Flow of Fluids and Bernoulli’s Equation

Introduction

• In the previous chapter we learn about body in static fluid

• In this chapter begins the study of fluid dynamics - moving fluids

• this chapter will be focusing on fluid dynamics in pipes of tubes

• Objective of this chapter is to be able to design and analyse the performance of pumped piping systems

• Application of the knowledge would be on the industrial (refinery plant) and products

Introductory Concepts

• Three measures of fluid flow rate are commonly used in fluid flow analysis:• Volume flow rate, Q is the volume of fluid flowing past a given section per

unit time

• Weight flow rate, W is the weight of fluid flowing past a given section per unit time

• Mass flow rate, M is the mass of fluid flowing past a given section per unit time

Introductory Concept

• Three kinds of energy possessed by the fluid at any point of interest within a fluid flow system:• Kinetic energy due to the motion of the fluid

• Potential energy due to the elevation of the fluid

• Flow energy, energy content based on the pressure in the fluid and its specific weight

• Bernoulli’s equation is the fundamental tool developed in this chapter for tracking changes in these three kinds of energy in a system (based on principle of conservation energy)

Introduction – application in cooling system of engine

Objectives

• Define volume flow rate, weight flow rate, mass flow rate and their unit

• Define steady flow and the principle of continuity

• Write the continuity equation, and use it it to relate the volume flow rate, area and velocity of flow between two points in a fluid flow system

• Describe five types of commercially available pipe and tubing

• Specify the desired size of pipe or tubing for carrying a given flow rate of fluid at a specified velocity

• State recommended velocities of flow and typical volume flow rates for various types of systems

Objectives

• Define potential energy, kinetic energy and flow energy

• Apply the principle of conservation of energy to develop Bernoulli’s equation and state the restrictions on its use

• Define the terms pressure head, elevation head, velocity head and total head.

• Apply Bernoulli’s equation to fluid flow systems

• Define Torricelli’s theorem, and apply it to compute the flow rate of fluid from a tank and the time required to empty a tank

Fluid Flow Rate and the Continuity Equation

• Fluid Flow Rate• As stated in introduction, the quantity of fluid flowing in a system

per unit time can be expressed in 3 different terms• Volume flow rate, Q is the volume of fluid flowing past a given section per

unit time

• Weight flow rate, W is the weight of fluid flowing past a given section per unit time

• Mass flow rate, M is the mass of fluid flowing past a given section per unit time

Volume Flow Rate

Q=Av

Where A is the area of the section and v is the average velocity of flow.

Units of Q can be derived from the formula:

Q = Av = m2 x m/s = m3/s

Weight Flow Rate

Weight flow rate, W is related to Q by:𝑊 = 𝛾𝑄

Where 𝛾 is the specific weight of the fluid

The SI units of W can be derived as follows:

𝑊 = 𝛾𝑄 =𝑁

𝑚3×𝑚3

𝑠= 𝑁/𝑠

Mass Flow Rate

Mass flow rate, M is related to Q by𝑀 = 𝜌𝑄

Where 𝜌 is the density of the fluid. The SI units based on the relationship:

𝑀 = 𝜌𝑄 =𝑘𝑔

𝑚3×𝑚3

𝑠= 𝑘𝑔/𝑠

In Summary

Conversion Factors for Volume Flow Rates

The Continuity Equation• The principle of continuity govern the method of

calculating the velocity of flow of a fluid in a closed pipe system

• Consider the pipe in figure, a fluid is flowing from section 1 to section 2 at a constant rate.

• constant rate - the continuity of fluid flowing past any section in a given amount of time is constant

• This type of flow is referred as steady flow

The continuity equation

• Based on the previous figure, should there is no fluid added, stored or removed between section 1 and 2, therefore• the mass of fluid flowing past section 2 in a given amount of time

must be the same as that flowing past section 1

• Hence:𝑀1 = 𝑀2

The Continuity Equation

• Because 𝑀 = 𝜌𝑄 = 𝜌𝐴𝑣

Hence, the continuity equation for any fluid:𝑀1 = 𝑀2

𝜌1𝐴1𝑣1 = 𝜌2𝐴2𝑣2This equation is a mathematical statement of the principle of continuity and is called the continuity equation

