PRESSURE
MEASUREMENT
Absolute Pressure, Gauge Pressure and Vacuum Pressure
Absolute Pressure
Reference to Zero Pressure
Gauge Pressure
Reference to local atmospheric Pressure (The word local is
where you are when you made the measurement)
Standard Pressure at sea level, Pstd= 14.7 psi or 101.3 kPa
Psi= Pound per square inch
It is always positive as it reference to zero ➢ The gauge pressure can be negative or positive.
pressure.
➢ Negative gauge pressure are sometime called as
Vacuum pressure.
➢ In a vacuum, the pressure is below atmospheric pressure, and
therefore, the gauge pressure would be negative
𝑝𝑔𝑢𝑎𝑔𝑒 = 𝑝𝑎𝑏𝑠 − 𝑝𝑎𝑡𝑚
𝑝𝑣𝑎𝑐𝑢𝑢𝑚 = 𝑝𝑎𝑡𝑚 − 𝑝𝑎𝑏𝑠
PRESSURE VARIATION
UNDER
DIFFERENT SCENARIOS
A)Pressure at any point in a fluid
Let us consider a point in a stationary fluid. Imagine a wedge shape fluid element, which is having ∆y, ∆z and ∆s as
shown in Fig.
P
z
∆s
∆z
y
x
Py
θ
∆y
Pz
We have pressure P acting on the hypotenuse side : acting perpendicular to the surface (normal stress)
Let the width of fluid element is unity
Let’s call pressure pointing in y direction Py and pointing in z direction Pz and the angle is theta
We are going to write summation of the forces. We start of with the Y direction. We are assuming a static fluid, no
acceleration, so the sum of forces is zero
P
Py
∆s
∆z
θ
∆y
Pz
෍ 𝐹𝑦 = 0
Py × ∆z × 1 − Psinθ × ∆s × 1 = 0
∆z = ∆s sinθ
∆y = ∆s cosθ
Py × ∆s sinθ × 1 − P × ∆s sinθ × 1 = 0
Py = P
Similarly σ 𝐹𝑧 = 0
(i)
Conclusion :Pressure at any
point in a fluid at rest is same
in all direction
(i)
Pz = P = Py
Pz = P
❑Pressure Variation in a Static Fluid Element (Hydrostatic Law)
All the arrows are point
in because pressure is a
compressive stress
Summation of force in x direction
𝑝+
𝛿𝑝 ∆𝑥
𝛿𝑥 2
∆𝑦∆𝑧 = 𝑝 −
𝛿𝑝 ∆𝑥
∆𝑦∆𝑧
𝛿𝑥 2
𝛿𝑝
=0
𝛿𝑥
Conclusion : Pressure must not
Vary on horizontal plane
Similarly in y direction,
𝛿𝑝
=0
𝛿𝑦
In z direction,
𝑝+
𝛿𝑝 ∆𝑧
𝛿𝑧 2
𝑝+
∆𝑥∆𝑦 + 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒇𝒍𝒖𝒊𝒅 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 = 𝑝 −
𝛿𝑝 ∆𝑧
𝛿𝑧 2
𝛿𝑝 ∆𝑧
𝑝+
𝛿𝑧 2
𝛿𝑝 ∆𝑧
𝛿𝑧 2
𝛿𝑝 ∆𝑧
𝛿𝑧 2
∆𝑥∆𝑦
𝛿𝑝 ∆𝑧
∆𝑥∆𝑦 + 𝜌 × 𝑔 × ∆𝑥∆𝑦∆𝑧 = 𝑝 −
𝛿𝑧 2
∆𝑥∆𝑦
∆𝑥∆𝑦 + 𝜌 × 𝑔 × 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑝 −
By solving,
𝛿𝑝
= −𝜌 × 𝑔
𝛿𝑧
Hence, pressure varies in direction of depth only,
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
𝑑𝑝 = −𝜌𝑔𝑑𝑧
∆𝑥∆𝑦
Considering two points in the liquid at vertical distance Z. Pressure difference between them can be calculated by
integrating above equation
𝑃2
𝑃2
න 𝑑𝑝 = − න 𝜌𝑔𝑑𝑧
𝑃1
𝑃1
𝑝2 − 𝑝1 = −𝜌𝑔 𝑍2 − 𝑍1
∆𝑝 = −𝜌𝑔∆𝑍
• This is the expression for pressure difference between two points in the static incompressible fluid.
•
In the Equation above, negative sign indicates that pressure decreases when we
moves upward however it increases when we move downward.
Pressure Does not vary in a horizontal plane as long as it is a continuous fluid
Hydrostatic-pressure distribution.
❑ Points a, b, c, and d are at equal depths in water and therefore have identical pressures.
❑ Points A, B, and C are also at equal depths in water and have identical pressures higher than a, b, c, and d.
❑ Point D has a different pressure from A, B, and C because it is not connected to them by a water path.
Conversion Unit for Pressure
101.3 kPa
14.7 psia
760 mm Hg
30 inches Hg
34 ft of Water
ALL ARE EQUAL
PRESSURE Measurement
1) If we simply punch a hole in the water pipe and insert a tube, water will rise up to certain in the tube
h
Piezometric tube or Piezometer
❑ The piezometric tube consists of a transparent tube made of glass or plastic that is open at the top and inserted
into the fluid whose pressure you want to measure.
❑ When the tube is inserted into the fluid, the fluid rises inside the tube to a certain level due to the pressure
exerted by the fluid. This rise in the fluid level is a result of the hydrostatic pressure at that particular point in
the fluid. The height of the fluid column inside the piezometric tube represents the pressure at that point in the
fluid.
Measure the pressure of groundwater in wells, including pumping wells.
U-tube manometers
❑ It consists of a U-shaped tube, typically made of glass or transparent plastic, filled with heavier fluid
(often mercury when we measure pressure of water, or water when we measure pressure of airduct)
PC + 𝜌1g h1- 𝜌2g Δh = 0
PATM
h1
PC = 𝜌2g Δh -𝜌1g h1
Δh
For the same Pressure 30psi, the height of fluid is 60 inches
of mercury 0r 5 ft
❑ Differential Manometer
• Differential manometers are used to measure the pressure difference between two points.
• A differential manometer is a glass U-tube connected at both ends to the points where the pressure is to be measured
and filled with a liquid having different specific gravity than the liquid of which pressure is to be measured.
✓ U-tube differential manometer:
Let,
h = Difference in mercury level in both limbs of U-tube
ℎ1 = Difference between center of pipe A and mercury level in connected
limb.
ℎ2 = Difference between center of pipe B and mercury level in connected
limb.
𝜌1 = Density of liquid in pipe A
𝜌2 = Density of liquid in pipe B
𝜌𝑚 = Density of manometric liquid(should be greater than 𝜌1 and 𝜌2 )
Therefore, manometric equation for this case will be
𝑃𝐴 + 𝜌1 𝑔ℎ1 + 𝜌𝑚 𝑔ℎ − 𝜌2 𝑔ℎ2 − 𝑃𝐵 = 0
So, pressure difference betn both pipes will be
𝑃𝐴 − 𝑃𝐵 = −𝜌1 𝑔ℎ1 − 𝜌𝑚 𝑔ℎ + 𝜌2 𝑔ℎ2
Let both the pipes are at same level as shown in figure, so for this condition
ℎ2 = ℎ1 + ℎ
• Therefore, the manometric equation will be,
𝑃𝐴 + 𝜌1 𝑔ℎ1 + 𝜌𝑚 𝑔ℎ − 𝜌2 𝑔(ℎ1 +ℎ) − 𝑃𝐵 = 0
𝑃𝐴 − 𝑃𝐵 = −𝜌1 𝑔ℎ1 −𝜌𝑚 𝑔ℎ + 𝜌2 𝑔(ℎ1 +ℎ)
= 𝒈𝒉𝟏 𝝆𝟐 − 𝝆𝟏 + 𝒈𝒉 𝝆𝟐 − 𝝆𝒎
• If liquid in both the pipes are same with density 1 = 2 = . So, the pressure difference
will be,
𝑃𝐴 − 𝑃𝐵 = 𝜌 − 𝜌𝑚 𝑔ℎ
HYDROSTATIC FORCES ON SURFACES
Introduction
❑Total pressure is defined as the force exerted by a static fluid on a surface either
plane or curved when the fluid is in contact with the surfaces .
❑Centre of Pressure : Point of application of total pressure on the surface
❑There are four cases of Submerged surfaces on which total pressure and centre of
pressure is to be determined. The submerged surfaces may be:
❑Horizontal plane surface
❑Vertical Plane surface
❑Inclined Plane surface
❑Curved Surface
Horizontal Plane Surface
❑ Consider a plane horizontal surface immersed in a static fluid
❑ As every point of the surface is of the same depth from the free surface of the liquid,
the pressure intensity will be the same on the entire surface and its equal to 𝜌𝑔ℎ
❑ Therefore the total hydrostatic force = pressure x area
𝐹=𝜌𝑔ℎ.𝐴
❑ For a horizontal plane surface, the center of pressure is located at the vertical centroid of the
submerged area.
❑ The centroid is the geometric center of the surface, and in the case of a simple geometric shape
like a rectangle or a circle, it corresponds to the center of the shape.
❑ hc =h = Depth of Centre of pressure from free surface
Inclined or vertical Plane Surface
𝑑𝐹 = 𝑃 𝑑𝐴 = ρ𝑔ℎ 𝑑𝐴
𝑑𝐹 = ρ𝑔 𝑦 𝑠𝑖𝑛θ 𝑑𝐴
Free surface
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
q
h
hC
𝐹𝑅 = න 𝑑𝐹 = ρ𝑔 𝑠𝑖𝑛θ න 𝑦𝑑𝐴
y
yC
X axis
C.g
dA
1𝑠𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓
𝑎𝑟𝑒𝑎 𝑤. 𝑟. 𝑡 𝑥𝑎𝑥𝑖𝑠
𝐴𝑙𝑠𝑜 𝑟𝑒𝑐𝑎𝑙𝑙 𝑡ℎ𝑎𝑡 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑎𝑛 𝑎𝑟𝑒𝑎
in y direction is
1
𝑦𝑐 = න 𝑦𝑑𝐴
𝐴
𝐹𝑅 = ρ𝑔 𝐴 𝑦𝑐 𝑠𝑖𝑛θ = ρ𝑔A hc
hc 𝑖𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑐𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑜 𝑡ℎ𝑒
𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑙𝑖𝑛𝑒𝑑 𝑝𝑙𝑎𝑛𝑐𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒,
Y axis
A is the area of the inclined plane surfaces
𝐹𝑅 = ρ𝑔 𝐴 𝑦𝑐 𝑠𝑖𝑛θ = ρ𝑔A hc
𝐼𝑓 𝑖𝑡 𝑖𝑠 𝑣𝑒𝑡𝑖𝑐𝑎𝑙 𝑆𝑢𝑟𝑓𝑎𝑐𝑒
𝑦=ℎ
𝜃 = 90 𝑑𝑒𝑔𝑟𝑒𝑒
Sin 90 degree=1
𝐹𝑅 = ρ𝑔A hc
Where does the force act ?
Pressure force is a normal stress therefore it will act normal.
Thus we know that Pressure force will act normal to the surface. But we have to find where
Sum moments about point O
i.e. the sum of all the individual forces
about point O, should be equal to
resultant force multiply by the distance
to where it act
Sum moments about point O
𝑦𝑅 𝐹𝑅 = න 𝑦 𝑑𝐹 = න 𝑦 𝑃𝑑𝐴
𝑦𝑅 𝐹𝑅 = න 𝑦 (𝜌𝑔 𝑦 sin θ ) 𝑑𝐴
O
𝑦𝑅 𝐹𝑅 = 𝜌𝑔 sin θ න 𝑦 2 𝑑𝐴
𝑦𝑅
Recall that 𝐼𝑥 = ‫ 𝑦 ׬‬2 𝑑𝐴
But also 𝐼𝑥 = 𝐼𝑥𝑐 + 𝐴𝑦𝑐 2
(𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑎𝑥𝑖𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚)
𝑦𝑅 𝐹𝑅 = 𝜌𝑔 sin θ (𝐼𝑥𝑐 + 𝐴𝑦𝑐 2 )
FR
𝑦𝑅 (𝜌𝑔 𝑦𝑐 𝐴 𝑆𝑖𝑛 θ) = 𝜌𝑔 sin θ (𝐼𝑋𝑐 + 𝐴𝑦𝑐 2 )
𝐼𝑥𝑐
𝑦𝑅 = 𝑦𝑐 +
𝑦𝑐 𝐴
𝑦𝑅 𝑖𝑠 𝑠𝑜𝑚𝑒𝑡𝑖𝑚𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐𝑝
In the centroid there is value of X and Y
In X direction, Similar Procedure gives
𝑋𝑅 = 𝑋𝑐 +
𝐼𝑥𝑦𝑐
𝑦𝑐 𝐴
In either of the centroidal axis are axis of symmetry , 𝐼𝑥𝑦𝑐 = 0, XR= Xc
In the centroidal axes of a section are symmetric, then the product of moment of inertia will
be zero. This is because symmetry about the symmetrical axis implies that the distribution of
area on one side of the axis is exactly balanced by the distribution on the other side
Example
y’
2m
O
hinge
The square flood gate (2m by 2m) is hinged along its
bottom as shown. Determine the moment at the hinge in
order to hold the gate steady.
First, find the resultant force:
FR = 𝜌𝑔ℎ𝐶 𝐴 = (1000)(9.8)(1)(2 × 2) = 39200(𝑁)
Then, determine the point of action:
1
3
(2)(2)
𝐼𝑥ො𝑥ො
1 4
12
y′=yC +
= (1) +
= 1 + = (𝑚)
𝐴𝑦𝐶
(2 × 2)(1)
3 3
As expected, it falls at a depth 2/3 of the total depth.
The holding moment (M) on the hinge O will be
4
෍ MO = 𝑀 − 𝐹𝑅 (2 − ) = 0,
3
x
𝑀 = 26133.33(𝑁. 𝑚)
y
Example (cont.)
