Chapter. 3 Elementary Fluid Dynamics:Bernoulli Equation

이동근ME Bldg. (517)

Tel. 051-510-2365Download at http://npt.pusan.ac.kr

Reference:1. Fluid Mechanics, Frank MWhite, 6th Edition, 2008, McGraw Hill2. Munson et al., Fundamentals of Fluid Mechanics, 8th Edition, 2006, John Wiley &Sons, Inc

1

3-2

Contents

Chap. 1 Introduction

Chap. 2 Fluid Statics

Chap. 3 Elementary Fluid Dynamics : the Bernoulli equation

- Newton's 2nd law; along a streamline or normal to a streamline

- Static, Stagnation, Dynamic, and Total Pressure

- Examples of Bernoulli equation

- Energy and Hydraulic Grade Lines

- Restrictions on use of the Bernoulli Equation

Chap. 4 Fluid Kinetics

Chap 5 Finite Control Volume Analysis

Chap. 6 Differential Analysis of Fluid Flow

Acceleration of Fluid Particle in a Curvilinear Motion

• To derive the acceleration vector in tangential and normal components, define the motion of a particle as shown in the figure.

• 𝑒𝑡 and 𝑒𝑡′are tangential unit vectors for the particle

path at P and P’. When drawn with respect to the same origin, ∆ 𝑒𝑡 = 𝑒𝑡

′− 𝑒𝑡 and ∆𝜃 is the angle between them.

3-3

The vector ∆ 𝑒𝑡/∆𝜃 has a magnitude of unity & a direction toward the O'

∆𝜃𝑂′

∆𝑒𝑡 = 2 sin(∆𝜃

2) lim

∆𝜃→0

∆𝑒𝑡∆𝜃

= lim∆𝜃→0

sin(∆𝜃/2)

∆𝜃/2=1

∴ lim∆𝜃→0

∆ 𝑒𝑡∆𝜃

= 𝑒𝑛 =𝑑 𝑒𝑡𝑑𝜃

• With the velocity vector expressed as 𝑣 = 𝑣 𝑒𝑡the particle acceleration may be written as

Acceleration of a particle along a streamline

where

After substituting,

where 𝜌 is the radius of curvature of the motion.

• Along a streamline,

3-4

𝑎 =𝑑 𝑣

𝑑𝑡=𝑑𝑣

𝑑𝑡 𝑒𝑡 + 𝑣

𝑑 𝑒𝑡𝑑𝑡

=𝑑𝑣

𝑑𝑡 𝑒𝑡 + 𝑣

𝑑 𝑒𝑡𝑑𝜃

𝑑𝜃

𝑑𝑠

𝑑𝑠

𝑑𝑡

𝑑 𝑒𝑡𝑑𝜃

= 𝑒𝑛; 𝜌𝑑𝜃 = 𝑑𝑠;𝑑𝑠

𝑑𝑡= 𝑣

𝑎 = 𝑎𝑡 𝑒𝑡 + 𝑎𝑛 𝑒𝑛 =𝑑𝑣

𝑑𝑡 𝑒𝑡 +

𝑣2

𝜌 𝑒𝑛

𝑎𝑡 =𝑑𝑣

𝑑𝑡=𝜕𝑣

𝜕𝑡+ 𝑣 ∙ 𝛻𝑣

=𝜕𝑣

𝜕𝑡+ 𝑣

𝜕𝑣

𝜕𝑠

← 𝑣 ∙ 𝛻𝑣 = 𝑣 𝑒𝑡 ∙𝜕𝑣

𝜕𝑠 𝑒𝑡 +

𝜕𝑣

𝜕𝑛 𝑒𝑛 = 𝑣

𝜕𝑣

𝜕𝑠

𝑎 =𝜕𝑣

𝜕𝑡+ 𝑣

𝜕𝑣

𝜕𝑠 𝑒𝑡 +

𝑣2

𝜌 𝑒𝑛

• For a steady state, 𝑎 = 𝑣𝜕𝑣

𝜕𝑠 𝑒𝑡 +

𝑣2

𝜌 𝑒𝑛

The Bernoulli Equation (BE)

