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GUJARAT TECHNOLOGICAL UNIVERSITY,
NOVEMBER , 2016
GOVERNMENT ENGINEERING COLLEGE,
PALANPUR
Subject :- Fundamental of Fluid Mechanics

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Submitted by :-
Bajariya Maheshkumar 150610122004
Kishankumar Nagar 150610122026
Patel Jigneshkumar 150610122033
Roy malay 150610122046
Solanki kalpesh 150610122048
BACHELOR OF ENGINEERING
In
MINING
Internal guide Head of department
Prof :- G. M. Savaliya Prof :- H. B. Patel

3. Fluid statics
Fluid Statics:- It is a study of a fluid at rest. In this fluid
there will be relative motion between adjacent or
neighbouring fluid layers. No shear force present force as
the fluid particles do not move respect to one another.
There velocity gradient is equal to zero.
Velocity gradient = change of velocity between two
adjecent layers
Distance between two layers
= du = 0
dy

4. Pressure variation in static fluid – hydrostatic
The hydrostatic law states that “the rate of increase of
Pressure in a vertically downward direction is equal to the
Weight density of fluid at that point.”
Consider a small element of fluid, vertical column of
constant cross sectional area dA, and totally surrounded b
fluid of mass density p .
p1 = Pressure on face AB,
dz = Height of fluid element,
p2 = Pressure on face CD
The forces acting on the fluid element are
i) Pressure force on face AB, p1dA, downward direction
ii) Pressure force on face CD, p2dA, upward direction

5. iii) Force due to weight of fluid element, W = mg = p.Vg
W = p(dA.dZ)g, downward direction.
iv) Pressure forces on surfaces AC and BD, but they are equ
and opposite to each other.
P1.dA
A B
C D
P2.d
dzCylindrical
Element of fluid
W
dA

6. Fluid is at rest the element must be in equilibrium and the sum of all
vertical forces must be zero.
p1dA – p2dA + W = 0
(p1 – p2)dA + p(dAdz)g = 0
(p2 – p1) – pgdz = 0
p2 – p1 = pgdz
dP = pgdz
dP = pg = w
dz
Where w = Weight density of fluid
Above equation states that “rate of increase of pressure in a vertica
direction is equal to weight density of the fluid at the point” and this
known as Hydrostatic law.
ʃ dp = ʃ pgdz
p = pgz
Where, p= pressure above atmospheric pressure, and z is the height o
point from free surfaces.

7. PASCAL’S LAW:- “The intensity of pressure at any point in a liq
at rest, is the same in all direction.”
Consider an arbitrary fluid element of wedge shape ABC in a fluid ma
rest . The width of the element perpendicular to the plane of paper is
Unity.

8. Let, px= Pressure acting on a face AB
py= Pressure acting on a face AC
pz= Pressure acting on a face BC
Px= Force on a face AB
Py= Force on a face AC
Pz= Force on a face BC
The forces acting on the element are
i) Force normal to the surface dye to fluid pressure
ii) Force due to weight of fluid mass in vertical direction
Px= px × area of face AB = px (dy × 1)
Py= py × area of face AC = py (dx × 1)
Pz= pz × area of face BC = pz (ds × 1)
Weight of element = mass of element × g
= p × volume ×g
= p×(½×AC×AB×1) × g
w= ½ p(dy×dx)g,
The element of the liquid is at rest, therefore sum of horizontal and
Vertical components of the forces equal to zero.

9. Force acting in X – direction , px × dy × 1 = pz × ds × sin(90-θ) × 1
px × py × 1 = pz ×dy × 1
px = pz
Force acting in Y – direction , py ×dx ×1 = pz ×ds ×cos(90-θ)× 1+(½dx×dy
py ×dx ×1 = pz ×dx ×1 (½dx×dy×1 is neglect
py = pz
px = py = pz
Hence , at any point in a fluid rest the pressure is exerted equally in
all directions.
Atmospheric pressure:- It is pressure exerted by the air on t
Surface of earth. The air is compressible fluid the density of air vary fro
time to time due to changes in its temperature, therefore atmospheric
pressure is not constant. The atmospheric pressure at sea level at 15Ċ is
101.3 KN/m2 in SI unit and 1.033 kgf/cm2 in MKS units.
1 atmospheric pressure = 101.3 KN/m2
= 1.033 kgf/cm2
= 760 mm of Hg
= 10.33 m of water

10. Gauge pressure:- Gauge pressure is measured with the help of
Pressure gauge. The atmospheric pressure is taken as datum. In this
Pressure, atmospheric pressure is considered zero, and this pressure is
Above the atmospheric pressure.
Vacuum pressure:- When pressure is below the atmospheric
pressure is called vacuum pressure. It is also known as negative gauge
pressure. It is measured by vacuum gauge.
Absolute pressure:- It is pressure which is measured with
reference to absolute vacuum pressure. It is independent of the chang
in atmospheric pressure. It is measured above the zero of pressure.
Mathematically, absolute pressure= atmo. Pressure + Gauge pressure
pabs = patm +pgauge
The Hydrostatic paradox:- We know the equation of press
at a point , p = pgh.
The intensity of pressure depends only on height of the column, and
density of fluid and it dose not depends on size of the column.

11. Consider four vessels all have the same base area A and are filled to the
same height h with the same liquid of density p. since weight of fluid
different in the four vessels.
We know, the forec on the base of vessel = p . A = p gh . A
For same base area A, same density p and same column height, the for
on the base of vessel is same for all four vessels and hence pressure
intensity exerted on the base are same for all four vessel.
Thus ,pressure intensity is independent on weight of fluid. This situa
is called Hydrostatic paradox.