45°
2m
y’
y
If the square gate is placed at an angle of 45° as
shown, recalculate the holding moment again.
CONTINUATION
HYDROSTATIC FORCES ON SURFACES……
Atmosphere
h1
Rectangular gate
Atmosphere
h2
hinge
Water
q
q = 45
h1= 1m
h2 = 3m
Width 1 m into the board
Weight of the Gate = 90 KN
Will the gate stay in place or fall ?
𝐹𝑅 = 𝜌𝑔𝐴 ℎ𝑐
𝑁
𝐹𝑅 = 9810 𝑚3
2.5 𝑚 1 ( 3 2 )
3
𝑆𝑖𝑛 45 = hypotenuse
, Sin 45 = 1/ 2
𝐹𝑅 = 104, 051 𝑁 = 104.1 𝐾𝑁
𝐼
𝑦𝑐 𝐴
𝑦𝑅 = 𝑦𝑐 + 𝑥𝑐
𝐹𝑅
𝑦𝐶 = 2.5 ( 2) = 3.54 m
𝑏𝑑3
𝐼𝑥𝑐 = 12 =
1 (3 2)3
= 6.36
12
𝑦𝑅 = 3.54 +
6.36
3.54 (3 2)
𝑦𝑅 = 3.54 + 0.424 = 3.96
Sum moment about Point O =0
=(1.5) (90) - (1.7) ( 104.1)
=-41.8 KN. m
W
R
O 1.5 m
=minus mean counter clockwise, so the gate will
stay in place
Hydrostatic Force on the Curved Surface
Not everything in the real world are plane surfaces, there are also Curved Surfaces
Curved Surface
Consider a curved surface AB submerged in the static fluid
Let dA be area of small strip at a depth h from the free surface
Normal elemental pressure force on the elemental strip,
𝑑𝐹=𝑝𝑑𝐴 =𝜌𝑔ℎ.d𝐴
𝑑𝐹v
Total pressure force on the curve surface should be ,
𝑑𝐹
A
𝑑𝐹h
𝑑𝐹= ‫𝑔𝜌 ׬‬h𝑑𝐴
But here the direction of forces on the small area is not the same
but varies from point to point
Hence integration for curved in this case is not possible
The problem can be solved by resolving the force dF into x and y
component.
B
The Resultant force is FR= (𝐹ℎ2 + 𝐹𝑣2 )
Rule 1: To find the horizontal Component FH
Project the curve surface into vertical plane and use the previous formula uses for vertical surfaces
ℎ𝐶
𝐹ℎ = 𝜌𝑔ℎ𝑐 𝐴
A
B
Also find Line of action same as before
Projected area
Rule 2: To find the Vertical Component FV
𝑭𝒗 = 𝝆𝒈𝑽
Where, 𝑉 = volume of fluid above the curve surface up to the free surface real or
imaginary
𝑭𝒗
volume
A
Line of Action: Forces goes through the centroid of the area.
B
It will not be in the middle, but somewhat towards the right
because there is more area on the right
The Resultant force is FR= (𝐹ℎ2 + 𝐹𝑣2 )
Compute the horizontal and vertical components of the total force acting on a curve surface
AB which is in the form of a quadrant of a circle of radius 2m. Take width of the gate as unity
𝜋𝑟 2
Quarter of a circle= 4
BUOYANCY
What is a Buoyant Force ?
❑ Consider a solid body of arbitrary shape submerged in a
liquid
h1
❑ Consider an elementary strip of this body.
❑ The pressure at the top of the body will be lesser as compare
to the pressure at its bottom
❑ Since the pressure at the bottom is more than at the top, it
will raise the object up. It’s weight will pull it down
❑ Pressure goes up as we goes down P= Po + 𝜌𝑔h,
is surface pressure at the top
where Po
Sum of the forces in the Z direction due to pressure in the elementary strip
dfZ= (Po + 𝜌𝑔h2) dA - (Po + 𝜌𝑔h1) dA
Pressure is a compressive stress, it will always point into the surface.
h2
dfZ= (Po + 𝜌𝑔h2) dA - (Po + 𝜌𝑔h1) dA
dfZ= 𝜌𝑔 (h2 - h1) dA
but (h2 - h1) dA = dv = differential volume
dfZ= 𝜌𝑔 dv
Integrate both side
𝑉
න dfZ = න 𝜌𝑔 dv = 𝜌𝑔 𝑉
0
𝐹𝐵 = 𝜌𝑔𝑉
Where V is the volume of liquid displaced by body.
It acts at the centroid of the displaced volume ( volume of liquid
pushed away by the object)
h1
h2
What is a Buoyant Force ?
When a body is wholly or partially submerged in the fluid, the upward force exerted by the
fluid on the body is called Buoyant force.
The phenomena is called Buoyancy
𝐹𝐵 = 𝜌𝑔𝑉
Where V is the volume of liquid displaced by body.
Example:
A circular log is found floating in a water. ¼ of its volume is above the water surface. Find the specific weight of the log
Sum of forces in y direction =0
Wt = FB
γwater
γlog * Volume of Log =γwater * Volume displaced
γlog = γwater * ( Voldisplaced / Vollog)
Wt
γlog = 9810 N/m3 * ¾ = 7351 N/m3
FB
Find the volume of water displaced and position of centre of Buoyancy for a wooden block of
width 2.5 m and depth of 1.5 m when it floats horizontally in water. The density of the
wooden block is 650 kg/m3 and its length is 6 m
❑ When a body (say a ship) float in water, it is acted upon by two forces:
1. Vertical downward force of gravity which is equivalent to the weight
of the body W acting at the centre of Gravity G
2. Upward vertical Buoyancy force FB which is equal to the weight of
water displaced by the immersed body which passes through the
centroid of displaced volume of water B (centre of Buoyancy)
Weight of water displaced by immersed body
❑ For a Floating body to be in Equilibrium
1. The Buoyancy force should be equal to the weight of the body
FB = W
2. Also the centre of Gravity and Centre of Buoyancy should lie on the
same vertical line called Centre line BG
❑ Now say a body undergoes an angular displacement say in the right
direction, then the volume displace is larger on the right side
❑ Because the volume displace is larger on the right side the center of
Buoyancy shift towards the right side from B to B’
❑ If we draw a vertical line from new center of Buoyancy.
❑Here the point of intersection of the vertical line through the new center
of Buoyancy B’ and center BG is termed as the MetaCenter of the body.
❑ The distance between the MetaCenter and the center of Gravity of the
floating body is called the MetaCentric height
MetaCentric height (GM) = BM- BG
❑ Depending upon the location of MetaCenter the floating body possesses
three types of Equilibrium:
1. Stable Equilibrium
2. Unstable Equilibrium
3. Neutral Equilibrium
Stable Equilibrium : When M is above G, then in this case, a restoring couple is
formed by the Buoyancy force and the weight of the body which tends to turn
the body to its original position
UnStable Equilibrium : When M is below G, then in this case, a overturning couple
is formed by the Buoyancy force and the weight of the body which tends to sink the
body from its original position
Neutral Equilibrium : When M coincide with G, then in this case, the line of action
of Buoyancy force and the weight of the body are collinear and passes through the
same point . Due to this the body neither return to its original position nor increase
its displacement further
REVISION IN THE FORM
OF MCQ
Q1. A fluid which has shear stress proportional to the rate of shear strain is called
a) Non- Newtonian Fluid
b) Ideal Fluid
c) Ideal Plastic Fluid
d) Newtonian Fluid
Answer : d) Newtonian Fluid
e.g. of Newtonian :Water, Oil, Air
e.g. of Non- Newtonian :Blood, Toothpaste , paint
Q2. The SI Unit of Kinematic Viscosity
a) 𝑚/𝑠 2
b) 𝑚2 /𝑠
c) 𝑚3 /𝑠
d) None of the above
Answer : b) 𝑚2 /𝑠
CGS = 𝑐𝑚2 /𝑠 = Stoke
1 Stoke = 10-4𝑚2 /𝑠
Q3. The SI Unit of Dynamic Viscosity
a) 𝑁. 𝑚2 /𝑠
b) 𝑁/𝑠 2 /𝑚
c) 𝑁. 𝑚/𝑠
d) 𝑁. 𝑠/𝑚2
a) Answer : d) 𝑁. 𝑠/𝑚2
= 𝑃𝑎. 𝑠
𝐶. 𝐺. 𝑆 = 𝑑𝑦𝑛𝑒. 𝑠/𝑐𝑚2 (𝑝𝑜𝑖𝑠𝑒)
0.1 𝑁. 𝑠/𝑚2 = 1 poise
Q4. The angle of contact between mercury and glass in case of capillary depression is
a) 600
b) 200
c) 900
d) 1280
Angle of contact: It is the angle between
the solid surface and the tangent to the
surface of water inside liquid.
For wetted surface, angle of contact remains less than 𝟗𝟎𝒐 .
Here the liquid will be spread when kept on solid surface.
Adhesion is more than cohesion . This type of substance are
called Hydrophilic
Answer : d) 1280
and for unwetted surface, angle of contact remains more
than 𝟗𝟎𝒐 . Rather than spreading it will accumulate like a
spherical. This kind of substance are called Hydrophobic.
Q5. With an increase in the radius of the tube, the rise of liquid in the tube due to
surface tension is :
a) Decrease
b) Increase
c) Remain Unchanged
d) None of the above
𝒉=
Answer : a) Decrease
𝟒𝝈𝒄𝒐𝒔𝜽
𝝆𝒈𝒅
Q6. Specific Volume and mass density is :
a) Directly Proportional
b) Inversely Proportional
c) Proportional to square
d) Proportional to square root
𝜌=
Answer : b) Inversely proportional
𝑚
𝑉
𝑉 1
=
𝑚 𝜌
Q7. The diameter of droplet is 0.075 mm. What is the intensity of the pressure
(N/sq.cm) developed in the droplet by surface tension of 0.000075 N/mm ?
a) 0.4
∆𝑃𝑅
2
2𝜎
𝑃=
𝑅
b) 0.6
𝜎 =
c) 0.8
d) 1
𝑃=
2×0.000075
0.0375
𝑃 = 4 × 10−3 𝑁/𝑚𝑚2
Answer : a) 0.4
𝑃 = 0.4 𝑁/𝑐𝑚2
Q8. For a floating body to be in stable equilibrium, its metacentre should be:
a) Below the Centre of gravity
b) Below the centre of Buoyancy
c) Above the centre of Buoyancy
d) Above the centre of gravity
Answer : d) Above the centre of gravity
Q9. The rise of liquid in manometers gives the :
a) Mercury level at the point
b) Discharge capacity at that point
c) Pressure head at that point
d) Density of water at that point
Answer : c) Pressure head at that point
Q10. The atmospheric pressure head is ____________ mm of mercury:
a) 9.81
b) 760
c) 10.33
d) 9810
101.3 kPa
14.7 psia
760 mm Hg
30 inches Hg
34 ft of Water
Answer : b) 760
1 atm= 760 mm of mercury
1 atm= 10.3 m of water
ALL ARE EQUAL
Q11. A differential manometer measures:
a) Absolute pressure at a point
b) Local atmospheric pressure
c) Difference in total energy between two points
d) Difference in pressure between two points
Answer : d) Difference in pressure between two points
Q12. Which is the simplest form of manometer used for measuring gauge pressures:
a) U-Tube manometer
b) Simple manometer
c) Differential manometer
d) Piezometer
Answer : d) Piezometer
Q13. Absolute Pressure is:
a) Atmospheric pressure + gauge pressure
b) Atmospheric pressure – gauge pressure
c) Vacuum pressure + gauge pressure
d) Vacuum pressure – gauge pressure
a) Answer : a) Atmospheric pressure + gauge pressure
𝑃𝐴 − 1000×9.81×50×10−3 −980× 9.81 × 50×10−3
+ 1000 × 9.81× 0.1 =𝑃𝐵
b) -10 N/m2
Answer d)
A wooden log of 0.6 m diameter and 5 m length is floating in a river water. Find the
depth of wooden log in water when the sp. gr. of the wooden log is 0.7
Given Data:
diameter, d = 0.6 m, L= 5m
Sp gravity of log =0.7, hence 𝜌𝑙𝑜𝑔 = 700 𝑘𝑔/𝑚3
B
- - -- ------ -- -
A
2𝜃
Specific weight of log, 𝛾𝑙𝑜𝑔 = 𝜌𝑙𝑜𝑔 × 9.81 = 700 × 9.81 N/𝑚3
Weight of the log = 𝛾𝑙𝑜𝑔 × volume of the log
𝜋
= 700×9.81× 4 × 0.6 2 × 5 = 9708 𝑁
Weight of the log = Weight of water displaced
9708 = 𝛾𝑤𝑎𝑡𝑒𝑟 × volume of the water which is displaced
9708
volume of the water which is displaced =9810 =0.98 96 𝑚3 = volume of the log below the water
C
- ------ --
0.6 m h
D
volume of the water which is displaced =0.98 96 𝑚3 = volume of the log below the water
Area (ADCA) × 5 =0.98 96 𝑚3
B
Area (ADCA) =0.1979 𝑚2
Area (ADCOA) + Area (AOC) =0.1979 𝑚2
(360−2𝜃)
1
𝜋r 2 360 + 2 𝑟𝑐𝑜𝑠𝜃2𝑟𝑆𝑖𝑛𝜃= 0.1979
𝜃 − 57.2 𝑆𝑖𝑛𝜃 𝐶𝑜𝑠𝜃 − 54. 01 = 0
By hit and trial method
𝜃= 71.50 (approximately , almost equal to +0.248)
ℎ = 𝑟 + 𝑟 cos 𝜃 = 0.3 + 0.3 𝐶𝑜𝑠 75 = 0.395 𝑚
- - -- ------ -- -
A
C
r
2𝜃
- ------ --
O
0.6 m h
D
𝜃
Area of a Sector=2𝜋 ×𝜋𝑟 2
rsin𝜃 C
A rsin𝜃
r
r Cos𝜃
𝜃 𝜃
O
r
Derivation of Expression for Metacentric Height (GM)
▪ Consider a body floating on the surface of a liquid having
the G and B as the center of gravity and center of
buoyancy as shown in Figure.
