CEEN598D Week 2 Statics - Inside Minesinside.mines.edu/~lbearup/CEEN598DF14/2.Notes.pdfPressure p is...

CEEN 598D– Fall 2015

Lindsay Bearup [email protected] Berthoud Hall 121

STATICS CEEN 598D: Fluid Mechanics for Hydro Systems


From your homework: What is bigger:

kg or a slug? lb or a N?


Fluid Statics •  Fluids that are in a hydrostatic condition

–  Hydrostatic Condition: means all fluid particles are in force equilibrium with one-another (i.e., at rest)

•  Calculate forces due to distributions of pressure

Pressure: normal force per unit area

•  Equilibrium of a fluid particle and surroundings –  Fluid particle: body of finite mass and internal structure

but negligible dimensions


Hydrostatic Condition


Fluid Statics For fluids at rest, only normal forces (non-shear forces) exist which act over the entirety of each boundary surface. These are called pressure forces. Pressure p is a scalar – we can check by calculating the pressure in every direction for an odd-shaped volume.

dAdF

AFp

A=

Δ

Δ=

→Δ 0lim


Pressure: Absolute, Gage and Vacuum

•  Pressure in a perfect vacuum is considered absolute zero.

•  All pressures relative to absolute zero are absolute pressures.

•  Sea level standard pressure is 101 kN/m2 or 101 kPa as an absolute pressure.


Measuring Pressure

•  Most pressure monitoring devices measure the difference in the relative pressure, with atmospheric pressure being the reference pressure. –  This is known as gage pressure. –  0 psi or 0 kPa gage pressure is atmospheric absolute

pressure ~ 101 kPa.

•  Absolute pressure can be less than atmospheric pressure, so the gage pressure is negative. –  This is called the vacuum pressure.


Pressure Transmission

•  Fluids act as a continuum where forces/deformations are translated throughout the fluid system.

•  Pascal’s Law: In a closed system, a pressure change at one point in the system will be transmitted throughout the entire system (Not instantaneously, however).

•  In watersheds, which is faster – fluid flow or pressure propagation? What are the implications of this?


Pressure Variation with Elevation

As you move down the water column, the pressure increases due to the weight of more and more overlying fluid. For water this is called hydrostatic pressure.

pA = 0 gage

A

z

zp p zγ= +


Pressure Variation with Elevation

Take an element of fluid:

A

Δz

zAW Δ−= γ

App )( Δ+−

pA

γγ

γ

γ

−=−=Δ

Δ

Δ−=Δ

+Δ+−Δ−==∑

dzdp

zp

pz

pAAppzAFz )(0

Fp

Fp Fw


Solve ODE for General Solution

Czp

dzdpdzdp

+−=

−=−=

γ

γγ

IF THE DENSITY IS CONSTANT •  Or, slightly rearranged, p/γ + z = constant. •  The quantity is called the piezometric head

H =p/γ + z

separate and

integrate

*Later we will see that H is the total head and also includes a velocity component, but for now our fluid is at rest.


Piezometric Head

1

2

z1

z2

22

11 zpzp

+=+γγ

Since this quantity is a constant in a static water body, we can apply this formula at any two points:


Piezometric Head

•  QUICK - What are the units of the piezometric head?

•  What is it measuring, or keeping track of?

p zγ+


Piezometric Head

•  The last term is the potential energy of the packet of water with volume, V.

•  Potential energy is a function of an objects position in space or arrangement of parts.

mgH =Vp+mgz


•  A decrease in elevation is accompanied by an increase in pressure. This is a simple conversion of energy.

•  Take a packet of water deeper in a pool and the potential energy has been converted to pressure energy. Water under more pressure has the ability to do more work than water under less pressure – recall the car jack!

mgH =Vp+mgzPiezometric Head


Let’s Summarize:

•  Piezometric Head

•  Hydrostatic Pressure

H =pγ+ z

zp p zγ= +pA = 0 gage

A

z


Converting One to the Other

Recall, for constant H:

pz = Hγ

But what does this all mean?