It is used to relate the fluid density 𝜌, flow area 𝐴 and velocity of flow 𝑣at 2 sections of the system in which there is steady flow (whether gas or liquid)

The Continuity Equation

• The equation is applicable for steady flow of either liquid or gas

• If the fluid is a liquid that can be considered incompressible, then the terms 𝜌1 and 𝜌2 in the equation are equal and they can be cancelled from the equation

• Hence the equation becomes 𝐴1𝑣1 = 𝐴2𝑣2𝑄1 = 𝑄2

This is the continuity equation for liquids

This equation states that for the steady flow, the volume flow rate is the same at any section

It can also be sure for gases at low velocity (<100m/s), with little error

Example

• In the figure, the inside diameters of pipe at section1 = 50mm and at section 2 = 100mm. Water 70oC is flowing with an average velocity of 8m/s at section 1. Calculate:• Velocity at section 2

• Volume flow rate

• Weight flow rate

• Mass flow rate

Solution

• Velocity at section 2𝐴1𝑣1 = 𝐴2𝑣2

𝑣2 = 𝑣1𝐴1𝐴2

𝐴1 =𝜋×502

4= 1963𝑚𝑚2 𝐴2 =

𝜋×1002

4= 7854𝑚𝑚2

𝑣2 = 81963

7854= 2𝑚/𝑠

Notice that for steady flow of a liquid, as the flow area increases, the velocity decreases with independent of pressure and elevation.

Solution

• Volume flow rate

𝑄 = 𝐴1𝑣1 = 1963𝑚𝑚2 × 8𝑚2 ×1𝑚2

1000𝑚𝑚2= 0.0157𝑚3/𝑠

Volume flow rate Q can be calculated either at section 1 or section 2 (principle of continuity)

Solution

• weight flow rate

At 70oC, sp. Weight of water is 9.59kN/m3

𝑊 = 𝛾𝑄 = 9.59𝑘𝑁/𝑚3 × 0.0157𝑚3/𝑠 = 0.151𝑘𝑁/𝑠

• Mass flow rate - At 70oC, density of water is 978kg/m3

𝑀 = ρ𝑄 = 978𝑘𝑔/𝑚3 × 0.0157𝑚3/𝑠 = 15.36𝑘𝑔/𝑠

Example

• At one section in an air distribution system, air at 14.7psi and 100oF has an average velocity of 1200ft/min and the duct is 12 in square. At another section, the duct is round with diameter of 18in and the velocity is measured to be 900ft/min.

Calculate:

(a) the density of the air in the round section

(b) the weight flow rate of air in pounds per hour. At 14.7 psia and 100oF, the density of air is 2.20 x 10-3 slugs/ft3 and the specific weight is 7.09 x 10-2lb/ft3

Solution

𝜌1𝐴1𝑣1 = 𝜌2𝐴2𝑣2Calculate density of air at section 2 (𝜌2)

𝐴2 = 12𝑖𝑛 × 12𝑖𝑛 = 144𝑖𝑛2 𝐴2 =𝜋 × (18𝑖𝑛)2

4= 254𝑖𝑛2

𝜌2 =𝜌1𝐴1𝑣1

𝐴2𝑣2=

2.20×10−3𝑠𝑙𝑢𝑔𝑠/𝑓𝑡 144𝑖𝑛2 (1200𝑓𝑡/min)

254𝑖𝑛2(900𝑓𝑡/𝑚𝑖𝑛)= 1.66 × 10−3𝑠𝑙𝑢𝑔𝑠/𝑓𝑡3

The weight flow rate can be found at section 1:

𝑊 = 𝛾1𝐴1𝑣1 = 7.09 ×10−2𝑙𝑏

𝑓𝑡3144𝑖𝑛2

1200𝑓𝑡

𝑚𝑖𝑛

1𝑓𝑡2

144𝑖𝑛260𝑚𝑖𝑛

ℎ= 5100𝑙𝑏/ℎ

Example

• At one section in an air distribution system, air at 101kPa and 37.8oChas an average velocity of 365.75m/min and the duct is 300mmsquare. At another section, the duct is round with diameter of450mm and the velocity is measured to be 274.32m/min.