• Let a small angular displacement θ has been given to the
body in the clock wise direction. The center of buoyancy
moves to 𝐵′ .
‘
• As the result of angular tilt, amount of water displaced by
body on the right side will increase by an amount of
volume of wedge 𝑄𝑂𝑄′
• Therefore, there will be a gain in buoyant force, let this
gain is 𝑑𝐹𝐵 .
• However, on left side the wedge, POP’ is coming out of
water so there will be loss in buoyant force in left side.
• 𝑑𝐹𝐵 in right side will be upward however in left side it will be downward. Therefore, the two forces will
form a couple against tilt.
❑ Consider the elemental strip of thickness 𝑑𝑥 on the wedge located at distance 𝑥 from the center of the body.
The height of the strip is 𝑥 × 𝜃 = 𝑥𝜃
Area of the strip = height × thickness= 𝑥𝜃 × 𝑑𝑥
Volume of the strip =area × L= 𝑥 × 𝜃 𝑑𝑥 × 𝐿
Therefore, the weight of strip will be
𝑑𝑊 = 𝜌 × 𝑔 × 𝑥 × 𝜃 × 𝑑𝑥 × 𝐿
If we consider a similar strip on the LHS, it will also have the
same weight but in the opposite direction
The two weight are acting in an opposite direction, hence they will
form a couple
‘
Moment of this couple = Weight of each strip × Distance
between these two weight
𝑑𝑀 = 𝜌𝑔𝑥𝜃𝐿𝑑𝑥 𝑥 + 𝑥
= 2𝜌𝑔𝑥 2 𝜃𝐿𝑑𝑥
Therefore, the total moment couple for whole wedge will be
= ‫ ׬‬2𝜌𝑔𝑥 2 𝜃𝐿𝑑𝑥
L
𝑥
𝜃
𝑥
• This moment will be equal to the moment due to force of buoyancy at B’. Therefore,
න 2𝜌𝑔𝑥 2 𝜃𝐿𝑑𝑥 = 𝐹𝐵 × 𝐵𝐵′
Since, 𝐵𝐵’ = 𝐵𝑀 × 𝜃
• Therefore
න 2𝜌𝑔𝑥 2 𝜃𝐿𝑑𝑥 = 𝐹𝐵 × 𝐵𝑀 × 𝜃
Since, 𝐿. 𝑑𝑥 = 𝑑𝐴 − 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑠𝑡𝑟𝑖𝑝. So,
න 2𝜌𝑔𝑥 2 𝜃dA = 𝐹𝐵 × 𝐵𝑀 × 𝜃
Since, 2‫ 𝑥 ׬‬2 𝑑𝐴 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑌𝑌 𝑎𝑥𝑖𝑠 = 𝐼𝑌𝑌
𝜌𝑔 × 𝐼𝑌𝑌 = 𝐹𝐵 × 𝐵𝑀
𝜌𝑔𝐼𝑌𝑌
𝐵𝑀 =
𝐹𝐵
Here force of buoyancy= 𝐹𝐵 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑏𝑜𝑑𝑦 = 𝜌𝑔𝑉′
• Therefore,
Therefore, metacentric height will be,
𝜌𝑔𝐼
𝐼
𝐵𝑀 =
=
𝜌𝑔𝑉′ 𝑉 ′
𝐼
𝐺𝑀 = 𝐵𝑀 − 𝐵𝐺 = ′ − 𝐵𝐺
𝑉
• Where I is the moment of inertia of horizontal cross section area of the body at the liquid surface
about its longitudinal axis.
A block of wood of specific gravity 0.7 floats in water. Find the metacentric height of
the block if its size is 2m × 1m × 0.8m
A rectangular pontoon is 5m long, 3m wide and 1.2m high. The depth of immersion
is 0.80m in sea water. If the CG is 0.6 m above the bottom of the pontoon, determine
the metacentric height. Density of sea water = 1025 kg/m3.
A
Top view
AG=0.6 m AB= 0.4 m ( half of the depth of Immersion)
Therefore BG= 0.2 m
Next we need to find Moment of Inertia about the Y-Y axis
.
REVISION-NUMERICAL
PROBLEMS
Question :GATE 2021 (1 mark)
A lake has maximum depth of 60 m. If the mean atmospheric pressure in the lake region is 91 kPa and
the unit weight of the lake water is 9790 N/m3, the absolute pressure (in kPa, round off to two decimal
places) at the maximum depth of the lake is _______.
Question :GATE 2021 (2 mark)
A cylinder (2.0 m diameter, 3.0 m long and 25 kN weight) is acted upon by water on one side and oil
(specific gravity = 0.8) on other side as shown in the figure.
The absolute ratio of the net magnitude of vertical forces to the net magnitude of horizontal forces
(round off to two decimal places) is_______
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑅𝑎𝑡𝑖𝑜 =
𝑁𝑒𝑡 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙
𝑁𝑒𝑡 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙
𝑁𝑒𝑡 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 = 𝐹𝑉𝑤 + 𝐹𝑉𝑂 − 𝐹𝑠𝑒𝑙𝑓𝑤𝑔𝑡
𝐹𝑠𝑒𝑙𝑓𝑤𝑔𝑡 = 25 𝐾𝑁 (𝐺𝑖𝑣𝑒𝑛)
𝜋 1 2
𝜋 1 2
Net vertical= 1000×9.8× 2 × 3 + 800×9.8× 4 × 3 – 25000= 39.72 KN
𝑁𝑒𝑡 𝐹𝐻 = 𝐹𝐻𝑤 − 𝐹𝐻𝑜
𝜌𝑤 𝑔ℎ1 𝐴1 − 𝜌𝑜 𝑔 ℎ2 𝐴2
Net horizontal= 1000×9.81× 2 × 2×3 − 800×9.81× 0.5 × 1×3 = 105. 95 𝐾𝑁
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑅𝑎𝑡𝑖𝑜 =
39.72
= 0.375
105.95
𝜋𝑟 2
Quarter of a circle= 4
BERNOULLI EQUATION
Bernoulli’s equation is of three form:
The first one is the pressure form of the Bernoulli’s equation
𝑃+
1
𝜌 𝑉 2 + 𝜌𝑔𝑧 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2
Unit of every term is Pa
𝑃: 𝑆𝑡𝑎𝑡𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
1
𝜌 𝑉 2 = 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
2
𝜌𝑔𝑧= Hydrostatic pressure
The second one is the head form of the Bernoulli’s equation (divided by rho g)
The third one (divided by rho). Not commonly used
Practical Applications of
Bernoulli's
Venturimeter
❑ A Venturimeter is a device used to measure the flow rate or discharge of the fluid through pipe
❑ It is based on the principle of Bernoulli’s equation
❑ It is consist of three part.
1. Converging section
2. Throat
3. Diverging section
❖Converging section
❑ The angle of convergence is approximately 210 ± 10 .
❑ This angle is steep enough to allow smooth acceleration but not too steep to cause
turbulence
❑ Approximate length is 2.7 (D-d), D is diameter of inlet section and d is diameter of
1
3
the throat ( d= 𝑡𝑜 D)
3
4
❖Throat
❑ Having minimum cross section area
❑ The length of the throat is approximately d (length is equal to its diameter).
❖Diverging section
❑ Diameter of the pipe changes from minimum to the diameter of the pipe
❑ The angle of divergence is approximately 50 to 70
• The divergence angle is much
smaller than the convergence
angle because the fluid is
decelerating as it moves through
the diverging section.
• A small divergence angle helps to
prevent flow separation and
minimize the formation of eddies
and vortices, which can cause
energy losses and lead to
inaccurate flow measurements.
❑ If the venturi's cross-sectional area increases too quickly (meaning the diverging angle is too wide), the fluid flow
can slow down and separate from the walls, and create areas of low pressure.
❑ These low-pressure areas cause the fluid to spin and swirl, forming eddies.
❑ These eddies disrupt the smooth, streamlined flow, leading to increased friction, energy loss, and inaccuracies
in measuring the flow rate.
Derivation of Expression for Rate of Flow through Venturimeter
Let a Venturimeter is fitted in a pipe
d1 = the diameter at section 1 or inlet
d2 = the diameter at section 2 or throat
P1 = pressure at section 1
P2 = pressure at section 2
V1 = velocity at section 1
V2 = velocity at section 2
Applying the Bernoulli's equation between section 1 and 2.
Since, the pipe is horizontal,
Let the pressure head difference between both limbs of differential manometer is therefore
ℎ = (𝑝1 −𝑝2 )/𝜌𝑔
𝑣22
𝑣1 2
ℎ= −
2𝑔
2𝑔
From Continuity Equation, 𝐴1 𝑣1 = 𝐴2 𝑣2
𝑣1 = (𝐴2 𝑣2 )/𝐴1
h
The discharge Q is under ideal conditions and its called theoretical discharge.
Actual discharge will be somewhat lesser than theoretical discharge (due to friction,
viscosity)
Actual discharge can be given as
where, 𝐶𝐷 is called coefficient of Venturimeter or coefficient of discharge.
Its value is always less than 1.
Now, if the differential manometer is used to measure the pressure difference between section 1
and section 2 then 𝒉 can be given as,
• Case 1: Manometric liquid is heavier than the liquid flowing in pipe
Let 𝑆𝑚 and 𝑆𝑝 are the specific gravity of liquids in manometer and in pipe respectively. Then
𝒉 can be given as,
𝑆𝑚
ℎ=𝑥
−1
𝑆𝑝
Where 𝑥 is the difference of manometer liquid column in differential manometer
Case 2: Manometric liquid is lighter than the liquid flowing in pipe
𝑆𝑚
ℎ =𝑥 1−
𝑆𝑝
The pressure difference isn't just due to the weight of the manometric liquid but is also offset by the presence of the
liquid in the pipe. By subtracting 1, you effectively correct for the pressure exerted by the liquid in the pipe, isolating
the contribution of the manometric liquid.
❑ Inclined and vertical Venturimeter
If the pipe in which discharge is to be measured is not horizontal then inclined or vertical
Venturimeter are used. Same Equation of discharge hold good in this case
𝑝1 𝑝2
ℎ=
−
+ 𝑧1 − 𝑧2
𝜌𝑔 𝜌𝑔
❖Advantages
❑ Head loss in pipe due to installation of venturi meter is small
❑ Due to small losses, the coefficient of discharge is high (generally from 0.95 to
0.98) .
❖Disadvantages
❑ Expensive
❑ Space requirement is more due to its long length
FLUIDS KINEMATICS
Definition
❑Kinematics is the branch of fluid mechanics which deals with
the motion of fluid particles without considering forces which
are responsible for the motion.
Methods of Describing Fluid Flow
❑Lagrangian Approach : The Lagrangian approach focuses on tracking individual
fluid particles as they move through space and time. In this approach, the fluid
particle is the point of interest, and its properties (such as velocity, acceleration, and
position) are followed as it moves along its trajectory.
❑Eulerian Approach :The Eulerian approach involves observing and analyzing the
flow properties of a fluid at specific points in space while keeping those points fixed.
Eulerian Approach : Looking at a fixed point in space
Lagrangian Approach: Tag a fluid particles
Types of fluid flow
1. Steady flow and Unsteady flow
Steady flow
• The flow in which the flow characteristic like velocity, pressure, density etc , at a point do not change with time is
called as steady flow
𝒅𝑽
𝒅𝒂
𝒅𝝆
= 𝟎;
= 𝟎;
= 𝟎;
𝒅𝒕
𝒅𝒕
𝒅𝒕
Unsteady flow
• If the flow in which the flow characteristic like velocity, pressure, density etc , at a point changes with time is called as
unsteady flow
𝒅𝑽
𝒅𝒂
𝒅𝝆
𝒅𝑷
≠ 𝟎;
≠ 𝟎;
≠ 𝟎;
≠ 𝟎;
𝒅𝒕
𝒅𝒕
𝒅𝒕
𝒅𝒕
2. Uniform and Non Uniform flow
Uniform flow
The flows, in which flow parameters such as velocity, acceleration, density and pressure at a given time does not
varies in space is called uniform flow,
𝒅𝑽 𝒅𝑽 𝒅𝑽
=
=
= 𝟎;
𝒅𝒙 𝒅𝒚 𝒅𝒛
Non-uniform Flow
The flows, in which the flow parameters at any given time changes with respect to space is known as non-uniform
flows.
𝒅𝑽
𝒅𝑽
𝒅𝑽
≠ 𝟎;
≠ 𝟎;
≠ 𝟎;
𝒅𝒙
𝒅𝒚
𝒅𝒛
Any of the four combination is possible
1. Steady uniform flow : Conditions do not change with position in the stream or with time. An example is the
flow of water in a pipe of constant diameter at constant velocity.
2. Steady non-uniform flow. Conditions change from point to point in the stream but do not change with time. An
example is flow in a tapering pipe with constant velocity at the inlet - velocity will change as you move along the
length of the pipe toward the exit
3. Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with
time. An example is a pipe of constant diameter connected to a pump pumping at a constant rate which is then switched
off.
4. Unsteady non-uniform flow. Every condition of the flow may change from point to point and with time at every
point. For example waves in a channel.
Laminar and Turbulent flow
Laminar flow
• Is defined as that type of flow in which the fluid particles move in a well defined path or
stream line and all the streamline are straight and parallel .
• The particle moves in lamina or layer gliding smoothly over the adjacent layer
• It is also sometime called streamline flow or viscous flow
Turbulent flows
• In turbulent flows, fluid particles moves in random and unpredictable manner.
Laminar and Turbulent flow are determined by a dimensionless number called Reynold Number
ρ 𝑉𝐿
Re=
µ
L in Reynold number is not length of the pipe. It is a characteristics length.
In case of pipe, this characteristic length is represent by diameter of the pipe.
For an open channel, the characteristic length often used is the hydraulic radius.