12. Manometer
 Manometer is a device used for measuring
the pressure at a point in a fluid by balancing
the column of fluid with the same column or
another of the fluid.
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13.  (1) Simple manometer:
 Piezometer
 U-tube manometer
 single column manometer
 Vertical single column manometer
 Inclined single column manometer
 (2) Differential manometer :
 U-tube differential manometer
 Inverted U-tube differential manometer
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Classification of
Manometers :

14. A piezometer is the simplest form of the manometer. It measures
gauge pressure only.
The pressure at any point in the liquid is
indicated by the height of the liquid in the
tube above that point, which can read on
the calibrated scale on glass tube.
The pressure at point A is given by;
𝑝 = 𝜌𝑔ℎ = 𝑤ℎ
∴ ℎ =
𝑝
𝜌𝑔
𝑝𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑
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1. Piezometer
:

15. It can be measure large pressure or vacuum pressure and gas pressu
∴ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑋𝑋 𝑖𝑛 𝑙𝑒𝑓𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
= 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑋𝑋 𝑖𝑛 𝑙𝑒𝑓𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
∴ 𝑝 + 𝜌1 𝑔ℎ1 = 𝜌2 𝑔ℎ2
∴ 𝑝 = 𝜌2 𝑔ℎ2 − 𝜌1 𝑔ℎ1
𝑁𝑜𝑤, 𝑝 = 𝜌𝑔ℎ, ℎ ℎ𝑒𝑎𝑑 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓
𝑤𝑎𝑡𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛,
𝜌𝑔ℎ = 𝜌2 𝑔ℎ2 − 𝜌1 𝑔ℎ1
∴ ℎ =
𝜌2
𝜌
ℎ2 −
𝜌1
𝜌
ℎ1
∴ ℎ = 𝑠2ℎ2 − 𝑠1ℎ1
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2. U-tube
Manometer :
x x

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3. Single column Manometer:
(A) Vertical single columnvmanometer
One of the limbs in double column manometer is converted into
reservoir having large cross sectional area (about 100 times) with
respect to the other limb.
∴ 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 ℎ𝑒𝑎𝑣𝑦 𝑙𝑖𝑞𝑢𝑖𝑑 𝑓𝑎𝑙𝑙 𝑖𝑛 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟
= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 ℎ𝑒𝑎𝑣𝑦 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑖𝑠𝑒 𝑖𝑛 𝑟𝑖𝑔ℎ𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
∴ 𝐴 × ∆ℎ = 𝑎 × ℎ2 → ∆ℎ =
𝑎 × ℎ2
𝐴
Pressure in left col. = pressure in right col.
∴ 𝑝 =
𝑎ℎ2
𝐴
𝜌2 𝑔 − 𝜌1 𝑔 + 𝜌2 𝑔ℎ2 − 𝜌1 𝑔ℎ1…………(i)
∴ 𝑝 = 𝜌2 𝑔ℎ2 − 𝜌1 𝑔ℎ1

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(B) Vertical single column manometer
It is modified of vertical column manometer. This manometer is
useful for the measurement of small pressure.
Here, ℎ𝑒𝑖𝑔ℎ𝑡 ℎ2 = 𝑙 𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑖𝑛 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞. 𝑖 ;
∴ 𝑝 =
𝑎𝑙 𝑠𝑖𝑛𝜃
𝐴
𝜌2 𝑔 − 𝜌1 𝑔 + 𝜌2 𝑔𝑙𝑠𝑖𝑛𝜃 − 𝜌1 𝑔ℎ1
𝑠𝑖𝑛𝑐𝑒, 𝑎 ≪ 𝐴, 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚;
∴ 𝑝 = 𝜌2 𝑔𝑙𝑠𝑖𝑛𝜃 − 𝜌1 𝑔ℎ1
∴ ℎ = 𝑠2 𝑙𝑠𝑖𝑛𝜃 − 𝑠1ℎ1

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4. U-tube differential manometer
It is used to measure pressure difference at two points
in a pipe or between two pipes at different levels.
Case 1 - U-tube upright differential manometer
connected at two points in a pipe at same level

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Case 2 - U-tube upright differential manometer
connected between two pipes at different levels
and carrying different fluids

20. It is used for low pressure difference.
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5. Inverted U-tube differential
manometer

21. Bourdon Tube

22. - The linkage is constructed so that the mechanism m
be adjusted for optimum linearity and minimum
hysteresis as well as compensate for wear which may
develop over a period of time.
- An electrical-resistance strain gauge may also be
installed on the bourdon-tube to sense thr elastic
deformation.

23. 4) Diaphragm and Bellows Gauges
● Represent similar types of elastic deformation
devices useful for pressure measurement applications.
● Architecture and operation:
Diaphragm gauge:
- Consider first the flat diaphragm subjected
to the differential pressure p1-p2 as shown in
figure .
- The diaphragm will be deflected in accordance with
this pressure differential and the deflection sensed
an appropriate displacement transducer.
- Various types of diaphragm gauge are shown figure 4.10

24. (a) Diaphragm and (b) Bellows
(a) (b)

25. Bellows Gauge:
- The bellows gauge is shown in figure (b).
- A differential gauge pressure force causes displacement
of the bellows, which may be converted to an electrical
signal or undergo a mechanical amplification to
permit display of the output on an indicator dial.
- Figure shows various types of bellows gauges.
● The bellows gauge is generally unsuitable for transient
measurements because of the larger relative motion and
mass involved.
● The diaphragm gauge which may be quite stiff, involves
rather small displacements and is suit for high frequency
pressure measurement.