(Chap.13 p.315)

The Mechanics of Nonviscous Fluids

A. Prof. Hamid NEBDIhbnebdi@uqu.edu.sa

Faculty of Science. Department of Physics. Room: 315 second floor. Phone: 3192

Umm Al-Qura University القرى أم معة جاFaculty of Scienceكـلية العلوم Department of Physics الفيزياء قسم

General Physics (For medical students): 403104-3 403104-3 : الطبية للفيزياء المدخل

Course’s Topics

- Introduction

13.1- Archimedes' Principle

13.2- The Equation of Continuity; Streamline Flow

13.3- Bernoulli's Equation

13.4- Static Consequences of Bernoulli's Equation

13.5- The Role of Gravity in the Circulation

13.6- Blood Pressure Measurements Using

the Sphygmomanometer

IntoductionIn this chapter, we:

- discuss fluids at rest and nonviscous (frictionless) fluids motion.

- develop an understanding of why an object may either sink or float in a fluid at rest?

- develop Bernoulli’s Equation, which puts work and energy concepts into a form suitable for fluids. Then, we can understand why fluids in connected containers tend to have the same surface levels?

And how fluids flow from one place to another?

On this discussion, the important condition is the assumption that the fluid is incompressible: a given mass of fluid always occupies the same volume though its shape my change.

13.1 Archimedes’ Principle

• An object floating or submerged in a fluid experiences an upward or buoyant force due to the fluid.

• Archimedes’ principle states that the buoyant force B on the object is equal to the weight of the displaced fluid: B = w0 = 0gV

What is the buoyant force ?• Consider a segment of fluid of volume V and density

ρo. The segment has a:

– Mass m0 = ρoV

– Weight w0= m0 g = ρo g V

• The segment is in equilibrium with the

surrounding fluid, thus the buoyant force B on the object is equal to the weight of the displaced fluid:

B = w0 = ρo g V (Archimedes’ Principle)

The buoyant force is the force exerted by the rest of the fluid to keep the segment at equilibrium.

Suppose now that an object of volume V and density ρ suspended by a string in the fluid (ρo).

Since it is equilibrium: T + B = -w

and T = w – B

or T = (ρ – ρo) g V

The tension in the string is reduced by the weight of displaced fluid.

Example 13.1:

A piece of metal of unknown volume V is

suspended from a string. Before

submersion the tension in the string is

10 N. When the metal is submerged in

water the tension is 8 N. The water

density is 103 kg/m3.

What is the density of the metal?

Before submersion the tension is : Ti = ρ g V

After submersion the tension is:

Solving for ρ:

Tf = (ρ – ρo) g V

fi

i

i

f

TT

T

T

T

0

0

ρ =5000 kg m-3

Solution 13.1:

V

Vs0

• An object less dense than a fluid will float partially submerged. If Vs is the portion of its volume V which is submerged, then the bouyant force is . This must equal the weight of the object, so:

• The ratio of the densities is equal to the fraction of the volume submerged.

sgV0

gVgVs 0

Example 13.2:

• The density of ice is 920 kg/m3 while

that of sea water is 1025 kg/m3.

What fraction of an iceberg is

submerged? Solution 13.2:The fraction is:

898.00

V

Vs

Almost 90 % of the iceberg is submerged.

13.2 The Equation of Continuity

• If the fluids enters one end of the channel such as a pipe or an artery at the flow rate Q1, it must leave the other end at a rate Q2, where

• The flow rate Q is the volume of the fluid flowing past a point in a channel per unit time:

Q1 = Q2 Equation of continuity

The S.I units of the flow rate Q is m3/s.

For example, if 1 m3/s enters, then 1m3/s must leave.

t

VQ

• Consider a section of a tube with a constant cross-sectional area A. In a time ∆t, the fluid moves a distance ∆x= v ∆t, and the volume of fluid leaving the tube is

∆V=A ∆x= A v ∆t

As ∆V= Q ∆t then Q=A vThe flow rate equals the cross-sectional area of the channel times the velocity of the fluid.

• For a channel whose cross section changes from A1 to A2

• The product of the cross-sectional area and the velocity of the fluid is constant

Q1 =Q2 A1 v1 = A2 v2

Example 13.4:

A water pipe leading up to a hose a radius of

1 cm. Water leaves the hose at a rate of 3

litres per minute.

(a)Find the velocity of the water in the pipe.

(b) The hose has a radius of 0.5 cm. What is

the velocity of the water in the hose.

Solution 13.4:

(a) The velocity (strictly speaking, the average velocity) can be

found from the flow rate and the area. The flow rate is the same

in the hose and in the pipe.

Using 1 litre=0.001 m3 and 1 min = 60 s, the flow rate is:

Q = ∆V/ ∆t = 0.003 m3/60s= 5x10-5 m3/s

We will call the velocity and area in the pipe v1 and A1, respectively.