The hydraulic radius (R) for an open channel is defined as the cross-sectional area of flow (A) divided by the
wetted perimeter (P
FOR PIPE FLOW
FOR OPEN CHANEL FLOW
Re < 2000, laminar
Re> 4000, Turbulent
Re < 500, laminar
Re> 2000, Turbulent
3. Compressible and incompressible flow
Compressible flow
• Compressible flows are the flows in which density varies considerably within flow field.
It means the density, 𝜌, is a function of space and time (𝑥; 𝑦; 𝑧; 𝑡) therefore it can not
be considered as constant.
Example : gases , vapour steam, changes in volume leads to change in density
Incompressible flows
• In a flow field, if density remains constant then the flow will be called as incompressible
flows. Incompressible flows should not be should not be confused with incompressible
fluids.
• Example water
One, two and three dimensional flow
• In real flows, flow parameters such as velocity, acceleration, density, pressure,
temperature etc., varies in all three dimensions and also with time therefore flow
parameters are function of x, y, z and t.
𝑉 = 𝑓 𝑥; 𝑦; 𝑧; 𝑡
Such flows are known as three dimensional flow.
Similarly, 𝑣 = 𝑓 𝑥; 𝑦; 𝑡 , for 2 D
𝑣 = 𝑓(𝑥; 𝑡) and 1D flow.
A 30 cm diameter pipe, conveying water, branches into two pipes of diameters 20 cm and 15 cm
respectively. If the average velocity in the 30 cm diameter pipe is 2.5 m/s. Find the discharge in
this pipe. Also determine the velocity in 15 cm pipe if the average velocity in 20 cm diameter pipe
is 2 m/s
GATE 2015 (1 mark)
A main channel of width 450 mm branches into two sub-channel having width 300 mm and
200 mm as shown in figure. If the volumetric flow rate (taking unit depth) of an in
compressible flow through the main channel is 0.9 m3/s and the velocity in the sub-channel of
width 200 mm is 3 m/s, the velocity in the sub-channel of width 300 mm is ________m/s.
Assume both inlet and outlet to be at the same elevation.
Water flows through a pipe AB 1.2 m diameter at 3 m/s and then passes
through a pipe BC 1.5 m diameter. At C, the pipe branches. Branch CD is 0.8 m
in diameter and carries one third of the flow in AB. The flow velocity in branch
CE is 2.5 m/s. Find the volume rate of flow in AB, the velocity in BC, the velocity
in CD and the diameter of CE
ORIFICE METER
❖Introduction
❑ An orifice meter is a device which is also used for measuring the discharge
through pipes.
❑ It works on the same principle as that of venturimeter
❑ It is a cheaper compared to venturimeter
❑ Finds application where space is limited and the accuracy is not that much
important
❖Construction
❑ An orifice meter consists of a flat circular plate with a circular hole called "orifice"
which is concentric with the pipe axis
❑ The orifice diameter (d) is generally kept at 0.5 times the diameter of the
pipe though it may varies from 0.4 to 0.8 times the diameter of the pipe (D)
❖Construction
❑ A differential manometer is connected at section a which is at a distance of
about 1.5 to 2 times the pipe diameter from the orifice plate
❑ Area and Position of position of vena contracta will change will shape of orifice. It
will vary between 0.5 to 0.6 times the diameter of the orifice on the
downstream side from the orifice plate.
❖Construction
❑ When the fluid passes through the orifice, it converge at the d/s side
❑ The minimum stream section occurs not in the plane of the orifice, but
somewhat downstream. This minimum cross section is called a vena contracta.
This is also the location of minimum pressure.
❑ The maximum possible
pressure difference exists
between the sections 1 and
2 ( vena contracta ), which
is measured by connecting a
differential manometer.
❖Construction
❑ The jet of the fluid coming out of the orifice expands from vena contracta to again
fill the pipe.
❑ Since there is an abrupt change in the cross sectional area of the flow
passage, so greater loss of energy is experienced compared to venturi meter
❑ The coefficient of discharge for orifices in
pipes varies from 0.60 to 0.80
Derivation of Expression for Rate of Flow through Orifice Meter
Apply Bernoulli Equation at section 1 and 2
h = Piezometric Differential head
From continuity equation,
𝐴1 𝑣1 = 𝐴2 𝑣2
𝐴2 𝑣2
𝑣1 =
𝐴1
Section 2 is at vena contracta then 𝐴2 will be of cross-section area at vena-contracta. Let 𝐴𝑜 is the area of hole on
orifice plate. Then, coefficient of contraction can be given as
𝐴2
𝐶𝑐 =
𝐴𝑜
So,
𝐴2 = 𝐴𝑜 𝐶𝑐
𝑣1 =
𝐴𝑜 𝐶𝑐 𝑣2
𝐴1
Now substituting the value of v1 in Equation (5.21).
𝑉2 =
𝐴𝑜 2 𝐶𝑐 2 𝑉22
2𝑔ℎ +
𝐴12
𝑉2 =
2𝑔ℎ + 𝑉12
𝑉2 =
𝐴𝑜 2 𝐶𝑐 2 𝑉22
2𝑔ℎ +
𝐴12
On simplification,
𝑉2 =
2𝑔ℎ
𝐴𝑜 2 2
1− 𝐴
𝐶𝑐
1
Discharge Q= 𝐴2 𝑉2
But we generally do not use Co-efficient of contraction, we use co-efficient of discharge and their relationship is
𝐶𝐶 = 𝐶𝑑
𝐴𝑜 2 2
1− 𝐴
𝐶𝑐
1
𝐴𝑜 2
1− 𝐴
1
Substituting the value of Cc and Solve for Q
𝑄=
𝐶𝑑 𝐴𝑜 𝐴1 2𝑔ℎ
𝐴12 − 𝐴2𝑜
𝑆𝑚
−1
𝑆𝑝
Where 𝑥 is the difference of manometer liquid column in differential manometer
ℎ=𝑥
Types of Orifice
Concentric Orifice Plate:
❑ This is the most common type of orifice plate and consists of a circular hole (orifice) that is centrally located in the pipe
and aligned with the pipe axis.
❑ It creates a symmetric flow pattern around the orifice.
❑ Concentric orifice plates are typically used for clean fluids, where the fluid can pass through the orifice without
any significant disturbances.
.
Eccentric Orifice Plate:
❑ In an eccentric orifice plate, the orifice is not located at the center of the pipe; it's offset from the pipe's axis.
❑ This design is used to reduce the effects of solids settling or accumulating at the bottom of the pipe.
Applications:
•Fluids with solids or debris that tend to settle at the bottom of the pipe.
•To prevent blockage or damage to the orifice by particulate matter.
•When accurate flow measurement is required despite the presence of solids.
Segmental Orifice Plate:
❑ A segmental orifice plate has a segment of the circular orifice removed, leaving a smaller, curved opening.
❑ This design is used to mitigate the effects of cavitation and erosion that can occur in high-velocity flows.
Applications:
•Fluids with high velocities or high Reynolds numbers.
•Fluids that have the potential to cause cavitation or erosion around the orifice.
•When the flow has impurities or solids that could damage a standard orifice plate.
PITOT TUBE
Example A vertical Venturimeter carries a liquid of relative density 0.9. The Venturimeter has
200 mm inlet and 100 mm throat diameter The pressure connection at the throat is
200 𝑚𝑚 above that at the inlet. If the actual rate of flow is 50 𝑙𝑖𝑡/sec and the 𝐶𝑑 = 0.98,
calculate the pressure difference between inlet and throat in 𝑁/𝑚2 .
Solution:
Given
Sp. gravity = 0.9
𝐷1 = 200 mm = 0.2 m
𝐷2 = 100 mm = 0.1 m
𝑍2 − 𝑍1 = 200 mm = 0:2 m
𝑄𝑎𝑐𝑡 = 50 lit=sec = 0.05 𝑚3 /𝑠
𝐶𝑑 = 0.98
Pressure difference:
FLOW IN PIPES
❑ Introduction
▪ Water naturally flows through open channels such as rivers and streams.
▪ For purposes like irrigation, water distribution is often managed through canals
▪ But many times we also need the help of Pipe ( e.g. For drinking water supply etc or waste water
transport or oil)
▪ Closed conduit is needed when no interaction with anything outside is desired
❑ Comparison Between Open Channel Flow and Pipe Flow
❑ OCF must have a Free Surface
❑Subjected to atmospheric pressure
❑Driving force is the gravity force
River
❑No free Surface
❑Water is flowing full and subjected to
hydraulic pressure
❑ The main driving force is pressure
difference although gravity may also be
contributing to the flow
Water Supply : Pipe Network
Hydraulic of Pipe
❑ When water flow through pipe, there will resistance offered by the pipe and viscosity of the fluid, in that
way some energy will be loss.
❑ Available energy will gradually reduce then if we do not design the pipe in a proper way, water may not
flow up to or desired point.
❑ When designing the pipe network, another important to note is the pressure that is been created inside the
pipe.
❑ If the pressure is very high, and if the pipe material is not sufficient, not of sufficient strength, then the pipe
may burst.
Pipe burst
Hydraulic Grade Line (HGL) and Energy Grade Line (EGL)
EGL
is
a
line
representing the total
head available to the
fluid
HGL
is
a
line
representing total head
minus the velocity
head
Piezometric Head
❑ Piezometric head is the sum of the pressure head and datum head.
❑ Piezometric head is important in the analysis of Pipe Flow.
❑ The diameter of the pipe at section 1 and section 2 will be same unless it’s a
converging or diverging pipe. That mean discharge flowing through the pipe is
constant, then velocity will also be constant.
❑ Therefore velocity between upstream point and downstream point may not have much
difference.
❑ So any energy difference in the pipe flow is because of Piezometric head
Dimensionless Number
❑ Incase of Open Channel Flow, Gravity is the dominant force, hence Froude number
❑ In pipe flow, Viscous force is Dominant, hence Reynold number
Reynold Experiment
Osborne Reynold in 1883 demonstrated that there are two types of flow through a simple Experiment
Reynold Number
ρ 𝑉𝐷
Re= µ
FOR PIPE FLOW
FOR OPEN CHANEL FLOW
Re < 2000, laminar
Re> 4000, Turbulent
Re < 500, laminar
Re> 2000, Turbulent
L in Reynold number is not length of the pipe. It
is a characteristics length.
In case of pipe, this characteristic length is
represent by diameter of the pipe.
What is the Significance of Reynold Number ?
❑ It is important to know whether the flow is laminar or turbulent, because the
resistance offered by the pipe to the flow incase of laminar flow will be different
from that of turbulent flow .
❑ When we say that resistance offered is different that mean the energy loss is
different
❑ So to calculate the energy loss, we have different equation for laminar and turbulent
flow.
Frictional Resistance- Recall
❑ In pipe flow, flow from one point to another depends on the difference of energy
level between the two points.
❑ The pipe boundary and fluid viscosity offer resistance to flow
❑ Fluid needs to work to overcome these resistance, thus energy got consumed
❑ Therefore estimating head loss in a pipe is very important
❑ Resistance offered to the flow depends on the types of flow
Fluid Friction for Laminar Flow
❑ Proportional to Velocity of Flow
❑ Independent of the Pressure ( fluid is always under pressure in pipe flow but friction loss is
independent of the amount of pressure in laminar flow)
❑ Proportional to the contact surface area (Longer pipes increase frictional loss because there is
more surface area in contact with the fluid, leading to more resistance due to viscous shear)
❑ Independent of the nature of Surface i.e. roughness ( In laminar flow, a layer is formed
on the boundary of the pipe which is at rest (no slip boundary condition). Water is
laminar flow is flowing layer by layer. So this make the situation as if water is flowing
over the static layer)
❑ Effected by variation of temperature (change of temperature leads to change of viscosity).
More temperature less viscous for Liquid
Fluid Friction for Turbulent Flow
❑ Proportional to Velocityn ( n varies from 1.71 to 2). n is often considered as 2 to be on safe
side. So we can say proportional to square of velocity
❑ Independent of the Pressure
❑ Proportional to the density of the fluid
❑ Proportional to the contact surface area
❑ Depend on the nature of Surface (water is not flowing in layer, the fluid particle are
always mixing, so the fluid contacting with the surface will again mix with other
layers). Incase of friction formula for Turbulent, roughness parameter will come.
❑ Effected slightly by variation of temperature (the viscosity become more significant when
the fluid is moving in layer)
❑ Consider a horizontal pipe of radius R.
❑Consider a fluid element of radius r.
❑Let length of the fluid element be dx.
.
Consider that this fluid element is steady state. Hence the acceleration is zero. That mean it is moving at a uniform
velocity
.
r
.r
.
dx y = R-r
.
