Hydraulics IM 2023 2024final Term

Module 5
Fluid Flow Concepts and Measurements
Engineers have developed ways to describe fluid flow patterns and to identify
important characteristics of the flow field. This terminology allows engineers to
communicate ideas essential to the design of systems such as bridge piers, air-
conditioning ducts, airfoils, and structures subjected to wind loads.
5.1. Flow Patterns
To visualize the flow field it is desirable to construct lines that show the flow
direction. Such a construction is called a flow pattern, and the lines are called
streamlines. The streamline is defined as a line drawn through the flow field in such a
manner that the local velocity vector is tangent to the streamline at every point along the
line at that instant. Thus the tangent of the streamline at a given time gives the direction
of the velocity vector. A streamline, however, does not indicate the magnitude of the
velocity. The flow pattern provided by the streamlines is an
instantaneous visualization of the flow field.
An example of streamlines and a flow pattern is shown
in the figure on the left for water flowing through a slot in the
side of a tank. The velocity vectors have been sketched at
three different locations: 𝑎, 𝑏, and 𝑐. The streamlines,
according to their definition, are tangent to the velocity
vectors at these points. This definition leads to the equation
for streamlines.
𝑑𝑢 𝑑𝑣 𝑑𝑤
= =
𝑢 𝑣 𝑤
where 𝑢, 𝑣, and 𝑤 are the velocity components in 𝑥, 𝑦,
and 𝑧 directions, respectively.
Also, the velocities are parallel to the wall in the wall region, so the streamlines
adjacent to the wall follow the contour of the wall. The generation of a flow pattern is a
very effective way of illustrating the geometric features of the flow field.
Whenever flow occurs around a body, part of it will go to one side and part to the
other as shown in the figure on the right above for
flow over an airfoil section. The streamline that
follows the flow division (that divides on the
upstream side and joins again on the downstream
side) is called the dividing streamline. At the
location where the dividing streamline intersects the
body, the velocity will be zero with respect to the
body. This is the stagnation point.
Pathline
The pathline simply is the path of a fluid particle as it moves through the flow
field. In other words, if a light were attached to a fluid particle, a time exposure
HYDRA 325 - HYDRAULICS 1

photograph taken of the moving light would be the pathline. Any particle of a stream of
fluid has at any given instant a certain velocity, 𝒗, which is a vector quantity and
therefore possesses both magnitude and direction. At the next instant forces acting on the
particle may cause it to have velocity which is different in amount and direction. The path
followed by a particle is called a path line.
Streakline
The streakline is the line generated by a tracer fluid, such as a dye, continuously
injected into the flow field at the starting point. Streakline concentrates on fluid particles
that have gone through a fixed station or point. At some instant of time the position of all
these particles are marked and a line is drawn through them. Such a line is called a
streakline.
5.2 Uniform and Non-Uniform Flow
Having introduced the general concepts of flow patterns, it is convenient to make

distinctions between different types of flows. These concepts can be best introduced by
expressing the velocity of the fluid in the form
𝒗 = 𝒗(𝒙, 𝒕)
where 𝒙 is the distance traveled by a fluid particle along a path, and 𝒕 is the time.
In a uniform flow, the velocity does not change along a fluid path.
𝝏𝒗
𝝏𝒙 = 𝟎
It follows that in uniform flow the fluid paths are straight and parallel, similar to
that of the figure on the right.
In nonuniform flow, the velocity changes along a fluid path.
𝝏𝒗
𝝏𝒙 = 𝟎
For the converging duct below, the magnitude of the velocity increases as the duct
converges, so the flow is nonuniform. For the vortex flow shown below, the magnitude of
the velocity does not change along the fluid path, but the direction does, so the flow is
nonuniform.
5.3 Steady and Unsteady Flow
Flows can be either steady or unsteady. In a steady flow, the velocity at a given
point on a fluid path does not change with time:
𝝏𝒗
=𝟎
𝝏𝒕
An unsteady flow exists if
𝝏𝒗
≠𝟎
𝝏𝒕

If the flow in the pipe changed with time due to a valve opening or closing, the flow
would be unsteady; that is, the velocity at any point selected on a fluid path would be
increasing or decreasing with time. Although unsteady, the flow would still be uniform.
Obviously, the pathline and streakline are very different. In general, neither pathlines
nor streaklines represent streamlines in an unsteady flow. Both the pathline and streakline
provide a history of the flow field, and the streamlines indicate the current flow pattern.
In steady flow the pathline, streakline, and streamline are coincident if they pass through
the same point.
5.4 Laminar and Turbulent Flow

Laminar flow is a well-ordered state of flow in which adjacent fluid layers move
smoothly with respect to each other. A typical laminar flow would be the flow of honey
or thick syrup from a pitcher. Laminar flow in a pipe has a smooth, parabolic velocity
distribution as shown in the figure below.
Turbulent flow is an unsteady flow characterized by intense cross-stream mixing. For
example, the flow in the wake of a ship is turbulent. The eddies observed in the wake
cause intense mixing. The transport of smoke from a smoke stack on a windy day also
exemplifies a turbulent flow. The mixing is apparent as the plume widens and disperses.
An instantaneous velocity profile for turbulent flow in a pipe is shown in the figure
above. A near uniform velocity distribution occurs across the pipe because the high-
velocity fluid at the pipe center is transported by turbulent eddies across the pipe to the
low-velocity region near the wall. Because the flow is unsteady, the velocity at any point
in the pipe fluctuates with time. The standard approach to treating turbulent flow is to
represent the velocity as a time-averaged average value plus a fluctuating quantity, 𝑢 =
𝑢̅ + 𝑢′ The time-averaged value is designated by 𝑢̅ the figure above. The fluctuation
velocity is the difference between the local velocity and the averaged velocity. A
turbulent flow is often designated as “steady” if the time-averaged velocity is unchanging
with time.
In general, laminar pipe flows are associated with low velocities and turbulent flows
with high velocities. Laminar flows can occur in small tubes, highly viscous flows, or
flows with low velocities, but turbulent flows are, by far, the most common.

5.5. Discharge, 𝑸
The discharge, 𝑄, often called the volume flow rate, is the volume of fluid that passes
through an area per unit time. For example, when filling the gas tank of an automobile,
the discharge or volume flow rate would be the gallons per minute flowing through the
nozzle.
Typical units for discharge are cubic feet per second (𝑓𝑡3/𝑠 or 𝑐𝑓𝑠), cubic feet per
minute (𝑓𝑡3/𝑚𝑖𝑛 or 𝑐𝑓𝑚), gallons per minute (𝑔𝑝𝑚), cubic meters per second (𝑚3/𝑠),
and liters per second (𝐿/𝑠).
Volumetric Flow Rate

The discharge or volume flow rate in a pipe is related to
the flow velocity and cross-sectional area. Consider the
idealized flow of fluid in a pipe as shown in the figure on the
right in which the velocity is constant across the pipe section.
Note that discharge refers to volume or volumetric flow
rate. The discharge or volume flow rate equation is defined by
𝑸 = lim ∆𝑉 𝑽
→ 𝑸 = = 𝑨𝒗
∆𝑡→0 ∆𝑡 𝒕
The discharge given above is based on a constant flow
velocity over the cross-sectional area. In general, the velocity
varies across the section such as in the figure on the right. The
volume flow rate through a differential area of the section is
𝑉 𝑑𝐴, and the volume flow rate is obtained by integration over
the entire cross-section:
𝑸 = ∫ 𝒗 ∙ 𝒅𝑨
𝑨
In many problems—for example, those involving flow in pipes—one may know
the discharge and need to find the mean (average) velocity without knowing the actual
velocity distribution across the pipe section. By definition, the mean velocity is the
discharge divided by the cross-sectional area.
𝑸
𝒗̅ =
𝑨
Mass Flow Rate,
𝒎̇
The mass flow rate, 𝒎̇ , is the mass of fluid passing through a cross-sectional area
per unit time. The common units for mass flow rate are kilograms per second (𝑘𝑔/𝑠),
pound mass per second (𝑙𝑏𝑚/𝑠), and slugs per second (𝑠𝑙𝑢𝑔𝑠/𝑠). Using the same
approach as for volume flow rate, the mass of the fluid is ∆𝒎 = 𝝆 ∙ ∆𝑽, where 𝝆 is the
average density of the fluid. The mass flow rate equation is
𝒎̇ = lim ∆𝑚
= 𝜌 lim ∆𝑉 → 𝒎̇ = 𝝆𝑸 = 𝝆𝑨𝒗
∆𝑡→0 ∆𝑡 ∆𝑡→0 ∆𝑡
The generalized form of the mass flow equation is
𝒎̇ = ∫ 𝝆𝒗 ∙ 𝒅𝑨

𝑨

5.6 The Continuity Equations
Several simplified forms of the continuity equation are used by engineers for flow
in a pipe. The equation is developed by positioning a control volume inside a pipe, as
shown in the figure below.
Mass enters through station 1 and exits through station 2. The control volume is
fixed to the pipe walls, and its volume is constant. If the flow is steady, then 𝒎𝑪𝑽 is
constant so the mass flow formulation of the continuity equation reduces to
𝒎̇ 𝟐 = 𝒎̇ 𝟏 → 𝝆𝟐𝑨𝟐𝒗𝟐 = 𝝆𝟏𝑨𝟏𝒗𝟏
For flow with a uniform velocity and density distribution, the continuity equation
above is applicable for steady flow in a pipe. If the flow is incompressible, then
𝑨𝟐𝒗𝟐 = 𝑨𝟏𝒗𝟏 → 𝑸𝟐 = 𝑸𝟏
Sample Problems
1. Compute the discharge of water through 2. The discharge of air through 24 in pipe
a 3 inches pipe if the mean velocity is is 8600 cfm. Compute the mean velocity.
8.5 ft/s.

3. What diameter of pipe in inches is 4. The pipe line shown with continuous
required to carry 10 gpm of gasoline at flow has a discharge of 0.25 m 3/s of
rate of 7 ft/s? water. Compute the velocity of each size
of pipe.

5. A circular chimney has a diameter of 6 m at the base and converges uniformly at a
height of 20 m. Coal gas having a specific weight of 4 N/m3 enters the chimney at the
bottom. Its specific weight increases uniformly until the top where it has a volume of
6 N/m3. Calculate the discharges in m3/s for every 5 m up the stack if the velocity at
the bottom is 8.20 m/s.

5.7 The Energy Equation
The energy equation involves energy, work, and power as well as machines that
interact with flowing fluids. These topics are introduced in this section.
When matter has energy, the matter can be used to do work. A fluid can have several
forms of energy. For example, a fluid jet has kinetic energy, water behind a dam has
gravitational potential energy, and hot steam has thermal energy. Work is force acting
through a distance when the force is parallel to the direction of motion. Work is done
when the piston exerts a pressure force that acts on the liquid over a distance. This force
produces a torque and work is given by the cross-product of the force and the distance; or
it could be obtained by the product of torque and the angular displacement.
𝑾=𝑭×𝒅
A. Work, Energy, and Energy Head

Work and energy both have the same primary dimensions, and the same units, and
both characterize an amount or quantity. For example, 1 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 is the amount of thermal
energy needed to raise the temperature of 1 𝑔𝑟𝑎𝑚 of water by 1°𝐶. Other common units
include the 𝐽𝑜𝑢𝑙𝑒 (𝐽), 𝑁𝑒𝑤𝑡𝑜𝑛 − 𝑚𝑒𝑡𝑒𝑟 (𝑁 − 𝑚), 𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡 − ℎ𝑜𝑢𝑟 (𝑘𝑊ℎ), 𝑓𝑜𝑜𝑡 −
𝑝𝑜𝑢𝑛𝑑 − 𝑓𝑜𝑟𝑐𝑒 (𝑓𝑡 − 𝑙𝑏𝑓), 𝑐𝑎𝑙𝑜𝑟𝑖𝑒 (𝑐𝑎𝑙), and the British thermal unit (𝐵𝑡𝑢).
Energy is defined as the ability to do work. Both energy and work are measured in
foot, lb or kN-m. The energy possessed by a flowing fluid consists of the kinetic and the
potential energy. Potential energy in fluids may on turn be subdivided into energy due to
position or elevation above a given datum plane, and energy due to pressure in fluid.
3 Forms of Energy
1. Kinetic energy is the ability of a mass to do work by virtue of its velocity
𝐾𝐸 = 1 𝑚𝑉2 1
𝑊 𝑉2
2 =
2𝑔
𝐾𝐸 𝑉 2
= = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 ℎ𝑒𝑎𝑑 𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑
𝑊 2𝑔
Here, 𝒎, 𝑾, and 𝒗 are the mass, weight, and
velocity of the object, respectively.
2. Elevation energy is manifested in a fluid by

virtue of its position or elevation with respect to
some arbitrarily selected horizontal datum
plane, considered in connection with the action
of gravity.
Consider a tank which is full of water. If a
hole at the bottom is opened, the fluid with flow by
virtue of the height of water above the hole.
Elev Energy = Wz
Elev Head = Z
Where:

Z = position of the fluid above (+) or below (-) the datum

3. Pressure energy
Consider a closed tank filled with a fluid which have a small opening at the top
without pressure at the top, the fluid practically would not floe. But with P, fluid will
flow.
𝑃
Pressure Energy = 𝑊
𝛾
𝑃
Pressure Head =
𝛾
Total energy of flow, E, is the sum of the kinetic and potential energy (total head).
𝐸=𝑉
2
𝑃
+ +𝑍
2𝑔 𝛾
5.8 Power
Power is the rate at which work is done.

