MEC 241 Chapter 2

Chapter 2 Pressure and Fluid Statics

2.0 Pressure

 Defined as a normal force exerted by a fluid per unit area.

F
P
A

Types of pressure:

 Pressure given a point is called absolute pressure (Pabs)

 Pressure measured by pressure measuring devices (calibrated to read zero in the

atmosphere) is called gage pressure (Pgage). Pgage can be positive or negative.

Pgage = Pabs – Patm (at sea level; Patm = 101.325 kPa)

 Pressure below atmospheric pressure is called vacuum pressure (Pvac).

Pvac = Patm – Pabs = -Pgage

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MEC 241 Chapter 2

2.1 Pressure at a point

Theory:

- Pressure at any point in a fluid is the same in all directions.

- Because it does not have a specific direction, only a specific magnitude, it is

considered a scalar quantity.

- Pressure same in all

directions.
- Does not have specific
direction.

Prove:

- The pressure acting at any point in a fluid is the same can be demonstrated by
considering a small wedge-shaped fluid element in equilibrium.

(1)

(2)

Where;

Substituting the geometric relations and dividing (1)

with yz and (2) with xy gives;
(3)

(4)

In equation (4), as z  0 , and the wedge becomes

infinitesimal, the fluid element shrinks to a point.

Combining both equations:

(5)

We can repeat the same analysis on different planes,

and still obtained the same results.

Thus it can be concluded that, pressure at a point in a

fluid has the same magnitude in all directions.
2
MEC 241 Chapter 2

2.2 Pressure variation in static fluid

- Pressure in a fluid does not change in the horizontal direction. This is proven below:

Consider a horizontal cylindrical fluid element with a cross-sectional area A, in a fluid

of density ρ, pressure P1 acting on the left side and P2 acting on the right side.

The fluid is in equilibrium so the sum of forces acting in the x-direction is:

F x  ma  0
 P1 A  P2 A  0
 P1  P2

- A consequence of the pressure in a fluid remaining constant in the horizontal direction

is that the pressure applied to a confined fluid increases the pressure throughout by
the same amount. This is called the Pascal’s Law:

F1 F2 F1 A1
P1  P2 →  → 
A1 A2 F2 A2

- A2/A1 is the area ratio. Application of the Pascal’s law can be seen in the figure
below.

Figure: The amount of force (F1) required

to lift the weight of the car can be
determined using the Pascal’s law

3
MEC 241 Chapter 2

- However, this is not the case in the vertical direction.

- Pressure increases with depth because more fluid is at rest on deeper layers.

Figure: The pressure of a fluid at

rest increases with depth.

- Pressure at various depth can be obtained using the equation:

Pbelow  Pabove  gh

Where:

gh  Pgage

For example:

Pbelow  Patm  gh

In summary:

4
MEC 241 Chapter 2

Example 1

Point A At this point, the oil is exposed to the atmosphere, and therefore

Point B The change in elevation between point A and point B is 3.0 m, with B lower than A.
To use Eq. (3–3) we need the specific weight of the oil:

Then we have

Now, the pressure at B is

Point C The change in elevation from point A to point C is 6.0 m, with C lower than A. Then,
the pressure at point C is

Point D because point D is at the same level as point B, the pressure is the same. That is, we
have

5
MEC 241 Chapter 2

Point E Because point E is at the same level as point A, the pressure is the same. That is, we
have

Point F The change in elevation between point A and point F is 1.5 m, with F higher than A.
Then, the pressure at F is

Air Pressure Because the air in the right side of the tank is exposed to the surface of the oil,
where pF = -13.2kPa the air pressure is also -13.2kPa or 13.2 kPa below atmospheric
pressure.

2.3 Manometric Pressure

 Standard technique for measuring pressure involves the use of liquid columns in
vertical or inclined tubes.

 Pressure measuring devices based on this technique are called manometers.

 Common types of manometers are:

i) U-Tube manometer

ii) Inclined-tube manometer

U-tube manometer

 A type of manometer which is widely used to measure pressure and pressure

difference between two points in a given system.

 Consists of a tube formed into the shape of a U.

Examples of U-tube manometer:

6
MEC 241 Chapter 2

 Many engineering problems and some manometers involve multiple immiscible fluids
of different densities stacked on top of each other. Such systems can be analysed
easily by remembering:

i. the pressure change across fluid column height, h is ΔP =ρgh

ii. pressure increases downward in a given fluid and decreases upward (i.e.,
Pbottom > Ptop)

iii. Two points at the same elevation in a continuous fluid at rest are at the same
pressure

 By remembering these steps, pressure at any point can be determined by starting with
a point of known pressure and adding (moving downwards) and subtracting (moving
upwards) ρgh terms as we move to the point of interest.