R
The forces acting on the fluid element are
1. Pressure force in both direction
2. Shear force
Shear force Fs = Shear stress  Contact Area (area where shear stress act= Circumference times length )
Fs = τ  2 π r dx
Fs =
2 π r dx τ
Fs =
2 π r dx τ
Pressure force = Pressure  Sectional Area
Say the Pressure is P
Pressure force at A,
FA = P πr
2
Pressure force at B, We can write in term of pressure gradient
𝑑𝑃
2
FB = (P + dx) π r
𝑑𝑥
Pressure is a compressive stress, always
act normal to the control volume
Equilibrium condition, summation of all force = 0
FA − FB - FS = 0
𝑑𝑃
2
2
π r P − (P +
𝑑𝑥)π r - 2 π r dx τ = 0
𝑑𝑥
πr P − πr P - πr
2
- πr
2
2 𝑑𝑃
𝑑𝑥
2 𝑑𝑃
𝑑𝑥
𝑑𝑥 - 2π r dx τ = 0
𝑑𝑥 – 2 π r dx τ = 0
2πrτ=-πr
2 𝑑𝑃
𝑑𝑥
2
-π
r
dP
τ=
2 π r dx
r −dp
τ=
………………..(1)
2 𝑑𝑥
Shear Stress diagram
❑ From Equation 1, shear stress varies linearly with r
❑ Also when r= 0, shear stress is 0
Velocity distribution diagram
❑ Because there is shear stress, we can apply Newton’s law of viscosity
τ = μ 𝑑𝑢
𝑑𝑦
❑ This equation is used for shear stress from the bottom surface of the pipe. But here we are talking about a fluid
element
So value of y = R- r
dy=-dr, R is constant
τ = μ - 𝑑𝑢
𝑑𝑟
………………..(2)
❑ Equate equations 1 and 2
❑ To find out velocity distribution we have to integrate with respect to r
………………..(3)
❑ To find out the value of C we need to apply boundary condition
Substitute the value of C in equation 3
This shows that velocity distribution is parabolic in nature
❑ Distribution of shear stress and velocity Profile in Laminar flow
Maximum and Average Velocity
❑ Discharge over the entire pipe
❑ To find pressure drop along a certain length of a pipe
2
1
R
L
x1
❑ From average velocity equation
x2
ഥ
𝑑𝑃 8μ𝑢
=
𝑅2
𝑑𝑥
Integrate between Section 1 and 2
-(P1 – P2) =
Take the negative to RHS
(P1 – P2) =
(P1 – P2) =
𝑅2
ഥ
8μ𝑢
𝑅2
ഥ
8μ𝑢
𝑅2
(P1 – P2) =
(P1 – P2) =
ഥ
8μ𝑢
(x1 – x2)
(x2 – x1)
L
ഥ
8μ𝑢
𝐷 2 /4
L
ഥL
32μ𝑢
𝐷2
(P1 – P2) 32μ𝑢ഥL
=h
=
f
2ρ g
𝐷
ρ𝑔
Hagen Poiseuille Equation
A pipe of 60 mm diameter and 450 m long slopes upwards at 1in 50. An oil of viscosity 0.9 Ns/m2 and specific gravity 0.9
is required to be pumped at the rate of 5 liters/sec.
i. Is the flow laminar ?
ii. What is the pressure difference required to maintain this condition ?
iii. What is the center line velocity and the velocity gradient at pipe wall
To find pressure difference.
Apply Bernoulli’s equation at section 1 and 2.
GATE QUESTION
Example (Adopted from FM White’s Fluid Mechanics)
An oil with ρ=900kg/m3 and μ=0.18kg/ms flows through an inclined pipe. Two sections, section 1 and
section 2 are 10 m apart. Assume steady laminar flow.
a)
Check whether flow is up or down.
b) Compute hf between 1 and 2.
c)
Compute the discharge Q.
d) Velocity, V.
e)
The Reynolds number.
Given following inputs: P1=350000Pa, Z1=0.0, P2=250000Pa, D=6cm.
Viscosity is Pa.s
1 Pa = 1 N/m² = 1 (kg·m/s²)/m² = 1
kg/m·s²
Z2=10sin40ο=6.43m
4
=
Derivation:
Q1. For a laminar flow through a circular pipe, prove that:
i)
ii)
The shear stress variation across the section of a pipe is linear
The velocity distribution is parabolic
Q2. Derive an expression for measuring discharge of fluid through a pipe using
Venturimeter
Q3. Derive an expression for measuring discharge of fluid through a pipe using
Orificemeter
Q4. Derive an expression for the depth of center of pressure from the free surface of a
liquid of an inclined plane submerge in a liquid.
Q5. Show that the average velocity is half of the maximum velocity for laminar flow in a
circular pipe by deriving an equation for velocity profile.
CONTINUATION ON PIPE FLOW
GENERAL EQUATION FOR HEAD LOSS IN PIPE
Darcy’s Weisbach Equation
❑ Consider a pipe whose length is
L, Diameter D
Head loss between two points 1 and 2
1
❑ Elevation at point 1 is Z1 and at
2 is Z2
❑ Weight of the fluid is acting in
vertical direction and denoted
by W
❑ If the angle is θ, then weight of
this fluid acting in direction of
flow is W cos θ
❑ So the forces is pressure force
on both end, shear stress acting
opposite to flow and W cos θ
𝑃1 𝑉12
𝑃2 𝑉22
ℎ𝑙 =
+
+ 𝑍1 −
+
+ 𝑍2
ρ𝑔 2𝑔
ρ𝑔 2𝑔
Consider Uniform section, V1= V2
Z 1- Z 2
Z1
θ
W
2
Z2
𝑃1
𝑃2
ℎ𝑙 =
+ 𝑍1 −
+ 𝑍2
ρ𝑔
ρ𝑔
𝑃1
𝑃2
ℎ𝑙 =
+ 𝑍1 −
+ 𝑍2
ρ𝑔
ρ𝑔
Applying force balance for Equilibrium
𝑃1 𝐴 − 𝑃2 𝐴 − τп𝐷𝐿 + 𝑊 𝑐𝑜𝑠θ = 0
п𝐷2
п𝐷2
𝑍1 − 𝑍2
𝑃1 − 𝑃2
− τп𝐷𝐿 + ρ𝑔
𝐿
=0
4
4
𝐿
Simplifying
𝑃1
𝑃2
4τ𝐿
+ 𝑍1 −
+ 𝑍2 =
ρ𝑔
ρ𝑔
ρ𝑔𝐷
4τ𝐿
ℎ𝑙 =
ρ𝑔𝐷
ℎ𝑙 ρ𝑔𝐷
4τ =
𝐿
ℎ𝑙 ρ𝑔𝐷
4τ =
𝐿
ρ𝑉 2
𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦
2
4τ
ℎ𝑙 ρ𝑔𝐷
=𝑓=
ρ𝑉 2
ρ𝑉 2
𝐿 2
2
𝑓𝐿𝑉 2
ℎ𝑙 =
2𝑔𝐷
IMPORTANT POINT
❑ This equation is valid for both laminar and turbulent flow
❑ To use this equation, the most important parameter is the value of f
❑ For Laminar Flow, both the equation should be equal . Therefore Darcy friction
factor is equal to
𝑓𝐿𝑉 2
32μ𝑉𝐿
ℎ𝑓 =
= 2
2𝑔𝐷
𝐷 ρg
64μ 64
𝑓=
=
ρ𝑉𝐷 𝑅𝑒
Thus in laminar flow, f is not influence by roughness of the pipe , but solely on Reynold number which depend on viscosity
of the fluid and velocity
❑Experiment on Friction factor was conducted
❑ Darcy, Nikuradse, Moody- All did experimentation to study various
parameters on friction factor
❑Nikuradse's did experimentation on pipe, artificially roughened by sand
grain. Grain diameter was taken as roughness
Nikuradse's diagram
❑Energy losses in Pipe are generally of two types:
1. Major losses and 2. Minor losses
❑ Total loss is the summation of major loss and minor loss
❑ Major loss is from friction loss
❑ Major loss is due to constant velocity. Thus it is also called as constant velocity head
loss
❑ Minor losses are due to change in Velocity and are called as variable velocity head
loss.
NOTE:
❑ If the pipe is long enough the minor losses can usually be neglected as they are much
smaller than the major losses
❑ Even though they are termed “minor”, the losses can be greater than the major
losses, for example, when there is a short pipe with many bends in it
❑ There are three types of forces that contribute to the total head in a pipe, which are
elevation head, pressure head, and velocity head.
❑ Minor losses are directly related to the velocity head of a pipe, meaning that the higher
the velocity head there is, the greater the losses will be
Minor Losses : Losses due to change in Velocity
❑ Loss at the exit of Pipe ( constant velocity head loss)
All other minor losses are due to variable velocity head loss
❑ Due to sudden enlargement
❑ Due to sudden Contraction
❑ Loss at the entrance of the pipe
❑ Loss at bends
❑ Loss due to various fitting (valve etc)
The loss at the exit of a pipe is considered as a
constant velocity head loss because the
velocity inside the pipe before the exit is
relatively uniform, and this velocity doesn't
suddenly change at the point of exit, even
though energy is dissipated
Expansion / Contraction Losses
❑ Expansions are defined as when the flow in a pipe goes from a small area to a larger
area and the velocity slows down (A1V1= A2V2)
❑ It is the exact opposite for contractions, the flow goes from a larger pipe to a smaller
one and the velocity increases.
❑ Sudden expansions and contractions are when the angle between the two pipe sizes is
equal to 90 degrees.
❑ Whenever there is rapid change in velocity, flow separation is taking place (separation of fluid taking place from the
pipe wall (green colour))
❑ Due to this separation eddies gets formed in this regions
❑ Eddies means the fluid particles started to swirl . Due to this some losses of energy is taking place
❑ Incase of sudden expansion and contraction, the diameter of the pipe is changing, hence there is changing in magnitude of
the velocity
❑ Incase of bend the diameter is the same, changes in velocity is due to change in direction
❑ Again due to change in velocity change separation is taking place
❑ Value of K can be find by conducting experiment in the lab
Control valve
Control
Valve
7m
a) Tank is full
b) Pipe material is Galvanized Iron
(surface roughness= 0.15)
c) Pipe is fitted with elbow joint and
a Gate/control valve at the end
d) When the gate valve is fully open,
water will discharge at the rate of
0.5 m3/s
e) Consider loss coefficient for bend
at the elbow as 0.9
f) Loss coefficient for fully open gate
valve is 0.2
g) Loss coefffient at the entrance is
0.5
Galvanized Iron
ሶ 0.5 𝒎𝟑 /s
𝑸=
4m
5m
Design the pipe diameter for a given
flow rate and given depth and height
9m
Assumption:
1. The tank is big so the velocity is zero at the tank compared to the
pipe
2. Flow is steady
3. Assume Kinematic viscosity at room temperature= 10-6 m2/s
4. Pipe diameter should be same through out
PIPE NETWORK ANALYSIS
ANALYSIS OF PIPE NETWORK: Hardy Cross Method
❑ In a water distribution system, pipe lines
conveying water generally branch up to
form loop of complex manner
❑ Determine the flow direction in all pipes by
simple observation is not possible
❑ There can be network having multiple loops
of common pipe
❑ The uncertainty about the flow directions
make the problem of flow analysis
complicated
❑ The Hardy Cross method provides a
systematic approach to analyze flow
distribution in these complex pipe networks
?
?
Basic of Pipe Network Net work Design
In any pipe network, the following two
conditions must be satisfied.
20
70
100
15
35
35
30
30
50
1. The continuity equation must be
satisfied at every node. The flow
entering in a junction or network
must be equal to the flow leaving
the same. So based on this
condition we assign logical value
and corrected it
Basic of Pipe Network Net work Design
2. Flow in each pipe must satisfy the head loss equation or the algebraic sum of the head loss around a closed
loop must be zero .
20
100
15
70
35
35
30
30
50
Head loss Formula:
𝑓𝐿𝑉 2
ℎ𝑙 =
2𝑔𝑑
𝑉=
ℎ𝑙 =
2𝑔
𝑄
п𝑑 2
4
𝑓𝐿𝑄2
2
п𝑑 2
4
20
𝑑
𝑓𝐿𝑄2
ℎ𝑙 =
п 2 5
2𝑔 4 𝑑
ℎ𝑙 = 𝐾𝑄2
❑ In order to satisfy the σ ℎ𝑙 = 0, flow is adjusted by
applying a correction factor ΔQ
100
15
70
35
35
30
30
Assumption:
clockwise --> positive
counter-clockwise --> negative
50
Calculation of Correction factor, ΔQ
Q0
Let the try discharge be represented by Q0
Let the correction = ΔQ
Revised or corrected Flow, Q
Q = Q0 - ΔQ
We know , ℎ𝑙 = 𝐾𝑄𝑛 = K (Q0 – ΔQ)n
෍ 𝐾𝑄0𝑛 − ∆𝑄 ෍ 𝐾𝑛 𝑄0𝑛−1 = 0
ℎ𝑙 = 𝐾 [ 𝑄0𝑛 − 𝑛 𝑄0𝑛−1 ∆𝑄 + 𝑛 𝑄0𝑛−2 ∆𝑄2 … … + ⋯ . . ]
As ΔQ is very small, ignore the higher power of ΔQ
ΔQ= σ
ℎ𝑙 = 𝐾 [ 𝑄0𝑛 − 𝑛 𝑄0𝑛−1 ∆𝑄]
For a loop, σ ℎ𝑙 = 0
σ 𝐾𝑄0𝑛
𝐾𝑛 𝑄0𝑛−1
Put n=2
෍ 𝐾𝑄0𝑛 − ෍ 𝐾𝑛 𝑄0𝑛−1 ∆𝑄 = 0
Modification (add or substrate) rule of thumb:
Clockwise= subtract with ΔQ
Counter-clockwise= add with ΔQ
σ 𝐾𝑄02
ΔQ= σ
2𝐾𝑄0
Checking after applying Correction
❑ In case of members, common to two loops, correction for each loop to be applied to this member
𝐾1
B
𝑄1
A
𝑄5
D
Say for loop ABD , correction is +∆𝑄′
Let the discharge in this loop be 𝑄1 , 𝑄2 𝑎𝑛𝑑 𝑄3 𝑎𝑛𝑑 𝐾 𝑖𝑠 𝐾1 , 𝐾2 𝑎𝑛𝑑 𝐾3
Correction for BD= 𝑄2 - ∆𝑄′ + ∆𝑄′′
C
𝐾2 𝑄2
𝐾3 𝑄3
Correction for AB= 𝑄1 - ∆𝑄′,
Correction for AD= 𝑄3 + ∆𝑄′
𝑄4
Say for loop BDC, correction is +∆𝑄′′
Correction for BC= 𝑄4 - ∆𝑄′′,
Correction for DC= 𝑄5 + ∆𝑄′′
Determine the distribution of flowrate?
σ 𝐾𝑄02
Basic EquationΔQ= σ 2𝐾𝑄
0
Loop 1
Loop 2
= 1 ∗ 602 + 3 ∗ 152 − 2
∗ 402 = 1075
= 2 ∗ 202 − 1 ∗ 552 − 3
∗ 152 = −2900
= 2 ∗ 1 ∗ 60 + 2 ∗ 3 ∗ 15
෍ 2𝐾𝑄0 + 2 ∗ 2 ∗ 40 = 370
= 2 ∗ 2 ∗ 20 + 2 ∗ 1 ∗ 55
+ 2 ∗ 3 ∗ 15 = 280
෍ 𝐾𝑄02
ΔQ
(L/s)
1075
= 2.91
370
−2900
= −10.31
280
Determine the distribution of flowrate?
