11-Nov-20

PP-207
FLUID MECHANICS
Lecture 4
(Part 1)

• The topics we will be covering are:

- Continuity Equation
- Pipes
- Bernoulli’s Equation

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FLUID FLOW RATE

The quantity of fluid flowing in a system per unit time can be expressed by three
different terms as:

Q The volume flow rate is the volume of fluid flowing past a section per unit time.

W The weight flow rate is the weight of fluid flowing past a section per unit time.

M The mass flow rate is the mass of fluid flowing past a section per unit time.

FLOW RATES

SYMBOL NAME DEFINITION SI UNITS U.S.

CUSTOMARY
SYSTEM UNITS
Q Volume flow 𝑄 = 𝐴𝑣 𝑚3 /𝑠 𝑓𝑡 3 /𝑠
rate
W Weight flow 𝑊 = 𝛾𝑄 𝑁/𝑠 𝑙𝑏/𝑠
rate 𝑊 = 𝛾𝐴𝑣
M Mass flow rate 𝑀 = 𝜌𝑄 𝑘𝑔/𝑠 𝑠𝑙𝑢𝑔𝑠/𝑠
𝑀 = 𝜌𝐴𝑣

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 CONTINUITY EQUATION

- The method of calculating the velocity of flow of a fluid in a closed pipe system
depends on the principle of continuity.

- Consider a pipe in which fluid is flowing from section 1 to section 2 at a constant

rate.

- Now, if there is no fluid added, stored or removed between these two sections, then
the mass of fluid flowing past section 2 in a given amount of time must be same as
that flowing past section 1.

- This can be expressed as,

𝑀1 = 𝑀2
𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2
This equation is a mathematical statement of principle of continuity.

- If the fluid in the pipe is a liquid (i.e. an incompressible liquid) then density
would be same:
𝜌1 = 𝜌2
So now the equation becomes,
𝐴1 𝑣1 = 𝐴2 𝑣2

𝑄1 = 𝑄2

-This is the continuity equation as applied on liquids; it states that for steady flow
the volume flow rate is the same at any section.

- It can also be used for gases at low velocity, i.e. less than 100 m/s with little error.

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EXAMPLE 6.4

The inside diameters of the pipe at sections 1 and 2 are 50 mm and 100 mm
respectively. Water at 70°C is flowing with an average velocity 8 m/s at section 1.
Calculate the following:

a) Velocity at section 2 100 𝑚𝑚

b) Volume flow rate
1 50 𝑚𝑚 2
c) Weight flow rate
𝑊𝑎𝑡𝑒𝑟 70℃
d) Mass flow rate 𝑣1 = 8𝑚/𝑠

𝑣2
𝑄
𝑊
𝑀

SOLUTION:
a)
𝐴1 𝑣1 = 𝐴2 𝑣2
𝐴1
𝑣2 = 𝑣
𝐴2 2
𝜋 50𝑥10−3 𝑚 2
𝐴1 = ⇒ 1.963 𝑥 10−3 𝑚2
4

𝜋 100𝑥10−3 𝑚 2
𝐴2 = ⇒ 7.854 𝑥 10−3 𝑚2
4

1.963 𝑥 10−3 𝑚2 m
𝑣2 = x 8
7.854 𝑥 10−3 𝑚2 s

𝒗𝟐 = 𝟐 𝒎/𝒔

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b)
d)
𝑄 = 𝐴𝑣
𝑄 = 𝐴1 𝑣1 M = ρ𝑄
𝑚 𝑘𝑔 𝑚3
𝑄 = (1.96 𝑥 10−3 𝑚2 )(8 ) M = 978 0.0157
𝑠 𝑚3 𝑠
𝑸 = 𝟎. 𝟎𝟏𝟓𝟕 𝒎𝟑 /𝒔
𝐌 = 𝟏𝟓. 𝟑𝟓 𝒌𝒈/𝒔
c)
𝑊 = 𝛾𝑄
𝑘𝑁 𝑚3
𝑊 = 9.59 0.0157
𝑚3 𝑠
𝑾 = 𝟎. 𝟏𝟓 𝒌𝑵/𝒔

Problem 6.41 (Robert L. Mott)

Figure 6.16 shows a fabricated assembly made from three different sizes of
standard steel tubing listed in Appendix G.2. The larger tube on the left carries
0.072 𝑚3 /𝑠 of water. The tee branches into two smaller sections. If the velocity in
the 50 − 𝑚𝑚 tube is 12 𝑚/𝑠, what is the velocity in the 100 − 𝑚𝑚 tube?

𝒗𝟏𝟎𝟎 =?