For a fluid of unit weight 𝛾 or W (N/m3) and moving at rate of Q (m3/s) with a
total energy of E (m), the power in N-m/s (Joule/s) or watts is
𝑃 = 𝑄𝑊𝐸 or
𝑃 = 𝑄𝛾𝐸
𝑄𝑊𝐸
Horsepower, 𝐻𝑝 =
746
1 hp = 746 watts
= 550 ft-lb/s
Sample Problems
1. A fluid is flowing in a pipe 8 inches in diameter with a mean velocity of 10 ft/s.
The pressure at the center of the pipe is 5 lb/in 2 and the elevation of the pipe
above the assumed datum is 15 ft. Compute the total head in feet if the fluid is
a. Water
b. Oil (s = 0.8)
c. Molasses (s = 1.5)
d. Gas (𝜔 = 0.040 lb/ft3)

2. The jet of water from a nozzle 3. At point A where the suction point
discharging into air has a diameter of leading to a pump is 4 ft below the
6 in. and a mean velocity of 120 ft/s. pump, an open manometer indicates
Compute the velocity head and the a vacuum of 7 in. of mercury. The
horsepower in the jet. pipe is 4 in. in diameter, and
discharge is 1.1 cfs of oil (s = 0.85).
Compute the total head at pt A with
respect to a datum at the pump.

4. A liquid of specific gravity 1.75 flows in a 6 cm horizontal pipe. The total energy at a
certain point in the flow is 80 J/N. The elevation of the pipe above a fixed datum is
2.6 m. If the pressure at the specified point is 75 kPa,
a. Determine the velocity of flow.
b. Determine the discharge.
c. Determine the power available at the specified point in watts.

5.9 Bernoulli’s Energy Theorem
Daniel Bernoulli, a Swiss mathematician, demonstrated a theorem with reference

to fluid motion under steady flow condition. He proved that in any fluid in motion where
friction is neglected, the total energy possessed by a given mass of the fluid is the same at
every point in the path of flow.
Bernoulli’s Energy Equation Without Head Lost:

E1 = E2
𝑉 2 𝑃 𝑉 2 𝑃2
1 2
1
+ + 𝑍1 = + + 𝑍2
2𝑔 𝛾 2𝑔 𝛾
Bernoulli’s Energy Equation With Head Lost:

E1 = E2 + HL
𝑉 2 𝑃
1 𝑉 2 𝑃2 2
1
+ + 𝑍1 = + + 𝑍2 + 𝐻𝐿

Presence of Pump
A pump is a mechanical device which when installed in the flow system will add
energy (output) to the system. It is usually used to raise water from a lower to a higher
elevation.
Bernoulli’s Equation from 1 – 2

E1 + HA= E2 + HL
𝑉12 𝑃 𝑉 2 𝑃2
1
2
+ + 𝑍1 + 𝐻𝐴 = + + 𝑍2 + 𝐻𝐿1−2
Presence of Turbine
A turbine or motor is a machine used to convert energy of flow into mechanical work.
Bernoulli’s Equation from 1 – 2

E1 = E2 + HE + HL
𝑉12 𝑃 𝑉 2 𝑃2
1
2
+ + 𝑍1 = + + 𝑍2 + 𝐻𝐸 + 𝐻𝐿1−2
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝐸𝑓𝑓𝑐𝑖𝑒𝑛𝑐𝑦 = 𝑥100%
𝑃𝑖𝑛𝑝𝑢𝑡

Sample Problems
1. The diameter of a pipe carrying oil 2. Water is discharged through a nozzle
with specific gravity = 0.80 changes having a diameter of the jet 100 mm
gradually from 100 mm at section 1 at a velocity of 60 m/s at a point 240
to 250 mm at section 2, section 1 m below the reservoir.
being 3 m lower than section 2 when a. Compute the total headloss.
0.25 m3/s is flowing. Find P 2 – P1. b. Compute the horsepower
Neglect head lost. produced by the jet.
c. Compute the power lost in
friction.

3. A 20 hp suction pump operating at 70% efficiency draws water from a suction line
whose diameter is 200 mm and discharges into air through a line whose diameter is
150 mm. The velocity in the 150 mm line is 3.6 m/s. If the pressure at pt A in the
suction pipe is 34 kPa below the atmosphere, where A is 1.8 m below B on the 150
mm line,
a. Fine the energy added by the pump
b. Determine the maximum elevation above B to which water can be raised
assuming a head loss of 3 m due to friction
c. Find the pressure at B

4. A turbine is installed as shown in the figure. The gage at Pt. 1 is 80 kPa while that at
Pt. 2 it is −46 kPa. If the rate of flow is 0.48 m3/s,
a. Compute the total head extracted by the turbine.
b. Compute the input horsepower of the turbine.
c. If the efficiency of the turbine is 82%, compute its output hp.
40 350 mm Ø
T
0 1
m 1.6 m
m
2

5. The 600 mm Ø pipe conducts water from reservoir A to a pressure turbine, which
discharges through another 600 mm Ø pipe into tail race B. The head loss from A to 1
is 5 times the velocity head in the pipe and from 2 to B is 0.2 times the velocity head
in the pipe. If the discharge is 700 li/sec
a. Compute the total headloss
b. Compute the energy given up by the water to the turbine
c. What horsepower is being given up by the water to the turbine
A LEl. 60 m
1 T 2 El. 15 m
B El. 0

Module 6
Fluid Flow Measurement
6. 1 Venturi Meter
Venturi meter is a device usually installed across a pipeline for the purpose of
measuring the quantity of fluid flowing in the line. It consists of converging tube which is
connected to the main pipe, a short section of uniform diameter known as the throat, and
a diverging tube which is again connected to the main pipe. The angle of divergence is
limited to about 6o to reduce the head lost caused by the turbulence as the velocity is
reduced.
Consider sections (1) and (2), neglecting friction losses,

Bernoulli’s Energy Equation between (1) and (2)
𝑉12 𝑃 𝑉22 𝑃2
1
+ + 𝑍1 = + + 𝑍2
Transposing, combining P.E. and K.E.
𝑉22 𝑃 𝑃2
𝑉1
2
1
− =( + 𝑍1) − ( + 𝑍2)
2𝑔 2𝑔 𝛾 𝛾
since Z1 = Z2, thus the equation can be reduced into
𝑽𝟐𝟐 𝑽𝟏𝟐 𝑷𝟏 𝑷𝟐
− = − (eq.1)
𝟐𝒈 𝟐𝒈 𝜸 𝜸
The Venturi Principle states that “the increase in kinetic energy per unit weight

is equal to the decrease in potential energy per unit weight.”

If the elevations and difference in pressure between pt. 1 and 2 are known, the
problem can be solved by relating velocities V1 and V2 by continuity equation.
𝐴1𝑉1 = 𝐴2𝑉2
𝐴2
𝑉1 = 𝑉2
1
𝐴2 𝐴
Let 𝑛 =
𝐴1
𝑽𝟏 = 𝒏𝑽𝟐 (eq. 2)
Substitute eq. 2 in eq. 1
𝑉2 (𝑛𝑉2) 𝑃 𝑃2
2 2 1 −
= 𝛾
2𝑔 − 2𝑔 𝛾
𝑽𝟐 = √ 𝟐𝒈 𝑷𝟏
(𝟏−𝒏 )
𝟐 ( 𝜸 − 𝑷𝜸𝟐 ) (eq. 3)
The theoretical values of Qt can be computed once V1 and V2 are known. The
actual discharge Q can be computed by multiplying the Qt by the coefficient of discharge
or meter coefficient C.
𝑄𝑡 = 𝐴2𝑉2
𝑸 = 𝑪𝑸𝒕 (eq. 4)
The actual discharge may be accomplished by series of observation, usually by
measuring the total amount of fluid passing through the device for a known period.

Sample Problems
1. A test to determine the discharge coefficient of a 5 cm by 1.25 cm Venturi meter,
the total weight of water passing through the meter in 5 minutes was measured as
3420 N. A mercury – water differential gage connected to inlet base and throat
showed an average deflection of 38 cm. Determine the meter coefficient.

2. A 150 mm diameter horizontal Venturi meter is installed in a 450 mm diameter
water main. The deflection of mercury in the differential manometer connected
from the inlet to the throat is 375 mm of mercury.
a. Determine the discharge neglecting headloss
b. Compute the discharge if the headloss from the throat is 300 mm of water
c. Determine the meter coefficient

3. A 30 cm by 15 cm Venturi meter is installed in a vertical pipe. The vertical
distance from the inlet to the throat is 30 cm. The flow is upward through the
meter containing a gage liquid (s = 1.50) attached to inlet base and throat, the
deflection being 80 cm. Determine the actual flow if C = 0.97.

4. A vertical venture meter, 150 mm in diameter is connected to a 300 mm diameter
pipe. The vertical distance from the inlet to the throat being 750 mm. If the
deflection of mercury from the inlet to the throat is 360 mm, determine the flow
of water through the meter if the meter coefficient is 0.69. Determine also the
headloss from the inlet to the throat.

6.2 Nozzle
A nozzle is a converging tube installed at the end of a pipe or hose for the purpose
of increasing the velocity of flow. It may be used, in particular to provide a high velocity
stream for firefighting or for measuring the flow of fluid in the pipe or hose into which it
is connected.
Two types in common use:
1. Converging part may be frustum of a cone
2. Or the inside be convex
Consider sections (1) and (2), neglecting 𝑃

𝑉1
2 𝑉22
1
friction losses, + =
2𝑔 𝛾 2𝑔
Bernoulli’s Energy Equation between 𝑽𝟏 𝑷𝟏
(1) and (2) 𝑽𝟐 = √ 2
𝟐𝒈 ( 𝟐 + ) (eq. 1)
𝑉1 𝑃1 𝟐𝒈 𝜸
𝑉12 + 𝑃1 + 𝑍 = 𝑉22 + 𝑃2 + 𝑍 if = + ;
2𝑔 𝛾 1
2𝑔 𝛾 2
2𝑔 𝛾
The lost can be expressed as a
since Z1 = Z2, thus the equation can be percentage of the velocity head in the jet
reduced into 𝑉2 = 𝑉𝑡 = √2𝑔𝐻
𝑃 𝑃2 𝑉 =𝐶 𝑉
𝑉1
2 1 𝑉2
2
+ = + 𝑎 𝑣 𝑡
2𝑔 𝛾 2𝑔 𝛾 𝑉𝑎 = 𝐶𝑣√2𝑔𝐻;
𝑃2
= 0, point 2 is exposed to the Where Cv = coefficient of velocity
𝛾
atmosphere 𝑸𝒂 = 𝑽𝒂𝑨𝟐
or 𝑸𝒂 = 𝑪𝒗𝑨𝟐√𝟐𝒈𝑯 (eq. 2)
Actually, there is a loss of head between these points and this is obtained by
considering actual conditions at the tip, that is
Head Loss = Initial Head - Actual Head

𝑉12 𝑃 𝑉𝑎2
1
𝐻𝐿𝑁 =
+ −
2𝑔 𝛾 2𝑔
From 𝑉𝑎 = 𝐶𝑣√2𝑔𝐻 → 𝑉𝑎2 = 𝐶𝑣22𝑔𝐻
𝑉𝑎2
𝐻= → actual head
𝐶𝑣2𝑔
𝑉𝑎2 𝑉𝑎2
𝑯𝑳

𝐻𝐿𝑁 = −
𝐶 2𝑔 2𝑔 𝑣
𝟐
𝑽𝒂 ( 𝟏 − 𝟏) (eq. 3) head loss due to nozzle
𝑵 = 𝟐𝒈
𝑪 𝒗𝟐

Sample Problems
1. A nozzle 10 cm by 4 cm is strained vertically upward. At a point 30 cm below the
tip, the pressure is maintained at 145 KPa. If the available power in the jet is 2.6
KW, find
a. The height through which the jet of water will rise, neglecting air friction
b. The value of Cv
c. The diameter of the jet 6m above the tip of the nozzle

2. Calculate the minimum power of the pump which will send the jet over the wall.
Neglect losses.

6.3 Pitot Tube
A tube with circular cross section bent in the shape of an L, with both ends open
was first used by a French scientist, Henri Pitot in 1732, it is a device used in measuring
the velocity fluid in an open channel.