NUMERICAL PROBLEMS ON FLOW
THROUGH PIPES
Determine the distribution of flowrate?
Answer: 24.35, 9.90, 25.65, 14.45, 10.55
Q1. Figure below shows a simple water distribution network in which Q and hf refers to discharge and
head loss respectively. Subscripts 1,2, 3,4, 5 designated respective values in pipe AB, BC, CD, DA and AC
respectively.
a) Find the Discharges Q B , Q 2 , Q 4 and Q 5
b) Head loss ℎ𝑓4 and ℎ𝑓5
c) Draw a neat final Sketch of the Network with flow directions
Q B =?
Q A = 20
Q1 = 30
ℎ𝑓1 = 60
𝐴
𝐵
Q2 = ?
ℎ𝑓2 = 40
Q 4 =?
ℎ𝑓4 =?
𝐶
𝐷
Q 3 = 40 ℎ𝑓3 = 120
Q D = 100
Q C = 30
Q2. A horizontal pipe line 40 m long is connected to a water tank at one end and discharge
freely into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is
150 m diameter and its diameter is suddenly enlarged to 300 mm. The height of water in the tank is 8m
above the centre of the pipe. Considering all losses of head which occur:
i. Determine the flow rate
ii. Draw the HGL and EGL Lines. Take f=0.01 for both sections of the pipe
Question :GATE 2020 (2 marks)
Question :GATE 2019 (2 marks)
Two identical pipes (i.e. having the same length , same diameter, and same roughness) are used to
withdraw water from a reservoir. In the first case, they are attached in series and also discharge
freely into the atmosphere. In the second case, they are attached in parallel and friction losses is
same in both cases. The ratio of the discharge in the parallel arrangement to that in the series
arrangement (round off to 2 decimal places ) is ____________________________
1st case: Series connection
Total head loss due to friction
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 ,
𝑓𝐿1 𝑉12 𝑓𝐿2 𝑉22
ℎ𝑙 =
+
2𝑔𝑑1
2𝑔𝑑2
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 , ℎ𝑙 =
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 , ℎ𝑙 =
𝑓 𝑄2
𝐿
𝑑1
𝐿
𝑑2
1
2
[
+
5
5]
п 2
2𝑔 4
𝑓 𝑄2
[
п 2
2𝑔 4
𝐿
𝐿
+
]
𝑑5
𝑑5
2𝑓 𝐿𝑄𝑠2
ℎ𝑙𝑠𝑒𝑟𝑖𝑒𝑠 =
12.1 𝑑 5
HEAD LOSS OF PIPE IN PARALLEL
𝑓𝐿1 𝑉12 𝑓𝐿2 𝑉22
ℎ𝑙 =
=
2𝑔𝑑1
2𝑔𝑑2
𝑓𝐿𝑉12
ℎ𝑙 =
=
2𝑔𝑑
𝑓𝐿𝑄 2
п𝑑
2
4
𝑓 𝐿𝑄 2
=
2 2
𝑔𝑑
12.1 𝑑 5
Question :GATE 2022 (2 marks)
As the pipe is in parallel , the head loss is both the pipes will be same
𝑓 𝐿𝑄2
𝑓 𝐿𝑄2
=
5
12.1 𝑑1 1
12.1 𝑑25 2
f and L are same
𝑄1
=
𝑄2
𝑑1
𝑑2
5/2
𝑄1
= 5.65
𝑄2
PROBLEM ON EQUIVALENT PIPE
Q2. A piping system consists of three pipes arranged in series; the length of the pipes are
1200 m, 750 m and 600 m and diameters 750 mm, 600 mm and 450 mm respectively.
a) Determine an equivalent diameter for the pipe, 2550 m long
b) Transform the system to an equivalent 450 mm diameter pipe
DIMENSIONAL ANALYSIS
Dimensional Analysis
❑ In many instances, the variables involved in physical phenomena are known,
while the relationship among the variables is not known. Such a relationship
can be formulated between a set of dimensionless groups of variables and the
groups numbering less than the variables. This procedure is called
dimensional analysis
❑ Sometimes, the experimental work in the laboratory is not only time-consuming, but also
expensive. So, the main goal is to extract maximum information from fewest experiments.
METHODS:
1. Rayleigh method
2. Buckingham pie’s theorem
Buckingham Π theorem
Buckingham Π theorem
❑ Step I: List out all the variables that are involved in the problem
❑ Step II: Express each variable in terms of basic dimensions.
❑ Step III: Decide the required number of pi terms: number of pi terms is equal to (n-m)
❑ Step IV: Select the number of repeating variables.
❑ Step V: Write the Pi term by combining repeating variables with each of the remaining variable
❑ Step VI: Solve the equation from step V
❑ Step VII: Checking of pi terms
❑ Step VIII: Final form of relationship among pi terms
Consider a steady flow of an incompressible
Newtonian fluid through a long, smooth walled,
horizontal circular pipe. Determine the pressure drop
per unit length of the pipe without the use of
experimental data
Step I: List out all the variables that are involved in the problem.
Typically, these variables are those that are necessary to describe the "geometry" of the system
(diameter, length etc.), to define fluid properties (density, viscosity etc.) and to indicate
the external effects influencing the system (force, pressure etc.).
• According to Step I, all the pertinent variables involved in the experimentation of pressure drop
per unit length of the pipe, must be noted and should be expressed in the form,
ΔP = f (D, ρ, µ, V)
Where D is the pipe diameter, ρ is the fluid density, µ is the viscosity of the fluid and V is the mean
velocity at which the fluid is flowing through the pipe.
Step II: Express each variable in terms of basic dimensions.
ΔP = f (D, ρ, µ, V)
ΔP= [M L-2 T-2]
D =[ L ]
ρ = [M L-3 ]
µ = [M L-1 T-1 ]
V= [L T-1]
Step III: Decide the required number of pi terms.
It can be determined by means of Buckingham pi theorem which indicates that the number of pi terms is equal
to (n-m), where n is the number of variables in the problem (determined from Step I) and m is the number
of reference dimensions required to describe these variables (determined from Step II).
• Since there are five variables (including the dependent variable) and three reference
dimensions. So, only two pi terms are required.
Step IV: Select the number of repeating variables.
❑ Don’t select dependent variable
❑ Variables should contain all m dimensions (M, L, T).
❑ Do not select dimensionless variables as repeating
ΔP = f (D, ρ, µ, V)
❑Since, there are three reference dimensions involved, so we need
to select three repeating variable.
❑We have no dimensionless variables
ΔP = f (D, ρ, µ, V)
❑In this case we choose D, ρ, V (should see L, M, T at least appear one time
from these 3)
D =[ L ]
Π1 = D V ρ
ρ = [M L-3 ]
V= [L T-1]
Π2 = D V ρ
Step V: Write the Pi term by combining repeating variables with each of the
remaining variable
❑ Essentially, the pi terms are formed by multiplying one of the non-repeating variables by the
product of the repeating variables each raised to an exponent that will make the combination
dimensionless.
Π1 = Da Vb ρc
Π2 = Dd Ve ρf
Tag the remaining variables
ΔP = f (D, ρ, µ, V)
Π1 = Da Vb ρc ΔP
Π2 = Dd Ve ρf µ
Step VI: Solve the equation from step V
Π1 = ΔP Da Vb ρc
=[M L-2 T-2] [ L ]a [L T-1]b [M L-3 ]c
Pi parameters are dimensionless parameters, therefore power on Mass, length and time should be zero
[M L-2 T-2] [ L ]a [L T-1]b [M L-3 ]c = [M0 L0 T0]
Solving
1+ c= 0 (For M)
-2+ a+ b-3c =0 (For L)
-2- b=0
a= 1, b=-2, c=-1
Π1 =
ΔP 𝑫
ρ𝑽𝟐
Π2 = µ Dd Ve ρf
[M L-1 T-1 ] [ L ]d [L T-1]e [M L-3 ]f = [M0 L0 T0]
Solving
1+ f= 0 (For M)
-1+ d+ e-3f =0 (For L)
-2- e=0
d= -1, e=-1, f=-1
Π2 =
µ
ρ𝑽𝑫
Step VII: Checking of pi terms
Make sure that all the pi terms must be dimensionless.
Step VIII: Final form of relationship among pi terms
µ
𝑫ΔP
=f(
)
ρ𝑽𝟐
ρ𝑽𝑫
SIMILITUDE
Similitude: Is the theory and art of predicting prototype performance from model
observation
❑ Model: Scaled version of the prototype
❑ Prototype: Actual Machine or structure
Note: A model need not to be always smaller than the prototype
In order that the results obtained in a model studies represent the
behaviour of a prototype, the following three similarities must be ensure
between a model and a prototype:
1.
Geometric Similarity
2. Kinematic Similarity
3. Dynamic Similarity
1. Geometric Similarity:
❑ For a geometric similarity to exist between a model and a prototype, all linear dimension of the model are
related to the corresponding dimensions of a prototype by a common scale factor
❑ Both model and prototype is of the same shape
❑ Included angle between the model and prototype must be same
Geometric Similarity:
𝐿𝑒𝑡 Lm = Length of model
Hm = height of model
Dm = Diameter of model
Am = Area of model,
Vm = Volume of model
Lp , Hp , Dp , Ap , Vp = Corresponding value of prototype
Then for Geometric Similarity, we must have
𝐿𝑚 𝐵𝑚 𝐻𝑚 𝐷𝑚
=
=
=
= 𝐿𝑟
𝐿𝑝
𝐵𝑝
𝐻𝑝
𝐷𝑝
Lr is called scale ratio or scale factor
Similarly, Ar = area ratio =
Am
= L2r
Ap
And , Vr = volume ratio =
Vm
= 𝑉𝑟3
Vp
Kinematic Similarity:
❑ In order to check kinematic similarity, geometric similarity must be first satisfied
❑ Kinematic similarity is the similarity of motion
❑ If at a corresponding points in the model and in prototype the velocity or acceleration ratio are
same and the velocity or acceleration vectors are in the same direction , then the two
flows are said to be kinematic similar
Distance should equal to linear scale
❑ Flow regime must be
the same (laminar, or
turbulent etc)
Kinematic Similarity:
𝐿𝑒𝑡 (V1 )m = Velocity of fluid at point 1 in the model
(V2 )m = Velocity of fluid at point 2 in the model
(a1 )m = acceleration of fluid at point 1 in the model
(a2 )m = acceleration of fluid at point 2 in the model
(V1 )p , (V2 )p , (a1 )p , (a2 )p = Corresponding value at a correspoding points of fluid velocity or acceleartion of prototype
Then for Kinematic Similarity, we must have
(V1 )m (V2 )m
=
= 𝑉𝑟
(V1 )p
(V2 )p
Vr is called scale ratio or scale factor
(a1 )m (a2 )m
Similarly, ar = acceleration ratio =
=
(a1 )p
(a2 )p
The direction of the velocities in the model and protype should be the same
Dynamic Similarity:
❑ Similarity of forces between model and prototype
❑ It is evaluate by equating dimensionless numbers of model and prototype
Forces:
1. Inertia Force
2. Viscous Force
3. Pressure Force
4. Surface Tension Force
5. Gravity Force
6. Compressible Force
Dimensionless Number:
1. Reynold Number
2. Euler’s number
3. Weber number
4. Froude’s number
5. Mach number
Inertia Force : Is the product of mass and acceleration of the flowing fluid
𝑣
𝑣
= 𝜌𝑙 3
𝑡
𝑡
𝑙
=𝜌𝑙 2 𝑡 𝑣 = 𝜌𝑙 2 v v = 𝜌𝑙 2 v 2
𝐹𝑖 = 𝑚𝑎 = 𝜌𝑉
Shear Force : Is the product of shear stress due to viscosity and surface area of the flow
𝑣
𝑣
𝐹𝑟 = 𝜏𝐴 = 𝜇 𝐴 = 𝜇 𝑙 2 = 𝜇𝑣𝑙
𝑙
𝑙
Pressure Force : Is the product of pressure intensity and cross sectional area of the flowing fluid
𝐹𝑝 = 𝑃𝐴 = 𝑃 𝑙 2
Surface tension Force : Is the product of surface tension and length of surface of the flowing fluid
𝐹𝑠 = 𝜎𝑙
Gravity Force : Is the product of mass and acceleration due to gravity
𝐹𝑔 = 𝑚𝑔
Compressibility Force :
𝐹𝑐 = 𝑘𝐴
𝐹𝑐 = 𝑘𝑙 2
Where k is the bulk modulus
Common Dimensionless Number
Name
Definition
Equation
Application
Reynold Number
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝐹𝑜𝑟𝑐𝑒
𝜌𝑉𝑙
𝜇
When viscous force
dominate (e.g Pressure
drop in a pipe)
Froude Number
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝐺𝑟𝑎𝑣𝑖𝑡𝑦 𝐹𝑜𝑟𝑐𝑒
𝑉
𝑙𝑔
When gravity force
dominate (open
channel)
Euler Number
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐹𝑜𝑟𝑐𝑒
𝜌
𝑉
𝑃
When pressure force
dominate (pump flow,
flow in venturi meter)
Weber Number
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒
Mach Number
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒
𝜌𝑙
𝜎
𝑉
𝑘/𝜌
𝑉
When surface force
dominate (e.g blood
flow, spray paint,
movement of water
into soil)
When Compressibility
flow dominates (Speed
of an aircraft, rocket)
An oil of specific gravity 0.92 and viscosity 0.03 poise is to be transported at the
rate of 2500 liters/second through a 1.2 diameter pipe. Tests were conducted on
a 12 cm diameter pipe using water at 20 degree Celsius. If the viscosity of water
at 20 degree Celsius is 0.01 poise. Find:
a) Velocity of flow in a model
b) Rate of flow in the model
A 1:12 scale model is to built to study the flow over
spillway. The flowrate in the prototype is 200 m3/s. The
flow rate in the model should be nearly equal to
A.0.40 m3/s
B.0.50 m3/s
C.0.70 m3/s
D.0.90 m3/s
Solution:
We know that Q= AV
𝑸𝑷
𝑨𝑷 𝑽𝑷
𝒍𝑷 𝟐 𝑽𝑷
𝒍𝑷
=
= 𝟐
=
𝑸𝒎 𝑨𝒎 𝑽𝒎 𝒍𝒎 𝑽𝒎
𝒍𝒎
𝟐
𝑽𝑷
𝑽𝒎
𝑸𝒓 = 𝒍𝒓 𝟐 𝑽𝒓 - (1)
❑ Equation 1 is true for any dynamic similar model
❑ However, the actual ratio depends on what kind of model it is. In this case we have flow over spillway which is dominate
by gravity so this is a Froude number model
𝐅𝐫𝐨𝐮𝐝𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐦𝐨𝐝𝐞𝐥 ∶ 𝑭𝒓𝒑 = 𝑭𝒓𝒎
𝑉
𝑔𝑙 𝑚
𝑸𝒓 = 𝒍𝒓 𝟓/𝟐
=
𝑉
𝑔𝑙 𝑝
𝑽𝑷
𝒍𝑷
=
𝑽𝒎
𝒍𝒎
𝑸𝒑
𝟏/𝟐
𝑽𝒓 = 𝒍𝒓 𝟏/𝟐 (𝟐)
𝟐𝟎𝟎
𝑸𝒎 = 𝟓/𝟐 = 𝟓/𝟐 = 𝟎. 𝟒𝟎
𝟏𝟐
𝒍𝒓
The spillway for a dam is 20m wide and is designed to
carry 125 m3/s . A 1:15 scale is to be constructed . The
flowrate in the model should be mostly nearly:
A.0.120 m3/s
B.0.143 m3/s
C.0.162 m3/s
D.0.185 m3/s
𝑸𝒓 = 𝒍𝒓 𝟐 𝑽𝒓
𝑸𝒓 = 𝒍𝒓 𝟓/𝟐
𝑸𝒎 =
𝑸𝒑
𝒍𝒓
=
𝟓/𝟐
𝟏𝟐𝟓
= 𝟎. 𝟏𝟒𝟑
𝟏𝟓𝟓/𝟐
NOTCHES & WEIRS
Weirs and Notches
❑ A notch means an opening provided in the side of a channel
❑ A weir is also a notch but it is made on a large scale.
