Fluid Statics-Application to manometers and barometers

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From the course by Georgia Institute of Technology

Fundamentals of Engineering Exam Review

219 ratings

Georgia Institute of Technology

Fundamentals of Engineering Exam Review

219 ratings

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.

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

Meet the Instructors

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

I want to continue my discussion of fluid of statics now by turning to applications

of the hydrostatic pressure equation to manometers and barometers.

Firstly, barometers are devices which have long been used to measure

atmospheric pressure.

And typically we do this by measuring the height of a column of a fluid,

which is typically mercury.

So a mercury barometer is shown here.

And we can relate the height of the mercury column to atmospheric pressure

by applying hydrostatics.

So, for example, at this point here the pressure there is

the same at those two points because the horizontal level is the same,

and remember the pressure on many horizontal plane is constant.

So therefore the pressure at this point is the same as the pressure here

which is atmospheric pressure.

And the pressure here is the pressure due to the height of this column of mercury

which is specific weight of mercury times h

plus the vapor pressure of the mercury above here.

And the vapor pressure of mercury is very low so we can effectively neglect it.

So basically the equation is that atmospheric pressure is

equal to the specific rate of mercury times the height of the fluid column.

To show what that would be for typical values,

we recognize that the specific gravity of mercury is about 13.6.

In other words, it's 13.6 times heavier than water, so

the density of mercury is 13.6 times the density of water which

is about 1000 kilograms per cubic meter which is 13600 kilograms per cubic meter.

Or the specific weight of mercury is the density times

the acceleration due to gravity which is about 133 times 10

cube newtons per cubic meter, or 133 kilonewtons per cubic meter.

Standard atmospheric pressure is approximately 101 kilopascals absolute.

So the height of the mercury column is atmospheric pressure

divided by the specific weight of mercury which is therefore 101 times 10 cubed

divided by 133 times 10 cubed is 0.759

meters, or, approximately 760 millimeters.

And, if we did the corresponding calculation assuming that this column here

was water, the corresponding height would be 34 feet which is obviously very

impractical which is why we typically use mercury as the fluid in a barometer.

And to illustrate the difference between the two, pressure due to a column

of 760 millimeters looks like this is approximately 34 feet tall.

Manometers are devices which are used to measure pressures of fluids by observing

the height of fluid columns.

Consider, for example, I have a pipe or a vessel here, and

I'm interested in the pressure in this pipe, which we'll denote by PA.

It's connected to a manometer, and where the manometer contains some

different fluid, the manometer fluid whose specific weight is gamma 2.

Now to analyze cases where we have columns of different densities,

let's consider a simplified example here.

So here I have an open column here which is filled with gasoline at the top and

it's overlaying an immiscible fluid, water, so the two fluids don't mix.

What is the pressure at the bottom of these two columns?

Well, to find that first of all we

apply hydrostatics through the gasoline from this point to this point.

And the difference in pressure between these two points, P1 minus P0 is

just the pressure which is due to this column of gasoline,

which is equal to minus gamma G times the height difference, Z1 minus Z2.

Similarly, I apply hydrostatics to the water column from the interface here

to the bottom of the water column and the equation there is the difference in

pressure, P2 minus P1 is minus gamma water times the height difference Z2 minus Z1.

Next I add these two equations together and

when I do that the pressure at the interface P1 cancels

out that leaves me with P2 minus P0 equals is to minus

gamma gasoline Z1- Z0 minus gamma water Z2 minus Z1.

Or rearranging, the pressure at the bottom, P2.

P2 is equal to the pressure at the top,

P 0, plus the pressures due to the two columns of liquids,

gamma G times Z node minus Z1 plus gamma water Z1 minus Z2.

In other words, very simply, all we do

is add the pressures due to each column of fluid when we have different fluids.

So, in this example, if the specific gravity of gasoline is 0.68,

I'll leave you to do these calculations such an example.

The pressure at the bottom here P2 is 6.31 psi.

In other words pounds per square inch gauge above atmospheric pressure.

If we have a single fluid, the equation is even more simple, and in this case,

the pressure here is just equal to the pressure due to the column of fluid here,

which is Gamma 1 times h 1 plus the pressure at the open surface here,

which is atmospheric pressure.

For high pressures or

gasses, we typically use a U-Tube manometer like on the previous slide.

Which looks like this.

And all of these manometer problems can be solved very simply,

by following a few general suggestions or rules that I'll give here.

And the suggestion is this: start at one end and write the pressure there.

For example, in this case PA.

And then, work your way through all of these different fluids to the other end

where we equate the pressures due to each fluid column.

And we add the pressure if the next meniscus is higher, subtract the pressure

if the next meniscus is lower and then finally add the pressure at the other end.