𝒘𝒂𝒕𝒆𝒓
𝟎. 𝟎𝟕𝟐 𝒎𝟑 /𝒔
dia
𝒅𝒊𝒏 = 𝟗𝟑 𝒎𝒎
thickness
𝒗𝟓𝟎 = 𝟏𝟐𝒎/𝒔

𝒅𝒊𝒏 = 𝒅𝒐𝒖𝒕 − 𝟐𝒕 𝒅𝒊𝒏 = 𝟏𝟒𝟗 𝒎𝒎

𝒅𝒊𝒏 = 𝟒𝟕 𝒎𝒎

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Solution:

Using continuity equation:

𝑄𝑡𝑜𝑡 = 𝑄1 + 𝑄2

𝑄𝑡𝑜𝑡 = 𝐴1 𝑣100 + 𝐴2 𝑣50

𝑚3 𝜋 2
𝜋 2
𝑚
0.072 = 𝑥 0.093𝑚 𝑣100 + 𝑥 0.047𝑚 12
𝑠 4 4 𝑠

𝑚3
0.072 = 6.8 𝑥 10−3 𝑚2 𝑣100 + 0.0208
𝑠

𝒗𝟏𝟎𝟎 = 𝟕. 𝟓 𝒎/𝒔

EXAMPLE

At one section in an air distribution system, air at 14.7 psia and 100°F has an average
velocity of 1200 ft/min and the duct is 12 in square. At another section, the duct is
round with a diameter of 18 in and the velocity is measured to be 900 ft/min.
Calculate the density of air in the round section and the weight flow rate of air in
pounds per hour. At 14.7 psia and 100°F the density of air is 2.20𝑥10−3 𝑠𝑙𝑢𝑔𝑠/𝑓𝑡 3
and specific weight is 7.09𝑥10−2 𝑙𝑏/𝑓𝑡 3 .

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 COMMERCIALLY AVAILABLE PIPE & TUBING

- Data given in Appendices in either U.S. units or metric units for actual outside
diameter, inside diameter, wall thickness and flow area for selected sizes and
types.

- You will notice that the dimensions are listed in inches (in) and millimeters
(mm)for outside diameter, inside diameter and wall thickness.

- The flow area are listed in square feet (𝑓𝑡 2 ) and square meters (𝑚2 ) to help in
maintaining consistent units in calculations.

- Data for inside diameter is also given.

- For many applications, codes and standards must be followed :

 International Organization for Standardization (ISO)

 British Standards (BS)
 German Standards (DIN)
 European Standards (EN)
 Japanese Standards (JIS)
 ASTM International

… and lots more!

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COMMERCIALLY AVAILABLE PIPE & TUBING

- Nominal Pipe Size (NPS) is a North American set of standards used to

designate pipe diameter and thickness.
- Pipe size is specified with two non-dimensional numbers: a nominal pipe
size (NPS) for inside diameter based on inches, and a schedule (Sched. or Sch.)
for wall thickness.

- Range of schedule numbers is from 10 to 160, with the higher numbers indicating a
heavier wall thickness.

- All schedules of pipe of a given nominal size have the same outside diameter, the
higher schedules have a smaller inside diameter.

- The most complete series available is of schedule 40 and 80.

- Birmingham Wire Gauge (BWG) is the dimensionless number which specifies the
thickness of a tube.

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 STEEL PIPE

- General purpose pipe lines.

- Standard sizes are designated by the Nominal Pipe Size (NPS) and schedule
number.

- Schedule numbers are related to the permissible operating pressure of the pipe
and to the allowable stress in the pipe.

- The most complete series of steel pipe available are Schedules 40 and 80.

- Data is given in Appendix F.

STEEL TUBING

- Standard steel tubing is used in hydraulics systems, condensers, heat

exchangers, engine fuel systems and industrial fluid processing systems.

- Sizes are designated by outside diameter and wall thickness in inches.

- Standard sizes from 1/8 in to 2 in for several wall thickness gauges are tabulated
in Appendix G.1.

- Other wall thicknesses are available.

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 COPPER TUBING

- Copper Development Association (CDA) develops standards for copper made

tubings made to U.S. Customary unit sizes.

- There are six types of CDA copper tubing offered.

1. Type K: water service, fuel oil, natural air and compressed air.
2. Type L: similar to Type K but with smaller wall thickness.
3. Type M: similar to Types K and L, but with smaller wall thicknesses; preferred
for most water services and heating applications at moderate pressures.
4. Type DWV: drain, waste and vent use in plumbing systems.
5. Type ACR: air conditioning, refrigeration, natural gas, LPG and compressed air.
6. Type OXY/MED: use for oxygen or medical gas distribution, compressed
medical air, and vacuum applications.

 DUCTILE IRON PIPE

- Water, gas and sewage lines are often made of ductile iron pipe because of its
strength, ductility and relative ease of handling.

- It has replaced cast iron in many applications.

- Standard fittings are supplied with pipe for convenient installation above or below
ground.

- Several classes of ductile iron pipe are available for use in systems with a range of
pressures.

- Appendix I lists the dimensions.

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 OTHER TYPES OF PIPE & TUBING

- Brass pipe is used for corrosive fluids, as is stainless steel.

- Other materials are aluminum, lead, tin, concrete and many other types of
plastics such as polyethylene, nylon and polyvinyl chloride.

- Appendix G.3 lists commercially available sizes of PVC pipes.