❑ Weirs can help raise the water level so that boats can pass through, and they can also reduce the flow of water
to prevent flooding
❑ Often made of metallic plate
❑ Used for measure discharge in small canal or
laboratory
❑ Often made of Concrete or masonry
❑ Used in rivers
Difference between Notches and Weirs
Notches
Weirs
• A notch may be define as an opening provided in
small channel, in such a way the liquid surface in
the tank or channel is below the top edge of the
opening
• A structure constructed across a river or
canal to store water on the upstream
side
• A notch is usually made of metallic plate
• A weir is made of cement concrete or
masonry
• A notch is usually used to measure small
discharge of small stream or canal
• A weir is used to measure large
discharge of rivers and canals
• Notches are small in size
• Weirs are larger in size
❑ Figure shows a channel in which a water is flowing
❑ There is an obstruction in the channel and in this
obstruction a cutoff has been made out of Width b
❑ The liquid flow out of that cut off says that it is till a
height H above the crest of the cutoff.
❑ Aim is to calculate the Flow rate Q based on the
value of H and b
❑ This is a typical picture of streamline for such a flow.
❑ If we analyze the flow like this, it will be complicated
❑ So we have to simplify thing but making some assumption
❑ Let us assume that there is no drawdown and the liquid is coming out horizontally
❑ Let us consider a single streamline within this flow from 1 to 2
❑ We also assume that the height of water in the channel is much larger than the height over the
obstruction. So the velocity at point 1 can be consider to be much lower compared to point 2.
❑ If the pressure is low at point 1, the vertical pressure can be considered as hydrostatic
❑ Since the pressure below and above the jet is atmospheric (depth is small ), so we assume that the pressure at
point 2 is atmospheric
Total Energy at 1 = Total Energy at 2
This is the velocity at Point 2.
It varies with y, the location of the point within the jet
The velocity is maximum when y= 0 , i.e. at the bottom most
point when the jet is coming out
This is minimum when y= H, we can say that the velocity is
close to zero
Now we know the velocity, we can find the discharge .
Q= AV
Let us find Q by considering the flow rate through this jet of
width b and height H
Consider a small strip of thickness dy and height y.
So discharge through this small strip dQ= V times the area ( b dy)
To find total discharge, integrate from y= 0 to y = H
So if I Know b and H which is the height of liquid above the crest, I can find Q
An obstruction is created and Vertical scale is erected upstream so that 0 coincides with the based of the cut off to
measure H
𝟐𝒃 𝟑/𝟐
𝑸 = 𝑪𝒅
𝑯
𝟐𝒈
𝟑
Since we made many assumption, the
actual discharge will be different from the
theoretical discharge hence we introduce
𝐶𝑑
Calibration Chart for Cd
H
P
b
B
V NOTCH
𝟖
𝜶
𝟓/𝟐
𝑸=
𝑪𝒅 𝑯
𝟐𝒈 𝒕𝒂𝒏
𝟏𝟓
𝟐
𝑯
𝑸=
𝟐𝒈𝒉 න 𝒃 𝒅𝒉
𝟎
h
b
H-h
𝜶
For a V-notch with an included angle q, put b = 2(H-h)tan(𝜶 /2)
𝑯
𝜶
𝑸 = 𝟐 𝟐𝒈 𝒕𝒂𝒏
න 𝑯 − 𝒉 𝒉𝟏/𝟐 𝒅𝒉
𝟐 𝟎
𝜶
𝑸 = 𝟐 𝟐𝒈 𝒕𝒂𝒏
𝟐
b/2
𝑯
𝟐
𝟐 𝟓/𝟐
𝟑/𝟐
𝑯𝒉 − 𝒉
𝟑
𝟑
𝟎
H-h
𝟖
𝜶
𝟓/𝟐
𝑸=
𝑪𝒅 𝑯
𝟐𝒈 𝒕𝒂𝒏
𝟏𝟓
𝟐
𝜶/𝟐
b/2 = (H-h)tan(𝜶 /2)
b = 2(H-h)tan(𝜶 /2)
H
Typical Discharge Coefficient graph for 90 degree notch
Trapezoidal weir:
❑ Trapezoidal weir is also called as Cippoletti weir. This is trapezoidal in shape and is the modification of
rectangular weir
❑ The sides are inclined outwards with a slope 1:4 (horizontal : vertical)
❑ In cippoletti weir both sides are having equal slope. So, we can divide the trapezoid into rectangle and triangle
portions.
❑ So, Total discharge over trapezoidal weir Q = discharge over rectangular weir + discharge over
triangular weir
𝟐𝒃 𝟑/𝟐
𝟖
𝜶
𝟓/𝟐
𝑸 = 𝑪𝒅
𝑯
𝟐𝒈 +
𝑪 𝑯
𝟐𝒈 𝒕𝒂𝒏
𝟑
𝟏𝟓 𝒅
𝟐
Types of Weirs/ Notches:
Weirs are classified according to:
1. Types of Weirs based on Shape of the Opening
•Rectangular weir
•Triangular weir
•Trapezoidal weir
2. Types of Weirs based on Shape of the Crest
•Sharp-crested weir
•Broad- crested weir
•Narrow-crested weir
•Ogee-shaped weir
Broad-crested weir:
•These are constructed only in rectangular shape and are suitable for the larger flows
.
•Head loss will be small in case of broad crested weir.
𝑸 = 𝑪𝒅 𝑳 𝟐𝒈 𝑯𝒉𝟐 − 𝒉𝟑
Ogee-shaped weir:
❑ Generally ogee shaped weirs are provided for the spillway of a storage dam.
❑ The crest of the ogee weir is slightly rises and falls into parabolic form.
❑ Discharge over ogee weir is also similar to flow over rectangular weir.
𝟐𝒃 𝟑/𝟐
𝑸 = 𝑪𝒅
𝑯
𝟐𝒈
𝟑
Narrow Crested weir:
Incase of Narrow crested weir , 2b < H
It is similar to rectangular weir or notch
𝟐𝒃 𝟑/𝟐
𝑸 = 𝑪𝒅
𝑯
𝟐𝒈
𝟑
Determine the height of rectangular weir of length
6 m to be built across a rectangular channel. The
maximum depth of water on the upstream side of
the weir is 1.8 m and discharge is 2000
liters/second. Take Cd =0.6
Find the discharge through a trapezoidal notch which is 1.2 m wide at the top and 0.5 m at the
bottom and is 0.4 m in height . The head of the water on the notch is 0.3 m.
Assume Cd for the rectangular portion= 0.62 and for the triangular portion = 0.60
VORTEX FLOW
Introduction
❑ When a fluid moves in a curve path, it is known as Vortex Motion
❑ There are two types of Vortex Motion:
1. Free Vortex motion
2. Forced Vortex motion
Free Vortex motion
❑ In a free vortex, fluid particles revolve around a central axis without any external forces driving the
rotation.
❑ The velocity of particles in a free vortex decreases as the distance from the center increases
1
𝑟
V is the tangential velocity and r is the radius from the centre
𝑉∝
❑ In a free vortex, vorticity is zero everywhere except at the very center, making it an irrotational vortex in most
of the flow field.
❑ Bernoulli’s equation is applicable as the flow field is mostly irrotational
❑ Example : Flow of water in a wash basin. Whirlpool in a river
Forced Vortex motion
❑ It is a type of vortex motion in which the fluid moves in a curved path by the action of some
external agency (Torque)
❑ Here, the tangential velocity increases with the radius, following 𝑽 = 𝒓𝝎 where ω is the angular
velocity.
❑ The forced vortex has non-zero vorticity throughout and is considered a rotational vortex., hence
Bernoulli’s equation is not applicable.
❑ Example : Water in a bucket when rotated with stick.
Generalized Equation of Vortex motion
𝑑𝑝
(𝑃 + 𝑑𝑟 𝑑𝑟) 𝑑𝐴
P𝑑𝐴
𝑑𝑟
When fluid flow between curve streamline , centrifugal forces
are set up and counter balance by the pressure force acting in
radial direction. To sustain this, the pressure must be higher on
the outer side (larger radius) and lower on the inner side
(smaller radius).
If the value of dr is very small, the area of both the upper and
lower curve is same and is denoted as dA
𝑟
𝑑𝐴 = 𝑟 𝑑θ ×(1)
dθ
PLAN
𝑑𝐴 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 × 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝐶
𝐼𝑛 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, ℎ𝑦𝑑𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤𝑖𝑙𝑙 𝑎𝑐𝑡
𝑑𝑝
= − 𝜌𝑔 ( 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑎𝑠 𝑤𝑒 𝑚𝑜𝑣𝑒 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡𝑡𝑜𝑚 𝑤𝑖𝑡ℎ
𝑑𝑧
Respect to h)
𝑚𝑉 2
𝐶𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 =
𝑟
𝑑𝑝
(𝑃 + 𝑑𝑟 𝑑𝑟)𝑑𝐴
𝑚𝑉 2
− P𝑑𝐴 = 𝑟
𝑑𝑝
𝑉2
𝑑𝑟 𝑑𝐴 = ρ𝑑𝐴 𝑑𝑟
𝑑𝑟
𝑟
𝑑𝑝
𝑉2
=ρ
𝑑𝑟
𝑟
Mass = density x volume
𝑑𝑝
𝑉2
=ρ
𝑑𝑟
𝑟
𝑑𝑝
= − 𝜌𝑔
𝑑𝑧
𝑑𝑝
𝑑𝑝
Total change in Pressure = 𝑑𝑟 𝑑𝑟 + 𝑑𝑧 𝑑𝑧
𝑉2
𝑑𝑝=ρ 𝑟 𝑑𝑟 − 𝜌𝑔𝑑𝑧
This is the general form of Vortex Flow
Forced Vortex
𝝎
𝑉2
𝑑𝑝=ρ 𝑑𝑟 − 𝜌𝑔𝑑𝑧
𝑟
2
Put 𝑽 = 𝒓𝝎
1
𝑟 2 𝝎2
𝑑𝑝=ρ
𝑑𝑟 − 𝜌𝑔𝑑𝑧
𝑟
𝑑𝑝=ρ𝑟ω2 𝑑𝑟 − 𝜌𝑔𝑑𝑧
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 1 𝑡𝑜 2
2
2
2
න 𝑑𝑝 = ρω2 න 𝑟𝑑𝑟 − 𝜌𝑔 න 𝑑𝑧
1
1
1
ρω2 2
𝑃2 − 𝑃1 =
𝑟2 − 𝑟12 − 𝜌𝑔(𝑧2 − 𝑧1 )
2
𝐼𝑓 𝑏𝑜𝑡ℎ 𝑝𝑜𝑖𝑛𝑡 𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑡𝑜𝑝 𝑙𝑒𝑣𝑒𝑙 𝑜𝑓 𝑣𝑜𝑟𝑡𝑒𝑥 , 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
ρω2
0= 2
𝑟22 − 𝑟12 − 𝜌𝑔(𝑧2 − 𝑧1 )
If we have a cylinder
filled with fluid and
rotate it, it will form a
vortex as shown
Rotate about an axis
with 𝝎
ρω2 2
𝑟1 − 𝑟22 = 𝜌𝑔(𝑧1 − 𝑧2 )
2
𝝎
ω2 𝑟12 − ω2 𝑟22
= 𝑧1 − 𝑧2
2𝑔
1
2
𝐼𝑓 𝑝𝑜𝑖𝑛𝑡 2 𝑖𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛, 𝑟2 = 0
ω2 𝑟12
=𝑍
2𝑔
𝑇ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑒𝑙𝑙𝑠 𝑚𝑒 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑚𝑜𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 𝑎𝑛𝑑 𝑎𝑛𝑦 𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑖𝑛𝑡
𝑻𝒉𝒖𝒔 𝒁 𝒗𝒂𝒓𝒊𝒆𝒔 𝒘𝒊𝒕𝒉 𝒓𝟐 . 𝑻𝒉𝒖𝒔 𝒔𝒉𝒂𝒑𝒆𝒅 𝒇𝒐𝒓𝒎 𝒊𝒏 𝒕𝒉𝒆 𝟑𝑫 𝒊𝒔 𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒍𝒐𝒊𝒅
𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒊𝒏 𝒂 𝒇𝒐𝒓𝒄𝒆𝒅 𝒗𝒐𝒓𝒕𝒆𝒙 𝒄𝒉𝒂𝒏𝒈𝒆𝒔 𝒘𝒊𝒕𝒉 𝒔𝒒𝒖𝒂𝒓𝒆 𝒐𝒇 𝒓𝒂𝒅𝒊𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝑍
Prove that in the case of a force vortex, the rise of the liquid level at the ends is equal to the fall of the liquid
level at the axis of rotation.