Then, with Q = A v, we have

v1=Q/A1=Q/r12 = (5x10-5m3/s)/(0.01m)2 = 0.159 ms-1

(b) The flow rate is constant, so A1 v1 = A2 v2, and the velocity v2 in

the hose is v2=v1 A1/A2=v1( r12/ r2

2 )=0.636 m.s-1

The water flows faster in the narrower channel.

13.3 Bernoulli’s Equation

Bernoulli’s equation can be used under the following conditions :

1- The fluid is incompressible; its density remains constant.2- The fluid is nonviscous (no mechanical energy is lost).3- The flow is streamline, not turbulent.4- The velocity of the fluid at any point does not change during the period of observation. (This is called the steady-state assumption.)

Bernoulli’s Equation

It states the consequences of the principle that the work done on a fluid as it flows from one place to another is equal to the change in its mechanical energy.

VvvK

yyVgU

VppxAppW

abا

ab

baba

)(2

1

)(

)()(

22

The work W done on the fluid must be set equal to the change in the potential energy ∆U plus kinetic energy ∆K.

22

2

1

2

1bbbaaa vgypvgyp

• The pressure plus the total mechanical energy per unit volume, , is the same everywhere in the flow tube.

• If v=constant then the pressure plus the potential energy per unit volume of the fluid is the same everywhere in the flow tube.

2

2

1vgyp

gy

bbaa gypgyp

13.4 Static consequences of Bernoulli’s Equation

- We first examine the implications of Bernoulli’s equation when the fluid is at rest, so v= 0 and P+ρgh = constant

- Fluid at Rest in a Container:From the figure find the pressure at a point B

in terms of the pressure at surface and the depth.

- Using Bernoulli’s equation we can write:PA+ρgyA = PB+ρgyB

At A the pressure is atmospheric pressure so PA=Patm and yA=h.

As d = h-yB then PB = Patm + gd.

This result shows that pressure at a depth d in a fluid at rest is equal to surface pressure plus the potential energy density change gd corresponding to this depth.

- This equation can also be interpreted as a statement of the force on a unit area at a distance d below the surface of the liquid. The force per unit area PB is the sum of two terms: Patm, the pressure due to the atmosphere, and gd , the pressure due to the weight of the liquid above the point B.

- Calculating P+ρgy at points B and D gives:

PB+ρgyB = PD+ρgyD

Or, since yB=yD , then PB = PD

Thus the pressure at the same depth at two places in a fluid at rest is the same. In particular, since points A and E are both at atmospheric pressure, the surface of the liquid is at the same height at both points. Thus the surfaces of liquids at rest in connected containers of any shape must be at the same height if they are open to the atmosphere.

Example 13.6:

• The pressure 1 m above a floor is measured

to be normal atmospheric pressure, 1.013 x

105 Pa. How much greater is the pressure at

the floor if the temperature is 0° C?Solution 13.6:

Here, d=1m. From Table 13.1 (p. 315) the density of air at atmospheric pressure and 0° C is

1.29 kg.m-3. ThusPB = Patm + gd

= 1.013 x 105 Pa + (1.29 kg.m-3)(9.8 m.s-2)(1m)

= 101312.64 Pa

The Manometer:- The open-tube manometer is a U-shaped tube used for measuring gas (or liquid if doesn’t mix with the manometer fluid) pressures. It contains a liquid that may be mercury or, for measurements of low pressures, water or oil.

- Measuring heights from the base of the U-tube, P+ρgy is P+ρgy1 at the left surface of the column and Patm+ρgy2 at

the right surface. Equating these: P+ρgy1 = Patm+ρgy2

or P =Patm+ρg(y2 –y1)=Patm+ρgh

Thus, a measurement of the height difference h of

the two columns determines the gas pressure P.- In the above equation, the pressure P is the absolute pressure.- From the same equation the difference, P-Patm, is the gauge pressure. The gauge pressure is then exactly equal to gh.

Blood Pressure Measurements by Cannulation:- In many experiments with anesthetized animals, the blood pressure in an artery or vein is measured by the direct insertion into the vessel of a cannula, which is a small glass or plastic tube containing saline solution plus an anticlotting agent.

- The saline solution, in turn, is in contact with the fluid in a manometer. It’s necessary to have the surface of contact between the saline solution and the manometer fluid either at the same level as the insertion point of the cannula or to correct for the height difference.

- Calculating P+ρg y at suitable points, it follows that the blood

pressure PB is given by : PB = Patm + ρ g h – ρs g h’

- Physiologists often use electromanometers in which the manometer fluid pushes against a membrane instead of rising up a tube.

- The flexing of the membrane is proportional to the pressure, and the extend that the membrane is flexed is translated into an electrical signal. This signal drives a recorder that automatically gives a continuous record of the blood pressure.