Example: multiple immiscible fluids of different densities stacked on top of each other

The pressure at the bottom (Point 1) can be determined by starting

from the free surface where the pressure is Patm, moving downwards until we point 1
at the bottom. This gives the equation:

Patm  1 gh1   2 gh2   3 gh3  P1

Example 1:

7
MEC 241 Chapter 2

 The pressure difference P1 – P2 can be obtained by starting at point 1 with P1 and

moving along the tube by adding or subtracting the ρgh terms until reaching point 2
and setting the result equals to P2.

Example 2

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MEC 241 Chapter 2

 Inclined tube manometer

Example:

9
MEC241 Fluid Statics

2.4 Fluid Statics

 Use to determine forces acting on floating or submerged bodies and the forces developed
by devices like hydraulic presses and car jacks.
 For problem analysis, only consider an element of the fluid – FBD is drawn to show the
forces acting on it due to solid boundaries or surrounding fluid.
 Engineering systems such as water dams, and liquid storage tanks requires the
determination of the forces acting on their surfaces using fluid statics.
 In this section, we will consider the forces acting on both plane and curved surfaces of
submerged bodies due to pressure.

2.4.1 Hydrostatic forces on submerged plane surfaces

 A plate, such as gate valve in a dam, the wall of a liquid storage tank, or the hull of a
ship at rest is subjected to fluid pressure distributed over its surface when exposed to a
liquid.
 For analysis purposes, we must be able to determine:
i) Resultant force (FR)
ii) Point of application (hp)
iii) Moment about a point (Mo) to determine external forces
 Submerged plane surfaces can be off three different orientations, horizontal, vertical,
and inclined.

i) Horizontal plane surface

FR  ghc A

hc  h

hp  hc

Where;

hc = vertical distance from free surface to centroid

area of the geometry
hp = vertical distance from free surface to center pressure
of the geometry
A = area of the geometry
MEC241 Fluid Statics

ii) Vertical plane surface

FR  ghc A

I xx
h p  hc 
hc A

Where;

hc = vertical distance from free surface to centroid

area of the geometry
hp = vertical distance from free surface to center pressure
of the geometry
A = area of the geometry
Ixx = 2nd moment area of the geometry

Inclined plane surface

FR  ghc A

hc  yc sin 

I xx sin 2 
hp  hc 
hc A
or
hp  y p sin 

Where;
hc = vertical distance from free surface to centroid
area of the geometry
hp = vertical distance from free surface to center pressure
of the geometry
yc = inclined distance from free surface to centroid
area of the geometry
yp = inclined distance from free surface to center pressure
of the geometry
A = area of the geometry
Ixx = 2nd moment area of the geometry
MEC241 Fluid Statics

 The values Ixx varies for different geometries, examples for centroid and the centroidal
moments of inertia for some common geometry are shown in the figure below.

Example: Hydrostatic forces on submerged plane surfaces

1. Submerged vertical plane surface

Determine the resultant force and the distance to the center of pressure due to the water acting
on the 1m by 2m rectangular area AB shown in the figure below.

The magnitude of the resultant force on the submerged plane is:

FR  ( ghc ) A  1000(kg / m 3 )  9.81(m / s 2 )  [1.22(m)  1(m)]  2(m 2 )  43,556kgm / s 2  43.556kN / m 2

MEC241 Fluid Statics

The distance to the center of pressure is:

I xx,c
h p  hc 
hc A

For rectangular shape, the 2nd moment of the area is:

bh 3 1(m)  2 3 (m 3 )
I xx,c    0.667 m 4
12 12

I xx,c 0.667 (m 4 )
 h p  hc   2.22(m)   2.37 m (from the free surface) or 0.15 below
hc A 2.22(m)  2(m 2 )
the center of gravity.

2. Submerged inclined plane surface

From the same figure above, determine the resultant force and distance to the center of
pressure due to the water acting on the 1.25 m by 2.0 m triangular area CD. The inclined
distance to the apex of the triangle is at C is 1.42 m.