 RECOMMENDED VELOCITY OF FLOW IN PIPE & TUBING

- Factors affecting the selection of a satisfactory velocity of flow in fluid systems are
numerous.

- Some important ones are the type of fluid, the length of the flow system, the type of pipe or
tube, the pressure drop that can be tolerated, the devices that maybe connected to the pipe or
tube, temperature, pressure and noise.

- We learned in continuity principle that as velocity of flow increases the area of the flow path
decreases, so, small tubes will cause high velocities and larger tubes will provide low
velocities.

- Later we will explain that the energy losses and the corresponding pressure drop increase
dramatically as the flow velocity increases.

- For this reason it is desirable to keep the velocities low.

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Pipe Selection Aid

- The resulting flow velocities from the recommended pipe sizes in last figure are
generally low for the smaller pipes and higher for the larger pipes, as shown in
the following data:

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- The U.S. Army Corps of Engineers manual Liquid Process Piping

recommends that for normal liquid service applications, the flow velocity
should be in the range of 𝟏. 𝟐 𝒎/𝒔 𝒕𝒐 𝟑. 𝟎 𝒎/𝒔 (𝟑. 𝟗 𝒇𝒕/𝒔 𝒕𝒐 𝟗. 𝟖 𝒇𝒕/𝒔).

EXAMPLE 6.6:

Determine the maximum allowable volume flow rate in 𝐿/min that can be carried
through a standard steel tube with an OD of 32 𝑚𝑚 and a 1.5 − 𝑚𝑚 of wall
thickness if the maximum velocity is to be 3.0 𝑚/𝑠.

SOLUTION
𝐴 = 6.605 𝑥 10−4 𝑚2 (from Appendix G.2)
So,
𝑄 = 𝐴𝑣
−4 2
𝑚 −3
𝑚3
𝑄 = 6.605 𝑥 10 𝑚 3 = 1.982 𝑥 10
𝑠 𝑠
𝑄 = 1.982 𝑥 10−3 𝑚3 60 𝑠 1𝐿 𝑳
−3 3 = 𝟏𝟏𝟗
𝑠 1 𝑚𝑖𝑛 10 𝑚 𝒔

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EXAMPLE 6.7:

Determine the required size standard Schedule 40 steel pipe to carry 192𝑚3 /𝑕 of
water with a maximum velocity of 6 m/s.

SOLUTION:
𝑄 = 𝐴𝑣 𝑄 = 192 𝑚3 1𝑕 𝑚3
= 0.0533
𝑄 0.0533 𝑚3 /𝑠 = 8.88 𝑥 10−3 𝑚2 𝑕 3600 𝑠 𝑠
𝐴= =
𝑣 6 𝑚/𝑠
 NPS 5
𝑄 0.0533 𝑚3 /𝑠
𝑣 = 𝐴 = 1.291 𝑥 10−2 𝑚2 = 4.13 𝑚/𝑠 𝒊𝒏 𝒓𝒂𝒏𝒈𝒆!

 NPS 4
𝑄 0.0533 𝑚3 /𝑠
𝑣 = 𝐴 = 8.213 𝑥 10−3 𝑚2 = 6.49 𝑚/𝑠 𝒕𝒐𝒐 𝒉𝒊𝒈𝒉 !

CONSERVATION OF ENERGY – BERNOULLI’S EQUATION

- There are 3 forms of energy which are always considered when analyzing a pipe
flow problem.

- Consider an element of fluid, that may be inside a pipe flow system. It would be
located at a certain elevation z, have a certain velocity v, and have a pressure p.

- The element of fluid would possess following forms of energy:

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1. POTENTIAL ENERGY:
due to its elevation, the potential energy of the element relative to some reference
level is,
𝑃𝐸 = 𝑤𝑧
2. KINETIC ENERGY
Due to its velocity, the kinetic energy of the element is
𝑤𝑣 2
𝐾𝐸 =
2𝑔
3. FLOW ENERGY
Sometimes called pressure energy or flow work, this represents the amount of work
necessary to move the element of fluid across a certain section against the pressure p.
𝑤𝑝
𝐹𝐸 =
𝛾

Figure above shows the element of fluid in the pipe being moved across a section. The
force on the element is 𝑝𝐴, where p is the pressure at the section and 𝐴 is the area of
the section. In moving the element across the section, the force moves a distance 𝐿
equal to the length of the element.

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Therefore the work done is:

𝑊𝑜𝑟𝑘 = 𝑝𝐴𝐿 = 𝑝𝑉
𝑤
𝑉=
𝛾
𝑝𝑤
𝑊𝑜𝑟𝑘 = 𝑝𝑉 =
𝛾
Which is called flow energy.

The total amount of energy of these three forms possessed by the element of fluid would be
the sum, called E:
𝐸 = 𝐹𝐸 + 𝑃𝐸 + 𝐾𝐸
𝑤𝑝 𝑤𝑣 2
= + 𝑤𝑧 +
𝛾 2𝑔
Each of these terms are expressed in units of energy.