Calculate:

(a) the density of the air in the round section

(b) the weight flow rate of air in pounds per hour. At 101kPa and 37.8oC, the density of air is 1133.834 x 10-3 kg/m3 and the specific weight is 1.114 x 10-2kN/m3

Commercially Available Pipe and Tubing

• Specifying piping and tubing for a particular application is the responsibility of the designer

• It has significant impact on the cost, life, safety and performance of the system

• Some of the codes that need to be follow in designing piping system:• British Standards

• European Standards

• ASTM International (ASTM)

Commercially Available Pipe and Tubing

• Steel pipe

• Steel tubing

• Copper Tubing

• Ductile Iron Pipe

• Plastic Pipe and Tubing

• Hydraulic Hose

Steel pipe

• For general-purpose pipe lines

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

• NPS is merely the standard designation and it is not used for calculation

• Schedule numbers are related to the allowable stress of the steel in the pipe, ranges from 10 – 160, with higher number indicating a heavier wall thickness.

Steel Pipe

• Nominal Pipe Sizes in Metric Units• Due to the long experience with manufacturing standard pipe according to

the standard NPS sizes and schedule numbers, they continue to be used often even when the piping system is specified in metric units.

• In such cases the DN set of equivalents has been established by the International Standards Organization (ISO)

• Symbol DN is used to designate the nominal diameter (diameter nominel) in mm.

• Example: DN 50-mm Schedule 40 steel pipe = 2-in Schedule 40 steel pipe

Steel Tubing

• Used in fluid power systems, condensers, heat exchangers, engine fuel systems, and industrial fluid processing systems.

• Standard sizes from 1/8in to 2in for several wall thickness

Copper Tubing

• 6 types of Copper Development Association (CDA) copper tubing offered:• Type K: Used for water service, fuel oil, natural gas and compressed air

• Type L: Similar to type K but with smaller wall thickness

• Type M: Similar to types K and L, but with smaller wall thicknesses; preferred for most water services and heating applications at moderate pressures.

• Type DWV: Drain, waste and vent uses in plumbing systems

• Type ACR: Air conditioning, refrigeration, natural gas, liquefied petroleum (LP) gas and compressed air.

• Type OXY/MED: Used for oxygen or medical gas distribution, compressed medical air, and vacuum applications. Available in sizes similar to Types K and L, but with special processing for increased cleanliness

Copper Tubing

• Available in either a soft, annealed condition or hard drawn

• Drawn tubing is stiffer and stronger, maintains a straight form, can carry higher pressures

• Annealed tubing is easier to form into coils and other special shapes

• Nominal or standard sizes for types K, L, M and DWV are all in 1/8 in less than the actual outside diameter

• The wall thicknesses are different for each type so that the inside diameter and flow areas vary

Ductile Iron Pipe

• Made for water, gas and sewage lines

• This is due its strength, ductility and relative ease of handling

• Has replaced cast iron for many applications

• Several classes of ductile iron pipe are available for use in systems with range of pressures

Plastic Pipe and Tubing

• Are being used in a wide variety of applications

• Due to its advantages:• Light weight• Ease of installation• Corrosion and chemical resistance• Very good flow characteristic

• Types of plastic used for pipe and tubing• Polyethylene (PE)• Cross-linked Polyethylene (PEX)• Polyamide (PA)• Polypropylene (PP)• Polyvinyl Chloride (PVC)

Polyamide (PA)

Cross-linked Polyethylene(PEX)

Polyvinyl Chloride (PVC)

Hydraulic Hose

• Reinforced flexible hose is used extensively in hydraulic fluid power systems and other industrial applications where flow lines must flex in service

• Hose materials include:• Butyl rubber

• Synthetic rubber

• Silicone rubber

• Thermoplastic

• Elastomers

• nylon

• Braided reinforcement may be made of:• Steel wire• Kevlar• Polyester• Fabric

• Industrial Application include:• Steam• Compressed air• Chemical transfer• Coolants• Heaters• Fuel transfer• Lubricants• Refrigerants• Power steering fluids

Steel wire braid

Kevlar braid

Recommended Velocity of Flow in Pipe and Tubing• Many factors affect the selection of a satisfactory velocity flow in fluid

systems

• The 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 connected to the pipe or tube (pumps, valves etc)• The temperature• The pressure • The noise

• As discussed in the continuity equation, the velocity of flow increases as the area of the flow path decreases

• Hence, smaller tubes will cause higher velocities

• Larger tubes will provide lower velocities

• As the flow velocity increases, the energy losses and pressure drop dramatically increases

• Due to this reason, it is desirable to keep velocities low.