13.5 The Role of Gravity in the Circulation

- Humans have adapted to the problems of moving blood upward a large distance against the force of gravity.

- Some animals die if held head upwards. The blood remains in the lower extremities, and the heart receives no blood from the venous system.

- The gauge pressure at the heart is PH

- The gauge pressure at the foot is PF

- The gauge pressure at the brain is PB

- If a person large arteries are cannulated. In the

reclining position, the pressures everywhere are

almost the same.

- The small pressure drop between the heart and

the feet or brain is due to the viscous forces.

- However, the pressures at the three points are

quite different in the standing position,

reflecting the large differences in their height.

- Since the viscous effects are small, we can use

Bernoulli’s equation:

P + ρgh + ½ ρ v2 = constant

- The velocities in the three arteries are small and

roughly equal, so the term ½ ρ v2 can be ignored.- The gauge pressures at the heart (PH), at the foot (PF),

and at the brain (PB) are related by the following

relation;PF = PH + ρ g hH = PB + ρ g hB,Where ρ is the density of blood.

- In discussions of the circulatory system, it’s convenient to measure pressure in kilopascals, kPa, where 1 kPa = 103 Pa is a multiple of the basic S.I. Pressure unit.

- Many discussions of the circulatory system use an older unit, the torr,

1 torr = 0.1333 kPa

- If hH =1.3 m and hB = 1.7 m.

With ρ =1.0595 x 103 kg/m3 , we find

PF – PH = ρghH = (1.0595 x 103 kg/m3) x

(9.8 m s-2) x (1.3 m) = 1.35 x 104 Pa = 13.5 kPa

PH is typically 13.3 kPa, so PF ~ 26.8 kPa.

- Similarly PB can be found as follows;

PB = PF – ρ ghB

PB = PH + ρ g (hH – hB)

- This explains why pressures in the lower and upper parts of the body are very different when the person is standing, although they are about equal when reclining.

- Blood returned to the heart, at least partially, by the pumping

action associated with breathing and by flexing of skeletal

muscle, as in walking.

- The importance of the role of gravity in the circulation is

illustrated by the fact that soldier who is required to stand at

strict attention may faint because of insufficient venous return.

- To regain consciousness, position has to be altered to horizontal,

then pressure is equalized.

Effects of Acceleration

- When a person in an erect position experiences an upward acceleration a, the

effective weight becomes m(g+a).

- Applying Bernoulli’s equation to the brain and heart with g replaced by (g+a):

PB + ρ(g+a)hB = PH + ρ (g+a)hH

or PB = PH - ρ (g+a) (hB – hH)

- Thus the blood pressure in the brains will be reduced even farther.

- It has been found that if a is 2 or 3 times g, a human will lose

consciousness because of the collapse of arteries in the brain. This

factor limits the speed with which a pilot can pull out of a dive.

- A releated experience is the feeling of light-headedness that

sometimes occurs when one suddenly stands up. Since muscular

movement is required to activate the venous return mechanism,

blood will tend to collect in the lower veins until normal activity is

resumed.

13.6 Blood Pressure Measurements Using the Sphygmomanometer

• The sphygmomanometer is an instrument to

measure the human’s blood pressure.

• In the sphygmomanometer the gauge

pressure in an air sack wrapped around the

upper arm is measured using a manometer.

• When using the sphygmomanometer,

why is the blood pressure usually

measured in the upper arm?

- (1) The upper arm of a human is at

about the same level as the heart,

blood pressure measurements made

there give values close to those near

the heart.

- (2) The fact that the upper arm

contains a single bone makes the

brachial artery located there easy to

compress.

- During a complete heart pumping cycle the pressure

in the heart and circulatory system goes through both

(1) a maximum (blood is pumped from the heart);

systolic, and (2) a minimum (as the heart relaxes

and fills with blood returned from veins); diastolic.

- The sphygmomanometer is used to measure these

extreme pressures (systolic & diastolic).

- The use of sphygmomanometer relies on the fact that

blood flow in the arteries is not always streamline.

- When the arteries are constricted and the blood flow

is noisy then it can be heard with a stethoscope.

1- The pressure in the sack is first increased until the artery

is closed.

2- Then the pressure in the sack is slowly reduced, while a

stethoscope is used to listen for noises in the artery

below the sack.

3- When the pressure is slightly below the maximum

pressure (systolic), the artery will open partially, so the

flow velocity is high and turbulent and therefore noisy.

4- The pressure in the sack is lowered the diastolic

pressure is measured.

The procedure of using the sphygmomanometer

- Blood pressures are presented as

Systolic / Diastolic

- Typical reading for normal blood

pressure for a healthy resting adult

are about120/80 torr and 16/11 kPa.

- The borderline for high blood pressure

(hypertension) is usually defined to

be 140/90 in torr and 19/12 in kPa.