Fluid elements used in Bernoulli’s

Equation

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Now consider the element of fluid in figure which moves from section 1 to a section 2.
The values of p, z and v will be different at the two sections. At section 1, the total
energy is,

𝑤𝑝1 𝑤𝑣12
𝐸1 = + 𝑤𝑧1 +
𝛾 2𝑔

At section 2, the total energy is,

𝑤𝑝2 𝑤𝑣22
𝐸2 = + 𝑤𝑧2 +
𝛾 2𝑔

If no energy is added to the fluid or lost between sections 1 and 2, then the principle
of conservation of energy requires that,

𝐸1 = 𝐸2

𝑤𝑝1 𝑤𝑣12 𝑤𝑝2 𝑤𝑣22

+ 𝑤𝑧1 + = + 𝑤𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

𝑝1 𝑣12 𝑝2 𝑣22
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

This is referred to as Bernoulli’s Equation.

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INTERPRETATION OF BERNOULLI’S EQUATION

- Each term of Bernoulli’s Equation resulted from dividing an expression for energy by
the weight of an element of the fluid.

- Therefore, it is appropriate to refer to the resulting forms as energy possessed by the

fluid per unit weight of fluid flowing in the system.

- The units for each term would be N-m/N or lb-ft/lb, leaving only a unit of height.

- Therefore, terms in Bernoulli’s are often expressed as “head” referring to a height

above a reference level.

- specifically,
𝑝
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡𝑕𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑕𝑒𝑎𝑑.
𝛾
𝑧 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡𝑕𝑒 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑕𝑒𝑎𝑑.
𝑣2
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡𝑕𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑕𝑒𝑎𝑑.
2𝑔

- The sum of three heads is called total head.

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Bernoulli’s Equation accounts for the changes in elevation head, and velocity
head between two points in a fluid flow system. It is assumed that there are no
energy losses or additions between the two points, so the total head remains
constant.

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 RESTRICTION ON BERNOULLI’S EQUATION

1. It is valid only for incompressible fluids.

2. There can be no mechanical devices between the two sections of interest that
would add energy to or remove energy from the system.

3. There can be no heat transferred to or out of the fluid.

4. There can be no energy lost due to friction.

EXAMPLE 6.9

Water at 10℃ is flowing from section 1 to section 2. At section 1, which is 25mm in

diameter, the gage pressure is 345 kPa and the velocity of flow is 3 m/s. Section 2
which is 50 mm in diameter, is 2 m above section 1. Assuming there are no energy
losses in the system, calculate the pressure 𝑝2 .

Data:
𝑊𝑎𝑡𝑒𝑟 10°𝐶
𝐷1 = 25 𝑚𝑚 𝑣1 = 3 𝑚/𝑠 𝑝1 = 345 𝑘𝑃𝑎 (𝑔𝑎𝑔𝑒)
𝐷2 = 50 𝑚𝑚
𝑧1 − 𝑧2 = 2 𝑚
𝑝2 =?

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SOLUTION:
Using Bernoulli’s between the two sections to find out the unknown,
𝑝1 𝑣1 2 𝑝2 𝑣2 2
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔
𝑝2 𝑝1 𝑣1 2 − 𝑣2 2
= + (𝑧1 −𝑧2 ) +
𝛾 𝛾 2𝑔
𝑣1 − 𝑣2 2
2
𝑝2 = 𝑝1 + 𝛾[(𝑧1 −𝑧2 ) + ]
2𝑔 1
𝑊𝑎𝑡𝑒𝑟 10°𝐶
𝛾 = 9.81 𝑘𝑁/𝑚3
For velocity at point 2, using continuity equation,
𝑄1 = 𝑄2
𝐴1 𝑣1 = 𝐴2 𝑣2
𝜋 −3 2
𝑣2 =
𝐴1
𝑣1 ⇒ 𝑣2 = 𝜋 4 (25 𝑥 10 ) 𝑥 3 𝑚 = 0.75 𝑚/𝑠
𝐴2 −3 2 𝑠
4 (50 𝑥 10 )

Now equation (1),

𝑣1 2 − 𝑣2 2
𝑝2 = 𝑝1 + 𝛾[(𝑧1 −𝑧2 ) + ]
2𝑔
𝑚 2 𝑚 2
9.81 𝑘𝑁 3 − 0.75
𝑠 𝑠
𝑝2 = 345 𝑘𝑃𝑎 + 3
[ 0−2 𝑚+ 𝑚 ]
𝑚 2 9.81 2
𝑠

𝒑𝟐 = 𝟑𝟐𝟗. 𝟔 𝒌𝑷𝒂

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PP-207
FLUID MECHANICS
Lecture 4
(Part 2)

EXAMPLE
Figure shows nozzle attached to a pipe that has diameter of 3 inch. Stream of water
exiting from the nozzle has a 2 inch diameter. If the pressure in pipe just ahead of
the nozzle is 150 psig, calculate volumetric flow rate through the nozzle in
gal/min.