• Because larger pipes and tubes are costly

Figure 6.3

• In general, the flow velocity is kept lower in suction lines providing flow into a pump to ensure proper filling of the suction inlet passage

• The lower velocity helps to limit energy losses in the suction line, keeping the pressure at the pump inlet relatively high to ensure that pure liquid enters the pump

• Lower pressures may cause a damaging condition, called cavitation

• Cavitation when occur can cause excessive noise, significantly degraded performance and rapid erosion of pump and impeller surfaces

The dark spots in the photo above (indicated by the ‘B’ arrows) are holes in the impeller. Bubbles form in the low pressure area at the center of the impeller (above, point A). These bubbles flow with the water out to where the pressure is higher. Here they implode and cause damage (above, points B).

• Note that in the figure, specifying one larger or one size smaller than indicated by the lines will not affect the performance of the system very much

• In general, the larger pipe size is favourable to achieve a lower velocity unless there are difficulties with space, cost or compatibilitywith a given pump connection.

Conservation of Energy – Bernoulli’s Equation

• Law of conservation of energy – energy can be neither created nor destroyed but it can be transformed from one form into another

• There are 3 forms of energy when analysing a pipe flow problem:• Potential energy

• Kinetic energy

• Flow energy

• Consider an element of fluid inside a pipe

• It is located at a certain elevation z, has a velocity v, and pressure p.

• The element of fluid possesses the following forms of energy:• Potential energy - due to its elevation (𝑤=weight, 𝑧=elevation):

𝑃𝐸 = 𝑤𝑧• Kinetic energy – due to its velocity

𝐾𝐸 = 𝑤𝑣2/2𝑔• Flow energy – sometimes called pressure energy or

flow work which represents the amount of work necessary to move the element of fluid across a certain section against the pressure, p.

𝐹𝐸 =𝑤

𝛾𝑝

• The total amount of energy possessed by the element of fluid:𝐸 = 𝐹𝐸 + 𝑃𝐸 + 𝐾𝐸

𝐸 =𝑤

𝛾𝑝 + 𝑤𝑧 +

𝑤𝑣2

2𝑔

• Consider the fluid element moves from section 1 to section 2 (figure)

• The values for p, z and v are different at the two sections.

𝐸1 =𝑤

𝛾1𝑝1 +𝑤𝑧1 +

𝑤𝑣12

2𝑔𝐸2 =

𝑤

𝛾2𝑝2 + 𝑤𝑧2 +

𝑤𝑣22

2𝑔

• If no energy is added to the fluid or lost between section 1 and 2, then the principle of conservation of energy :

𝐸1 = 𝐸2

𝑝1

𝛾1+ 𝑧1 +

𝑣12

2𝑔=

𝑝2

𝛾2+ 𝑧2 +

𝑣22

2𝑔(Bernoulli’s equation)

Interpretation of Bernoulli’s Equation

• Each term in Bernoulli’s equation is one form of energy possessed by the fluid per unit weight of fluid flow in the system

𝑝

𝛾is called the pressure head

𝑧 is called the elevation head𝑣2

2𝑔is called the velocity head

Sum of these is total head

• As the fluid moves from 1 – 2 (figure):• The total head remains at constant level (no energy added or lost)

• Velocity reduce due to increment of flow area

• Pressure head increases because velocity head decreases (so that total head is constant)

• Elevation is increased hence increment in elevation head

• In summary, Bernoulli’s equation accounts for the changes in elevation head, pressure 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

Restrictions on Bernoulli’s Equation

• It is valid only for incompressible fluids because the specific weight of the fluid is assumed to be the same at the two sections of interest

• There can be no mechanical devices between the two sections of interest that would add energy to or remove energy from the system, because the equation states that the total energy in the fluid is constant

• There can be no heat transferred into or out of the fluid

• There can be no energy lost due to friction

Restrictions on Bernoulli’s Equation

• In reality no system satisfies all the restrictions

• However, there are many systems for which only a negligible error will result when Bernoulli’s equation is used

• The use of this equation allows a rapid calculation if only rough estimation is required

Application of Bernoulli’s equation -Procedure1. Decide which items are known and what is to be found

2. Decide which two sections in the system will be used when writing Bernoulli’s equation. One section is chosen for which many data values are known. The second is the section which something is to be calculated

3. Write Bernoulli’s equation for the two selected sections in the system. It is important that the equation is written in the direction of flow. That is, the flow must proceed from the section on the left side of the equation to that on the right side.

