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Fluid Properties-Stresses Viscosity

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Fundamentals of Engineering Exam Review

Georgia Institute of Technology

4.7 (237 ratings) | 32K Students Enrolled

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

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PD

Dec 09, 2018

These are best videos for the Working professional who can not devote much time reading the test material. Nicely explained !

DY

Jan 12, 2016

This is a fantastic course that helped me prepare for the EIT exam. Thanks for providing this and it is greatly appreciated.

From the lesson

Fluid Mechanics

This module reviews the basic principles of fluid mechanics particularly the topics covered in the FE Exam. It first discusses what a fluid is and how it is distinguished from a solid, basic characteristics of liquids and gases, and concepts of normal and shear forces and stresses. The major fluid properties are then discussed. Next fluid statics are addressed: pressure variation in homogeneous and stratified fluids and application to manometers; forces on submerged plane surfaces and buoyancy forces on fully and partially submerged objects.Flowing fluids are then covered. This includes the equations for conservation of mass (the continuity equation) and energy (the Bernoulli equation). These are then applied to velocity and flow measuring devices: the Pitot tube, and Venturi and orifice meters.The final topic is similitude and dimensional analysis. This includes concepts of fundamental dimensions and dimensional homogeneity, the Buckingham Pi theorem of dimensional analysis, and the conditions for complete similitude between a full-scale prototype flow situation and a small scale model.In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 6 hours | Difficulty Level: Medium

Fluid Properties- Introduction6:29

Fluid Properties-Density and Pressure10:37

Fluid Properties-Stresses Viscosity15:46

Fluid Properties-Surface tension10:21

Fluid Properties-Units and other properties4:16

Fluid Statics- Introduction and Pressure Variation12:13

Fluid Statics-Application to manometers and barometers14:55

Fluid Statics-Forces on submerged plane surfaces10:19

Fluid Statics-Forces on submerged plane surfaces continued9:28

Fluid Statics-Buoyancy and stability9:12

Continuity and Energy Equations: Continuity and mass conservation7:45

Continuity and Energy Equations: Energy equation9:12

Continuity and Energy Equations: Energy equation examples9:24

Flow Measurement-Pilot tubes10:51

Flow Measurement-Venturi meter4:27

Flow Measurement-Orifice meter9:17

Flow Measurement-Dimensions and units, Pi theorem12:24

Flow Measurement-Similitude9:26

Flow Measurement-Similitude examples13:06

Taught By

Dr. Philip Roberts
Professor

Transcript

In this segment, I'm going to continue our discussion of fluid properties,

particularly looking at the idea of tangential stresses and

the related property of viscosity.

In the manual, we have definitions of stresses which are given here.

The stress is force per unit area, and

they decompose this into a normal and tangential component.

So if we have some surface here,

let's suppose we have a force at some arbitrary angle to that surface,

we can represent that in terms of a tangential stress and a normal stress.

Stress is decomposed like this, the normal component, the magnitude of the normal

component is just the pressure, and the tangential component is the sheer stress.

And in the handbook, the normal stress is denoted by tau subscript m but

that's not a normal convention because what it is is simply the pressure, or

simply the magnitude of the pressure.

And also somewhat unfortunately they use the symbol uppercase P

to denote pressure which is not commonplace.

Normally we denote pressure by lowercase p.

The tangential ex, stress they denote by tau subscript t,

but again, this is somewhat non-conventional and

typically we simply denote the tangential stress by tau.

And in a particular class of fluids, which will be most important to us,

the relationship between the shear stress and the velocity

is given by this expression, tau is equal to mu, dv by dy,

where v is the velocity, at any height y for, above a boundary.

So dv by dy is called the velocity gradient, and

any fluid which obeys this equation is called a Newtonian fluid,

after Sir Isaac Newton who first wrote the equation in this form in the 17th century.

And in this equation, mu is a fluid property

which is called the dynamic or the absolute viscosity,

which has fundamental dimensions of force times time per area.

So, for example, in metric units the dimensions of viscosity would be Newtons

second per square meter, or in English units pound force second per square foot.

