Medicine

Part b of the figure shows the streamlines of simple flow

Theme: Basic biorheology and gemodynamics

Amount of hours: 6 hours

Place of leadthrough: physical laboratory.

 

І. Practical work – 9:00-12:00 (3 hours)

Method of implementation of practical work:

Theoretical information

Here we survey modern methods for measuring the hydrodynamic properties of biological macromolecules and for calculating the hydrodynamic properties of useful models for biomolecules.

Hydrodynamic properties are quantities like sedimentation and diffusion coefficient, rotational relaxation time, intrinsic viscosity, and viscoelastic relaxation time. They enable one to determine how rapidly molecules translate and rotate in solution, and how they influence the steady state and time-dependent viscosity of solutions. These measured properties, in turn, when interpreted in terms of suitable models, give us information about the molecular weight, size, hydration, shape, exibility, conformation, and degree of association of biological macromolecules.

Like so much of science, hydrodynamics has a long history, but has been dramatically hanged by recent advances in experiment, theory, and computation. Classic textbooks tell of the experimental work of Fick in 1855 on diffusion, Svedberg in the 1920s on sedimentation, and Poiseuille in 1846 on viscosity. Theories for simple shapes such as spheres and ellipsoids were worked out by Stokes (1847), Einstein (1906), and Perrin (1936). However, recent experimental developments such as dynamic laser light scattering, nanosecond fluorescence depolarization measurements, fluorescence recovery after photoblea hing,.uorescence correlation spectroscopy, pulsed field gradient NMR, and improved analytical ultracentrifuge design have made measurements of hydrodynamic properties more convenient, precise, and refined. At the same time, development of new theoretical and computational tools have enabled calculation of the properties of complex, realistic molecular models.

Fluids In motion

Part b of the figure shows the streamlines of simple flow. These lines show the path a tiny particle in the fluid follows as it moves along the pipe. Flow such as this is termed laminar flow.

The fluid velocity is lower in the large cross-sectional region. Part c of the figure shows what happens if the flow becomes too swift past an obstruction. The smooth flow lines no longer exist. As the fluid rushes past the obstacle, it starts to swirl in erratic motion. No longer can one predict the exact path a particle will follow. This region of constantly changing flow lines is said to consist of turbulent flow.

We saw in the last section that friction effects occur in a flowing fluid. This effect is described in terms of the viscosity of the fluid Viscosity measures how much force is required to slide one layer of the fluid over another layer. Substances which do not flow easily, such as thick tar or syrup, have large viscosity. Substances (like water) which flow easily have small viscosity

 

The Equation of Continuity

Up to now, we have studied only fluids at rest. Let us now study fluids in motion, the subject matter of hydrodynamics. The study of fluids in motion is relatively complicated, but the analysis can be simplified by making a few assumptions. Let us assume that the fluid is incompressible and flows freely without any turbulence or friction between the various parts of the fluid itself and any boundary containing the fluid, such as the walls of a pipe. A fluid in which friction can be neglected is called a nonviscous fluid. A fluid, flowing steadily without turbulence, is usually referred to as being in streamline flow. The rather complicated analysis is further simplified by the use of two great conservation principles: the conservation of mass, and the conservation of energy. The law of conservation of mass results in a mathematical equation, usually called the equation of continuity. The law of conservation of energy is the basis of Bernoulli's theorem.

Let us consider an incompressible fluid flowing in the pipe of figure 1. At a particular instant of time the small mass of fluid , shown in the left-hand portion of the pipe will be considered.

                     (1)

Because the pipe is cylindrical, the small portion of volume of fluid is given by the product of the cross-sectional area A1 times the length of the pipe containing the mass, that is,

                                (2)

The length  of the fluid in the pipe is related to the velocity of the fluid in the left-hand pipe. Because the fluid in  moves a distance  in time, . Thus,

                                       (3)

Substituting equation 3 into equation 2, we get for the volume of fluid,

                                                            (4)

Figure 1. The law of conservation of mass and the equation of continuity.

 

Substituting equation 4 into equation 1 yields the mass of the fluid as

             (5)

We can also express this as the rate at which the mass is flowing in the left-hand portion of the pipe by dividing both sides of equation 5 by . Thus

            (6)

                  (7)

Equation 7 is called the equation of continuity and is an indirect statement of the law of conservation of mass. Since we have assumed an incompressible fluid, the densities on both sides of equation 7 are equal and can be canceled out leaving. When the cross-sectional area of a pipe gets smaller, the velocity of the fluid must become greater in order that the same amount of mass passes a given point in a given time. Conversely, when the cross-sectional area increases, the velocity of the fluid must decrease.