𝑉12 𝑃 𝑉22 𝑃2
1
+ + 𝑍1 = + + 𝑍2
since Z1 = Z2=V2= 0 (at stagnation point 2, V = 0),
thus the equation can be reduced into
𝑃 𝑃2
𝑉1
2 =
1
𝛾
2𝑔 + 𝛾
𝑃2 𝑃1
𝑉1 = √ 2𝑔 ( − )
𝛾 𝛾
𝑃2 𝑃1
But 𝐻 = −
𝛾 𝛾
𝑉𝑡 = 𝑉1 = √2𝑔𝐻
𝑉𝑎 = 𝐶𝑡𝑉𝑡
𝐶𝑡 = coefficient of Pitot Tube

Pitot tube in a pipe
1 2
Sg gr = S1
y
Sg gr = S2
𝑃1𝑆1
+ 𝑦𝑆 + ℎ𝑆 − (ℎ + 𝑦)𝑆 = 𝑃2𝑆1
1 2 1
𝛾 𝛾
𝑃2𝑆1 − 𝑃1𝑆1
= 𝑦𝑆 + ℎ𝑆 − (ℎ + 𝑦)𝑆
1 2 1
𝛾
(𝑃2 − 𝑃1)𝑆1
= ℎ(𝑆2 − 𝑆1)
(𝑃2 −𝛾𝑃1) ℎ(𝑆2 − 𝑆1)
=
𝛾 𝑆1
(𝑃2 − 𝑃1)
𝑉 = 𝐶 √2𝑔
𝑡
𝛾
ℎ(𝑆2 − 𝑆1)
𝑉 = 𝐶𝑡 √2𝑔
𝑆1

Sample Problem
1. A pitot static tube (C=1.0) is used to measure air speeds. With water in the
differential manometer and a gage difference of 75 mm, calculate the air speed
𝑘𝑔
using 𝜌𝑎𝑖𝑟 = 1.16 .
𝑚3
2. A pitot static tube (C=1.0) is is placed at a point where the velocity is 2.0 m/s.
The specific gravity of fluid is 2.0, and the upper portion of the manometer
contains air (γ_air=12 N/m^3). Compute for the value of h (m).

Module 7
Flow in Orifice
7.1 Steady Flow in Orifice
Orifice is an opening with a closed perimeter through which fluid flows. The
primary purpose of this opening is to measure and control the fluid flow. The upstream
face maybe rounded or sharp. An orifice with prolonged side such as piece of pipe,
having a length of about three times its diameter is called a tube. According to shape,
orifice maybe circular, square or rectangular in cross-section. A circular sharp-crested
orifice is the most commonly used for metering because of the simplicity of its design
and construction.

𝑉 2 𝑃
1 𝑉 2 𝑃2
2
1
+ + 𝑍1 = + + 𝑍2
Where V1= 0, P1 and P2 = 0 (exposed to atmosphere, Z1 = H, and Z2 = 0 (datum)
𝑽𝟐 = √𝟐𝒈𝑯 (eq 1)
A. Coefficient of Velocity
Experimentally, it is found out that the actual mean velocity of the jet from a
sharp-edged orifice is little less than the theoretical velocity. This difference is attributed
to the fluids viscosity which was neglected in deriving the formula above, to correct for
the head losses initially neglected, a correction factor known as coefficient of velocity Cv
must be multiplied to the theoretical velocity.
𝑽𝒕 = 𝑽𝟐 = √𝟐𝒈𝑯
𝑉𝑎 = 𝐶𝑣𝑉𝑡
𝑽𝒂 = 𝑪𝒗√𝟐𝒈𝑯
The discharge through an orifice is therefore:
𝑄𝑡 = 𝐴𝑉𝑡
𝑸𝒕 = 𝑨√𝟐𝒈𝑯

𝑸𝒂 = 𝑪𝑨√𝟐𝒈𝑯

B. Coefficient of Contraction
In the figure shown, the individual particles of the mass approaching the orifice
follow converging paths. Due to the inertia of the particles getting close to the inner wall,
they cannot make abrupt changes in their direction as they get to the opening. This causes
them to follow curvilinear paths affecting a contraction of the jet up to a section, say B-C,
at which point the stream paths are assumed to be parallel, and the pressure having a
value equal to that of the surrounding medium.
The section B-C where the contraction of the jet ceases is called the vena
contracta and its distance from the inner wall of the orifice is approximately ½ of the
diameter of the orifice. Moreover, the unit less ratio of the area of the vena contracta to
the area of the orifice is known as the coefficient of contraction, Cc.
𝒂
𝑪𝒄 =
𝑨
C. Coefficient of Discharge
The theoretical discharge is the product of the area of the orifice and the
theoretical velocity.
𝑸𝒕 = 𝑨𝑽𝒕
Furthermore, the real or actual discharge is obtained at the issuing jet (Vena
contracta) and represented as the product of the area of the contracted section and the
actual velocity.
𝑄𝑎 = 𝐴𝑉𝑎
𝑄𝑎 = 𝐶𝑐𝐴(𝐶𝑣√2𝑔𝐻)
𝑄𝑎 = 𝐶𝑐𝐶𝑣𝐴√2𝑔𝐻
Where
C = coefficient of discharge, C = CcCv
A = cross-sectional area of orifice

D. Submerged orifice
A submerged orifice is the one with downstream force submerged as shown. If the
surface is subject to atmospheric pressure, and neglecting velocity of approach, the head
H is given by
𝑯 = 𝒉𝟏 − 𝒉𝟐 = 𝒉
l.s
h
h1 h2
If there is a pressure on top of the liquid surface, convert it into equivalent height
if the liquid in the container.
E. Horizontal orifice discharging upward
HL
h
H
𝐻𝐿 = ℎ − 𝐻

Sample Problems
1. Solve for the theoretical velocity of the jet, the actual velocity and the discharge
of the orifice considering the following:
a. P1 = P2 = 0
h = 5 m of oil P1
Do = 5cm
Cc = 1. 0 (rounded)
Cv = 0.97 P2
h
b. P1 = P2 = 0
h = 5 m of water V2
Do = 5cm
Cc = 0.62 (sharp edged)
Cv = 0.98
c. P1 = 70 kPa
P2 = 0
h = 1.6 m (liquid s = 3.0)
Do = 7.5 cm
Cc = Cv = 0.95 (rounded)

2. The horizontal orifice is 7.5 cm in diameter with Cc = 0.63, Cv = 0.98. When h =
2.30 m, compute the height to which the jet will rise above the orifice. What is the
diameter of the jet 1 m above the orifice? Neglect air friction.

3. Water flows from a tank through 200 mm and 125 mm diameter orifices with a
total drop in the water surface of 12.0 m. The coefficient of discharge for each
orifices is C = 0.65. assuming the flow to be uniform and neglecting velocity of
approach, determine the following:
a. Height h1
b. Height h2
c. Maximum discharge
h1
h2 12.0 m
200 mm ∅ 125 mm ∅

4. A jet is issued from the side of a tank under a constant head of 3 m. The side of
the tank has an inclination of 1H to 1V. The total depth of water in tank is 6.70 m.
Neglecting air resistance and assuming Cv = 1.0, determine the following:
a. The maximum height to which the jet will rise
b. The point it strikes a horizontal plane 1.20 m below the bottom of the
tank, and
c. The velocity of the jet as it strikes the ground

5. An orifice at the side of the tank is located 1 m above the bottom of the tank
which is resting on the ground. The jet of water strikes a distance of 2.75 m
horizontally away from the orifice with C v = 0.98. The height of the tank is 4 m
and it is filled with water 2 m depth and on top of it is another liquid having a
depth of 1 m.
a. Determine the velocity of the jet.
b. Determine the equivalent constant head of water that causes flow out of
the tank.
c. Determine the specific gravity of the liquid.

7.2 Unsteady Flow in Orifices: Discharge under a Falling or Rising Head
𝑄𝑖 ≠ 𝑄𝑜
Case 1: Qi = 0, the surface drops and the volume decreases

Rate of change in the volume is the outflow, Qo
𝑑𝑉𝑜𝑙 1 ℎ1
𝐴𝑑ℎ
= 𝑄𝑜 = 𝐶𝐴𝑜√2𝑔ℎ 𝑡= ∫
𝑑𝑡 ℎ1/2
𝐶𝐴𝑜√2𝑔 ℎ2
𝑑𝑉𝑜𝑙
𝑑𝑡 = Sides of the tank are vertical, A is
𝐶𝐴𝑜√2𝑔 constant
ℎ
ℎ2 < ℎ < ℎ1 𝟐𝑨
𝒕= {𝒉𝟏 𝟏/𝟐 𝟏/𝟐
− }
𝒉𝟐
𝑪𝑨𝒐√𝟐𝒈
1 ℎ1
𝑑𝑉𝑜𝑙
𝑡 = 𝐶𝐴 2𝑔 ∫
√ 1/2
𝑜 ℎ2 ℎ
Case 2: when 𝑸𝒊 ≠ 𝑸𝒐 at time t = 0

a. 𝑄𝑖 < 𝑄𝑜, the liquid surface drops
𝑑𝑉𝑜𝑙 𝒉𝟏 𝑨𝒅𝒉
= 𝑄𝑜 − 𝑄𝑖 𝒕=∫
𝑑𝑡
ℎ1
𝑑𝑉𝑜𝑙 𝒉𝟐 𝑪𝑨𝒐√𝟐𝒈𝒉 − 𝑸𝒊
𝑡=∫
𝐴 = 𝑓(ℎ)
ℎ2 𝐶𝐴𝑜√2𝑔ℎ − 𝑄𝑖
b. 𝑄𝑖 > 𝑄𝑜, the liquid surface rises and the volume increases
𝑑𝑉𝑜𝑙
= 𝑄𝑖 − 𝑄𝑜
𝑑𝑡
𝒉𝟐

𝑨𝒅𝒉
𝒕=∫
𝒉𝟏 𝑸𝒊 − 𝑪𝑨𝒐√𝟐𝒈𝒉

Unsteady Flow in a submerged orifice
Unsteady flow problem in a submerged orifice usually requires the determination

of the time for the equalization of the surfaces in the two tanks, that is, when the flow
through the orifice ceases or when the surfaces reach the same elevation, using the free-
flow type.
ℎ𝑜 𝑑𝑉𝑜𝑙
𝑡=∫
0 𝐶𝐴𝑜√2𝑔ℎ
ℎ𝑜 = ℎ1 − ℎ2 at t = 0
𝑨 𝟏𝑨 𝟐
𝒅𝑽𝒐𝒍 = 𝒅𝒉
𝑨𝟏 + 𝑨𝟐
A1 and A2 – cross sectional areas of the chambers
A1 A2
dVol1 Vlost
dh1
H2 H1
dVol2 Vgain
dh2
Vlost = Vgain
𝐻1
𝐴2𝑑ℎ2
𝑡=∫
𝐻2 𝑄𝑜
1 𝐴1𝐴2
𝑡= 𝐻1 −1/2
∫ 𝐴1 + 𝐴2 𝐻 𝑑ℎ
𝐶𝐴𝑜√2𝑔 𝐻2
𝟐 𝑨𝟏𝑨𝟐
𝒕 = 𝑪𝑨 𝟐𝒈 𝑨 + (√𝑯𝟏 − √𝑯𝟐)
𝑨
𝒐√ 𝟏 𝟐

Sample Problems
1. A vertical cylindrical tank has an orifice in its side at a point 0.60 m above the
bottom. If the discharge reduces the depth of water in the tank from 6 m to 4 m in 78
seconds, what is the time required to reduce the depth from 3.6 m to 1.6 m?
2. A tank is in the shape of a frustum of a cone having its bases horizontal and the axis
vertical. The tank is 3 m high and filled with water. It has a diameter of 2.5 m at the
top and 1 m at the bottom. What is the time required to empty the tank through a
sharp-edged orifice 7.5 cm square? C = 0.60.

3. A hemispherical shell with the base horizontal and uppermost is filled with water. If
the radius is 2.5 m, determine the time required to empty the container through the
sharp-edged 15 cm diameter orifice (C=0.60) located at the lowest point.

4. A sharp-edged orifice 10 cm in 5. A sharp-edged orifice 10 cm in
diameter in the side of the tank diameter in the side of the tank
having a horizontal cross-section 2 m
having a horizontal cross-section 2 m
square discharges water under a
square discharges water under a
constant head. The rate by which the
head is kept constant is suddenly constant head. The rate by which the
changed from 0.023 m3/s to 0.034 head is kept constant is suddenly
m3/s. How long will it take after the decreased from 0.034 m3/s to 0.023
change occurs for the head over the m3/s. How long will it take after the
orifice to be 2 m? Use C = 0.60. change occurs for the head to drop
by 1m? Use C = 0.60.