4. Be explicit when labelling the subscript for the pressure head, elevation head and velocity head terms in Bernoulli’s equation. You should note where the reference points are on a sketch of the system.

5. Simplify the equation, if possible, by cancelling terms that are zero or those that are equal on both sides of the equation.

6. Solve the equation algebraically for the desired term.

7. Substitute known quantities and calculate the result, using consistent units throughout the calculations.

Example

Water at 10oC is flowing from section 1 to section 2. At section 1, which is 25mm in diameter, the gage pressure is 345kPa and the velocity of flow is 3.0m/s. Section 2, which is 50mm in diameter, is 2.0m abovesection 1. assuming there are no energy losses in the system, calculate the pressure p2.

• Known term

Section 1:

Diameter = 25mm, hence area= 𝐴2 = 𝜋(25/2)2= 491𝑚𝑚2

Gage pressure, 𝑝1 = 345𝑘𝑃𝑎

Velocity of flow, 𝑣1 = 3.0𝑚/𝑠

Section 2:

Diameter = 50mm, area=𝐴2 = 𝜋(50/2)2= 1963𝑚𝑚2

𝑧2 − 𝑧1 = 2.0𝑚

𝑝1

𝛾1+ 𝑧1 +

𝑣12

2𝑔=

𝑝2

𝛾2+ 𝑧2 +

𝑣22

2𝑔(red-coloured terms are known)

𝑣2 are determined from continuity equation:𝐴1𝑣1 = 𝐴2 𝑣2

𝑣2=𝐴1𝑣1

𝐴2

Substitute into the Bernoulli’s equation:

𝑝1𝛾1

+ 𝑧1 +𝑣1

2

2𝑔=𝑝2𝛾2

+ 𝑧2 +

𝐴1𝑣1𝐴2

2

2𝑔Rearrange the equation:

𝑝2 = 𝛾2[𝑝1𝛾1

+ 𝑧1 +𝑣1

2

2𝑔− 𝑧2 +

𝐴1𝑣1𝐴2

2

2𝑔]

γ2 = γ1, therefore,

𝑝2 = 𝛾1[𝑝1𝛾1

+ 𝑧1 +𝑣1

2

2𝑔− 𝑧2 +

𝐴1𝑣1𝐴2

2

2𝑔]

𝑝2 = 𝑝1 + 𝛾1[𝑧1 − 𝑧2 +𝑣1

2

2𝑔+

𝐴1𝑣1𝐴2

2

2𝑔]

𝑝2 = 345𝑘𝑃𝑎 + 9.81𝑘𝑁/𝑚3[(−2𝑚) +3𝑚/𝑠2

2(9.81𝑚𝑠2

)+

(491𝑚𝑚2)(3𝑚𝑠)

1963𝑚𝑚2

2

2(9.81𝑚𝑠2

)]

Solve the equation:

𝒑𝟐 =329.6kPa

Tanks, Reservoirs and Nozzles Exposed to the Atmosphere• Figure shows a fluid flow system

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

• Note that the surface of the tank and the free stream of fluid exiting the nozzle are not confined by solid boundaries and exposed to the atmosphere

• Hence, the pressure at this section is zero gage pressure

Tanks, Reservoirs and Nozzles Exposed to the Atmosphere• Hence, we can use the following rule:

When the fluid at a reference point is exposed to the atmosphere, the pressure is zero and the pressure head term can be cancelled from Bernoulli’s equation

Tanks, Reservoirs and Nozzles Exposed to the Atmosphere• The tank from which the fluid is being drawn can be assume to be

quite large compared to the size of the flow area inside the pipe

• Due to v=Q/a, the velocity at the surface of such a tank will be very small

• In term of velocity head, the velocity v is squared hence producing much less value for v<1.0 (e.g 0.52 = 0.025)

• Hence, we can adopt the following rule:

The velocity head at the surface of a tank or reservoir is considered to be zero and it can be cancelled from Bernoulli’s equation

When Both Reference Points Are in the Same Pipe• At points B-E, several points are inside the pipe, which has a uniform

flow area.