Now a Newtonian fluid follows this equation, however,

more generally, other types of materials follow some kind of relationship where tau

is some unknown function of dv by dy, and

dv by dy is the rate of shearing strain, or the velocity gradient in the flow.

So generally speaking, different materials will have curves which have different

shapes if we plot the shear stress towel versus dv by dy.

And in this graph then the simple fluid, the Newtonian fluid,

has a straight line where the slope of this line is the viscosity mu.

We have other types of materials here, for example this one here,

which is called a shear thickening fluid, where the gradient, or

mu, increases as the strain rate increases.

In other words, the viscosity increases.

The sort of opposite of that is the so-called shear thinning fluid,

where the viscosity decreases as the strain rating increases.

Then we have some materials which follow this curve here, where it's,

it behaves essentially like a solid for slow rates of for

small stresses, but then begins to flow like a fluid.

And the material which follows that curve is called a Bingham Curve or

a Bingham Plastic or an Ideal Plastic.

And these materials are discussed in this sidebar in the FE Handbook.

However, we don't need to worry about all of these materials.

The only types of fluids that we will be concerned with here

are Newtonian fluids which follow this simple equation here and

fortunately, all of the fluids that we're interested in obey this relationship.

For example, oil, water, gasoline, air, et cetera, are all

Newtonian fluids so we don't have to worry about the more complex fluid behaviors.

A closely related property is the kinematic viscosity,

which we give a symbol mu, the greek letter mu, and the kinematic viscosity

is the ratio of the dynamic viscosity divided by the density of the fluid.

This is called the kinematic viscosity and

it has fundamental, fundamental dimensions of length squared by time.

For example, either meters squared per second or feet squared per second.

Some other, some other important behavior of fluids are that, for

liquids, the viscosity decreases with temperature.

In other words, it becomes more runny, if you like.

But conversely, for gases, as we heat up gases,

the viscosity increases as the temperature increases.

So some examples, water flows in a pipe,

the viscosity is 8 times 10 to the minus 4 Newtons per square meter,

and the density is 998 kilograms per cubic meter.

The kinematic viscosity is mostly nearly which of these alternatives.

So to find out, we simply apply our definition of kinematic viscosity,

mu is equal to mu over rho, is equal to 8 times 10 to the minus 4

divided by the density we take as 998 is equal to 0.8 times 10 to the minus 6, and

the answer is A is the closest answer.

Remembering that a Newton is the same as kilogram times a meter per second squared.

Now a common problem that comes up in viscosity is this one,

where we have a plate which is flowing, moving over a stationary

boundary at a constant speed and the lower boundary is fixed.

The gap in between here is filled with some fluid, for example oil, or water or

whatever and a typical question is what is the force required to

keep the plate moving at a constant speed and what is the sheer stress in the fluid.

Very important properties of real fluids,

in other words fluids which have viscosity is that they stick to a solid boundary.

So the velocity of the fluid at the lowest boundary here

is the same as the velocity of the plate which is zero.

Similarly, at the upper boundary here, the fluid is sticking to the plate so

the velocity of the fluid there is equal to

the velocity of the plate which is capital V.

It turns out in this case that

the distribution of velocity in the fluid, in the gap, is linear.

It increases linearly from zero at the lower plate

up to the moving plate velocity at the upper plate.

So, in that case, the sheer stress,

which is given by our Newtonian formula, tau is equal to mu du by dy.

Because the velocity increases linearly du by dy is constant,

so tau is given by this symbol expression, is equal to mu u over b.

The shear stress, tau, is force divided by area,

so conversely, the total force on the upper plate is equal to tau times area.

So the total force to keep the plate moving is equal to tau times A.

An example on that, we have a plate which is moving like that, and we're given that

the plates are spaced 5 millimeters apart and the space between them is filled with

oil with a specific gravity of 0.92, and kinematic viscosity as shown.

The upper plates is moving at a speed of 1.0 meter per second,

the shear stress on the upper plate is most nearly which of these alternatives?

So, we just apply the equations from the previous slide.

The velocity distribution is linear, so tau is equal to mu u over b.

But, in this case, we're given the kinematic viscosity so

we have to calculate the dynamic viscosity.