Inclined distance from the free surface to the centroid of the geometry is:

2 2
yc  1.42(m)  d  1.42(m)   2(m)  2.75m
3 3

Vertical distance from the free surface to the centroid of the geometry is:

The magnitude of the resultant force on the submerged plane is:

hc  yc sin   2.75(m) sin 45  1.945m

The magnitude of the resultant force on the submerged plane is:

 2.0(m)  1.25(m) 
FR  ( ghc ) A  1000(kg / m 3 )  9.81(m / s 2 )  1.945(m)     23,826 kgm / s 2  23.826 kN / m
 2.0 ( m) 

For triangle shape, the 2nd moment of the area is:

bh 3 1.25(m)  2 3 (m 3 )
I xx,c    0.278m 4
36 36

Vertical distance from the free surface to the center of pressure is:

I xx sin 2  0.278 sin 2 45

hp  hc   1.945   2m
hc A 1.945(1.25)
MEC241 Fluid Statics

 To determine external forces subjected on the plane surface, a summation of moment

(∑M=0) equation of forces about a fix point must be obtained.

Example:

1. A 6-m height, 6-m wide rectangular plate blocks the end

of a 5-m deep fresh water channel, as shown in the figure
given. The plate is hinged about the horizontal axis along
its upper edge through a point A and is restrained from
opening by a fixed ridge at point B. Determine the
hydrostatic force subjected on the plate AB, and the force
exerted by the ridge on the plate.

Solution:

Determine the hydrostatic force subjected on the plate AB, and the force exerted by the ridge
on the plate.

Hydrostatic force on plate AB

FR  ( ghc ) A  (1000)(9.81)(2.5)(6  5)  735.9 N

Center of pressure, hp;

I xx,c 62.5
hp  hc   2.5   3.33m (from free surface)
hc A (2.5)(30)

Force subjected on plate AB by ridge at B

M A 0

FR (d )  Fridge (d )  0

 FR (d )  Fridge (d ); 735.9(1  3.33)  Fridge (6) ; Fridge  531.07 N

MEC241 Fluid Statics

The 280-kg, 6-m wide rectangular gate shown in the

figure is hinged at B and leans against the floor at A
making an angle of 45° with the horizontal. The gate
is to be opened from its lower edge by applying a
normal force at its center. Determine force F required
to open the water gate.

Determine the hydrostatic force subjected on the gate AB,

3
hc  0.5   2m
2

FR  ( ghc ) A  (1000)(9.81)(2)(6  4.243)  499.5kN

Determine the vertical distance from the free surface to the center pressure of the plate AB
(hp)

ab3 6(4.243)3
I xx    38.193m4
12 12

I xx sin 2  38.193(sin 2 45)

hp  hc   2  2.375m
hc A 2(25.458)

Determine the inclined distance from the free surface to the center pressure of the plate AB
(yp)

hp 2.375
yp    3.359 m
sin  sin 45

Determine the force F required to open the water gate

(CCW ) M B  0
 F (l1 )  FR (l2 )  W cos (l1 )  0
FR (l2 )  W cos (l1 )
F
l1

l1 = 4.243/2 = 2.1215 m

l2 = 3.359 – 0.5/sin 45°= 2.652 m

FR (l2 )  W cos  (l1 ) 1,324,674  4120 .55

F   626.3kN
l1 2.1215
MEC241 Fluid Statics

2.4.2 Hydrostatic forces on submerged curved surfaces

 For submerged curved surface, determination of the resultant hydrostatic force, FR

requires the integration of the pressure forces that change direction along the curve
surface.

 The easiest way to determine the resultant hydrostatic force, FR acting on the curved
surface is to determine the horizontal (FH) and vertical (FV) forces separately.

 The horizontal (FH) and vertical (FV) forces can be obtained by using ∑F = 0 with
respect to their direction. Therefore, the horizontal (FH) and vertical (FV) forces
acting on the submerged curved surface are:

 The magnitude of the resultant hydrostatic force, FR acting on the curve surface is:

FR  ( FH ) 2  ( FZ ) 2

 and the tangent of the angle it makes with the horizontal is:

a) Fluid above curve surface b) Fluid below curve surface

FH  Fx  ghc A FH  Fx  ghc A
FV  Fy  gV FV  Fy  gV
Fy  ghA
MEC241 Fluid Statics

Example

a) fluid above curve surface

A 4-m long curved gate is located in the side of a reservoir

containing water as shown in figure. Determine the magnitude
of the horizontal and vertical components of the force of the
water on the gate and the resultant force with its point of
application.

