressurePressure


NR = Pas1008.2

)m104)(ms1099.1)(Kgm10059.1(2rv23

1233 3

−

−−−−

××××

=ηρ

= 80.9

This value is much less than 2000, so the flow is laminar .

In turbulent flow, some energy is dissipated as sound and some as heat. The noise associated with turbulent flow in arteries facilitates blood pressure measurements, as described in chapter four, in addition, it makes possible the detection of some heart abnormalities by the use of stethoscope.

Flow resistance: The flow resistance Rf is defined, in general, as the ratio of pressure drop to the flow rate of any kinds. That is,

Rf = QPΔ (11)

When the flow is laminar, we can compare this with the Poiseuille’s law. Then we get:

4f rL8R

πη

= (12)

From definition the units of flow resistance is Pascal-second per cubic meter, Pa s m−3. We will use (KPa s m−3). In physiology texts the unit of flow resistance is

1 torr s cm-3 = 1.33×105 KPa s m−3 The flow resistance in a large artery is smaller than that in the small one. This explains why the pressure drop in large areteries is small. The following example will clarify this fact.

Example (7):

The aorta of an average adult human has a radius 1.3×10−2m. What are the resistance and pressure drop over a 0.2m distance, assuming a flow rate of 10−4m3s-1 and η for blood is 2.0844×10-3 Pas.

Solution:


Rf = 42

3

4 )m103.1()m2.0)(Pas1008.2( 8

rL8

−

−

××

π=

πη

= 3.72×104 Pasm−3 = 37.2 KPasm−3 ΔP = RfQ

= (37.2 KPasm−3)(10-4 m3 s-1 )= 0.00372 K Pa This is very small compared to the total pressure drop in the whole body system which is about 13.3 Kpa. Most of the flow resistance and pressure drops occur in the smaller arteries and vascular beds of the body.

The flow resistance of a collection of arteries, such as the mesenteric bed of the dog, can be measured or calculated. The calculation can be done by considering, for example, that all arteries of a given size are in parallel. Also each artery carries its equal share of the total flow. If the flow through one aretery of resistance Rf1 is Q1, then from Eq. (11).

1f

1 RPQ Δ

=

Where ΔP is the pressure across all arteries. If there are N identical arteries, the total flow Q = Q1 + Q2 + …. Qn or

1fRPNQ Δ

=

If we define the equivalent flow resistance Rp of this arrangement by

Q = pRPΔ , then

Rp = N

R1f (13)

Example (8): If the radius of a single capillary is 4×10-6 m and its length is 10-3 m, then calculate the net resistance of the 4.73×107 capillaries in the mesenteric vascular bed of a dog if they are assumed to be in parallel. (ηblood = 2.084×10−3 Pas).

Solution:


46

33

4f )m104()m10)(Pas10084.2(8

rL8R

1 −

−−

×π×

=πη

=

= 2.073×1016 Pa sm-3 = 2.073×1013 KPasm-3

Since there are N = 4.73×107 capillaries in parallel, so their effective resistance is

7

313f

f 1073.4KPasm10073.2

NR

R 1

××

==−

= 4.38×105 KPasm-3

Blood flow in blood vessels Blood is a viscous liquid contains many different constituents, including red blood cells, white blood cells, platelets, and proteins. Accordingly, its flow through the body blood vessel is laminar but in special conditions are turbulent. This turbulent flow generally occurred at the heart valve edges, during high body activity and in case of the presence of certain obstruction inside the arteries.

As the blood goes from the aorta into the smaller areteries and areterioles with greater total corss-sectional areas the velocity of the blood decreases as shown in Fig. (9).

Fig. (9). The change in velocity (solid lines) and the cross-sectional area (dashed curve) during the flow of blood in the circulatory system.

It shows how the velocity of the blood is related in an inverse way to the total cross-sectional area of the vessels carrying the blood. Thus if a certain artery is branched into, n, arterioles of equal cross-sectional

Area

Vena cava

CapillariesAorta

3 cm2

velocity

0.3 ms−1

0.05 ms−1

600cm2

Area


area, then according to Poiseuille’s law and in order to keep the pressure gradient constant, the area of the artery should be n . To clarify this information, let us consider the following example.

Example (9): A blood artery of radius r1 is branched to n arterioles each of equal radius r2. What is the relation between the areas and velocity of the artery and arterioles.

Solution:

The relation between the areas;

Since A1v1 = nA2v2 then from Poiseuile’s law

L8

PrnL8Pr 4

24

1η

Δπ=

ηΔπ that is

π r 41 = nπ r 4

2 , r 21 = n r 2

2 that is A1 = n A2 (14) The relation between velocities

Since Q1 = Q2 and A1 = n A2, Eq. 14, then

A1v1 = nA2v2 , n A2v1 = nA2v2,

that is v2 = n

v1 (15)

Eq. 15 indicates that the velocity in the arterioles is much less than that in the artery. For example if the velocity in the aorta is about 0.3 ms−1, then by the use of Eq. 15, the velocity in the ten equal branched arterioles will be 0.094 m s-1.

