Fluid Mechanics Hydrostatics
Fluid Mechanics Hydrostatics
Fluid Mechanics Hydrostatics
ENGINEERING STUDENTS
Notes on
-FLUID MECHANICSHydrostatics
The nature of Fluids and the study of Fluid
Mechanics, Forces due to Static Fluids,
Buoyancy and stability Density.
Prepared by Tshuma T
F
A
Fluid Properties
Other fluid properties are specific weight, density, specific gravity,
surface tension and viscosity.
It is also important in determining the character of the flow of fluids and
the amount of energy that is lost from a fluid flowing in a system.
Similarly
SI Unit Prefixes
Table 1.1 shows the SI unit prefixes.
We can also compute the mass of an object if we know its weight. For
example, assume that we have measured the weight of a valve to be
8.25 N. What is its mass?
F
A
The figure below shows the pressure acting uniformly in all directions on
a small volume of fluid.
The bar is another unit used by some people working in fluid mechanics
and thermodynamics.
The bar is defined as 105 Pa or 105 N/m2.
Another way of expressing the bar is 1 bar = 100 x 103 N/m2, which is
equivalent to 100 kPa.
Example 1.2
The figure below shows a container of liquid with a movable piston
supporting a load. Compute the magnitude of the pressure in the liquid
under the piston if the total weight of the piston and the load is 500 N and
the area of the piston is 2500 mm2.
It is reasonable to assume that the entire surface of the fluid under the
piston is sharing in the task of supporting the load. The second of Pascals
laws states that the fluid pressure acts perpendicular to the piston. Then,
we have
The standard unit of pressure in the SI system is the N/m2 called the pascal
(Pa) in honor of Blaise Pascal. The conversion can be made by using the
factor 103 mm = 1 m. We have
Example 1.3
A load of 200 pounds (lb) is exerted on a piston confining oil in a circular
cylinder with an inside diameter of 2.50 inches (in). Compute the pressure
in the oil at the piston.
To use Eq. p
F
, we must compute the area of the piston
A
Although the standard unit for pressure in the U.S. Customary System is
pounds per square foot (lb/ft2), it is not often used because it is
inconvenient. Length measurements are more conveniently made in inches,
and pounds per square inch (lb/in2) abbreviated psi, is used most often for
pressure in this system. The pressure in the oil is 40.7 psi. This is a fairly low
pressure; it is not unusual to encounter pressures of several hundred or
several thousand psi.
m /V
where V is the volume of the substance having a mass m.
The units for density are kilograms per cubic meter in the SI system and
slugs per cubic foot in the U.S. Customary System.
w @ 4C
w @ 4C
where the subscript s refers to the substance whose specific gravity is being
determined and the subscript w refers to water.
Example 1.5
Calculate the weight of a reservoir of oil if it has a mass of 825 kg.
We have
Example 1.6
If the reservoir from Example Problem 1.5 has a volume of 0.917m3,
compute the density, the specific weight, and the specific gravity of the oil.
Example 1.7
Glycerine at 20C has a specific gravity of 1.263. Compute its density and
specific weight.
Example 1.8
A pint of water weighs 4.632 N. Find its mass.
Because w = mg, the mass is
Example 1.9
One gallon of mercury has a mass of 51.2 kg. Find its weight. Write
This is correct, but the units may seem confusing because weight is normally
expressed in N. The units of mass may be rewritten as Ns2/m, and we have
Surface Tension
Surface tension acts somewhat like a film at the interface between the
liquid water surface and the air above it.
Surface tension is also the reason that water droplets assume a nearly
spherical shape.
The movement of liquids within small spaces depends on this capillary
action.
Wicking is the term often used to describe the rise of a fluid from a
liquid surface into a woven material.
Fig. 4.1
In each case, the fluid exerts a force on the surface of interest that acts
perpendicular to the surface, considering the basic definition of
pressure, p = F/A and the corresponding form, F = pA.
We apply these equations directly only when the pressure is uniform
over the entire area of interest.