Solution

FH  Fx  ghc A
FH  (1000 )(9.81)(6  1.5)(3  4)  882 kN

FV  Fy  gV
Fy  ghA  (1000 )(9.81)(6)(3)(4)  706.32kN
1
FV  (1000 )(9.81)(  (3) 2 (4)  277.37 kN
4
FV  983.69 kN

FR  FH  FV2  1321 .2kN

2

FV
  tan 1  48.1
FH

b) Fluid below curve surface

Gate AB in the figure given is a quarter circle 10 m wide into

the paper and hinged at B. Determine the horizontal, vertical
components of the hydrostatic force, and resultant force against
the dam and the center of pressure where the resultant strikes
the dam.
MEC241 Fluid Statics

Solution

FH  Fx  ghc A
FH  (1000 )(9.81)(8 / 2)(8  10)  3.139 MN

F 0
 FV  Fy  gV  0
FV  Fy  gV
Fy  ghA  (1000 )(9.81)(8)(8)(10)  6.278 MN
1
gV  (1000 )(9.81)(  (8) 2 (10)  4.931MN
4
FV  1.347 MN

FR  FH  FV2  3.415 MN
2

FV
  tan 1  23.22
FH

 Similar, to determine external forces subjected on the plane surface, a summation of

moment (∑M=0) equation of forces about a fix point must be obtained.

Example

A 4-m long quarter-circular gate of radius 3 m and of

negligible weight is hinged about its upper edge A as
shown in the figure. The gate controls the flow of
water over the ledge at B, where the gate is pressed
by a spring. Determine the minimum spring force
required to keep the gate closed when the water level
rises to A at the upper edge of the gate.
MEC241 Fluid Statics

Solution

FH  Fx  ghc A
FH  (1000 )(9.81)(3 / 2)(4  3)  176.58kN

F 0
 FV  Fy  gV  0
FV  Fy  gV
Fy  ghA  (1000 )(9.81)(3)(3)(4)  353.16kN
1
gV  (1000 )(9.81)(  (3) 2 (4)  277.37 kN
4
FV  75.79 kN

FR  FH  FV2  192.16kN
2

FV
  tan 1  23.22
FH

The minimum spring force needed is determined by taking a moment about the point A
where the hinge is, and setting equal to zero.

M A 0
FR ( R)(sin 90   )  Fspring ( R)  0

Solving for Fspring and substituting, the spring force is determined to be

Fspring  FR (sin 90   )  (192.16)(sin 90  23.22)  176.59kN

MEC241 Fluid Statics

2.5 Buoyancy and Stability

1. A body in fluid, whether floating or submerged, is buoyed up by a force equal to the

weight of the fluid displaced.

2. The buoyant force, FB for a submerged body acts vertically upward through the centroid of
the displaced volume and can be defined as:
FB   f gVbody

3. The buoyant force, FB for a floating body can be defined as:

FB  W
 f gVsubmerged   body gVtotal
Vsubmerged  body
 
VTotal f

Example 1

A cube 80 mm on a side made of a rigid foam material and floats in water with 60 mm of the
cube below the surface is shown in the Figure below. Calculate the magnitude and direction
of the force (Fe) required to hold it completely submerged in glycerine, which has a specific
gravity of 1.26.
MEC241 Fluid Statics

Solution

- 1st determine the buoyant force (Fb) for the cube in (a);

F v 0
Fb  W  0
Fb  W  (  fluid g )Vcube   fluid Vcube _ submerged
Fb  9.81(kN / m 3 )  0.000304 (m 3 )  3.77 kN

- Next, determine the force (Fe) required to hold the cube submerged in (b);

F v 0
Fb  Fe  W  0
Fe  Fb  W  (  glycerine g )Vcube _ total  3.77(kN )   glycerineVcube _ total  3.77(kN )
Fe  12.36(kN / m 3 )  0.512(m 3 )  3.77(kN )  2.56kN ()

Example 2

A freshly cut log as shown in the figure given floats with ¾ of its volume submerged under
the water. Determine the density of the log.

Solution:

For floating bodies;

FB  W
 f gVlog_ ubmerged   log gVlog_ total

Simplifying and rearranging the equation:

3 
 Vloq _ submerged   V 
 log   fluid     fluid  4   (1000 ) 3   750 kg / m 3
 V   V  4
 log_ total   
 
MEC241 Fluid Statics

2.5.1 Stability

 A body in a fluid is considered stable if it will return to its original position after being
rotated a small amount about the horizontal axis.

For floating bodies such as ships, Stability is easily understood

stability is an important by analyzing a ball on the
consideration for safety. floor.

 Stability of a body can be analysed whether it is submerged or floating.

 The condition for stability of bodies completely submerged in a fluid is that the centre of
gravity of the body must be below the centre of the buoyancy.

 The centre of the buoyancy of a body is at the centroid of the displaced volume of the
fluid, and it is through this point that the buoyant force acts in a vertical direction.