In summary the gradual decrease in the velocity of blood when streaming in the circulatory system is due to the blood viscosity and the increase in the total cross-sectional areas. Such decrease represents a dissipation in power and hence energy. This fact could be clarified, if we rewrite the expression of power as


vFtx.F

timeworkP =

ΔΔ

== (16)

Where Δx is the change in the blood flow distance under the force F. v is the resultant average velocity of blood flow. Since F equals the difference in pressure ΔP time the cross-section A, then Eq. (16) becomes

P = F v = ΔPA v = ΔP πr2 v = ΔPQ (17)

where Q is the rate of flow.

Equation 17 indicates that the power dissipation in the circulatory system leads also to the gradual decrease of pressure in the blood vessels as shown in Fig. 10. It shows the gradual decrease in blood pressure streaming through arteries, arterioles, capillaries, and veins during a certain period of time.

Fig. (10). The variation of pressure throughout the circulatory system at a certain

period of time where the pressure gradient is constant. Example (10)−1

1. b)

23

133

2 )104()ms1099.1)(m1.0)(Pas1008.2(8

rvL8P −

−−−

×××

=η

=Δ

=2.07 Pa = 0.00207 KPa

Pressure (mmHg)

Arteries

Arterioles

Capillaries

veins

Time


2. a liquid, and then define the Reynold’s number. 3. Use the relation between the blood pressure and its rate of

flow for normal aorta and obstructed one to compare between their behaviors.

4. Define the flow resistance of a certain liquid flowing in a vessel.

5. A stream of fluid has its cross-sectional area halved in a certain region. What its average velocity ?

6. The radius of water pipe decreases from 0.2 to 0.1 m. If the average velocity in the wider portion is 3 ms−1, find the average velocity in the narrower region.

7. A blood vessel of radius r splits into four vessels, each with radius r/3. If the average velocity in larger vessel is v, find the average velocity in each of the smaller vessels.

8. Bernoulli’s equation implies that the work done on the fluid flowing in a tube must equal its change in potential and kinetic energy.

a) If dissipative forces are present, how does that affect the pressure changes as a fluid moves along the tube?

b) Are dissipative forces more or less significant as the flow velocity increases? Explain.

9. During a whole-blood transfusion, the needle is inserted in a vein where the pressure is 2000 Pa. At what height must be the blood container be placed, relative to the vein, so that the blood just enters the vein?

10. A venture tube has a radius of 1 cm in its narrower portion and 2 cm in its wider sections. The velocity of water in the wider sections is 0.1 ms−1. Find (a) the pressure drop; (b) the velocity in the narrow portion.

11. Two arteries have the same flow rate, but one has twice the radius of the other. What is the ratio of the pressure drops over a given distance.

12. A blood vessel of radius 10−3 m has a pressure gradient ΔP/L of 600 Pa m−1 (Assume laminar flow). (a) What is the flow rate of blood at 37oC in the vessel? (b) What is the maximum velocity of the blood in the vessel?

13. The pressure drop along a length of a horizontal artery is 100 Pa. The radius of the artery is 0.01 m, and the flow is laminar. (a) What is the net force on the blood in this portion of the


artery? (b) If the average velocity of the blood is 1.5×10-

2ms−1, find the power expended in maintaining the flow. 14. The radius of an artery is increased by a factor of 1.5. (a) If

the pressure drop remains the same, what happens to the flow rate? (b) If the flow rate stays the same, what happens to the pressure drop? (Assume laminar flow).

15. Define the type of flow of a liquid has a Reynolds number less than 2000.

16. A blood vessel of radius R branches into several vessels of smaller radius r. If the average fluid viscosity in the smaller vessel is half that in the large vessel, how many vessels of radius r must be there.

17. A small artery has a length of 1.1×10−3 m and a radius of 2.5×10−5 m.

a) Calculate its resistance. b) If the pressure drop across the artery is 1.3 KPa,

what is the flow rate? 18. A glass tube has a radius of 10−3 m and a length of 0.1 m.

a) What is the resistance to the flow of a liquid with a viscosity of 10−3 Pas?

b) If the pressure drop along the tube is 103 Pa, what is the flow rate?

19. A glass plate 0.25 m2 in area is pulled at 0.1 ms-1 across a large glass plate that is at rest. What force is necessary to pull the upper plate if the space between them is 0.03 m thick and filled with (a) Water with η = 1.005×10-3 Pas.

(b) oil with η = 0.01 Pas? (c) why might oil be preferred to water as a lubricant?

20. The average flow rate of blood in the aorta is 4.2×10−6 m3s-1. The aorta is 1.3×10−2 m in radius

(a) What is the average blood velocity in the aorta? (b) What is the pressure drop along 0.1m of the aorta? (c) What is the power required to pump blood through this

portion of the aorta? 21. When a calf is at rest, its heart pumps blood at rate of

6×10-5m3s−1. The pressure drop from the arterial to venous system is 12 KPa. (a) What is the flow resistance of its circulatory system? (b) How much work does the heart do in pumping the blood? (c) An experimental artificial heart


powered by an electrical pump is implanted in place of animal’s heart. If the pump has an efficiency of 50 percent, how much electrical power does it require?

22. a) Calculate the flow resistance of a typical human capillary 2×10−6 m in radius and 10-3 m long.

b) Estimate the number of capillaries in a human using this result, given that the net flow rate through aorta is 9.7×10−5m3s-1 and that the pressure drop from the arterial system to the venous system is 11.6 KPa. Assume that all the capillaries are in parallel and that 9% of the pressure drop occurs in the capillaries.

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