An example is when the fluid is a gas for which we consider the pressure
to be equal throughout the gas because of its low specific weight.
Fig.4.2
Example 4.1
If the cylinder in the figure above has an internal diameter of 51 mm and
operates at a pressure of 2070 kPa (gage), calculate the force on the ends of
the cylinder.
Notice that gage pressure was used in the calculation of force instead of
absolute pressure. The additional force due to atmospheric pressure acts on
both sides of the area and is thus balanced. If the pressure on the outside
surface is not atmospheric, then all external forces
must be considered to determine a net force on the area.
Fig 4.3
Example 4.2
If the drum in the figure above is open to the atmosphere at the top, calculate
the force on the bottom.
To use we must first calculate the pressure at the bottom of the drum and the
area of the bottom:
Would there be any difference between the force on the bottom of the drum
in Fig. 4.3 and that on the bottom of the cone-shaped container in Fig. 4.4?
Fig. 4.4
The force would be the same because the pressure at the bottom is dependent
only on the depth and specific weight of the fluid in the container. The total
weight of fluid is not the controlling factor.
Comment: The force computed in these two example problems is the force
exerted by the fluid on the inside bottom of the container. Of course, when
designing the support structure for the container, the total weight of the
container and the fluids must be considered. For the structural design, the
cone-shaped container will be lighter than the cylindrical drum.
Rectangular Walls
Fig 4.5 below shows the rectangular walls.
They are walls which exposed to a pressure varying from zero on the
surface of the fluid to a maximum at the bottom of the wall.
The force due to the fluid pressure tends to overturn the wall or break it
at the place where it is fixed to the bottom.
The actual force is distributed over the entire wall, but for the purpose
of analysis it is desirable to determine the resultant force and the place
where it acts, called the center of pressure.
Fig 4.5
Fig 4.6 below shows the vertical rectangular wall.
Fig.4.6
The total resultant force can be calculated from the equation
where pavg is the average pressure and A is the total area of the wall.
But the average pressure is that at the middle of the wall and can be
calculated from the equation
Example 4.4
In Fig. 4.6, the fluid is gasoline and the total depth is 3.7 m. The wall is
12.2 m long. Calculate the magnitude of the resultant force on the wall
and the location of the center of pressure.
Step 1
FR h / 2 A
45.14m2 557.0kN
3
m
2
Step 2
Step 3
The force acts perpendicular to the wall at the center of pressure as shown in
Fig. 4.6.
Example 4.5
Figure 4.7 shows a dam 30.5 m long that retains 8 m of fresh water and is
inclined at an angle of 60. Calculate the magnitude of the resultant force on
the dam and the location of the center of pressure.
Fig. 4.7
Step 1
To calculate the area of the dam we need the length of its face, called L in Fig.
4.7:
Example 4.5
Step 1
Now we can calculate the resultant force:
Step 2
The center of pressure is at a vertical distance of
from the bottom of the dam, or, measured from the bottom of the dam along
the face of the dam, the center of pressure is at
Fig.4.8
The standard dimensions and symbols used in the procedure described
later are shown in the figure and defined as follows:
Below are the procedure for computing the force on a submerged plane
area:
1. Identify the point where the angle of inclination of the area of interest
intersects the level of the free surface of the fluid. This may require the
extension of the angled surface or the fluid surface line. Call this point S.
2. Locate the centroid of the area from its geometry.
3. Determine hc as the vertical distance from the level of the free surface
down to the centroid of the area.
4. Determine Lc as the inclined distance from the level of the free surface
down to the centroid of the area. This is the distance from S to the
centroid. Note that hc and Lc are related by
where is the specific weight of the fluid. This equation states that the
resultant force is the product of the pressure at the centroid of the area and
the total area.