6. Two vertical circular cylindrical tanks are connected near the bottom by a short tube
having a cross-sectional area of 78.5 cm2. The inside diameters of the tanks are 3 m
and 1.5 m respectively. The tanks contain oil. With the valve connecting tube closed,
the oil surfaces are 4 m in the larger tank and 0.60 m in the smaller tank above the
tube. With C = 0.78, find the time required for the surfaces to equalize following a
quick opening of the valve. Assume the properties of the tube to be similar to that of
an orifice.

7. Two vertical cylindrical tanks 1 and 2 having diameters 2 m and 3 m respectively are
connected with a 200 mm diameter tube at the lower portion and having C = 0.60.
When the tube is closed, the water surface in tank 1 is 5 m above tank 2. How long
will it take after opening the tube for the water surface in tank 2 to rise by 1 m?

Module 8
Weirs
Weirs are overflow structure which are built across an open channel for the
purpose of measuring or controlling the flow of liquids.
According to shape, weirs may be rectangular, triangular, trapezoidal, circular,
parabolic, or any other regular form. According to form, weirs may be sharp-crested or
broad crested.
The flow over a weir may either be free or submerged. If the water downstream
from the weir is lower than the crest, the flow is free, but if the downstream surface is
higher than the crest, the flow is submerged.
8.1 Steady Flow in weirs

A. Rectangular weir
H V1
L h V2
dh
0
d
P
𝑑𝑄𝑡 = 𝑑𝐴 𝑉𝑡
𝑑𝑄𝑡 = 𝐿 𝑑ℎ 𝑉2
𝑉 12
𝑑𝑄𝑡 = 𝐿 𝑑ℎ √2𝑔 (ℎ + )
2𝑔
V1 = Va = velocity of approach
𝐻
𝑉𝑎2 1/2
𝑄𝑡 = √2𝑔𝐿 ∫ (ℎ + 𝑑ℎ
)
0 2𝑔
Assuming Va is constant
2 𝑉2 3/2 3/2
𝑎 𝑉𝑎2
𝑄𝑡 = √2𝑔𝐿 [(𝐻 + ) −( )
]
3 2𝑔 2𝑔
𝑄𝑎 = 𝐶𝑄𝑡
2 3/2 3/2
𝑉2 𝑉𝑎2
𝑄 = 𝐶√2𝑔𝐿 [(𝐻 𝑎 − ( ) ]
𝑎 3 2𝑔 ) 2𝑔
+2
𝐶𝑤 =
𝐶√2𝑔 3 𝟑/𝟐 𝟑/𝟐
𝑽 𝟐 𝑽
𝟐
𝑸𝒂 = 𝑪𝒘 𝑳 [(𝑯 + 𝒂
𝟐𝒈 )

𝒂
−( ) 𝟐 𝒈
𝐻 2
If the ratio is sufficiently small, V
is negligible; 𝑉𝑎 ≈ 0
a
𝑃 2𝑔
𝟑/𝟐
𝑸𝒂 = 𝑪𝒘𝑳𝑯

Cw Formulas (Metric)
1. Francis Formula
𝐻
Cw = 1.84; < 0.40
𝑃
2. Rehbock and Chow Formula
𝐻
𝐶𝑤 = 1.8 + 0.22
3. Bazin Formula 𝑃
0.02161 2
𝐶𝑤 = 0.5518 (3.248 ) [1 + 0.55 (𝐻 ) ]
+ 𝐻 𝑑
Contracted Rectangular Weir
The effective length of L of a contracted weir is given by:
L L
L’ L’
One-end Contraction Two-end Contraction

(N = 1) (N = 2)
𝐿 = 𝐿′ − 0.1𝑁𝐻
Where: L’ = measured length of crest
N = number of end contraction (1 or 2)
H = measured head

Sample Problems
1. Find the width in meters of the channel at the back of a suppressed weir using the
following data:
Head, H = 28.5 cm
Depth of water, d = 2.485 m
Discharge, Q = 0.84 m3/s
Consider velocity of approach and use Francis formula.

2. A rectangular channel 6.10 m wide has a depth of 90 cm of water flowing with a
mean velocity of 0.75 m/s. Determine the height of standard sharp-crested weir that
will increase the depth in the channel of approach to 1.5 m.

3. The discharge from a 150-mm diameter orifice under a head of 3.05 m and coefficient
of discharge C = 0.60 flows into a regular channel and over a rectangular suppressed
weir. The channel is 1.83 m wide and the weir has a height of 1.50 m and the length
is
0.31 m. Determine the depth of water in the channel. Use Francis formula and neglect
velocity of approach.

4. A rectangular sharp-crested weir with end contractions is 1.4 m long. How high
should it be placed in a channel to maintain an upstream depth of 2.35 m for a
flow of 400 li/sec?

B. Triangular Weir (V-Notch)
For very low heads, a V-notch weir should be used if accuracy of measurement is
required. The vertex angle Ɵ of a V-notch weir is between usually between 10° to 90° but
rarely larger.
L
Integrate:
𝐿 𝐻
L/2 𝑄𝑡 = √2𝑔 ∫ (𝐻ℎ1/2 − ℎℎ1/2) 𝑑ℎ

x 𝐻 0
x/2 x/2 h 5/2 𝐻
2𝐻ℎ3/2 − 2ℎ ]
dh 𝑄𝑡 = √2𝑔 5
𝐻 3 0
Ɵ/2 H–h [ 3/2
2𝐻𝐻 2𝐻 5/2
H
𝐿 − ] − 0}
Ɵ 𝑄𝑡 = √2𝑔 3 5
{[ 𝐻
𝑄 = 𝐿 √2𝑔 ( 4 𝐻5/2)
𝑡 𝐻 15
4
𝑄𝑡 = √2𝑔
𝑑𝑄 = 𝑉 𝑑𝐴 𝐿𝐻3/2 15
𝑡 𝑡
𝑄𝑎 = 𝐶𝑄𝑡
Neglecting velocity of approach 𝟒
𝑉𝑡 = √2𝑔ℎ 𝑸𝒂 = 𝑪√𝟐𝒈 𝑳𝑯𝟑/𝟐
𝟏𝟓
𝑑𝐴 = 𝑥 𝑑ℎ
By similar triangles 𝜃 𝐿/2
𝐿 𝑥 tan =
= 2 𝐻
𝐻 𝐻−ℎ 𝜃
𝐿 𝐿 = 2𝐻 tan
𝑥 = ( 𝐻 − ℎ) 4 2 𝜃
𝐻
𝐿 𝑄𝑎 = 𝐶√ 2𝑔 (2𝐻 tan ) 𝐻3/2
𝑑𝐴 = (𝐻 − ℎ) 𝑑ℎ 15 𝜃 5/22
𝐻 𝑄 = 8
𝑑𝑄𝑡 = √2𝑔ℎ 𝐿 (𝐻 − ℎ) 𝑑ℎ 𝑎 𝐶√2𝑔 tan 𝐻
𝐻 15
8 2
𝐿
𝑑𝑄 = √2𝑔 ℎ1/2(𝐻 − ℎ) 𝐶𝑤 =
𝐶√2𝑔 15 𝟓/𝟐
𝑑ℎ 𝜽
𝐻
𝐿
𝑑𝑄𝑡 = √2𝑔 (𝐻ℎ1/2 − 𝑸𝒂 = 𝑪𝒘 𝐭𝐚𝐧 𝑯
𝟐
ℎℎ1/2) 𝑑ℎ
𝐻 For standard 90° weir, Cw = 1.4
𝑸𝒂 = 𝟏. 𝟒𝑯𝟓/𝟐

Sample Problems
1. Determine the discharge of water over 60° triangular weir if the measured head is
19 cm. Assume C=0.60.
2. The flow in a rectangular channel varies from 225 liters per second to 350 li/s,
and it is desired to regulate the depth by installing standard 90° V-notch weir at
the end. How many weirs are needed to regulate the variations in depth to 60 mm?

C. Trapezoidal Sharp Crested Weir
The discharge from a trapezoidal weir is assumed the same as that from a
rectangular weir and a triangular weir in combination.
b L b
Ɵ/2 H Ɵ/2
𝑸 = 𝑪𝒘𝟏𝑳𝑯𝑏𝟑/𝟐 + 𝑪𝒘𝟏𝒁𝑯𝟓/𝟐 𝜃
where 𝑍 = , substituted for tan
𝐻 2
Cipolletti Weir
4
4 H
Ɵ = 75.96°
1
1 𝛼 = 14.04°
Ɵ
L
𝟑
𝑸 = 𝟏. 𝟖𝟓𝟗𝑳𝑯 (Metric)
𝟐

Sample Problems
1. A trapezoidal weir having side slope 2. The discharge over a trapezoidal
of 1H to 2V discharges 50 m3/s weir is 1.315 m3/s. The crest length
under a constant head of 2m. Find is 2 m and the sides are inclined at
the length of the weir assuming C = 75°57’49” with the horizontal. Find
0.60. the head on the weir in meters.

3. Determine the discharge of the weir having a head of 0.30 m in liters/sec.
a. If a 90° triangular weir is used.
b. If a trapezoidal weir with sides inclined 14.04° with the vertical and a length of
crest of 2 m.
c. If a contracted rectangular sharp crested weir 2 m long is used.

8.2 Unsteady Flow in Weirs (Variable Head)
Reservoir or tank with constant

water surface area, As
H
Weir with
varying head
𝑯𝟏
𝒕=∫ 𝑨𝒔𝒅𝑯
𝑯𝟐 𝑸𝒐
If the flow is through a suppressed rectangular weir
𝑯𝟏
𝑨𝒔𝒅𝑯
𝒕=∫
𝑪𝒘 𝑳𝑯𝟑/𝟐
𝑯𝟐
𝑨 𝒔 𝑯𝟏
𝒕= −𝟑/𝟐
𝑪𝒘𝑳 ∫𝑯𝟐 𝑯 𝒅𝒉
𝑨𝒔 𝑯𝟐
𝒕= [−𝟐𝑯−𝟏/𝟐]
𝑪𝒘𝑳 𝑯𝟏
𝟐𝑨𝒔 𝟏 𝟏
𝒕= [ − ]
𝑪𝒘𝑳 √𝑯𝟐 √ 𝑯𝟏
where Cw = weir factor

L = crest length
As = constant water surface are of reservoir or tank
H1 = initial head
H2 = final head

Sample Problems
1. A spillway controls a reservoir 4.6 2. A rectangular suppressed weir of

hectares in area. The permanent crest length 1 m is constructed or cut at
is at elevation 75 m. If water can be the top of a tall rectangular tank
drawn from elevation 76.5 to having a horizontal section 20 m by
elevation 75.5 m in 42 minutes, find 20 m. If the initial head over the weir
the length of the spillway in meters. is 1 m, compute the time required to
Use Francis Formula neglecting discharge 72 m3 of water. Use
velocity of approach. Francis Formula.

3. A V-notch is located or cut at one end of a tank having a horizontal square section 10
m by 10 m. If the initial head on the weir is 1.20 m and it takes 375 seconds to
discharge 100 m3 of water, what could have been the vertex angle of the weir. Use C
= 0.60.

Module 9
Closed Conduit Flows
A conduit is any pipe, tube, or duct that is completely filled with a flowing fluid.
Examples include a pipeline transporting liquefied natural gas, a microchannel
transporting hydrogen in a fuel cell, and a duct transporting air for heating of a building.
A pipe that is partially filled with a flowing fluid, for example a drainage pipe, is
classified as an open-channel flow and will be analyzed in the next module.
Pipes are closed conduits through which fluids or gasses flows. In hydraulics, pipes
are referred to as conduits (usually circular) which flow full. Conduits flowing partially
are called open channels. Fluid flow in pipes maybe steady or unsteady. In steady flow,
there are two types of flow that exist, they are called laminar and turbulent flow.
9.1. Laminar and Turbulent Flows

Flow in a conduit is classified as being either laminar or turbulent, depending on
the magnitude of the Reynolds number. The original research involved visualizing flow
in a glass tube as shown in Figure 1 below. Reynolds in the 1880s injected dye into the
center of the tube and observed the following:
 When the velocity was low, the streak of dye flowed down the tube with little
expansion, as shown in Figure 1 (b). However, if the water in the tank was
disturbed, the streak would shift about in the tube.
 If velocity was increased, at some point in the tube, the dye would all at once mix
with the water as shown in Figure 1 (c).
 When the dye exhibited rapid mixing, shown in Figure 1 (c), illumination with an
electric spark revealed eddies in the mixed fluid as shown in Figure 1 (d).
Figure 9.1. Reynolds' experiment: (a) Apparatus, (b) Laminar flow of dye in tube, (c) Turbulent flow of dye in tube,
and (d) Eddies in turbulent flow.
The flow regimes shown in Figure 9.1 are laminar flow Figure 1 (b) and turbulent flow
Figures 1 (c) and (d). The flow is said to be laminar if when the particles move along a
straight, parallel paths in layer of laminae. The flow is said to be turbulent if when the
particles move in an irregular path.