𝜋
𝐴1 = (3 𝑖𝑛. )2 = 7.07 𝑖𝑛.2
4
𝜋
𝐴2 = (2 𝑖𝑛. )2 = 3.14 𝑖𝑛.2
4

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Solution:

Using Bernoulli’s equation,

𝑝1 𝑣1 2 𝑝2 𝑣2 2
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

𝑝1 𝑣1 2 𝑣2 2
+ = → (1)
𝛾 2𝑔 2𝑔

Using Continuity Equation to find velocities,

𝐴1 𝑣1 = 𝐴2 𝑣2
𝐴1
⇒ 𝑣2 = 𝑣
𝐴2 2
7.07 𝑖𝑛.2
⇒ 𝑣2 = 𝑣 = 2.25𝑣1
3.14 𝑖𝑛.2 1

𝛾𝑤𝑎𝑡𝑒𝑟 = 62.4 𝑙𝑏/𝑓𝑡 3

Now substituting values in equation (1),

𝑝1 𝑣1 2 𝑣2 2
+ =
𝛾 2𝑔 2𝑔
𝑝1 𝑣1 2 𝑣2 2
+ =
𝛾 2𝑔 2𝑔 𝑝1 = 150 𝑙𝑏 122 𝑖𝑛.2 𝑙𝑏
= 21600
21600 2
𝑙𝑏
𝑖𝑛.2 12 𝑓𝑡 2 𝑓𝑡 2
𝑓𝑡 𝑣1 2 (2.25𝑣1 )2
+ =
2(32.2 𝑓𝑡2 )
62.4 𝑙𝑏/𝑓𝑡 3 2𝑔
𝑠
𝑣1 = 74.1 𝑓𝑡/𝑠
We know, 𝑄 = 𝐴𝑣
𝑓𝑡 1𝑓𝑡 2
⇒ 𝑄 = 7.07 𝑖𝑛.2 74.1
𝑠 144 𝑖𝑛.2

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𝑓𝑡 3
𝑄 = 3.64
𝑠
Convert it to gal/min,

3.64𝑓𝑡 3 7.48 𝑔𝑎𝑙 60 𝑠

𝑄= = 𝟏𝟔𝟑𝟎 𝒈𝒂𝒍/𝒎𝒊𝒏
𝑠 1 𝑓𝑡 3 1 𝑚𝑖𝑛

TANKS, RESERVOIRS & NOZZLES EXPOSED TO THE ATMOSPHERE

When the reference point is exposed to the The velocity head at the surface of the tank or
atmosphere, the pressure is zero and the pressure reservoir is considered to be zero and it can be
head term can be cancelled from Bernoulli’s cancelled from Bernoulli’s equation.
equation.

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 TANKS, RESERVOIRS & NOZZLES EXPOSED TO THE ATMOSPHERE

- Figure shows a fluid system in which siphon draws fluid from a tank or a reservoir
and delivers it through a nozzle at the end of the pipe.

- Note that the surface of the tank (Point A) and the free stream of fluid exiting the
nozzle (Point F) are not confined by the solid boundaries and are exposed to the
prevailing atmosphere. So the pressure at those sections would be zero gage
pressure.

- When such points are used as reference points in Bernoulli’s equation, the pressure
head terms will be zero and can be cancelled.

- The tank from which the fluid is being drawn can be assumed to be quite
large compared to the size of the flow area inside the pipe

- Now, because of 𝑣 = 𝑄/𝐴, the velocity at the surface of such a tank will be
very small and then we compute velocity head through this velocity, we
square it and so it will be a very very small value

- For these reasons, we consider the velocity head at the surface of a tank or a
reservoir to be very nearly zero and we cancel it out from the equation

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TANKS, RESERVOIRS & NOZZLES EXPOSED TO THE ATMOSPHERE

When the two points of reference for Bernoulli’s When the two points of reference for
equation are both inside a pipe of the same size, the Bernoulli’s equation are both at the same
velocity head terms on both sides of the equation are elevation, the elevation head terms 𝒛𝟏 𝒂𝒏𝒅 𝒛𝟐
equal and can be cancelled. are equal and can be cancelled.

WHEN REFERENCE POINTS ARE IN THE SAME PIPE

When the two points of reference for Bernoulli’s equation are both inside a pipe of
the same size, the velocity head terms on both sides of the equation are equal and can
be cancelled.

WHEN ELEVATION IS EQUAL AT BOTH REFERNCE POINTS

When the two points of reference for Bernoulli’s equation are both at the same
elevation, the elevation head terms 𝒛𝟏 𝒂𝒏𝒅 𝒛𝟐 are equal and can be cancelled.

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EXAMPLE 6.10

Figure shows a siphon that is

used to draw water from a
swimming pool. The pipe that
makes up the siphon has an
inside diameter of 40 mm and
terminates with a 25 mm
diameter nozzle. Assuming that
there are no energy losses in the
systems, calculate the volume
flow rate through the siphon and
the pressure at points B, C, D
and E.