𝝎
Initial water level is AB
Provide angular velocity to the cylinder
𝑦1
B 𝑦
2
A
Because of this angular velocity, paraboloid will form
ℎ
Volume of initial Liquid = 𝜋𝑅 2 (ℎ + 𝑦2 )
Volume of Liquid during rotation = 𝜋𝑅2 ℎ + 𝑦2 + 𝑦1
𝑅
− 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑙𝑜𝑖𝑑
1
Volume of Liquid during rotation = 𝜋𝑅2 ℎ + 𝑦2 + 𝑦1 − 2 𝜋𝑅2 (𝑦1 + 𝑦2 )
Volume of Liquid during rotation = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑏𝑒𝑓𝑜𝑟𝑒 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛
1 2
2
2
𝜋𝑅 ℎ + 𝑦2 = 𝜋𝑅 ℎ + 𝑦2 + 𝑦1 − 𝜋𝑅 (𝑦1 + 𝑦2 )
2
(𝑦1 +𝑦2 ) −
1
𝑦 + 𝑦2 = 𝑦2
2 1
𝑦1 𝑦2
+ = 𝑦2
2
2
𝑦1 = 𝑦2
Impulse- Momentum Equation
BASIC OF OPEN CHANNEL
FLOW
❑ Open channel flow is a type of flow with a free surface open
to atmosphere
❑ Driven by Gravity force
❑ The basic force balance is between the gravity and friction
Hydraulic Grade Line (HGL) and Energy Grade Line (EGL)
EGL
is
a
line
representing the total
head available to the
fluid
HGL
is
a
line
representing total head
minus the velocity head
In open channel flow,
HGL coincide with free
surface
Driving force is gravity,
pipe flow driving force is
the pressure difference
CLASSIFICATION OF OCF:TYPE 1
❑ OCF can be manmade or natural
Natural: They are generally very irregular in shape. Example: Rivers
Artificial : They are usually designed with regular geometric shapes. Example:
Irrigation canals, culverts,
CLASSIFICATION OF OCF:TYPE 2
❑ OCF can be rigid boundary and mobile boundary channels
Mobile boundary: composed of loose sedimentary particles moving under the action of flowing water
Rigid boundary : immovable bed and sides (lined canal)
CLASSIFICATION OF OCF:TYPE 3
❑ OCF can be prismatic or non prismatic channel
Prismatic: A channel in which the cross sectional shape, size and the bed slope are constant is
termed as Prismatic channel. All artificial channels are usually prismatic. The rectangular,
trapezoid, are the most commonly used shapes of prismatic channels.
Non prismatic : All natural channels generally have varying cross section and consequently are
known as Non-prismatic channels
Types of flow in open channel
❖ According to the characteristics of the flow with respect to time and place, different
categories can be set
A. Steady flow:- Here the criterion is time. A flow can be said steady if the fluid
characteristics like velocity, pressure density, depth of flow doesn’t change or if it can be
assumed constant between the time of consideration
V
= 0,
t
p
=0
t
and
y
=0
t
B. Unsteady flow:- Here the fluid characteristics vary with time such that
y
p
V
and
0
0
 0,
t
t
t
,
C. Uniform flow:- A space as a criterion is used . Open channel flow is said to be uniform if the
depth of flow, velocity remains constant or the same at every section of the channel
Uniform flow may be steady or unsteady, depending on whether or not the depth changes with
time
y
V
=0
= 0, and
s
s
D. Non uniform flow:- In case when the velocity, depth of flow in a channel changes with
space:
y
V
0
 0, and
s
s
E. Steady uniform flow
F. Unsteady uniform flow
G. Unsteady uniform flow
H. Unsteady non uniform flow
Non-Uniform Flow is divided into Rapidly varying flow (R.V.F) and
Gradually Varied Flow (G.V.F)
In the case of R.V.F, the depth of flow rapidly changes over a smaller length of the channel. It rises up
suddenly for a short length and settles back.
While in a G.V.F, the depth of flow changes gradually over a longer length of the channel.
Manning’s Equation
❑
The basic force balance is between the gravity and friction
❑
Friction force comes from the perimeter of the channel
One of the important term in Manning formula is hydraulic radius, Depth is the depth of water not depth of
channel ( wetted perimeter )
A high hydraulic radius value indicates that the channel contains a lower volume of contact fluid and a
greater cross-sectional area.
These conditions result in increased flow velocity and capacity, as well as improved channel efficiency.
For prismatic channel it is easy to calculate the area
For natural divide it into trapezoidal and triangle . Area is summation of all the areas.
Manning formula is used to estimate the average velocity of water flowing in an open channel
Manning equation can be also used for overland flow
1 2Τ3 1ൗ
𝑉= 𝑅 𝑆 2
𝑛
V is average velocity
n is manning roughness parameters ( it will tell us how rough or smooth the channel is. For smooth n
is small)
S is the bed slope or slope of energy grade line
R is the hydraulic radius
Typical Contour of Velocity (m/s) in Open Channel Section
No slip boundary condition:
Velocity near to the
boundary is Zero
CONCEPT OF ECONOMIC
CROSS SECTION
ECONOMIC SECTION
❑ If we know the discharge, we have decided the lining material and shape of the channel plus we have
the slope of the topography, then our next target is to design the channel with most economic section.
❑ Such channel shape is called as the most economic section
❑ Most economic section is also called as Best hydraulic section or efficient section
❑ Let’s say we are lining the channel, so to reduce the cost the perimeter should be minimum. In that
process we achieve another objective as resistance to flow comes from the perimeter.
❑ If the length of the perimeter is less, resistance is less, discharge is more
❑ Thus in case of lined canal our objective is to design the Best hydraulic section or most economical
section i.e both from economic and discharge point of view
To maximize the discharge, we need to minimize the perimeter
For a Rectangular channel,
P= B+ 2y
Also, A= By
𝐴
Thus B=
𝑦
(1)
(2),
y
Put the value of B in (1)
𝐴
P= + 2y
𝑦
For minimum P
𝑑𝑃
=0
𝑑𝑦
B
For minimum P
𝑑𝑃
=0
𝑑𝑦
P=
A
1
−𝑦 2
𝐴
+ 2y
𝑦
+2 =0
𝐴
=2
𝑦2
Also, A= By (from equation 2)
𝐵𝑦
=2
𝑦2
B= 2𝑦
This is the expression for most economic section in case of rectangular channel
Best Hydraulic Radius for rectangular channel
𝐴
𝐵𝑦
2𝑦 𝑦
2𝑦 2
𝑦
R= =
=
=
=
𝑃
𝐵+2𝑦 2𝑦+2𝑦
4𝑦
2
1.The most economical section of a rectangular channel is one where the width of the channel
is twice the depth of flow and
2.The most economical section of a rectangular channel is one which has hydraulic radius
equal to half the depth of flow.
BEST ECONOMIC SECTION FOR Trapezoidal Section
𝐴 = 𝐵 + 𝑧𝑦 𝑦
𝐴
𝐵=
− 𝑧𝑦
𝑦
(1)
1
(2)
𝑃 = 𝐵 + 2𝑦 1 + 𝑧 2
(3)
Substitute the value of B from 2
𝐴
𝑃=
− 𝑧𝑦 + 2𝑦 1 + 𝑧 2
𝑦
For most economic section,
𝑑𝑃
=0
𝑑𝑦
0=
𝐴
− 𝑧 + 2 1 + 𝑧2
2
−𝑦
𝐴
+ 𝑧 = 2 1 + 𝑧2
2
𝑦
y
z
B
𝐴
+ 𝑧 = 2 1 + 𝑧2
2
𝑦
Replace the value of A from 1
𝐵 + 𝑧𝑦 𝑦
+ 𝑧 = 2 1 + 𝑧2
2
𝑦
Multiply throughout by y
(B + zy) + 𝑧𝑦 = 2 𝑦 1 + 𝑧 2
𝑦 1 + 𝑧2
1
=2 (B + 2zy)
zy
zy
1
y
z
B
𝑦2 + 𝑧 2𝑦2
𝑦 1 + 𝑧2
Length of the inclined side=
1
2
Top width
Homework Assignment:
1. Best Economic section for triangular and circular section
2. Rayleigh method from Dimension Analysis
FLOW THROUGH
ORIFICES
AND MOUTHPIECES
ORIFICES
❑ An orifice is a small opening of any cross section (circular, triangular , rectangular etc) on the side of the
tank or bottom of the tank through which the fluid is flowing.
❑ The typical purpose of an orifice is the measurement of discharge.
Mouthpiece
A mouthpiece is an attachment in the form of a small tube or pipe fixed to the orifice (the length of pipe
extension is usually 2 to 3 times the orifice diameter) and is used to increase the amount of discharge.
❑ Mouthpiece is also used for measuring discharge
Classification of Orifice
1. Based on size
2. Based on shape
3. Based on shape of entrance or upstream edge
4. Based on discharge condition
Based on size
1. Small orifice : H > 5d
2. Large orifice : H ≤ 5𝑑
H is from centre of orifice to free surface of the liquid
d is diameter of orifice
Based on shape
1. Circular
2. Rectangular
3. Square
4. Triangular
Based on shape of entrance or upstream edge
a) Sharp edge orifice
b) Bell mounted orifice
c) Square orifice
Based on discharge condition
Derivation of Expression for Velocity and
Discharge Through the Orifice
Flow through small orifice ( head is more than 5 times the diameter)
Consider two points 1 and 2. Point 1 is inside the tank and point 2 is at the vena-contracta
Let the flow is steady and at a constant head
FLOW THROUGH A LARGE ORIFICES
If the head of liquid is less than 5 times the depth of the orifice, the orifice is called large orifice
Incase of small orifice the velocity in the entire cross- section of the jet is considered to be constant
and discharge can be calculate as :
𝑄 = 𝑐𝑑 𝑎 2𝑔ℎ
Incase of large orifice, the velocity is not constant over the entire cross section of the jet and hence
Q cannot be calculate using the formula above
FLOW THROUGH A LARGE ORIFICES
Consider a large rectangular orifice in one side of the tank discharging freely into the atmosphere under constant head H
TIME REQUIRED FOR EMPTYING A TANK THROUGH AN ORIFICE AT ITS BOTTOM.
Consider a tank, of uniform cross-sectional area, containing some liquid, and having an orifice at its bottom
Let A= Cross sectional area of the tank
a = area of the orifice
H1= Initial height of the liquid
H2= Final Height of the liquid
T= Time taken in seconds required to bring the level from H1 to H2.
TIME REQUIRED FOR EMPTYING A TANK THROUGH AN ORIFICE AT ITS BOTTOM.
Let A= Cross sectional area of the tank
a = area of the orifice
H1= Initial height of the liquid
H2= Final Height of the liquid
T= Time taken in seconds required to bring the level from H1 to H2.
Let at some instant the height of a liquid be h above the orifice and the liquid fall by an amount dh in small
time interval dt
The volume of liquid that has pass the tank in small time interval dt
𝑑𝑞 = −𝐴 𝑑ℎ
(𝑖)
During this small time dt , there will also small change in fluid height (dh) within the tank/reservoir, the height
will drop (negative sign)
TIME REQUIRED FOR EMPTYING A TANK THROUGH AN ORIFICE AT ITS BOTTOM.
We know that theoretical velocity through the orifice is
𝑉=
2𝑔ℎ
Discharge through the orifice in a small interval of time dt
𝑑𝑞 = 𝐶𝑑 𝑎 2𝑔ℎ 𝑑𝑡
(𝑖𝑖)
Equating (i) and (ii)
−𝐴 𝑑ℎ = 𝐶𝑑 𝑎 2𝑔ℎ 𝑑𝑡
𝑑𝑡 =
−𝐴 𝑑ℎ
𝐶𝑑 𝑎 2𝑔ℎ
=
1
−2
−𝐴 ℎ 𝑑ℎ
𝐶𝑑 𝑎 2𝑔
Time taken (T) to lower the level from H1 to H2 is by integrating the above equation from H1 to H2
Find the discharge through a rectangular orifice 2.0 m wide and
1.5 m deep fitted to a water tank. The water level in the tank is
3.0m above the top edge of the orifice. Take Cd = 0.62
The head of water over an orifice of diameter 40mm is 10 m.
Find the actual discharge and actual velocity of the jet at venacontracta. Take Cd =0.6 and
Cv =0.98
Classification of Mouthpiece
The mouthpieces can be classified based on the following categories:
1.
Classification based on the position of the mouthpiece
2.
Classification based on the shape of the mouthpiece
3.
Classification based on the nature of discharge at the outlet of the mouthpiece
Based on Position of Mouthpiece
1 Internal Mouthpiece : Internal mouthpieces are pipes fixed inside the tank or the vessel to
measure the rate of flow.
2. External Mouthpiece External mouthpieces are pipes that are fixed externally and project out
of the vessel or tank walls.