Fluid Mechanics (1TV024)! Hydromechanics

Fluid Mechanics (1TV024)!
!
and!
!
Hydromechanics (1HY125)!
!
Spring 2014!
Fluid Mechanics
!
!Hydromechanics
!!
1TV024, 5 hp !
!
!1HY125, 5hp!
!
Instructor:!
Chris Hieronymus!
Office: Geocentrum Dk255!
Phone: 471 2383!
email: [email protected]!
!
Literature:!
Engineering Fluid Mechanics!
by Clayton T. Crowe, Donald F. Elger, Barbara C. Williams and
John A. Roberson!
9th edition, 2010 (alternatively 8th ed., 2005)!
Wiley & Sons, Inc., Hoboken, NJ, USA!
!
Mandatory components of the course:!
Grades will be based on final examination.!
Attendance at Lab Exercises towards end of term is mandatory. !
Lab report to be handed in is also mandatory.!
Further Information:!
Additional information regarding schedule, assignments, etc. can be
found at!
http://studentportalen.uu.se!
Past student evaluations of this course are also on studentportalen.!
The exam will be closed-book. In other words, you cannot use your
book or your notes during the exam. However, you may (and should!)
prepare a cheat sheet to bring to the exam. One page (A-4, doublesided) of equations and whatever you think might be useful. (A list of
equations will also be provided with the exam. But I think you ll learn
more if you make your own.)!
Grading:!
!
The goals of this course are quantitative.!
At the end of the course the student is expected to be able
to solve specific problems of fluid mechanics. Examples of
past examinations are given at
http://studentportalen.uu.se.!
!
GradingScale:!
!
80%– 100% !
!5!
60%– 79.9% !
!4!
40%– 59.9% !
!3!
0%– 39.9% !
!U!
!
!
!
Course Outline!
•  Fluid Statics (Chapter 3)!
•  Flowing Fluids and Pressure Variation (Chapter 4)!
Chapter 3!
!
FLUID STATICS!
•  Control Volume Approach and Continuum Principle
(Chapter 5)!
•  Momentum Principle (Chapter 6)!
•  Energy Principle (Chapter 7)!
•  Surface Resistance (Chapter 9)!
•  Flow in Conduits (Chapter 10)!
•  Varied Flow in Open Channels (Chapter 15)!
Fluid Mechanics, Spring Term 2014!
Shear Forces!
Normal Forces!
(pressure)!
where F is a force normal to area A!
Flow of an
unconfined viscous
fluid down an
incline.!
•  Flowing viscous fluid exert shear forces.!
•  Static fluids only exert normal forces.!
•  Moving fluids (dynamics) will be covered later.!
Pressure is a scalar quantity!
Figure 3.1 (p. 31)!
Force balance in the z-direction:!
Force balance in the x-direction:!
Vertical force
on ΔA!
From last slide:!
Divide through by
Vertical force
on lower
boundary!
Total weight of wedge
element!
= specific weight!
For your Culture (i.e., not required for this course…)!
to get !
=!
Now shrink the element to a point: !
This can be done for any orientation α, so!
It is possible to have different normal stresses.!
Consider a small cubic fluid element that is part of a
larger fluid mass:!
Different normal forces in one (coordinate) orientation
are equal to shear forces in another orientation.!
(cont d:)!
!
So what is the pressure then?!
!
•  Pressure is the average of the normal forces acting
at a point.!
•  Differences between normal forces are due to fluid
motion.!
Fy!
Pressure Transmission!
In this case, if the force vectors are
equal in magnitude, then!
Fx!
-Fx!
Hydraulic Lift!
!
p = 0!
In a closed system, pressure changes from one point
are transmitted throughout the entire system!
(Pascal s Law).!
-Fy!
Absolute Pressure, Gage Pressure, and Vacuum!
Figure 3.4 (p. 36)!
Example of
pressure relations!
Pressure Variation with Elevation!
Static fluid:!
All forces must
balance as there are
no accelerations.!
!
Look at force balance
in direction of !`
Figure 3.5 (p. 37)!
•  Pressure in a vacuum is p = 0.!
•  Absolute pressure is referenced to perfect vacuum.!
•  Gage pressure is referenced to another pressure,
typically atmospheric pressure (most gages measure
relative pressures). !
Pressure Variation for a Uniform-Density Fluid!
From figure, note that!
The pressure-elevation relation derived on the previous slide,!
Shrink cylinder to
zero length:!
is perfectly general (applies also to variable γ).!
!
But if γ is constant, the above equation is easy to integrate:!
(from previous slide)!
The quantity
!
or!
is known as the piezometric pressure and!
is called the piezometric head.!
Example: What is the water pressure at a depth of 35 ft?!
For an incompressible fluid, γ is constant.!
Pressure and elevation at one point can thus be related
to pressure and elevation at another point:!
for!
With the information given,
all we can calculate is the
pressure difference between
points 1 and 2.!
!
!
or!
(Do yourself a favor and work in SI-units!)!