A. Reynold’s Number
Reynold’s number, which is dimensionless, is the ratio of the inertia force to

viscous force. For pipes flowing full,
𝑽𝑫𝝆 𝑽𝑫
𝑹𝒆 = =
𝝁 𝒗
where
V = mean velocity in m/s
D = pipe diameter in meter
v = kinematic viscosity of the fluid in m2/s
𝜇
=𝜌
µ = absolute or dynamic viscosity in Pa-s
For non-circular pipes, use D = 4R, then the formula becomes,
𝟒𝑽𝑹
𝑹𝒆 =
𝒗
where
R = hydraulic radius = A/P
A = cross sectional area of the pipe in m2
P = perimeter in m
B. Shearing Stress in Pipes

Shear stress at the pipe wall
𝝉𝒐 𝒇𝑽𝟐𝝆 or 𝝉𝒐 𝜸𝒉𝒇𝑫
= 𝟖 = 𝟒𝑳
Shear velocity or friction velocity
𝜏𝑜 𝑓
𝑣 =√ = 𝑣√
𝑠
𝜌 8
where 𝝉𝟎 is the shear stress at wall of pipe (in 𝑃𝑎 or 𝑙𝑏/𝑓𝑡2),

𝒇 is the dimensionless friction factor from Darcy-Weisbach,
𝒗 is the average velocity on the conduit (in 𝑚/𝑠 or 𝑓𝑡/𝑠),
𝝆 is the fluid’s mass density (in 𝑘𝑔/𝑚3 or 𝑠𝑙𝑢𝑔/𝑓𝑡3),
𝜸 is the fluid’s unit or specific weight (in 𝑁/𝑚3 or 𝑙𝑏/𝑓𝑡3),
𝒉𝒇 is the head loss due to friction (in 𝑚 or 𝑓𝑡),
𝑫 is the cross-sectional diameter of the pipe (in 𝑚 or 𝑓𝑡),
𝑳 is the length of the pipe (in 𝑚 or 𝑓𝑡), and
𝒗𝒔 is the shear velocity or friction velocity.

9.2 Head Losses in Pipes
Head losses in pipes maybe classified into two: the major head loss, which is
caused by pipe friction along straight section of pipe of uniform diameter and uniform
roughness, and minor head loss, which are caused by changes in velocity or direction of
flow and are commonly expressed in terms of kinetic energy.
A. Major Head loss, hf

There are three primary formulas used in quantifying the major head loss or pipe
friction head loss on pipes: (1) Darcy-Weisbach Formula, (2) Manning’s Formula, and
(3) Hazen Williams Formula.
1. Darcy-Weisbach Formula
𝒉𝒇 𝟐
𝒇𝑳 𝑽𝟐 or 𝒉𝒇 = 𝟎.𝟎𝟖𝟐𝟔𝒇𝑳𝑸
= 𝑫 𝑫𝟓
𝟐𝒈
where
f = friction factor
L = length of pipe, m
D = pipe diameter, m
V = mean velocity of flow, m/s
Q = discharge, m3/s
For non-circular pipes, use D = 4R.

Value of f:
For Laminar Flow:
𝟔𝟒
𝒇=
𝑹𝒆
For Turbulent Flow:

Colebrook Equation
𝟏 𝜺 𝟗. 𝟑𝟓
= 𝟏. 𝟏𝟒 − 𝟐. 𝟎 𝐥𝐨𝐠 [ + ]
√ 𝒇 𝑫 𝑹𝒆 √ 𝒇
Colebrook-White Equation
𝟏 𝜺 𝟐. 𝟓𝟏
= −𝟐 𝐥𝐨𝐠 [ + ]
√𝒇 𝟑. 𝟕𝑫 𝑹𝒆√𝒇
where 𝒇 is the dimensionless friction factor from Darcy-Weisbach,
𝜺 is the absolute pipe roughness (in 𝑚𝑚 or 𝑖𝑛) obtained from the figure
below,
𝑫 is the cross-sectional diameter of the pipe (in 𝑚𝑚 or 𝑖𝑛), and
𝑹𝒆 is the Reynolds Number.

Moody used the Colebrook-White formula to generate a design chart similar to that
shown in Figure 9.2 below. This chart is now known as the Moody diagram for
commercial pipes.

Figure 9.1. Resistance coefficient 𝑓 versus 𝑅𝑒 for sand roughened pipe
2. Manning Formula
𝟏 𝟐⁄ 𝟏⁄
𝟑𝑺 𝟐
𝑽=
where: 𝑹 𝒏
n = roughness coefficient
R = hydraulic radius, m
𝒉𝒇
𝑺 = , slope of the energy grade line
𝑳
then
𝒉 𝟔.𝟑𝟓𝒏𝟐𝑳𝑽𝟐 𝟏𝟎.𝟐𝟗𝒏𝟐𝑳𝑸𝟐
𝒇
= 𝟒 or 𝒉𝒇 = 𝟏𝟔⁄
⁄
𝑫 𝟑 𝑫 𝟑
3. Hazen Williams Formula

𝑸 = 𝟎. 𝟐𝟕𝟖𝟓𝑪𝟏𝑫𝟐.𝟔𝟑𝑺𝟎.𝟓𝟒
where:
C1 = Hazen William coefficient
D = pipe diameter, m
S = slope of the energy grade line
= hf/L
Q = discharge, m3/s
Solving for hf, the formula becomes,
𝟏𝟎. 𝟔𝟕𝑳𝑸𝟏.𝟖𝟓
𝒉𝒇 =
𝑪𝟏.𝟖𝟓𝑫𝟒.𝟖𝟕 𝟏

B. Minor Head Loss
Minor Head Loss due to Sudden Contraction

The loss of head, hc, caused by a reduction in the cross sectional area of the
stream which results to an increase in the velocity of flow. The contraction maybe sudden
or maybe tapered. A special case of this is the loss of the head at the entrance to a pipe
from a reservoir. The head loss is given by the formula
𝑽𝟐
𝒉 𝒄 = 𝑲𝒄
𝟐𝒈and Kc is the minor loss coefficient due
where V is the velocity in the smaller pipe
to sudden contraction.
Figure 9.3. Illustration of a Sudden Contraction as a Pipe Connection
Values of Kc: Sudden Contraction

Velocity RATIO
V2 (m/s) D2/D1
0.00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.60 0.49 0.49 0.48 0.45 0.42 0.38 0.28 0.18 0.07 0.03
1.50 0.48 0.48 0.47 0.44 0.41 0.37 0.28 0.18 0.09 0.04
3.00 0.47 0.46 0.45 0.43 0.40 0.36 0.28 0.18 0.10 0.04
6.00 0.44 0.43 0.42 0.40 0.37 0.33 0.27 0.19 0.11 0.05
12.00 0.38 0.36 0.35 0.33 0.31 0.29 0.25 0.20 0.13 0.06
Figure 9.4. Illustration of a Sudden Contraction as Water Flows from a Large Tank into a Pipe
Type of Connection (Entrance) 𝑲𝒄

Flush Connection 0.50
Projecting Connection 1.00
Rounded Connection 0.05

Minor Head Loss due to Sudden or Gradual Enlargement
The loss of head he, caused by an increase in cross sectional area of the stream
which results to a decrease in the velocity of flow. The enlargement may either be sudden
or gradual. A special case of this is the head loss at the outlet end of a pipe where it
discharges into a reservoir. The head loss is given by the formula
𝑽𝟐
𝒉 𝒆 = 𝑲𝒆
𝟐𝒈and Ke is the minor loss coefficient due
to sudden or gradual expansion.
Figure 9.5. Illustration of a Sudden Expansion as a Pipe Connection
Values of Ke: Sudden Enlargement

Velocity RATIO
V1 (m/s) D1/D2
0.00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.60 1.00 1.00 0.96 0.86 0.74 0.60 0.44 0.29 0.15 0.04
1.50 0.96 0.95 0.89 0.80 0.69 0.55 0.41 0.27 0.14 0.04
3.00 0.91 0.89 0.84 0.76 0.65 0.52 0.39 0.26 0.13 0.04
6.00 0.86 0.84 0.80 0.72 0.62 0.50 0.37 0.24 0.12 0.04
12.00 0.81 0.80 0.75 0.68 0.58 0.47 0.35 0.22 0.11 0.03
Values of Ke: Gradual Enlargement

RATIO
Ɵ D1/D2
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
5° 0.04 0.04 0.04 0.04 0.04 0.04 0.03 0.02 0.01
15° 0.16 0.16 0.16 0.16 0.15 0.15 0.13 0.10 0.06
30° 0.49 0.49 0.48 0.48 0.46 0.43 0.37 0.27 0.16
55° 0.64 0.63 0.63 0.62 0.60 0.55 0.49 0.38 0.20
60° 0.72 0.72 0.71 0.70 0.67 0.62 0.54 0.43 0.24
Value of K from the tank of pipe (exit only) is taken as 1.0

Minor Head Loss due to Obstructions
The loss of head ho, caused by obstruction such as gates, valves, and any other
pipe fittings which produce a change in cross-sectional area in the pipe or in the direction
of flow. The head loss is given by the formula
𝑽𝟐
𝒉 𝒐 = 𝑲𝒐
𝟐𝒈
Threaded Pipe Fittings
𝑲𝒐
(Obstruction)
Globe valve – wide open 10.0
Angle valve – wide open 5.0
Gate valve – wide open 0.2
Gate valve – half open 5.6
Minor Head Loss due to Bends

The loss of head hb, due to bends or curves in the pipe, which produced a change
in the direction of flow. The head loss is given by the formula
𝑽𝟐
𝒉 𝒃 = 𝑲𝒃
𝟐𝒈and Kb is the minor loss coefficient due
to sudden or gradual expansion.
Threaded Pipe Fittings (Bends) 𝑲𝒃

90° miter bend
Without vanes 1.1
With vanes 0.2
Interpolate for values

90° smooth bend of 𝒓/𝒅
1 0.35
2 0.19
4 0.16
6 0.21
8 0.28
10 0.32
Return Bend 2.2
Straight-through flow 0.4
Tees Side-outlet flow 1.8
90° elbow 0.9
45° elbow 0.4

Sample Problems:
1. A pipe has a diameter of 20 mm and has a length of 80 m. A liquid having a

kinematic viscosity of 4 x 10-5 m2/s is flowing through the pipe at a velocity of 3 m/s.
a. Compute the Reynolds number
b. Compute the friction factor
c. Compute the head loss of the pipe

2. The head loss of a 120 m length of 3. Oil having a specific gravity of 0.869
tube having a diameter of 10 mm is and dynamic viscosity of 0.0814 Pa-s
30 m. Oil flows through a tube at a flows through a cast iron pipe at a
Reynolds number of 1800. velocity of 1 m/s. The pipe is 50 m
a. Compute the friction factor long and 150 mm in diameter.
b. Compute the kinematic a. Find the head lost due to
viscosity of the fluid in m2/s friction, and
c. Compute the discharge in b. The shearing stress at the
li/min walls of the pipe

4. In a syringe, the drug had a density of 900 kg/m3 and an absolute viscosity of 0.002
Pa-s. The flow through the needle is 0.4 mL/s. Neglect head loss in the larger cylinder
a. Determine the Reynolds number through the needle
b. Determine the head loss
c. Determine the required force, F, to produce the given flow
d2 = 10 mm
d1 = 0.25 mm
Q F
20 mm 30 mm

5. The pipe flow in the figure is driven by the pressurized air in the tank. Assuming f =
0.014 and the flow rate of 13.60 li/s
a. Find the velocity of the water in the pipe
b. Find the head loss in the pipe neglecting minor losses
c. Find the gage pressure needed to provide a flow rate of 13.60 li/s
20 m
P 50 mm Ø
100 m
10 m
40 m

6. There is a leak in a horizontal 300-mm diameter pipeline. Upstream from the leak two
gages 600 m apart showed a difference of 140 KPa. Downstream from the leak two
gages 600 m apart showed a difference of 126 KPa. Assuming f = 0.025, how much
water is being lost from the pipe.