 

Bernoullis equation

Bernoulli's theorem is a fundamental theory of hydrodynamics that describes a fluid in motion. It is really the application of the law of conservation of energy to fluid flow. Let us consider the fluid flowing in the pipe of figure 1. The left-hand side of the pipe has a uniform cross-sectional area ,which eventually tapers to the uniform cross-sectional area A2 of the right-hand side of the pipe. The pipe is filled with a nonviscous, incompressible fluid. A uniform pressure p\ is applied, such as from a piston, to a small element of mass of the fluid Am and causes this mass to move through a distance Ax\ of the pipe. Because the fluid is incompressible, the fluid moves throughout the rest of the pipe. The same small mass Am, at the right- hand side of the pipe, moves through a distance

Figure 2.  Bernoulli's theorem.

Bernoulli's theorem. It says that the sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any one location of  the fluid is equal to the sum of the pressure, the potential energy per unit  volume, and the kinetic energy per unit volume at any other location in  the  fluid, for a nonviscous, incompressible fluid in streamlined flow.

Since this sum is the same at any arbitrary point in the fluid, the sum itself must therefore be a constant. Thus, we sometimes write Bernoulli's equation in the equivalent form

The Flow of a Liquid Through an Orifice

Let us consider the large tank of water shown in figure 2. Let the top of the fluid be location 1 and the orifice be location 2. Bernoulli's theorem is

But the pressure at the top of the tank and the outside pressure at the orifice are both p0, the normal atmospheric pressure. Also, because of the very large volume of fluid, the small loss through the orifice causes an insignificant vertical motion of the top of the fluid Bernoulli's equation becomes

The pressure term p0 on both sides of the equation cancels out. Also h2 is very small compared to h1 and it can be neglected, leaving

Solving for the velocity of efflux, we get

Figure 3.

Viscosity is a measure of a fluid's resistance to flow. It describes the internal friction of a moving fluid. A fluidwith large viscosity resists motion because its molecular makeup gives it a lot of internal friction. A fluid with low viscosity flows easily because its molecular makeup results in very little friction when it is in motion.

In everyday terms (and for fluids only), viscosity is "thickness" or "internal friction". Thus, water is "thin", having a lower viscosity, while honey is "thick", having a higher viscosity. Put simply, the less viscous the fluid is, the greater its ease of movement (fluidity).

Viscosity describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. For example, high-viscosity felsic magma will create a tall, steep stratovolcano, because it cannot flow far before it cools, while low-viscosity mafic lava will create a wide, shallow-sloped shield volcano.

With the exception of superfluids, all real fluids have some resistance to stress and therefore are viscous. A fluid which has no resistance to shear stress is known as an ideal fluid or inviscid fluid. In common usage, a liquid with the viscosity less than water is known as a mobile liquid, while a substance with a viscosity substantially greater than water is simply called a viscous liquid.

The study of flowing matter is known as rheology, which includes viscosity and related concepts.

Laminar shear of fluid between two plates. Friction between the fluid and the moving boundaries causes the fluid to shear. The force required for this action is a measure of the fluid's viscosity. This type of flow is known as a Couette flow.

Laminar shear, the non-constant gradient, is a result of the geometry the fluid is flowing through (e.g. a pipe).

 

In general, in any flow, layers move at different velocities and the fluid's viscosity arises from the shear stress between the layers that ultimately opposes any applied force. The relationship between the shear stress and the velocity gradient can be obtained by considering two plates closely spaced at a distance y, and separated by a homogeneous substance. Assuming that the plates are very large, with a large area A, such that edge effects may be ignored, and that the lower plate is fixed, let a force F be applied to the upper plate. If this force causes the substance between the plates to undergo shear flow with a velocity gradient u/y (as opposed to just shearing elastically until the shear stress in the substance balances the applied force), the substance is called a fluid.

Newtons equation for fluids

The applied force is proportional to the area and velocity gradient in the fluid:

where  - is the proportionality factor called dynamic viscosity.

Poiseuille formula                           

Hydraulic resistence

 

Reynolds number       

In 1851, George Gabriel Stokes derived an expression, now known as Stokes' law, for the frictional force – also called drag force – exerted on spherical objects with very small Reynolds numbers (e.g., very small particles) in a continuous viscous fluid. Stokes' law is derived by solving the Stokes flow limit for small Reynolds numbers of the generally unsolvable Navier–Stokes equations:[1]

F_d = 6 \pi\,\mu\,R\,v_s\,

where:

·        Fd is the frictional force – known as Stokes' drag – acting on the interface between the fluid and the particle (in N),

·        μ is the dynamic viscosity (N s/m2),

·        R is the radius of the spherical object (in m), and

·        vs is the particle's settling velocity (in m/s).