And we're given the specific gravity, therefore,

we have to calculate the density.

So, firstly, the fluid density is the specific gravity multiplied by the density

of water, is equal to 0.92 times 1,000, or 920 kilograms per cubic meter.

The dynamic viscosity, mu, is mu times rho,

is equal to 0.524 Newton second per square meter.

And now we can plug into our formula for sheer stress, tau is equal to mu u over b,

is that calculating that is equal to 104.8 Newtons per square meter, or Pascals.

So the answer is B, 105 Newtons per square meter.

Now very similar problems to this are, are given sometimes like this,

in this case, we have a round shaft which is being pulled through a bearing

at a constant speed with a small gap in between.

So these kinds of problems are very similar, here's a specific one.

We have a 25 millimeter diameter solid shaft,

pulled through a cylindrical bearing.

The gap between, the bearing and the shaft is 0.3 millimeters, and

it's filled with an oil whose viscosity is given, and specific gravity is given.

What is the force needed to pull the shaft at a constant speed

of three meters per second?

Which of these alternatives?

So for these problems, we assume that the behavior

of the fluid in the gap here is the same as the behavior of the fluid for

the flat plate being pulled over a fl-, over a moving plate.

A moving plate being pulled over a stationary plate.

So that the velocity distribution in the gap here is again linear

which is like this.

So all of the same equations apply, so tau again

is equal to mu u over b and the procedures are the same for the last problem.

First calculate the density, it's 910 kilograms per cubic meter.

Then the dynamic viscosity,

the kinematic viscosity multiplied by the density is 0.728.

Then we can calculate the sheer stress, mu u over b is equal to,

in this case, 7,280 Newtons per square meter.

And the last step is that the force required

is the shear stress multiplied by the area over which it's applied.

And in this case, the area is the curved surface of the cylindrical shaft,

so the surface area is the circumference, which is pi d, multiplied by the length.

So, plugging in the numbers, that is 0.039 square meters,

and the force required is tau times A, is equal to 284 Newtons,

and the final answer, then, is C.

Similar kinds of problems look like this, commonly comes up

where we have a block which is sliding at a steady speed down an incline.

And in this case, and the gap in between the block and the plate is filled with

some oil, questions would be typically, what is the steady state speed aquired.

And in this case we just do a force balance, the force is acting on the block

ate the weight of the block which is acting straight downwards, and

the component of this weight force in the down incline direction is w sine theta.

And this is balanced by the friction force

due to the shear stress in the oil which is tau times A.

So a simple force balance gives us w sign theta is equal to tau times A, where

tau we compute in exactly the same way as we did in the previous two examples.

Now, that linear velocity distribution is a particular distribution which applies

only to the problem of, a moving plate flowing or moving over a stationary plate.

For flows which are driven by pressure drops,

which we'll look at later, the velocity distribution is a little bit more complex.

And in particular, for laminar flows which again, we'll look at later,

the velocity distribution is typically parabolic like this, with a zero velocity

at the plates or the boundaries and a maximum velocity on the center line.

So here's a typical problem, the velocity distribution in the oil that

is flowing in the gap between two parallel plates is given by this expression,

u is equal to 3V over 2, 1 minus y over h squared,

where V is the maximum velocity that occurs on the center line.

Y is the height measured from the center line and h is the half width of the plate?

We're given that the distance h is 2 inches maximum velocity

is 2 feet per second and dynamic viscosity is 0.04.

Question is, what is the shear stress on the bottom wall?

Is it most nearly which of these alternatives.

So, to solve this problem,

we apply our basic relationship between shear stress and the velocity profile.

Tau is equal to mu du by dy, but, in this case,

because the velocity is varying, in a nonlinear manner,

we have to differentiate this expression with respect to y du by dy.

And the answer is minus 3 Vy over h squared, or

tau at any point in this gap is equal to minus 3 mu V y over h squared.

On the bottom wall in particular, which is what we're looking at,

y is equal to minus h, so the shear stress on the bottom wall is three mu V over h.

Plugging in the numbers, that is equal to 14.4 pounds force per square foot,

so the answer is A, 14.4 pounds per square foot.

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