€

Δp = −γΔz2

21

1 zpzp+=+

γγ


Measuring Pressure: Bourdon-Tube Gage

•  Measures pressure by sensing deflection of a coiled hollow tube

Pressure in tube balanced by spring rate of the tube


Problem

For the closed tank with Bourdon-type gages tapped into it, what is the specific gravity of the oil and the pressure reading on gage C?


Problem: Solution Part 1

Solve for γoil assuming γair is small (true value is 11.8 N/m3)

€

Δp = −γΔz

€

pa − pb = −γΔz

€

50000 N/m2 − 58530 N/m2 = −γ oil (1.0 m)

€

−8530 N/m3 = −γ oil

€

γ oil = 8530 N/m3

From Converting One to the Other Slide:


€

pc = 58530 N/m2 + 8530 N/m3(0.5 m) + 9800 N/m3(1.0 m) = 72595 N/m2

Problem: Solution Part 2

Solve for pc by stepping from point C to point B: Pressure must be same at interface (bottom of oil) Substitute for pbottom_oil

€

pbottom _ oil − pc = −γwaterΔz

€

pb − pbottom _ oil = −γ oilΔz

€

pc = pb + γ oil (0.5m) + γwater(1.0m)

For water at 10◦C


Measuring Pressure: Aneroid Barometer

http://usatoday30.usatoday.com/weather/waneroid.htm


Measuring Pressure: Barometer

•  Used to measure atmospheric pressure

•  Most common types: mercury barometer and aneroid barometer

€

patm = γ Hgh + pv ≈ γ Hgh


Measuring Pressure: Piezometer

•  Vertical tube in which liquid (flowing or not) rises due to pressure in the pipe.

Where would you see something like this?


Measuring Pressure: Pressure Transducers

•  Converts pressure to electrical signal


Measuring Pressure: Manometer

Pressure device that uses specific weight (or density) of a fluid to measure pressure.


Manometry

•  Pressure and elevation heads are transferrable in static liquids.

•  Allows us to construct very simple pressure gauges called manometers.

•  Attach a fluid-filled tube to the system in question and measure the height of the fluid.


U-Tube Manometer


hpp mγ+= 12

23 pp =

lhplpp

m γγ

γ

−+=

−=

04

34

Start at the open end and go around the horn:

U-Tube Manometer


There’s almost no need for a general formula, but if you wish to, define a break point at every fluid change and find the subsequent elevation around a bend that is across from the break (if it exists). Then, •  By the way, either pstart or pend must be known the solve

the other. •  If you only care about the pressure difference in a pipe,

then you can use a differential manometer.

∑∑ −+=ups

jjdowns

iistartend hhpp γγ

U-Tube Manometer


U-Tube Manometer with Two Fluids

1

2

3 4 h

γm

γ

5

6

Go round the horn in your head. •  From 1 to 2, as much pressure

is added as is lost eventually going from 5 to 6 (those segments cancel out).

•  Pressures at 3 and 4 are equal.

•  From 2 to 3 add γmh and from 4 to 5 we lose γh.

So:

)(16

16

γγ

γγ

−=−

−+=

m

m

hpphhpp


Uniform Pressure Distribution

•  Uniform pressure distribution –  For a panel or a plane or flat surface of arbitrary

shape the description of the pressure along all points along the surface is called a pressure distribution.

–  If this pressure is the same at every point, it is called a uniform pressure distribution.


AF pdA pA= =∫


Hydrostatic Pressure Distribution

When the pressure distribution is produced by a fluid in hydrostatic equilibrium the pressure distribution is called a hydrostatic pressure distribution.


Another Representation

( sin )

sin

sin

A

A

A

c

c

F pdA

y dA

ydA

y A

p A

γ θ

γ θ

γ θ

=

=

=

=

=

∫

∫

∫

pressure at the centroid

To find the total force on the plane, integrate pdA

Integral of distance times area is the first moment of the area.