• Under the conditions of steady flow assumed for these problems, the velocity will be the same throughout the pipe.

• Hence, the following rule applies when steady flow occurs:

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

When Elevations Are Equal at Both Reference PointsWhen the two points of reference for Bernoulli’s equation are both at the same elevation, the elevation head terms are equal and can be cancelled

Example

• Figure above shows a siphon that is used to draw water from a swimming pool. The tube that makes up the siphon has an ID of 40mm and terminates with a 25mm diameter nozzle. Assuming that there are no energy losses in the system, calculate the volume flow rate through the siphon and the pressure at points B-E.

• Known value:

ID 40mm of pipe = appendix J

At nozzle = 25mm diameter

Pressure at A and F = 0Pa due to exposed to atmosphere

Velocity at A is almost zero because surface area of the pool is relatively large compared to the pipe

Hence, by analysing point A-F𝑝𝐴𝛾𝐴

+ 𝑧𝐴 +𝑣𝐴

2

2𝑔=𝑝𝐹𝛾𝐹

+ 𝑧𝐹 +𝑣𝐹

2

2𝑔

𝑧𝐴 = 𝑧𝐹 +𝑣𝐹

2

2𝑔

𝑣𝐹 = 2𝑔(𝑧𝐴 − 𝑧𝐹)

From figure, 𝑧𝐴 − 𝑧𝐹 = 1.2 + 1.8 = 3.0𝑚

𝑣𝐹 = 2𝑔(3) = 7.67𝑚/𝑠

To solve for volume flow rate Q:𝑄 = 𝐴𝐹𝑣𝐹 = 3.77 × 10−3𝑚3/𝑠

To solve for pressure at points B-E, first we solve points B:𝑝𝐴𝛾𝐴

+ 𝑧𝐴 +𝑣𝐴

2

2𝑔=𝑝𝐵𝛾𝐵

+ 𝑧𝐵 +𝑣𝐵

2

2𝑔

Elevation head s are cancelled because point A and B are at the same elevation

By using Q value determine previously for the whole system, we can calculate 𝑣𝐵.

𝑣𝐵 =𝑄

𝐴𝐵= 3.77 × 10−3/(1.257 × 10−3𝑚2)= 3m/s

Hence, solve pB:

𝑝𝐵 = 𝛾𝐵𝑣𝐵

2

2𝑔= −4.5𝑘𝑃𝑎

(negative sign indicate that pB is 4.5kPa below atmospheric pressure)

Solve for points C-E

pC=-16.27kPa

pD=-4.5kPa

pE=24.93kPa

Summary of the example

1. The velocity of flow from the nozzle, and therefore the volume flow rate delivered by the siphon depends on the elevation difference between the free surface of the fluid and the outlet of the nozzle (analysis points A-F)

2. The pressure at point B is below atmospheric pressure even though it is on the same level as point A which is exposed to the atmosphere. In calculation at point B, the pressure head at B is decreased by the amount of the velocity head. This is because some of the energy is converted to kinetic energy, resulting in a lower pressure at B.

Summary of the example

3. When steady flow exists, the velocity of flow is the same at all points where the tube size is the same (analysis B-C-D).

4. The pressure at point C is the lowest in the system because point Cis at the highest elevation. (conversion of potential – flow energy)

5. The pressure at point D is the same as that at point B because both are on the same elevation and the velocity head at both points is the same.

6. The pressure at point E is the highest in the system because point E is at the lowest elevation.

Venturi Meters and Other Closed Systems with Unknown Velocities• Venturi meter is a device that can be used to measure

the velocity of flow in a fluid flow system. Analysis of this device is based on the application of Bernoulli’s equation.

• The reduced-diameter section at B causes the velocityof flow to increase there with a corresponding decrease in the pressure. It will be shown that the velocity of flow is dependent on the difference in pressure between points A and B.