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• Points A & B
𝑝𝐴 𝑣𝐴 2 𝑝𝐵 𝑣𝐵 2
+ 𝑧𝐴 + = + 𝑧𝐵 +
𝛾 2𝑔 𝛾 2𝑔

𝑣𝐵 2
𝑃𝐵 = −𝛾
2𝑔

Solution:

• Points A & F
𝑝𝐴 𝑣𝐴 2 𝑝𝐹 𝑣𝐹 2
+ 𝑧𝐴 + = + 𝑧𝐹 +
𝛾 2𝑔 𝛾 2𝑔
2
𝑣𝐹
𝑧𝐴 = 𝑧𝐹 +
2𝑔

𝑣𝐹 = (𝑧𝐴 − 𝑧𝐹 )(2𝑔)

2𝑥9.81𝑚
𝑣𝐹 = 3𝑚 − 0
𝑠2
𝒗𝑭 = 𝟕. 𝟔𝟕 𝐦/𝐬

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• Using the definition of volume flow rate,

𝑸 = 𝑨𝒗

𝜋 𝑚
𝑄= 𝑥 25 𝑥 10−3 2 𝑚2 7.67
4 𝑠

𝒎𝟑
𝑸 = 𝟑. 𝟕𝟕 𝒙 𝟏𝟎−𝟑
𝒔

• Calculating the velocity inside the siphon, using,

𝑄 = 𝐴𝑣

𝑄 = 𝐴𝑠𝑖𝑝𝑕𝑜𝑛 𝑣𝐵

𝜋
(40 𝑥 10−3 )2 𝑚2
𝑣𝐵 = 4
3.77 𝑥 10−3 𝑚3 /𝑠

𝒗𝑩 = 𝟑 𝒎/𝒔

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• Points A & B
𝑝𝐴 𝑣𝐴 2 𝑝𝐵 𝑣𝐵 2
+ 𝑧𝐴 + = + 𝑧𝐵 +
𝛾 2𝑔 𝛾 2𝑔

𝑣𝐵 2
𝑝𝐵 = 𝛾 −
2𝑔

𝑚 2
𝑘𝑁 3 𝑠
𝑝𝐵 = 9.81 3 − 𝑚
𝑚 2 𝑥 9.81 2
𝑠

𝒑𝑩 = −𝟒. 𝟓𝟎 𝒌𝑷𝒂

• Points A & C
𝑝𝐴 𝑣𝐴 2 𝑝𝐶 𝑣𝐶 2
+ 𝑧𝐴 + = + 𝑧𝐶 +
𝛾 2𝑔 𝛾 2𝑔

𝑣𝐶 2
𝑝𝐶 = 𝛾[(𝑧𝐴 − 𝑧𝐶 ) − )]
2𝑔

𝑚 2
𝑘𝑁 3𝑠
𝑝𝐶 = 9.81 3 (0 − 1.2 𝑚) − 𝑚
𝑚 2 𝑥 9.81 2
𝑠

𝒑𝑪 = −𝟏𝟔. 𝟐𝟕 𝒌𝑷𝒂

𝒑𝑫 = 𝒑𝑩

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• Points A & E
𝑝𝐴 𝑣𝐴 2 𝑝𝐸 𝑣𝐸 2
+ 𝑧𝐴 + = + 𝑧𝐸 +
𝛾 2𝑔 𝛾 2𝑔

𝑣𝐸 2
𝑝𝐸 = 𝛾[(𝑧𝐴 − 𝑧𝐸 ) − )]
2𝑔

𝑚 2
𝑘𝑁 3
𝑝𝐸 = 9.81 3 (3𝑚 − 0) − 𝑠
𝑚 𝑚
2 𝑥 9.81 2
𝑠

𝒑𝑬 = 𝟐𝟒. 𝟗𝟑 𝒌𝑷𝒂

Answers:

𝒎𝟑
𝑸 = 𝟑. 𝟕𝟕 𝒙 𝟏𝟎−𝟑 𝒔
𝒗𝑭 = 𝟕. 𝟔𝟕 𝐦/𝐬
𝒗𝑩→𝑬 = 𝟑 𝒎/𝒔
𝒑𝑩 = −𝟒. 𝟓𝟎 𝒌𝑷𝒂
𝒑𝑪 = −𝟏𝟔. 𝟐𝟕 𝒌𝑷𝒂
𝒑𝑫 = 𝒑𝑩
𝒑𝑬 = 𝟐𝟒. 𝟗𝟑 𝒌𝑷𝒂

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Example Problem 6.11:

The venturi meter as shown in figure

carries water at 60℃. The specific
gravity of the gage fluid in the
manometer is 1.25. Calculate the
velocity of flow at section A and the
volume flow rate of water.