If the particles are falling in the viscous fluid by their own weight due to gravity, then a terminal velocity, also known as the settling velocity, is reached when this frictional force combined with the buoyant force exactly balance the gravitational force. The resulting settling velocity (or terminal velocity) is given by:[2]

v_s = \frac{2}{9}\frac{\left(\rho_p - \rho_f\right)}{\mu} g\, R^2

where:

·        vs is the particles' settling velocity (m/s) (vertically downwards if ρp > ρf, upwards if ρp < ρf ),

·        g is the gravitational acceleration (m/s2),

·        ρp is the mass density of the particles (kg/m3), and

·        ρf is the mass density of the fluid (kg/m3).

 

 

QUESTION 1

Flow rate. What is the flow rate of water in a pipe whose diameter d is 10.0 cm when the water is moving at a velocity of 0.322

SOLUTION

The cross-sectional area of the pipe is

A==7.85 ×10 m

The flow rate is = v= (1.00 × 10 kg/m)(7.85 ×10m)(0.322 m/s)= 2.53 kg/s

Thus the 2.35 kg of water flow through the pipe per second.

 

QUESTION 2

Applying the equation of continuity. In question 1 the cross-sectional area A WAS 7.85 ×10m and the velocity vwas 0.322 m/s .If the diameter of the pipe to the right is 4.00cm, find the fluid in the right –hand pip

SOLUTION

The cross-sectional area of the right-hand side of the pipe is

A===1.26 × 10m.

The velocity of the fluid on the right-hand side v,

v=v == 2.01 m/s

The fluid velocity increase more six times when it flowed through the constricted pipe.

 

QUESTION 3

Applying Bernoulli’s theorem. In figure 13.10, pressure p= 2.94  N/m,whereas the velocity of the water is v= 0.322m/s .The diameter of the pipe at location 1 is 10.00cm and it is 5.00m above the ground .If the diameter of the pipe at location 2 is 4.00 cm,and the pipe is 2.00m above the ground , find the velocity of the water v at the position 2, and the pressure pof the water at the position 2.

SOLUTION

The area A is

A=== 7.85

Whereas the area A is

A=== 1.26

The velocity at location 2 is found from the equation of continuity

v =v==2.01 m/s

The pressure at location 2 is found from rearranging Bernoulli’s equation as

p= p +  +

=2.94 ++-

 =2.94 =3.04

 

QUESTION 4

When the velocity increases pressure, the pressure increases. In example 13.16 associated with figure 13.9, the velocity v in area A was 0.322 m/s and the velocity v in Awas found to be 2.01 m/s .If the pressure in the left pipe is 2.94 Pa, what is the pressure p in the constricted pipe?

SOLUTION

The pressure p+=2.94=

=2.94 =9.7

Thus, the pressure of the water in the constricted portion of the tube has decreased to 9.7

 

QUESTION 5

A venturi meter .A Venturi meter reads height of h=30.0cm and h=10.0cm .Find the velocity of flow v in the pipe .The area A and the area of A.

SOLUTION

The velocity of flow vin the main pipe is given by

v==0.322m/s

 

QUESTION 6

How much water will flow in 30s through 200mm of capillary tube 1.5 i.d, if the pressure differential across the tube is 5centimeter of mercury? The viscosity of water is 0.801 Cp and  for mercury is 13600 kg/m.

SOLUTION

We shall make use of Poiseuille’s law with

p = 6660N/m

Then           Q=

In 30s, the quantity that would flow out of the tube is (5.2cm

 

QUESTION 7. Applying of the Bernoulli theorem for analysis of an aneurysm. Let's esteem artery a dia d1= 2,5 cm (one of arteries of abdominal cavity), which has a puff-up (aneurysm) a dia 5 cm (fig. 8). Knowing, that the average speed of a blood in norm compounds v1= 30sm * c-1, and the relative blood pressure is equal 120 torr (we shall remind, that 760 torr = 105 Pa), we shall calculate pressure P2 in an aneurysm. How will the aneurysm develop? At calculus’s we shall accept, that the artery is posed horizontally.

SOLUTION

The applying of a Bernoulli relation to a horizontal lease A1 A2, gives

The condition of a continuity of current results equals

        or      

whence

*

The substitution of numerical values gives magnitude, equal approximately 42 Pa, or

The overpressure is insignificant as contrasted to absolute (760 + + 120 = 880 torr), however, it tends even more to expand puff-up, that in turn results in further pressure increase and so on. If, as it frequently happens, the walls of artery are resized by the pathological process, the breaking of an aneurysm and internal hemorrhage, quite often resulting death, is possible.

 

 QUESTION 8. Clottage, arterial noise. In case of partial clogging of artery it is usually said, that clottage takes place (formation an atheromatosis plaque). On carotid artery, which average diameter dt = 1 cm (fig.), the blood circulates with an average speed v1=20cM x c-1.