Example 3.3: What is the gage pressure at point 3 ?!
Pressure Measurements!
Two step solution:!
1)  Calculate!
2)  Calculate !
Figure 3.7 (p. 42)!
Figure 3.6 (p. 41)!
Piezometer or !
simple manometer!
(relative to atmospheric pressure at point 1)!
U-tube manometer!
!
Better for higher pressures.!
Possible to measure pressure
in gases.!
Find pressure at center of pipe:!
The complete path from
point 1 to point 2 may
include several U-tubes.!
!
In general:!
Can start either at open end
or inside pipe.!
!
!
Here we start at open end:!
p at
open
end!
From example 3.7!
Change
in p from
1 to 2!
Change
in p from
3 to 4!
p in
pipe!
Example 3.8: Find the change in piezometric pressure
and in piezometric head between points 1 and 2.!
Differential Manometer!
Figure 3.12 (p. 47)!
Used for measuring pressure differences between
points along a pipe.!
( from
)!
Hydrostatic Forces on Plane Surfaces!
The
to give!
cancel out
(piezometric pressure)!
(piezometric head)!
The white area AB
in the figure is a
plane of irregular
shape.!
Line A-B is an edge
view of that area.!
!
What is the net force
due to pressure
acting on the sloping
plane AB?!
Fig. 3.16!
First, note that hydrostatic pressure increases along y as!
(since y is not vertical) !
This figure is absolutely
awful!
!
Line AB represents the
true location of the
surface.!
!
The white surface is not
drawn in its actual
location.!
Line 0-0 is horizontal; the white area has been rotated about
axis A-B from its proper location. In other words, the apparent
depth of the white area within the fluid is not as it appears.!
From the definition of pressure:!
or!
so that the total force on a plane area A is!
F =
)
F =
Z
Z
p dA
A
y sin ↵ dA
A
or, since γ and sin α are constants!
F = γ sin α
The first moment of an area (that is, the mean of y over an
area A) is defined as!
so that the total force can be written as!
!
y dA
A
Vertical Location of Line of Action of Resultant
Hydrostatic Force!
In English: We wish to represent the distributed pressure
force by an equivalent point force. Where (in the vertical)
does that force act?!
where p̄ is the pressure at the centroid of the area.!
The boxed equation is known as the hydrostatic force
equation.!
We have thus replaced an integral involving a variable
pressure by a constant resultant pressure:!
Similar problem: 2 weights on
a beam supported at ycp!
So, for the moment about a point at ycp we have!
The book just refers to the parallel axis theorem to write!
Mathematically, I think it is easy to see that using!
But with
and
we get!
The integral on the right-hand side is the second moment of
the area (about point y=0):!
!
!
!
If we can evaluate this integral, then the problem is solved!!
Notice that the last term is zero because!
As an aside, you may recall that!
Identifying
as the 2nd moment about y=0!
and
as the 2nd moment about!
!
(= moment of inertia about point
)!
we have thus proved the parallel-axis theorem:!
!
The moment of inertia of an object
about an axis through its center of
mass Icm is the minimum moment
about any axis in that direction.!
!
The moment about any other
parallel axis is equal to Icm plus the
moment of inertia about distance d
of the entire object treated as a
point mass located at the center of
mass.!
Our system of pressures has nothing to do with rotations, but
the equations are of the same form…!
We thus have:!
Back to the problem at hand:!
Recall from a few slides ago that!
or:!
so that!
Why is this result useful for us?!
!
is something that is unique to a given area; in other
words, can be tabulated for different shapes.!
!
!
is not unique but depends on choice of!
!
y=0. So it cannot be tabulated.!
!
Example 3.10: !
or!
or!
Note that at great depth
, the difference between
the centroid and the center of pressure gets very small.!
Now calculate the slant distance between
ȳ and !
Find the normal force
required to open the
elliptical gate if it is
hinged at the top.!
!
First find Ftotal, the total
hydrostatic force acting
on the plate:!
The slant distance to the hinge is 8m x 5m/4m = 10m, and the
slant distance from the hinge to the centroid is 2.5m. Hence, !
With
The two moments about the hinge must add to zero:!
(Appendix p. A-5) we get!
Hydrostatic Forces on Curved Surfaces!
We could integrate the vector forces along segment AB, but it
is often easier to find equivalent forces on a free body as
illustrated above.!
FAC acts at the center of pressure as from previous section,!
FCB acts at centroid of area CB, and W acts at the center of
mass of the free body ABC.!
Example 3.11:!
Find magnitude and line of
action of equivalent force F.!
!
Force balance in x and y:!
The line of action of the horizontal force is!
Where we just read
directly off the figure.
!
The line of action for the vertical force can be found by
summing the moments about C (or any other point…)!
(notice that we could add a constant to every x-coordinate
since
)!
From Appendix p. A-1 (Figure A.1):!
The complete result is summarized below:!
Distance from C to centroid is:!
So that xcp is found to be!