7. What commercial size of a new cast iron pipe shall be used to carry 4,490 gpm with a
lost head of 10.56 feet per mile.
a. Using f = 0.019
b. Using n = 0.014
c. Using C1 = 130

8. In a pipe carrying 57 li/s of water, determine the head loss and the pressure difference
resulting from a change of section:
a. due to sudden contraction from 20 cm to 15 cm
b. due to sudden enlargement from 15 cm to 20 cm

9.3 Reservoir Problems
A. Pipe Discharging from a Reservoir
LS hc
EG
H hf (a – b) hf
HG 𝑉2
2𝑔
(a) (b)
Horizontal Pipe
If friction is neglected the velocity at the discharge end of the pipe shown in the
figure will be the same as the theoretical flow velocity from an orifice,
𝑽 = √𝟐𝒈𝑯
For a pipe having a uniform diameter, the head loss between any two sections, say
a and b, is measured by the height difference hf (a – b), of the liquid in the piezometers
tapped at these sections.
The energy gradient shows a drop at the entrance from the tank to the pipe due to
sudden contraction loss, hc. Since the velocity of the liquid mass entering the pipe
increases, a corresponding drop in the hydraulic gradient equal to the velocity head in the
2
pipe, 𝑉 , is shown. The total head H is
2𝑔
𝑽𝟐
𝑯 = 𝒉𝒄 + 𝒉𝒇 +
𝟐𝒈 between the reservoir and the
which is easily verified by writing the energy equation
discharge of the pipe.
Assuming turbulent flow conditions and using the Darcy-Weisbach equation, H

may be written as
𝑽𝟐 𝒇𝑳𝑽
𝟐
𝟐
𝑯 = 𝑲𝒄 + +𝑽
𝟐𝒈 𝑫𝟐𝒈 𝟐𝒈
or
𝟐
𝑯=𝑪𝑽
𝟐𝒈
where
𝒇𝑳
𝑪 = 𝑲𝒄 + +𝟏
𝑫

Sample Problems:
1. A horizontal pipe 20 cm in diameter 2. A 20 mm diameter commercial steel
and 30 m in length discharges water pipe, 30 m long is used to drain an
from a reservoir. If the entrance is oil tank. Determine the discharge
sharp-cornered, what is the when the oil level in the tank is 3 m
difference between the water surface above the exit of the pipe. Neglect
in the reservoir and the discharge end minor losses.
of the pipe if the flow is 140 liters/s? a. Assume f = 0.12
Assume f = 0.021. b. Assume n = 0.011
c. Assume C1 = 100

3. The water system in a suburban area consists of an old 20 cm pipe line 750 m long
which conveys water from a pump to a reservoir whose water surface is 105 m higher
than the pump. Water is pumped at the rate of 70 li/s. Neglecting minor losses
a. Determine the head loss added by the pump using the 20 cm pipeline.
Assume f = 0.033.
b. If the old pipe is replaced with a new 25 cm diameter pipe, compute the
energy supplied by the pump. Assume f = 0.022.
c. Determine the horsepower saved by replacing the old pipe with a new 25 cm
pipe.

B. Pipe Between Two Reservoirs
hc
LS 𝑉2
2𝑔 EG hf
HG
he
h
h2
V
1· ·2
The flow analysis of the pipe system follows:

At the pipe entrance, as the velocity changes from zero in the reservoir to the
2
𝑉
mean value V in the pipe, the pressure head h is changed to the velocity head . This is
2𝑔
shown as a drop in the hydraulic gradient. In the short distance between the contracted
stream and the section where normal flow starts, the hydraulic gradient has a
corresponding rise. As for the energy gradient, there is a drop at the entrance hc due to
contraction. From the entrance to the discharge end there is a further drop in the energy
gradient and hf attributed to pipe losses. Upon reaching this end the energy gradient again
drops by he due to the enlargement loss. All in all, the total energy lost is
𝑯 = 𝒉𝒄 + 𝒉𝒇 + 𝒉𝒆
and if 𝑉2 𝑉2
ℎ = 0.50 and = 1.00 , then H becomes
ℎ
𝑐 2𝑔 𝑒 2𝑔
𝒇𝑳𝑽𝟐 𝟐
𝑯= + 𝟏. 𝟓𝟎 𝑽
𝑫𝟐𝒈 𝟐𝒈
utilizing the Darcy-Weisbach equation to express hf.
Sample Problems
1. A new cast-iron pipe 30 cm in diameter connects two reservoirs. It is 30 m long with
both ends being sharp-cornered and submerged. Determine the difference in the water
elevations if the discharge is 450 liters/s. Assume f = 0.019.

2. A wood stave pipe 120 m long is to carry 3400 li/sec across a ravine. Water enters
one end of the pipe from an open flume, and discharges at the other end into another
flume. If the difference in the flume is to be 1 m,
a. Determine the necessary diameter of pipe, assuming well designed
transition (Kc = 0.1, Ke = 0.2) and neglecting the effect of velocity in the
flume.
Use f = 0.013.
b. Compute the velocity of water in the pipe,
c. Compute the minor losses.

3. Two reservoirs A and B have elevations of 250 m and 100 m respectively. It is
connected by a pipe having a diameter of 250 mm and a length of 100 m. A
turbine is installed at a point in between A and B. The discharge flowing in the
pipeline is 150 li/s. If C = 120,
4. Determine the head loss due to friction
5. Determine the head extracted by the turbine
6. Compute the power generated by the turbine

C. Branching Pipes
Consider three reservoirs A, B, and C shown in the figure below connected to
common junction P’ by pipes 1, 2, and 3 with friction losses h f1, hf2, and hf3 respectively.
Assume that the pipes sufficiently long (≥ 1000 D) so the minor losses may be taken
negligible. By reason of continuity the flows into and out of junction P are equal. And, as
indicated by the piezometric tube at the junction, the pressure head pp/w is common to all
three pipes.
The main objective is to locate the position (elevation) of the energy at the
junction P. This position represents the water surface of an imaginary reservoir at P’. The
difference in elevation between this surface and the reservoir surface represents the head
lost in the pipe leading to the corresponding reservoir.
Case 1: Given the discharge in one of the pipes, find the elevation of the other reservoir
or may be the diameter or length of one of the pipes.
Procedures:
1. Given the flow in one of the pipe leading to or flowing out from a reservoir of the
other reservoir of known elevation, solve for its head loss hf.
2. Determine the elevation of the energy grade line at the junction of the pipes (P’)
by adding or subtracting the head loss in the pipe from the elevation of the water
surface in the reservoir.
3. Draw a line from P’ to the surface of the other reservoir. These lines represents
the EGL (energy grade lines) of each pipe. The difference in elevation between P’
and the surface of the reservoir is the head loss in the pipe.
4. Solve for the required variable.

Case 2: Given all the pipe properties and elevation of all reservoirs, find the flow in each
pipe which can be solved by trial and error.
Procedures:
1. Given all the elevation of reservoirs and pipe properties, determine the direction
of flow of each pipe. The highest reservoir always has an outflow and the lowest
always has an inflow.
2. The flow in the 2nd pipe can be determined by assuming that Q 2 = 0 or hf2 = 0.
This condition makes hf1 = H1 and hf3 = H2, which in actuality is not true. Q1 and
Q2 may be obtained from either Darcy-Weisbach or Manning equation by
assuming trial values of f or n (if these are not given)
3. If Q1 > Q3, the flow in pipe 2 is directed toward reservoir B. If Q 1 < Q3, the flow
in pipe 2 is directed away from reservoir B. Whatever the result may be, the
piezometric surface is correspondingly corrected: above reservoir B if Q 1 > Q3, or
below reservoir B if Q1 > Q3.
4. Finally, the continuity of flow equation and the head loss relations are written
based on the result obtained from 2 and 3.

Sample Problems
1. A 1,200 mm diameter concrete 1,800 m long carries 1.35 m 3/s from reservoir A,
whose water surface is at elevation 50 m, and discharges into two concrete pipes,
each 1,350 m long and 750 mm in diameter. One of the 750 mm diameter pipe
discharges into reservoir B in which the water surface is at elevation 44 m.
Determine the elevation of the water surface of reservoir C into which the other
750 mm diameter pipe is flowing. Assume f = 0.02 for all pipes.

2. Determine the flow into or out of each reservoir in the pipe system shown in the
figure. Use n = 0.11 for all pipes.

3. The turbine shown in the figure is located in the 350 mm-diameter line. If the turbine
efficiency is 90%, determine its output in kilowatts.

4. Determine the flow in each of the pipes shown the figure. Assume f = 0.02 for all
pipes.

9.6 Pipe Systems in Series
LS hc1
hf1
he1
hf2 hc2 hf3 H
he2
LS
V1
V2 V3
H = hc1 + hf1 + he1 + hf2 + hc2 + hf3 + he2
If minor losses are considered negligible,

H = hf1 + hf2 + hf3
Equivalent Length of Pipes

𝒉𝒇𝒆 = 𝒉𝒇
𝒇
𝑳𝒆 = 𝑳 𝑫𝒆)𝟓
𝒇𝒆 (
𝑫
where fe and De are the properties of the standard pipe, while L, f, and D are those of the
given pipe whose equivalent length is Le.
for minor losses:
𝑲𝑫
𝑳𝒆 =
𝒇

Sample Problems:
1. The compound pipe in series of the previous figure has the following properties:
L1 = 400 m, D1 = 30 cm, f1 = 0.020
L2 = 600 m, D2 = 60 cm, f2 = 0.019
L3 = 500 m, D3 = 45 cm, f3 = 0.020
If the flow through the system is 230 liters/s, determine the head lost. Check by the
equivalent length method.

2. Three concrete pipes are connected in series. If the flow in the pipe is 0.10 m3/s,
L1 = 360 m, D1 = 200 mm, f1 = 0.0248
L2 = 300 m, D2 = 150 mm, f2 = 0.0242
L3 = 600 m, D3 = 250 mm, f3 = 0.0255
a. Determine the total head loss of the pipe
b. Determine the length of a 200 mm diameter equivalent pipe
c. Determine the diameter of a 1,260 m long equivalent pipe with f = 0.0255

3. A series arrangement for a horizontal with details of each pipe is shown below.
The pressure head at point A is 200 m and the discharge through the system is
𝑉𝑐2
0.179 m /s. The energy loss at the contraction BC is given as 0.3
3
and the
2 2𝑔
(𝑉𝐷− 𝑉𝐸)
energy loss at the expansion DE is
2𝑔
Pipe Diameter (m) Length Friction factor, f
AB 0.300 61 m 0.020
CD 0.150 30.5 m 0.015
EF 0.300 30.5 m 0.020
a. Calculate the elevation of energy gradient at B

b. Calculate the elevation of energy gradient at D
c. Calculate the elevation of energy gradient at F

9.7 Pipe System in Parallel
Top View
Pipe 1 Q1
A B
QA Pipe 2 Q2 QB
Pipe 3 Q3
hf
A
Elevation B
Q A = Q 1 + Q2 + Q3 = Q B
Since the pressure at A or B is the same at both ends of the three pipes, then the head loss
in each pipe is the same, or
hf = hf1 = hf2 = hf3
If the pipeline before and after the junctions A and B, respectively, are included in the
system, them the total loss of head will be
H = hfA + hf1 + hfB
where hf1 may be replaced either by hf2 or hf3. The losses hfA and hfB are those pipes before
and after the junctions, respectively.

Sample Problems:
1. The total discharge into the pipe system of the figure below is 440 liters/s.
Determine the flow division into each pipe. Use n = 0.020 for all pipes. The
lengths and diameters of the pipes are
L1 = 600 m, D1 = 30 cm
L2 = 500 m, D2 = 25 cm
L3 = 800 m, D3 = 40 cm
Pipe 1 Q1
A B
QA Pipe 2 Q2 QB
Pipe 3 Q3

2. The total head lost from A to E in the figure shown is 20 m. Find the discharge in
each pipe assuming f = 0.02 for all pipes.
L1 = 450 m
L2 = 650 m L5 = 540 m
D1 = 600 mm
D2 = 500 mm D5 = 600 mm
A B D E
L3 = 360 m L4 = 480 m
C
D3 = 450 mm D4 = 450 mm

9.8 Water Hammer
Water hammer (or hydraulic shock) is the momentary increase in pressure, which
occurs in a water system when there is a sudden change of direction or velocity of the
water.
When a rapidly closed valve suddenly stops water flowing in a pipeline, pressure
energy is transferred to the valve and pipe wall. Shock waves are set up within the
system. Pressure waves travel backward until encountering the next solid obstacle (or
change in density), then forward, then back again. The pressure wave’s velocity is equal
to the speed of sound; therefore it “bangs” as it travels back and forth, until dissipated by
friction losses.
Anyone who has lived in an older house is familiar with the “bang” that resounds
through the pipes when a faucet is suddenly closed. This is an effect of water hammer.
A less severe form of hammer is called surge, a slow motion mass oscillation of
water caused by internal pressure fluctuations in the system. This can be pictured as a
slower “wave” of pressure building within the system. Both water hammer and surge are
referred to as transient pressures.
If not controlled, they both yield the same results: damage to pipes, fittings, and valves,
causing leaks and shortening the life of the system. Neither the pipe nor the water will
compress to absorb the shock.
Critical time (tc) of closure of a valve is equal to 2L/c, where L is the length of
the pipe in the upstream of the valve up to the reservoir, and c is the velocity of sound in
fluid.
If the closure time of a valve is less than tc the maximum pressure difference
developed in the downstream end is given by 𝜌𝑐𝑉; where V is the velocity in the
pipeline.
Water hammer pressures are quite large. Therefore, engineers must design piping
systems to keep the pressure within acceptable limits. This is done by installing an
accumulator near the valve and/or operating the valve in such a way that rapid closure is
prevented. Accumulators may be in the form of air chambers for relatively small systems,
or surge tanks. Another way to eliminate excessive water hammer pressures is to install
pressure-relief valves at critical points in the pipe system.
Valve closure:
 Water in close proximity to the valve is brought to rest.
 Sudden change of velocity in the water mass cause a local
pressure increase, DP
 Water column is compressed and the pipe walls expands
slightly.
 Both compression and expansion help provide a little extra
volume, allowing water enter the section continuously until it
comes to a complete stop.
 A wave of increased pressure propogates up the pipe toward
reservoir, the entire pipe is expanded and the water column within
is compressed by the increased pressure.
t = L/C  The entire water column within the pipe comes to a

complete halt. (V=0)
2nd period (EGL in the pipe > EGL in reservoir)