7. Calculate Ic the moment of inertia of the area about its centroidal axis.
8. Calculate the location of the center of pressure from
Notice that the center of pressure is always below the centroid of an area that
is inclined with the horizontal. In some cases it may be of interest to calculate
only the difference between Lp and Lc from
Example 4.6
The tank shown in Fig. 4.8 contains a lubricating oil with a specific gravity of
0.91. A rectangular gate with the dimensions B=1.2 m and H=0.6 m is placed in
the inclined wall of the tank (=60).The centroid of the gate is at a depth of
1.5 m from the surface of the oil. Calculate (a) the magnitude of the resultant
force FRon the gate and (b) the location of the center of pressure.
Therefore, we have
Then we have
FR 0 hc A
8.92kN
1.5m 0.72m2 9.63kN
3
m
Ic
0.02m4
1.73m
Lc A
1.73m 0.72m2
This means that the center of pressure is 0.016 m (or 16 mm) below the
centroid of the gate.
Be sure you understand how the dimension Lp is drawn from the reference
line.
Center of Pressure
The center of pressure is that point on an area where the resultant force
can be assumed to act so as to have the same effect as the distributed
force over the entire area due to fluid pressure.
A more convenient expression can be developed by using the transfer
theorem for moment of inertia from mechanics.
The following equation:
then becomes
Rearranging gives:
Then
Example 5.1
A cube 0.50 m on a side is made of bronze having a specific weight of
86.9kN/m3. Determine the magnitude and direction of the force required to
hold the cube in equilibrium completely submerged (a) in water and (b) in
mercury. The specific gravity of mercury is 13.54.
Consider part (a) first. Imagine the cube of bronze submerged in water. Now
do Step 1 of the procedure.
On the assumption that the bronze cube will not stay in equilibrium by itself,
some external force is required. The objective is to find the magnitude of this
force and the direction in which it would actthat is, up or down.
Now do Step 2 of the procedure before looking at the next panel.
The free body is simply the cube itself. There are three forces acting on the
cube in the vertical direction, as shown in Fig. 5.2: the weight of the cube w,
acting downward through its center of gravity; the buoyant force Fb acting
upward through the centroid of the displaced volume; and the externally
applied supporting force Fe .
Part (a) of Fig. 5.2 shows the cube as a three-dimensional object with the three
forces acting along a vertical line through the centroid of the volume. This is
the preferred visualization of the free-body diagram. However, for most
As a part of Step 4, solve this equation algebraically for the desired term. You
should now have
There is another unknown on the right side of Eq. (53). How do we calculate
Fb ?
RESULTS- PART A
Notice that the result is positive. This means that our assumed direction for Fe
was correct. Then the solution to the problem is that an upward force of 9.63
kN is required to hold the block of bronze in equilibrium under water.
What about part (b) of the problem, where the cube is submerged in mercury?
Our objective is the same as beforeto determine the magnitude and
direction of the force required to hold the cube in equilibrium.
Now do Step 2 of the procedure.
Either of two free-body diagrams is correct as shown in Fig. 5.3, depending on
the assumed direction for the external force Fe. The solution for the two
diagrams will be carried out simultaneously so you can check your work
regardless of which diagram looks like yours and to demonstrate that either
approach will yield the correct answer.
Now do Step 3 of the procedure.
The following are the correct equations of equilibrium. Notice the differences
and relate them to the figures:
Because the magnitudes of w and Fb are the same for each equation, they can
now be calculated.
As in part (a) of the problem, the weight of the cube is
Notice that both solutions yield the same numerical value, but they have
opposite signs. The negative sign for the solution on the left means that the
assumed direction for Fe in Fig. 5.3(a) was wrong. Therefore, both approaches
give the same result.
RESULTS PART B
The required external force is a downward force of 5.74 kN. How could you
have reasoned from the start that a downward force would be required?
Items c and d of Step 4 of the procedure suggest that the specific weight of the
cube and the fluid be compared. In this case we have the following results:
COMMENT
Because the specific weight of the cube is less than that of the mercury, it
would tend to float without an external force. Therefore, a downward force, as
pictured in Fig. 5.3(b), would be required to hold it in equilibrium under the
surface of the mercury.