• Therefore, a differential manometer is convenient to use.

Example

• The venture meter shown in figure carries water at 60oC. The inside dimensions are machined to the sizes shown in the figure. 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.

𝑝𝐴𝛾+ 𝑧𝐴 +

𝑣𝐴2

2𝑔=𝑝𝐵𝛾+ 𝑧𝐵 +

𝑣𝐵2

2𝑔

• Rearrange the equation𝑝𝐴 − 𝑝𝐵

𝛾+ 𝑧𝐴 − 𝑧𝐵 =

𝑣𝐵2 − 𝑣𝐴

2

2𝑔

𝑧𝐴 − 𝑧𝐵 = −0.46𝑚

Pressure head Velocity head

• To solve 𝑝𝐴−𝑝𝐵

𝛾, we need to write the equation

for the manometer. Write from A:𝑝𝐴 + 𝛾 𝑦 + 𝛾 1.18 − 𝛾𝑔 1.18 − 𝛾 𝑦 − 𝛾 0.46 = 𝑝𝐵

• As given, sg of gage fluid = 1.25, hence𝛾𝑔 = 𝛾𝑤𝑎𝑡𝑒𝑟 × 𝑠𝑔 = 9.81 × 1.25 = 12.26𝑘𝑁/𝑚3

• Cancel out 𝛾 𝑦 , 𝑝𝐴 − 𝑝𝐵 = 𝛾 0.46 − 1.18 + 𝛾𝑔 1.18

• Rearrange equation to be equal to 𝑝𝐴−𝑝𝐵

𝛾𝑝𝐴 − 𝑝𝐵

𝛾= 0.46 − 1.18 +

𝛾𝑔 1.18

𝛾= 0.78𝑚

• Solve for velocity head term:

Using continuity equation𝐴𝐴𝑣𝐴 = 𝐴𝐵𝑣𝐵

𝑣𝐵 =𝐴𝐴𝑣𝐴𝐴𝐵

Area at section A and B are calculated based on the diameter given

𝑣𝐵 =(7.069 × 10−2)𝑣𝐴(3.142 × 10−2)

= 2.25𝑣𝐴

𝑣𝐵2 − 𝑣𝐴

2 = 2.25𝑣𝐴2 − 𝑣𝐴

2 = 4.06𝑣𝐴2

• Input all the results into Bernoulli’s equation:𝑝𝐴 − 𝑝𝐵

𝛾+ 𝑧𝐴 − 𝑧𝐵 =

𝑣𝐵2 − 𝑣𝐴

2

2𝑔

0.78 + −0.46 =4.06𝑣𝐴

2

2 9.81

Solve for 𝑣𝐴,𝑣𝐴 = 1.24𝑚/𝑠

Volume flow rate:

𝑄𝐴 = 𝐴𝐴𝑣𝐴 = 8.77 × 10−2𝑚3/𝑠

Torricelli’s Theorem

• In the siphon analysed previously, 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 classic application of this is in the flow of water from tank.

• Fluid is flowing from side of tank through a smooth rounded nozzle

• The velocity of flow from the nozzle can be determined using Bernoulli’s equation between a reference point on the fluid surface and a point in the jet issuing from the nozzle

𝑝1𝛾+ 𝑧1 +

𝑣12

2𝑔=𝑝2𝛾+ 𝑧2 +

𝑣22

2𝑔

• However, p1=p2=0 and v1 is approximately zero

0 + 𝑧1 + 0 = 0 + 𝑧2 +𝑣2

2

2𝑔

• Then solving for v2 gives

𝑣2 = 2𝑔(𝑧1 − 𝑧2)

Let h= (𝑧1 − 𝑧2)

𝑣2 = 2𝑔(𝑧1 − 𝑧2) (Torricelli’s Theorem)

• This equation is called Torricelli’s theorem in honour of Evangelista Torricelli who discovered it in 1645

Example• For the tank shown in the figure, compute the velocity of flow from

the nozzle for a fluid depth h of 5m

• Solution:

Directly apply Torricelli’s theorem:

𝑣2 = 2𝑔(ℎ)

𝑣2 = 2(9.81)(5) = 9.91𝑚/𝑠