 The halted water in the pipe flows back into the reservoir as
soon as the pressure wave reached the reservoir.
 Decreased pressure wave travels downstream toward the
valve.
 During this period water behind the wave front moves in the
upstream direction as soon as the column decompresses.
 By the time, decreased pressure wave
arrives at the valve, the entire water column of water
within the pipe is in motion in the upstream direction.
 Pressure at the valve to drop below the normal static pressure.
 Pressure at the valve to drop below the normal static pressure.
t = 2L/C
3rd osciallation period (Pressure at the valve < normal static
pressure)
 This period begins as a wave of negative pressure
propagates up the pipe toward reservoir.
 As the instant negative pressure reaches the reservoir the
water column within the pipe again comes to a complete standstill.
 EGL of the reservoir > EGL of the pipe water flows into the
pipe
4th osciallation period (Pressure at the valve < normal static

pressure)
 This period is marked by a wave of normal static
pressure moving downstream toward the valve.
 The water mass behind the wave front also moves in the
downstream direction.
 As the pressure wave arrives the valve, the entire pipe
returns to original EGL and water is moving downstream.
4th osciallation period-continue (Pressure at the valve < normal

static pressure)
 For an instant the conditions throughout the pipe are somewhat similar to the
conditions at the time of valve closure except the water velocity.
 Water velocity has been reduced because of the energy losses.
 Another cycle starts
 Four sequential waves travels up & down the pipe in exactly the same manner
as the first cycle .
 Pressure waves are smaller in magnitude.
 Pressure-wave oscillations continues until finally the waves die out
completely.

1. Celerity, c
a. For rigid pipes 6. Time of travel of the pressure wave
𝐸𝐵 up and back (round trip), t
𝑐=√ 2𝐿
𝜌 𝑡=
𝑐
𝑃 = 𝜌𝑐𝑉 𝐸𝐵
𝑐=√
b. For non-rigid pipes 𝜌
𝐸𝐵
𝑐=√ 2. Change in pressure, ∆𝑃
𝜌 (1 𝐸𝐵𝐷) ∆𝑃 = 𝜌𝑐𝑉
𝐸𝑡
+
𝑃 = 𝜌𝑐𝑉
3. Mach Angle 𝜶
2. Time of Closure, t 𝑐𝑒𝑙𝑒𝑟𝑖𝑡𝑦 𝑐
𝑆𝑖𝑛𝛼 = =
2𝐿 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑉
𝑡= 𝑉
𝑐 𝑀𝑎 =
𝟐𝑳 𝑐
3. When 𝒕 >
𝒄
𝑃 2𝐿 Where:
𝑃𝑚𝑎𝑥 = P = water hammer pressure
𝑐 V = velocity of water
4. Rise in pressure when the valve EB = bulk modulus of water
is suddenly closed 𝜌 = mass density of water
∆𝑃 = 𝜌𝑐𝑉 c = celerity
𝐸𝐵 E = bulk modulus of steel
𝑐=√ D = diameter of pipe
𝜌 t = thickness of pipe
L = length of pipe in meter
5. Velocity of sound through water, 𝑉𝑝 ∆𝑃 = change in pressure
𝐸𝐵 𝑉𝑝= velocity of sound
𝑉 =√ Ma = Mach angle
𝑝
𝜌

Sample Problems:
1. The elasticity and dimension of the pipe are such that the celerity of the pressure is
970 m/s. Suppose the pipe has the length of 1,600 m and a diameter of 1.2 m and an
initial flow of 0.85 m/s.
a. Find the water hammer pressure for the instantaneous valve closure.
b. How much time should be allowed for closing a valve to avoid water hammer?
c. Find the appropriate water hammer pressure of the valve if it is closed in 4 sec.

2. A non-rigid steel pipe of 60 cm in diameter is to carry a maximum static head of 350
m of water and additional pressure due to water hammer when quick closing valve
stops a flow of 0.85 m3/s. The pipe has a thickness of 18 mm and a bulk modules EB
= 2,068.5 MPa, E = 200,000 MPa for steel.
a. Compute the celerity of the pressure wave.
b. Compute the total pressure on the pipe after the closure of the valve.
c. Compute the maximum stress acting on the pipe.

3. A valve is suddenly closed in a 200 mm Ø pipe. The increase in pressure is 700 kPa.
Assuming that the pipe is rigid and the bulk modulus is 2.07 x 109 N/m2.
a. Compute the celerity of the pressure wave
b. Compute the velocity of flow
c. If the length of the pipe is 650 m long, compute the water hammer pressure at the
valve if it is closed in 3 sec.

4. Water from a reservoir flowing through a pipe having a dimeter of 600 mm with a
velocity of 1.75 m/s is completely stopped by a closure of a valve situated 150 m
from the reservoir. Assume that the pressure increases at a uniform rate and there is
no damping of the pressure wave. The pipe has a thickness of 20 mm and bulk
modulus of water is 2.2 x 109 N/m2 and modulus of elasticity of steel is 1.4 x 10 11
N/m2.
a. Compute the celerity of the pressure wave
b. Compute the equivalent bulk modulus of water.
c. Compute the pressure of the pipe after the closure of the valve

Module 10
Open Channel
In general, the term open channel is used to define all natural streams and
artificial canal having surfaces exposed to the pressure of the atmosphere. All forms of
closed conduits which flow only partially full are also in the category of open channels.
Unlike in pipes flowing full and flowing under pressure, the flow in an open channel
depends upon the slope of the streambed as the slope of water surface.
In an open channel flow, the hydraulic grade line is coincident with the stream
surface since the pressure at the surface is atmospheric.
The flow in open channels may either be uniform or non-uniform.
The main types of open channels are natural streams or rivers; artificial canals or
flumes; sewers, tunnels and pipelines not flowing full.
10.1 Uniform Flow

Essential properties of uniform flow in an open channel are
a.
uniform depth, or d1 = d2 = depth at any section
b.
uniform area, A1 = A2 = area at any section
c.
uniform velocity, or V1 = V2 = velocity at any section
d.
equal slopes of energy gradient, streambed and water surface, or S = So = Sws,
ℎ𝑓
where 𝑆 = is the slope of the energy gradient, So is the slope of the streambed
𝐿
and Sws is the slope of the water surface or the hydraulic gradient
1 2
𝑉12 EGL
ℎ𝐿 = 𝑆𝐿
2𝑔 Slope = S
Sws 𝑉22
𝑑1 2𝑔
Streambed, Slope = So
𝑆𝑜𝐿 𝑑2 A
Flow Section
A. Specific Energy
The energy equation between sections 1 and 2 gives
E1 = E 2 + h f
𝑉1 2
Where: 𝐸 = + 𝑑 , the total energy per unit weight at section 1
1 2𝑔 1

2
𝐸2 = 𝑉2 + 𝑑
2𝑔 2 , the total energy per unit weight at section 2
hf is the frictional less between two section

L is the length of the channel
B. Chezy Formula
In the figure above, the head lost between any points in the channel is
hL = SL
where S is the slope of the energy grade line and L is the length or run. The head loss
balances the loss in height in the channel.
For a given channel shape and bottom roughness and constants are denoted by C,
𝑽 = 𝑪√𝑹𝑺 C = Chezy coefficient
𝑸 = 𝑨𝑪√𝑹𝑺 R = 𝐴, hydraulic radius (m),
𝑃
Where: V = mean velocity of flow (m/s) S = slope of the energy grade line
Values of C (SI Units)

a. Kutter’s Formula c. Bazin Formula
𝟏 𝟎.𝟎𝟎𝟏𝟓𝟓 𝟖𝟕
𝒏
+ 𝟐𝟑 + 𝑺 𝑪=𝟏+ 𝒎
𝑪= 𝒏
𝟏+ √𝑹
√𝑹 (𝟐𝟑 +)
𝟎.𝟎𝟎𝟏𝟓𝟓
d. Powell Equation
𝑺 𝑪 𝜺
b. Manning Fomula 𝑪 = −𝟒𝟐 𝒍𝒐𝒈 (𝑹 + 𝑹)
𝒆
𝑪=𝟏 𝟏
𝑹𝟔 Where n = roughness coefficient
𝒏
𝑽=𝟏 𝟐 𝟏 m = Basin coefficient
𝑹𝟑𝑺𝟐 R = hydraulic radius
𝒏
𝟏 𝟐 𝟏 𝜀 = roughness, m
𝑸 = 𝑨 𝑹𝟑𝑺𝟐 Re = Reynold’s number
𝒏
S = slope of EGL
C. Boundary Shear Stress, 𝝉𝒐

The average boundary shear stress, 𝜏𝑜, acting over the wetted surface of the channel is
given by
𝝉𝒐 = 𝜸𝑹𝑺
Where 𝛾 = unit weight of the liquid
R = hydraulic radius
S = slope of the EGL
D. Normal Depth
The normal depth, dn, is the depth at which uniform flow will occur in an open channel.
Normal depth may be determined from Chezy formula with S = So.

Sample Problems
1. A trapezoidal channel has a bottom width of 6 m and side slopes of 2 horizontal to
1 vertical. If the depth of flow is 1.2 m and the flow is 20.40 m3/sec
a. Compute the specific energy
b. Compute the slope of the channel if n = 0.014
c. Compute the average shearing stress at the boundary
d. Determine the value of n in the Kutter formula
e. Determine the value of m in the Bazin formula

2. A 600-mm radius sewer pipe is laid on a slope of 0.001 and has a roughness
coefficient of n = 0.012, was found to be 7/8 full. Determine the discharge
through the pipe.

3. A rectangular, concrete channel, 15 4. Water flows in a triangular V-notch
m wide is to carry water at a rate of steel channel, with vertex of angle of
22 m3/s. If the channel slope is 60°, at a velocity of 1.2 m/s. Find the
0.00025, determine the normal depth normal depth if the channel is laid on
of flow. Use roughness coefficient of a slope of 0.0017. Use n = 0.014.
n = 0.013.

E. Most Efficient Section (MES)
Also known as the most economical sections, these are sections which, for a given slope
S, channel cross-sectional area A, and roughness n, the rate of discharge is a maximum.
Of all canal shapes, the semicircular open channel is the most efficient. For wooden
flumes the rectangular shape is usually employed. Canals excavated in earth must have a
trapezoidal cross section, with the slope less than the angle of repose of the bank
material.
To derive the proportions for most efficient sections, minimize the perimeter with the
cross-sectional area constant.
Rectangular Section
b = 2d
R = d/2 d
b
Trapezoidal Section
x
R = d/2
x = 2y Ø
Ø d
Ø = 30° y
Triangular Section
A = d2
Ø = 90°
d
Ø
Circular Sections
A circular channel will have its maximum discharge when the depth of flow, d, is 93.8%
of the diameter D, and the velocity is a maximum when the depth is 82% of the diameter.
D
d

Sample Problems
1. A rectangular canal, 6.5 m wide and 1.4 m deep lined with smooth stone (n =
0.013) has a hydraulic slope of 0.001. What savings in earth excavation and lining
per meter length of canal could have been effected by using the best proportion of
rectangular canal section but adhering to the same discharge and slope?

2. A trapezoidal flume of most efficient proportion has a base width of 1.5 m. Its full
discharge is 3 m3/s. If the same material were used for a most efficient rectangular
section, by how much would the discharge be decreased in m3/s?

3. A triangular channel with most 4. Determine the maximum flow
efficient proportion discharges through a 1.2 m diameter
water at the rate of 1 m 3/s. concrete culvert which is laid on
Assuming n = 0.018 and a slope of 0.09. Use n = 0.013.
S = 0.0021 calculate the normal
depth of flow in meters.