Or, another way

where h=y sinθ for constant γ and θ. The integral is the first moment of the area with respect to the x-axis and can be defined as:

where yc is the y-coordinate of the centroid of the area A.

sin sinA A A A

F pdA hdA y dA ydAγ γ θ γ θ= = = =∫ ∫ ∫ ∫

€

ydAA∫ = ycdA


where hc is the vertical distance from the fluid surface to the centroid of the area.

sin c cF y A h Aγ θ γ= =

Or, another way


Example Concrete (23.6 kN/m3) poured into a vertical form 2.44m high by 1.22m by x = 0.5 m thick exerts what force?

2

23

1.22sin 90 1

2.44 1.22 3

23.6 1.22 3 85.7

( 19,000 )

cy m

A m m mkNF m m kNmlb

=

° =

= × =

= × × =

≈

If the thickness x doubles, how does the force change? x

2.44

m


Line of Action

•  In the last slides, we calculated the point at which the effective pressure is calculated (i.e., the centroid) to find the magnitude of the force on a submerged plane.

•  Is this the point at which the force appears to be applied?


Example

yc

x

2.44

m

yR

1.22

m

1.63m

0.81m

For the previous example, the resultant force for a triangular distribution (incompressible fluid) will act h/3 up from the bottom.


•  Coordinate of resultant force can be determined by a summation of moments around x-axis.

•  Since F=γΑycsinθ we can cancel the RHS

2

xAR

c c

y dA Iyy A y A

= =∫

2sinR A AFy ydF y dAγ θ= =∫ ∫

Moments

Second moment of the area OR Moment of inertia


Parallel Axis Theorem: Used to determine the moment of inertia about any axis given the body’s moment of inertia about a parallel axis through the centroid and the perpendicular distance between the axis. where Ixc is the second moment of the area with respect to an axis passing through the centroid and parallel to the x-axis. Inserting into previous equation:

So, resultant force does not pass through the centroid but always below it since Ixc/ycA > 0.

Parallel Axis Theorem

€

Ix = Ixc + Ayc2

xcR c

c

Iy yy A

= +


Let’s go over one more time!

•  The position of the “line of action” of a hydrostatic force lies below the centroid of a plane surface.

•  This location is called the center of pressure. Call the distance along the plane surface, yR (R for resultant).

2( sin ) sinRA A A A

y F ydF ypdA y y dA y dAγ θ γ θ= = = =∫ ∫ ∫ ∫


•  Insert the parallel axis theorem:

•  The above expression is finding the point at which the

total pressure force appears to be applied. •  Insert the magnitude of the total force:

2sin ( )R xc cy F I y Aγ θ= +

2

2

sin sin ( )R c xc c

xc c xcR c

c c

xcR c

c

y y A I y AI y A Iy yy A y AIy yy A

γ θ γ θ= +

+= = +

− =

sin cF y Aγ θ=


First, the magnitude of the hydrostatic force is , where pc is the pressure at the centroid, which is at a depth of 10m.

cF p A=

3

6

10 9810 2 2.51.541 10

F m N m m mN

π= × × × ×

= ×

Example What is the force F required to open the gate?


The slant distance is calculated using similar triangles (no need to know α): The sum of torque moments around the hinge is zero:

The slant distance to depth ratio maintains the ratio 4:5.

mymy 5.12

45

10=∴=

mmm

ya

AyIyycp 125.0

)5.12(4)5.2(

4

22

====−

5 4 Fpush

F

€

Fpush × 5m − Fp × (2.5 + 0.125)m = 0

Fpush =1.541×106N 2.625m5m

= 8.09 ×105N


Problem

Given: A 4-m diameter circular gate is located in the inclined wall of a large reservoir containing water (γ=9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter and the water depth is 10m above the shaft. Find: a)  The magnitude and location of the resultant force

exerted on the gate by the water. b)  The moment that would have to be applied to the shaft

to open the gate.

CEEN598D Week 2 Statics - Inside Minesinside.mines.edu/~lbearup/CEEN598DF14/2.Notes.pdfPressure p is...

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