For simplicity, it is possible to accept, that the artery is posed horizontally, the gravity of a blood is peer to density of water (103 kg x m-3), and the overpressure inside an unclogged lease of artery

 

Fcompounds P1-P2 = 100 torr (P0 — pressure outside).

1°. What is the minimum diameter d2, at which blood flow is still possible (let's consider, that the

fractionally clogged lease of artery is cylinder)?

2°. What will happen, if diameter becomes less, than d2?

SOLUTION

1 . Let's examine dots a1  and A2, posed on the same horizontal, in which artery diameter is equal accordingly dt, (when absence of a pathology) and d2 (partly clogged artery). Assume that P1,v1  and P2, v2 are pressure and speed of blood in these dots; then, according to the Bernoulli theorem,

Blood flow will exist till P2 is bigger then outside pressure P0 or

Considering that artery is undeformable and blood is incompressible, we shall write a condition of flow conservation:

       or    

The condition of existence of blood flow will be rewritten as

       or     

The substitution of numerical values gives d2 >= 2 mms.

2°. When d2 becomes less than this magnitude, the artery flattens under outside pressure. The pressure pi prolongs thus to increase owing to incessant heart activity (which thus works in conditions of heightened load); blood starts to leak by jerks, and with the help of a stethoscope the discontinuous noise telling about failure of circulation is being listened.

 

ІІ. Seminar discussion of practical work –12:30 - 14:00

 

 

 

Independent work

1.     Oil flows through a 4cm i.d pipe at an average velocity of 2.5m/s.Find the flow in  and

Answer: 3.14

2.     The velocity of glycerin in a 5cm i.d. pipe is 0.54. Find the velocity in a 3cm i.d pipe that connects with it, both pipes flowing full?

Answer: 1.50

3.     How long will it take for 500cm of water to flow through a 15cm long, 3mm i.d pipe, if the pressure differential across the pipe is 4kPa? The viscosity is 0.80Cp

Answer: 7.5s

4.     A molten plastic flows out of a tube of8cm long at a rate of 13when the pressure differential between the two ends of the tube is 18cm of mercury .Find the viscosity of the plastic i.d of the tube is 1.30mm. The density is 13.6

Answer: 0.097

5.     Calculate the theoretical velocity of efflux of water from an aperture that is 8m below the surface of water in a large tank, if an added pressure of 140kPa is applied to the surface of the water.

Answer: 21

6.          A ¾ n. pipe is connected to a ½-in. pipe. If the velocity of the fluid in the ¾-in. pipe is 2.00ft/s. what is velocity in the ½-in. pipe ? How much water flows per second from the ½-in. pipe ?

7.          A 2.50-cm pipe is connected to a 0.900-cm pipe. If the velocity of the fluid in the 2.50-cm. pipe is 1.50 /s what is the velocity in the 0.900-cm pipe ? How much water flows per second from the 0.900-cm pipe ?

8.          A duct for a home air conditioning unit is 10.0 in. in diameter. If the duct is to remove the air in a room 20.0 ft by 24.0 ft by 8.00 ft high every 20.0 min, what must be velocity of the air in the duct be ?

9.          A duct for a home air conditioning unit is 35 cm. in diameter. If the duct is to remove the air in a room 9.0 m by 6.0 m by 3.0 m high every 15 min, what must be velocity of the air in the duct be ?

10.     Water enters the house from a main at a pressure of 1.5 x 105 Pa at a speed of 40 cm/s in a pipe 4.0 cm in diameter. What will be the pressure in a 2 cm pipe located on the second floor 6.0m high when no water is flowing from the upstairs pipe ? When the water starts flowing, at what velocity will it emerge from upstairs pipe ?

11.     A can of water 30.0 cm high sits on a table 80.0 cm high. If the can develops a leak 5.00 cm from the bottom, how far away from the table will the water hit the floor ?

12.     The Vehicle Assembly Building at the Kennedy Space Center is 160 m high. Assuming the density of air to be a constant, find the difference in atmospheric pressure between the ground floor and the ceiling of the building.

13.     If the height of a water tower is 20.0 m, what is the pressure of the water as it comes out of a pipe at the ground ?

14.     A can 30.0 cm high is filled to the top with water. Where should a hole be made in the side of the can such that the escaping water reaches the maximum distance x in the horizontal direction ? ( Hint: calculate the distance x for values of h from 0 to 30.0 cm in steps of 5.00 cm)

III. Testing of knowledges of students – 14:15-15:00.

Author:                         as. Bagriy-Zayats O.A.

as. Palasiyk B.M.,

It is revised on meeting of department

___________2012___             Protocol №_1_