And if you follow these simple rules, you can solve any complex manometer problem.

In this case, we have PA, write the pressure at the one end

PA and then go from here to this meniscus.

This meniscus here is lower than this so

it's going to be negative and the hydrocenic pressure due to each column

Is just the specific weight multiplied by the height difference.

So it's minus gamma 1 times h1.

Next, we go through the blue fluid here, the manometer fluid,

to here passing straight through the 0.3 because that's irrelevant.

This meniscus is higher than this, so

it's positive gamma two times the height difference, which is h2.

And then finally adding the pressure at the open end here

which is atmospheric pressure.

So this is the equation that would apply to that particular example.

Now, let me do some numerical problems.

So, in this case, we have a closed tank, which is filled with water,

above here we have air and the pressure gauge is reading 7 psi.

If the specific weight of the water is 62.4 pounds per cubic foot,

then the height, h, of the water column is most nearly which of these alternatives.

So here's the solution.

We apply that simple rule that I just applied.

So start from this point and write the pressure there P air.

Go through the water column from here up to here

and this meniscus is higher than the free surface here so

that is equal to gamma times the height difference which is h minus 2, and

then add the pressure at the free surface there which is atmospheric pressure.

So this is our equation.

So rearranging, in this case, we want to solve this for h.

So we have h is equal to 2 plus P air over gamma.

The pressure of the air here were given a 7 pounds per square inch and

convert that to pounds per square foot multiply by 144.

So therefore, h is equal 2 plus 1008

divided by 62.4 is equal to 18.2 feet.

So the closest answer is C, 18.2 feet.

Another example, a little bit more complicated,

the closed tank contains a liquid with a vapor above it.

The liquid density here is 800 kilograms per cubic meter and

the vapor density is negligible.

The pressure at the top of the liquid in the tank is 120 kilopascals absolute,

atmospheric pressure is 101 kilopascals.

Specific gravity is given and density of water is given.

Question is, the reading h here on the mercury manometer

is most nearly which of these alternatives?

So again the solution, just follow those simple rules that I just gave you.

Start at this point here, and

the pressure at this point which I'll call Pv is equal to, so

first we go through the liquid here all the way through here to here.

Just go through the continuous liquid and

this meniscus is lower than this meniscus so it's going to be negative.

And the hydrostatic pressure is minus gamma l, the specific

weight of the liquid, multiplied by this height difference which is one meter.

Then we go through the mercury column from here to here.

This meniscus is higher so it's going to be positive and

the hydrostatic pressure is gamma mercury, gamma hG, times the height of h.

And then finally, add on the atmospheric pressure at the free end.

So, rearranging this equation for h, which is what we're trying to find, we get this.

And remembering that the specific weight gamma of mercury is

equal to the specific gravity SG multiplied by the density, HG,

we have this equation, And now, we can plug in the numbers.

The pressure here is 120 kilopascals, 120 times 10 cubed plus, etc.

And plugging in the numbers, the answer is h = 0.201 m or 201 mm.

So the best answer is D, 200 millimeters.

Notice in this problem that I work this through in absolute pressures

because we gave the atmospheric pressure here as an absolute pressure.

You can work these problems as either gauge pressures or absolute pressures.

You will get the same answer.

However, you must be careful not to mix the two.

Work only in absolute pressures or only in gauge pressures.

The final example, we have an inverted U-Tube manometer

containing an oil which has a specific gravity of 0.9.

The pressure difference between the two pipes here, A and B,

is minus five kilopascals, and

the specific weight of water is 9800 newtons per cubic meter.

The reading age here where age is the difference between these two

levels is most nearly which of these alternatives?

So again follow same procedure.

We start here at the left-hand side.

So I write the pressure here as Pa, then go through here to this meniscus.

Okay.

So, that is equal to the pressure due to this column of fluid here

which is positive because this meniscus is higher.

Specific way to times the height which is 0.2.

Then I go through the oil here, from here to here.

This meniscus is lower so it's going to be negative.

And the pressure is gamma o, specific weight of the oil times height times h.

Finally, I go from here to here.

Again this meniscus is lower so it's going to be negative here.

Specific weight of the water times 0.3, and

finally plus the pressure at the other end which is PB.

So rearranging,

I get the pressure difference PB minus PB is given by this expression.

And, again, I'm interested here in the barometer reading h.

Which is equal to this so rearranging and I get this expression.

Plugging in the numbers, PA minus PB is equal to we're given minus

five kilopascals minus five times ten cubed plus the specific weight of water

which is 9800 newtons per cubic meter multiplied by the height which is 0.1.

And the answer is 0.456 meters or 456 millimeters.

So the closest answer is A which is 460 millimeters.

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