 The weight of the body acts vertically downwards through the centre of gravity.

An immersed neutrally buoyant body is (a) stable if the center of gravity G is directly
below the center of buoyancy B of the body, (b) neutrally stable if G and B are coincident,
and (c) unstable if G is directly above B.
MEC241 Fluid Statics

 The condition for stability of bodies floating in a fluid is that the centre of gravity of
the body must be above the centre of the buoyancy.

From the figure above, A floating body is (a) stable if the body is bottom-heavy and thus the
center of gravity G is directly above the centroid B of the body, or (b) if the metacenter M is
above point G. However, the body is (c) unstable if point M is below point G.

Steps for evaluating the stability of floating bodies

1. Using the principles of buoyancy, determine the position of the floating body.

2. Locate the centre of buoyancy, Cb. This is done by computing the distance from a
reference axis to Cb. This distance is noted with ycb. Usually, the bottom of the object
is taken as the reference axis.

3. Locate the centre of gravity, CG. This is done by computing the distance (ycg) from the
same reference axis.

4. Relative to the front view, determine the shape of the area at the fluid surface and
compute the smallest moment of inertia Iyy for that shape.

5. Compute the displaced volume of the shape, Vd

I yy
6. Compute BM 
Vd

7. Compute

ymc  ycb  BM

* If ymc > ycg, the body is stable

If ymc < ycg, the body is unstable

MEC241 Fluid Statics

The locations for each abbreviation are illustrated in the figure below:

BM
Ymc

Reference axis Ycb Ycg

The conditions for stability of bodies in a fluid can be summarized as follow:

 Completely submerged bodies are stable if the centre of gravity is below the centre of
buoyancy.

 Floating bodies are stable if the centre of gravity is below the metacentre.

Example 1

The figure given below shows a flatboat hull that when fully loaded weighs 150 kN.
Determine whether the boat is stable in fresh water. Note the location of the centre of gravity,
cg.
MEC241 Fluid Statics

Solution:

1. Calculate the depth of the submergence or draft (X)

of the boat.

- Equation of equilibrium;

F v  0  Fb  W
 Fb  W

- Submerged volume (Vd);

Vd  B  L  X

- Buoyant force (Fb);

Fb  (  fluid g )Vd   fluid  ( B  L  X )

- From equation of equilibrium we had:

Fb  W
Fb   fluid  ( B  L  X )
  fluid  ( B  L  X )  W
W 150(kN )
X    1.06m
 fluid  ( B  L) 9.81(kN / m 3 )  2.4(m)  6.0(m)

Meaning that, the flatboat floats, with 1.06 m of the boat is submerged under water.

The center of buoyancy (cb) is at the centre of the displaced volume of water. In this case as
shown in the figure below is on the vertical axis of the boat at a distance of 0.53 m from the
bottom. That is half of the draft (X).

ycb = 0.53 m

Because the center of gravity is above the

center of buoyancy, we must locate the
metacenter to determine whether the boat is
stable.

-
I
BM 
Vd
Vd  B  L  X  2.4(m)  6.0(m)  1.06(m)  15.26m 3
MEC241 Fluid Statics

The moment of inertia Iyy is determined about the axis y-y from the first figure above;

LB3 6.0(m)  2.43 (m3 )

I yy    6.91m 4
12 12

6.91(m 4 )
 BM   0.45m
15.26(m 3 )

y mc  ycb  BM  0.53(m)  0.45(m)  0.98m

Is the boat stable?

Yes it is. Because the metacenter is above the center of gravity (ymc > ycg) as shown in the
figure.

Example 2

Consider a wooden cylinder (SG = 0.6) 1 m in diameter and 0.8 m long. Would the cylinder
this cylinder be stable if placed to float with its axis vertical in oil (SG = 0.85)?

1. Using the principle of buoyancy (for floating body);

FB  W
 f gVsubmerged   cylinder gVtotal
(0.85)(R 2 h)  (0.6)(R 2 H )
h  0.565 m

This means that the cylinder floats, with 0.565 m of the cylinder submerged under water.

2. Centre of buoyancy is at the centre of the displaced volume;

ycb  h / 2  0.565 / 2  0.282m

3. Locate the metacentre height;
I
BM 
Vd
Vd  R 2 h   (0.5) 2 (0.565)  0.444m 3
R 4  (0.5) 4
I   0.0491m 4
4 4
I 0.0491
 BM    0.111m
Vd 0.444
y mc  ycb  BM  0.282  0.111  0.393m
Knowing that ycg = 0.4 m, therefore the boat is slightly unstable because, ymc < ycg.