𝛾𝐻2𝑂 = 𝟗. 𝟔𝟓 𝒌𝑵/𝒎𝟑 𝐴𝑝𝑝𝑒𝑛𝑑𝑖𝑥 𝐴

𝑘𝑁
𝛾𝑓 = 1.25 9.81 = 𝟏𝟐. 𝟐𝟔 𝒌𝑵/𝒎𝟑
𝑚3

Solution:

𝑝𝐴 𝑣𝐴 2 𝑝𝐵 𝑣𝐵 2
+ 𝑧𝐴 + = + 𝑧𝐵 +  (1)
𝛾 2𝑔 𝛾 2𝑔

Using manometer equation to solve for pressures,

𝑝1 = 𝑝2
𝛾𝐻2 𝑂 1.18 + 𝑦 + 𝑝𝐴 = 𝛾𝑓 1.18 𝑚 + 𝛾𝐻2 𝑂 𝑦 + 0.46 + 𝑝𝐵

𝑝𝐴 − 𝑝𝐵 = 𝛾𝑓 1.18 𝑚 + 𝛾𝐻2 𝑂 0.46 + 𝛾𝐻2 𝑂 (𝑦) − 𝛾𝐻2 𝑂 1.18 − 𝛾𝐻2 𝑂 𝑦

𝒑𝑨 − 𝒑𝑩 = 𝟕. 𝟔𝟑 𝒌𝑷𝒂
𝝅
Using Continuity Equation for velocities, 𝑨𝑨 = (𝟑𝟎𝟎 𝒙 𝟏𝟎−𝟑)𝟐 = 𝟕. 𝟎𝟕𝒙𝟏𝟎−𝟐 𝒎𝟐
𝟒
𝐴𝐴 𝑣𝐴 = 𝐴𝐵 𝑣𝐵 𝝅
𝑨𝑩 = (𝟐𝟎𝟎 𝒙 𝟏𝟎−𝟑 )𝟐 = 𝟑. 𝟏𝟒𝟐𝒙𝟏𝟎−𝟐𝒎𝟐
𝑣𝐵 = 2.25 𝑣𝐴 𝟒

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Equation (1)
𝑣𝐵 2 − 𝑣𝐴 2 𝑝𝐴 − 𝑝𝐵
⇒ = + (𝑧𝐴 − 𝑧𝐵 )
2𝑔 𝛾
Substituting values,

(2.25 𝑣𝐴 )2 −𝑣𝐴 2 7.63 𝑘𝑃𝑎

𝑚 = + (0 − 0.46 𝑚)
2(9.81 2 ) 9.65 𝑘𝑁/𝑚3
𝑠
Solving for 𝑣𝐴 ,
𝒎
𝒗𝑨 = 𝟏. 𝟐𝟒
𝒔
Now the volume flow rate,
𝑄 = 𝐴𝐴 𝑣𝐴
𝑸 = 𝟖. 𝟕𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝟑 /𝒔

TORRICELLI’S THEOREM

- In the siphon analyzed example problem

it was observed that the velocity of flow
from the siphon depends on the elevation
difference between the free surface of the
fluid and the outlet of the siphon.

- A classical example is shown in the

figure where fluid is flowing from the
side of the tank through a smooth,
rounded nozzle.

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- To determine the velocity of flow from the nozzle,

write Bernoulli’s equation between a reference point on
the fluid surface and a point in the jet issuing from the
nozzle:

𝑝1 𝑣12 𝑝2 𝑣22
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

- But 𝑝1 = 𝑝2 = 0 and 𝑣1 is approximately zero. Then,

solving for 𝑣2 gives
𝑣2 = 2𝑔(𝑧1 − 𝑧2 ) 𝑜𝑟

𝑣2 = 2𝑔𝑕
This Equation is called Torricelli’s theorem.

- Another interesting application of Torricelli’s

theorem is shown in figure in which a jet of fluid
is shooting upward.

- If no energy losses occur, the jet will reach

height equal to the elevation of the free surface
of the fluid in the tank and at this height the
velocity in the stream is zero.

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- Lets apply Bernoulli’s in such situation,

𝑝1 𝑣12 𝑝2 𝑣22
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔
Simplifying,
𝑣2 = 2𝑔𝑕

Now apply Bernoulli’s at points 2 & 3

𝑝2 𝑣22 𝑝3 𝑣32
+ 𝑧2 + = + 𝑧3 +
𝛾 2𝑔 𝛾 2𝑔

Again, 𝑝2 = 𝑝3 = 0, then solving for 𝑣3

𝑣3 = 𝑣22 + 2𝑔(𝑧2 − 𝑧3 )

Substitute 𝑣2 = 2𝑔𝑕 and (𝑧2 − 𝑧3 )=-h. Then,

𝑣3 = 2𝑔𝑕 + 2𝑔(−𝑕) = 0

This result verifies that the stream just reaches the height
of the free surface of the fluid in the tank.

To make the jet go higher, a greater pressure can be

developed above the fluid in the reservoir or a pump can
be used to develop a higher pressure.

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EXAMPLE
For the tank shown in figure, compute the velocity of flow from the nozzle and the
volume flow rate for a change of depths from 3 m to 0.5 m in steps of 0.5m. The
diameter of the jet at the nozzle is 50 mm.