This example problem is concluded.
Example 5.2
A certain solid metal object has such an irregular shape that it is difficult to
calculate its volume by geometry. Use the principle of buoyancy to calculate its
volume and specific weight.
First, the mass of the object is determined in the normal manner to be 27.2 kg.
Then, using a setup similar to that in Fig. 5.4, we find its apparent mass while
submerged in water to be 21.1 kg. Using these data and the procedure for
analyzing buoyancy problems, we can find
the volume of the object. (W=mg)
Fig.5.4
Now apply Step 2 of the procedure and draw the free-body diagram of the
object while it is suspended in the water.
The free-body diagram of the object while it is suspended in the water should
look like Fig. 5.5. In this figure, what are the two forces Fe and w?
Fig.5.5
We know that w=266.8 N, the weight of the object in air, and Fe=207 N, the
supporting force exerted by the balance shown in Fig. 5.4. Now do Step 3 of
the procedure.
Using Fv = 0, we get
Our objective is to find the total volume V of the object. How can we get V into
this equation?
We use this equation from Step 4a,
Example 5.3
A cube 80 mm on a side is made of a rigid foam material and floats in water
with 60 mm of the cube below the surface. Calculate the magnitude and
direction of the force required to hold it completely submerged in glycerine,
which has a specific gravity of 1.26.
Complete the solution before looking at the next panel.
Fig. 5.6
From Fig. 5.6(a), we have
Example 5.4
A brass cube 152.4 mm on a side weighs 298.2 N. We want to hold this cube in
equilibrium under water by attaching a light foam buoy to it. If the foam
weighs 707.3 N/m3, what is the minimum required volume of the buoy?
Complete the solution before looking at the solution.
Calculate the minimum volume of foam to hold the brass cube in equilibrium.
Notice that the foam and brass in Fig. 5.7 are considered as parts of a single
system
and that there is a buoyant force on each. The subscript F refers to the foam
and the subscript B refers to the brass. No external force is required.
Fig.5.7
The equilibrium equation is
This means that if 0.029 m3 of foam were attached to the brass cube, the
combination would be in equilibrium in water without any external force. It
would be neutrally buoyant.
Buoyancy Materials
The design of floating bodies often requires the use of lightweight
materials that offer a high degree of buoyancy.
The buoyancy material should typically have the following properties:
1. Low specific weight and density
2. Little or no tendency to absorb the fluid
3. Compatibility with the fluid in which it will operate
4. Ability to be formed to appropriate shapes
5. Ability to withstand fluid pressures to which it will be subjected
6. Abrasion resistance and damage tolerance
The weight of the body acts vertically downward through the center of
gravity.
Fig. 5.8
Because their lines of action are now offset, these forces create a
righting couple that brings the vehicle back to its original orientation,
demonstrating stability.
Fig.5.9
If the cg is above the cb, the couple created when the body is tilted
would produce an overturning couple that would cause it to capsize.
Solid objects have the cg and cb coincident and they exhibit neutral
stability when completely submerged, meaning that they tend to stay in
whatever position they are placed.
Fig.5.10
In part (a) of the figure, the floating body is at its equilibrium orientation
and the center of gravity (cg) is above the center of buoyancy (cb).
A vertical line through these points will be called the vertical axis of the
body.
Figure 5.10(b) shows that if the body is rotated slightly, the center of
buoyancy shifts to a new position because the geometry of the displaced
volume has changed.
The buoyant force and the weight now produce a righting couple that
tends to return the body to its original orientation.
Thus, the body is stable.
A floating body is stable if its center of gravity is below the metacenter.
The distance to the metacenter from the center of buoyancy is called
MB and is calculated from
In this equation, Vd is the displaced volume of fluid and I is the least
moment of inertia of a horizontal section of the body taken at the
surface of the fluid.
If the distance MB places the metacenter above the center of gravity, the
body is stable.
Compute MB = I/V .
d
Compute y
mc
= y + MB.
cb
If y
If y
mc
mc
mg
mg