F. Alternate Stages of Flow
The channel shown in the figure below carries water at a depth of d and a mean velocity
of V.
𝑉2
The total specific energy in the channel is 𝐻 = + 𝑑
2𝑔
solving for V, 𝑉 = √2𝑔(𝐻 − 𝑑)

and the discharge is, 𝑄 = 𝐴√2𝑔(𝐻 − 𝑑)
If the equation will be plotted (as shown in the figure at the right side), it can be seen that
when 𝑑 = 0, 𝑄 = 0 and when 𝑑 = 𝐻, 𝑄 = 0 and by substituting values of d in terms of H
a curve is established.
EGL d
𝑉2
2𝑔 Upper stage, FN < 1.0
subcritical FN = 1.0 Qmax

H depth
d Lower stage, FN > 1.0 critical depth, dc
supercritical
depth
0 Q
Variation of discharge with
depth for constant energy
It appears in the curve that, within limits (from 0 to H), there are two depth at which any
given discharge will flow with the same energy content. These two depths are called
alternate stages, and are spoken as the tranquil or upper stage and the rapid or lower
stage.
On the upper stage, the Froude number FN < 1, while on the lower stage FN > 1.0.
G. Froude Number
The ratio of the inertia force gravity and For rectangular channel, L= depth of
is given by the expression. flow d
𝑽 𝑽
𝑭𝑵 = 𝑭𝑵 =
√𝒈𝑳 √𝒈𝒅
H. Critical Depth, dc
From the figure shown above that there is a certain depth dc, that for a given total specific
energy H, the discharge is maximum. This depth is called the critical depth and is defined
as the depth at which for a given total head, the discharge is maximum, or conversely, the

depth at which for a given flow, the specific energy is minimum. Its value can be obtained
by differentiating the following equation:
𝑸 = 𝑨√𝟐𝒈(𝑯 − 𝒅)
Critical Depth on Rectangular Section
𝒒 = 𝒅√𝟐𝒈(𝑯 − 𝒅)
Where:
q = unit flow in m3/s per meter width of canal
𝑄
𝑞 = = 𝑉𝑑
�
�
Q = total flow in m3/s
b = channel width in m
𝟐 𝟑
𝒅𝒄 = 𝑯 𝒐𝒓 𝑯 = 𝒅𝒄
𝟑 𝟐
𝟑 𝒒𝟐
𝒅𝒄 = √
𝒈
replacing q=Vd
𝑽
= 𝟏(Froude number, F)
√𝒈𝒅
Critical Slope
The slope required to give uniform flow at critical depth is known as the critical slope S.
The equation for critical slope for a wide rectangular channel is
𝒏𝟐𝑽𝟐
𝑺𝒄 =
𝟒⁄
𝑹 𝟑
Critical Depth on Any Section

B
𝑸𝟐𝑩 𝑸 =𝑨 𝟐 𝟑
= 𝟏 𝒐𝒓 𝒈 𝑩
𝒈𝑨𝟑 dc
Where A and/or B, if variable must be expressed Area = A
terms of dc in
The critical velocity V, in irregular channel can be
taken by replacing 𝑄 = 𝐴𝑐𝑉𝑐
𝑽𝒄 = √𝒈𝑨𝒄
𝑩𝒄

Sample Problems
1. Water is flowing at a depth of 1.5 m in a 3-m wide rectangular channel having n =
0.013 and S = 0.0009.
a. What is the stage of flow?
b. Determine the critical depth
c. Determine the critical slope

2. A triangular flume having side slope of 1:1 carries water at the rate of 4 m3/s. The
bottom of the flume is on a slope of 0.004 with n = 0.013.
a. Determine the stage of flow
b. What is the critical depth of the given flow

3. If water flows full in trapezoidal canal having a base width of 1.8 m, width at the top
of 2.8m and a depth of 1.2 m. Assume S = 0.002 and n = 0.012.
a. Compute the rate of flow in the canal
b. Compute the critical depth
c. Determine the type of flow

I. Irregular Open Channel
For natural streams which are most often irregular in shape, the same fundamental
principles of open-channels may also be applied with the consideration of certain
additional factors such as:
a. Variation of slope and channel section,
b. Non-uniformity of flow
c. Variation of roughness coefficient.
However, an appropriate solution is possible by breaking down irregular section into

simple elementary sections and independently analyzing each divided section. In the
figure below is the section of a natural stream in flood stage. It is divided into two
sections by the dotted line ab: the main section and the overbank section. Obviously,
under low tide the overbank section does not have any flow. Moreover, it will have a
higher coefficient of roughness due to the possible presence of vegetation or small trees.
a Overbank section (os)
b
Main section (ms)
The total discharge through the irregular channel is

Q = Qms + Qos
where
𝑄 =𝐴 1
𝑚𝑠 𝑚𝑠 𝑅𝑚𝑠 2/3𝑆 𝑚𝑠 1/2
𝑛𝑚𝑠
𝑄 =𝐴 1
𝑜𝑠 𝑜𝑠 𝑅𝑜𝑠 2/3𝑆 𝑜𝑠 1/2
𝑛𝑜𝑠
by the Manning equation.
In the computation of the wetted perimeter, the dividing line ab must not be
included since no appreciable shear is occurring along this line. The values of the
roughness coefficients nms and nos, and the slope Sms and Sos are ordinarily taken as
average values within a certain reach of the stream.

Sample Problem
1. A flood occurs in a main channel having a trapezoidal section (side slope on both
sides: 2H to 1V) and base width of 12 m. The depth of flow in this section is 3.60
m and the flood spills out over an almost horizontal plane on the side of the main
channel. The width of the flood plain is 60 m with an overflow depth of 1 m. If n
= 0.025 for the main channel and two times as large for the overflow section,
estimate the discharge if the slope for both is 0.00030.

10.2 Non-Uniform Flow
Uniform flow rarely occurs in natural streams because of changes in depth, width,
and slope along the channel. The Manning equation for uniform flow can be applied to
non-uniform flow with accuracy dependent on the length of reach L taken. Thus a long
stream should be divided into several reaches if varying length such that the change in
depth is roughly the same within each reach.
1 2
𝑉12
2𝑔
EGL ℎ𝐿 = 𝑆𝐿
Slope = S
𝐻1 Sw.s. 𝑉22
𝑑1 2𝑔
Streambed,
𝐻2
Slope = So 𝑑2
𝑆𝑜𝐿
Section 1 L Section 2
V1, R1, V2, R2, S2
S1
𝑯𝟐 − 𝑯𝟏
𝑳= ̅
𝑺𝒐 − 𝑺
𝑺𝟏 + 𝑺𝟐
̅𝑺 = 𝟐
Where H = specific energy

𝑉2
𝐻= +𝑑
2𝑔
L = length of reach
So = slope of channel bed
S1 and S2 = slope of EGL at sections 1 & 2 respectively, computed using
Manning’s formula
If the slope of EGL is not known, the slope 𝑆̅, may be approximated as
𝒏 𝟐 𝑽̅ 𝟐
̅𝑺 =
𝑹̅ 𝟒 / 𝟑
̅ ̅
Where 𝑉 and 𝑅 are the averages of the velocities and hydraulic radii at the two end
sections.

Sample Problems
1. A rectangular flume of planed timber (n = 0.012) is 1.5 m wide and carries 1.70 m 3/s
of water. The bed slope is 0.00060. At a certain section the depth is 1 m. Find the
distance to the section where the depth is 0.75 m.

2. A small stream has a trapezoidal cross section with base width of 12 m and side slope
2 horizontal: 1 vertical in a reach of 8 km. During a flood, high water levels recorded
at either ends of the reach are as below:
Section Elevation of bed Water surface elevation Remarks
Upstream 100.20 102.70 Manning’s n = 0.030
Downstream 98.60 101.30
Estimate the discharge in the stream.

3. A reinforced concrete drainage outfall 2 m in diameter was constructed with a
uniform slope of 2 per thousand and discharges to the sea as shown in the figure.
During the recent heavy downpour, the peak flow in the drainage pipe was 2.05 m 3/s.
Assuming steady flow and n = 0.015, determine the distance between manholes 1 and
2.
L
MH #1 MH #2 MH #3
d1 = 1.25 m d2 = 1.50 m

A. Hydraulic Jump
It is an abrupt rise in water surface which results from retarding water flowing at lower
stage. The change in stage is from a depth less than the critical depth to a one greater than
critical depth but due to loss of head in the jump, the total energy after the jump is less
than the alternate stage before the jump.
EGL
HL
𝑉2
2
𝑉1
2 2𝑔
F
2𝑔 V2
V1 F2
d2 m
F1 dc
m d1
1 2
For rectangular sections
𝒒𝟐 𝒅𝟏𝒅𝟐(𝒅𝟏 + 𝒅𝟐)
𝒈 = 𝟐
(𝒅𝟐 − 𝒅𝟏)𝟑
𝑯𝑳 =
𝟒𝒅𝟏𝒅𝟐
For non-rectangular section
𝑸𝟐 = 𝒉𝟏𝑨𝟏 − 𝒉𝟐𝑨𝟐
𝒈 𝟏 𝟏
𝑨𝟐 − 𝑨𝟏
𝑽 𝟏𝟐 𝑽 𝟐𝟐
𝑯𝑳 = ( + 𝒅 𝟏) − ( + 𝒅 𝟐)
𝟐𝒈 𝟐𝒈
h1 – centroid of section 1 below the water surface
h2 – centroid of section 2 below the water surface

Sample Problems
1. If a discharge of 1 m3/s per meter 2. Water is flowing in a 3 m wide
width of the channel has a velocity rectangular channel at a depth of 360
of 4 m/s, to what depth could it mm. A hydraulic jump occurs and
possible jump? Check the critical the depth of water downstream from
depth. the jump is 1300 mm. Determine the
discharge.

3. A hydraulic jump occurs in a trapezoidal channel with side slope of 1:1 and base
width of 4 m. If the upstream depth is 1.0 m and the downstream depth is 2.0 m,
compute the discharge and the power lost in the jump

4. A spillway 1.5 m deep goes over a dam 30 m high. Using a weir factor Cw of
1.92, determine
a. The depth of flow at the foot of the spillway and after the jump
b. Velocity of flow after the jump
c. The energy after the jump
d. The head loss in the jump

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Module 5

Fluid Flow Concepts and Measurements

5.1. Flow Patterns

HYDRA 325 - HYDRAULICS 1

5.2 Uniform and Non-Uniform Flow

Having introduced the general concepts of flow patterns, it is convenient to make

5.3 Steady and Unsteady Flow

HYDRA 325 - HYDRAULICS 2

5.4 Laminar and Turbulent Flow

HYDRA 325 - HYDRAULICS 3

Volumetric Flow Rate

HYDRA 325 - HYDRAULICS 4

HYDRA 325 - HYDRAULICS 5

HYDRA 325 - HYDRAULICS 6

HYDRA 325 - HYDRAULICS 7

HYDRA 325 - HYDRAULICS 8

A. Work, Energy, and Energy Head

2. Elevation energy is manifested in a fluid by

HYDRA 325 - HYDRAULICS 9

HYDRA 325 - HYDRAULICS 10

Power is the rate at which work is done.

HYDRA 325 - HYDRAULICS 11

HYDRA 325 - HYDRAULICS 12

HYDRA 325 - HYDRAULICS 13

Daniel Bernoulli, a Swiss mathematician, demonstrated a theorem with reference

Bernoulli’s Energy Equation Without Head Lost:

Bernoulli’s Energy Equation With Head Lost:

HYDRA 325 - HYDRAULICS 14

Bernoulli’s Equation from 1 – 2

Bernoulli’s Equation from 1 – 2

HYDRA 325 - HYDRAULICS 15

HYDRA 325 - HYDRAULICS 16

HYDRA 325 - HYDRAULICS 17

HYDRA 325 - HYDRAULICS 18

HYDRA 325 - HYDRAULICS 19

Consider sections (1) and (2), neglecting friction losses,

HYDRA 325 - HYDRAULICS 20

HYDRA 325 - HYDRAULICS 21

HYDRA 325 - HYDRAULICS 22

HYDRA 325 - HYDRAULICS 23

HYDRA 325 - HYDRAULICS 24

HYDRA 325 - HYDRAULICS 25

HYDRA 325 - HYDRAULICS 26

Consider sections (1) and (2), neglecting 𝑃

Head Loss = Initial Head - Actual Head

HYDRA 325 - HYDRAULICS 27

HYDRA 325 - HYDRAULICS 28

HYDRA 325 - HYDRAULICS 29

HYDRA 325 - HYDRAULICS 30

Consider sections (1) and (2), neglecting friction losses,

HYDRA 325 - HYDRAULICS 31

HYDRA 325 - HYDRAULICS 32

HYDRA 325 - HYDRAULICS 33

7.1 Steady Flow in Orifice

Consider sections (1) and (2), neglecting friction losses,

HYDRA 325 - HYDRAULICS 34

HYDRA 325 - HYDRAULICS 35

HYDRA 325 - HYDRAULICS 36

E. Horizontal orifice discharging upward