SOLUTION
nozzle dia = 50 mm
50𝑥10−3 2
Area of nozzle = 𝜋( ) = 1.96𝑥10−3 𝑚2
4

𝒎
Depth, h (m) 𝒗𝟐 ( 𝒔 ) 𝑸(𝒎𝟑 /𝒔)
3.0 7.67 1.51𝑥10−2
2.5 7.00 1.38𝑥10−2
2.0 6.26 1.23𝑥10−2
1.5 5.42 1.07𝑥10−2
1.0 4..43 0.87𝑥10−2
0.5 3.13 0.61𝑥10−2

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6.67E:
For the tank shown in figure, calculate the volume flow rate of water from the nozzle.
The tank is sealed with a pressure of 20 psig above the water. The depth h is 8 ft.

𝛾 = 62.4 𝑙𝑏/𝑓𝑡 3
𝜋
𝐴2 = (0.25 𝑓𝑡)2 = 0.049 𝑓𝑡 2
4
20 lb 144 𝑖𝑛2 lb
𝑝2 = = 2880 2
𝑖𝑛 2
1 𝑓𝑡 2 𝑓𝑡

3 in 1 𝑓𝑡. = 0.25 𝑓𝑡
𝑑=
12 𝑖𝑛.

𝑝1 𝑣12 𝑝2 𝑣22
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

2880 𝑙𝑏/𝑓𝑡 2 𝑣22

+ 𝑧1 − 𝑧2 =
62.4 𝑙𝑏/𝑓𝑡 3 2 𝑥 32.2 𝑓𝑡/𝑠 2

𝑓𝑡
⇒ 𝑣2 = 59
𝑠
Now calculating the volume flow rate,

𝑓𝑡
𝑄 = 0.049 𝑓𝑡 2 𝑥 59
𝑠
𝑸 = 𝟐. 𝟗 𝒇𝒕𝟑 /𝒔

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6.66M:
For the system shown in figure, calculate:
a) The volume flow rate of water from the nozzle
b) The pressures at points A and B

Solution:

𝑝1 𝑣12 𝑝2 𝑣22
+ 𝑧1 + = + 𝑧2 +
𝛾 2𝑔 𝛾 2𝑔

𝑣2 = 2𝑔(𝑧1 − 𝑧2 )

𝑚
𝑣2 = 2 9.8 (3)
𝑠2
𝜋
𝐴𝑛𝑜𝑧𝑧𝑙𝑒 = (35 𝑥 10−3 )2 = 9.64 𝑥 10−4 𝑚2
4
𝑣2 = 7.67 𝑚/𝑠 𝜋
𝐴𝑝𝑖𝑝𝑒 = (100 𝑥 10−3 )2 = 7.85 𝑥 10−3 𝑚2
4
𝑚 𝑘𝑁
𝑄 = 9.64 𝑥 10−4 𝑚2 7.67 𝛾𝑜𝑖𝑙 = 0.85 9.81 = 8.34𝑘𝑁/𝑚3
𝑠 𝑚3
𝑄 = 7.38 𝑥 10−3 𝑚3 /𝑠

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𝑝1 𝑣12 𝑝𝐴 𝑣𝐴2
+ 𝑧1 + = + 𝑧𝐴 +
𝛾 2𝑔 𝛾 2𝑔

For velocity using continuity principle,

𝑄
𝑄 = 𝐴𝐴 𝑣𝐴 or ⇒ 𝑣𝐴 = 𝐴
𝐴
𝑚
𝑣𝐴 = 0.94
𝑠
Substituting in Bernoulli’s to find out pressure,
𝑝𝐴 𝑣𝐴2
𝑧1 = + 𝑧𝐴 +
𝛾 2𝑔
𝑚 2
8.34𝑘𝑁 0.94
𝑝𝐴 = 4𝑚 − 𝑠
𝑚3 𝑚
2 𝑥 9.8 2
𝑠
𝒑𝑨 = 𝟑𝟑 𝒌𝑷𝒂

𝑝1 𝑣12 𝑝𝐵 𝑣𝐵2
+ 𝑧1 + = + 𝑧𝐵 +
𝛾 2𝑔 𝛾 2𝑔

Substituting in Bernoulli’s to find out pressure,

𝑝𝐵 𝑣𝐵2
𝑧1 = + 𝑧𝐵 +
𝛾 2𝑔
𝑚 2
8.34𝑘𝑁 0.94 𝑠
𝑝𝐵 = 3
3𝑚 − 𝑚
𝑚 2 𝑥 9.8 2
𝑠
𝒑𝑩 = 𝟐𝟒. 𝟔𝟒 𝒌𝑷𝒂

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Home Tasks:

6.56M:
Figure shows a heat exchanger in which each of the two 6-in Schedule 40 pipes
carries 450 L/min of water. The pipes are inside a rectangular duct whose inside
dimensions are 200 mm by 400 mm. Compute the velocity of flow in the pipes.
Then, compute the required volume flow rate of water in the duct to obtain the
same average velocity.

6.53M:

From the list of standard tubing in given Appendix, select the smallest size which
would carry 2.80 L/min of oil with a maximum velocity of 0.3 m/s

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6.65M:
For the system shown in figure,
calculate:
a) The volume flow rate of water
from the nozzle.
b) The pressure at point A.

20