3-5

𝜃

𝜃

𝛿𝑧

Free body diagram Assumptions: Steady state, Inviscid, incompressible flow

𝛿𝐹𝑠,𝑖 = 𝛿m 𝑎𝑠

1) Stream-wise direction along a streamline

= 𝜌𝛿∀ 𝑣𝜕𝑣

𝜕𝑠

𝛿𝐹𝑠,𝑖 = 𝛿𝑊𝑠 + 𝛿𝐹𝑝𝑠

𝛿𝑊𝑠 = −𝛿𝑊 sin 𝜃 = −𝛾𝛿∀ sin 𝜃

𝛿𝐹𝑝𝑠 = 𝑝 − 𝛿𝑝𝑠 𝛿n𝛿𝑦 − 𝑝 + 𝛿𝑝𝑠 𝛿n𝛿𝑦

= −2𝛿𝑝𝑠𝛿n𝛿𝑦 = −𝜕𝑝

𝜕𝑠𝛿s𝛿n𝛿𝑦

= −𝜕𝑝

𝜕𝑠𝛿∀

𝛿𝐹𝑠,𝑖 = − 𝛾 sin 𝜃 +𝜕𝑝

𝜕𝑠𝛿∀

The Bernoulli Equation (BE)

3-6

𝜃

𝜃

𝛿𝑧

Free body diagram Assumptions: Steady state, Inviscid, incompressible flow

1) Stream-wise direction along a streamline

𝜌𝑎𝑠 = 𝜌𝑣𝜕𝑣

𝜕𝑠= − 𝛾 sin 𝜃 +

𝜕𝑝

𝜕𝑠

← sin 𝜃 =𝑑𝑧

𝑑𝑠

𝜌𝑣𝜕𝑣

𝜕𝑠= −𝛾

𝑑𝑧

𝑑𝑠−𝜕𝑝

𝜕𝑠

비압축성

𝜕 𝜌𝑣2/2

𝜕𝑠= −

𝜕

𝜕𝑠𝑝 + 𝛾𝑧

∴𝜕

𝜕𝑠

𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧 = 0

Integrate against 𝑑𝑠along a streamline

The Bernoulli Equation (BE)

3-6

Assumptions: Steady state, Inviscid, incompressible flow

1) Stream-wise direction along a streamline

𝜕

𝜕𝑠

𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧 = 0

Integrate against 𝑑𝑠 along a streamline

𝑑 = 𝛻 ∙ 𝑑 𝑟 at steady state

𝛻 =𝜕( )

𝜕𝑠𝑑𝑠 +

𝜕( )

𝜕𝑛𝑑𝑛 +

𝜕( )

𝜕𝑦𝑑𝑦

where 𝑑 𝑟 is the integration path

along a streamline

𝑑 = 𝛻 ∙ 𝑑 𝑟 =𝜕( )

𝜕𝑠𝑑𝑠

1

2 𝜕

𝜕𝑠

𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧 𝑑𝑠 = 0

1

2

𝑑𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧 = 0

𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧

1

= 𝜌𝑣2

2+ 𝑝 + 𝜌𝑔𝑧

2

This is a sort of energy equationTotal energy is conserved along a streamline when the fluid inviscid and incompressible and at steady state

𝑣2

2𝑔+

𝑝

𝜌𝑔+ 𝑧

1

= 𝑣2

2𝑔+

𝑝

𝜌𝑔+ 𝑧

2

The Bernoulli Equation (BE)

3-6

Assumptions: Steady state, Inviscid, incompressible flow

2) Normal direction across a streamline

𝛿𝐹𝑛,𝑖 = 𝛿𝑊𝑛 + 𝛿𝐹𝑝𝑛 = 𝛿𝑚 𝑎𝑛 = 𝛿𝑚𝑣2

𝑅

𝛿𝑊𝑛 = −𝛿𝑊 cos 𝜃 = −𝛾𝛿∀ cos 𝜃 = −𝛾𝛿∀𝑑𝑧

𝑑𝑛

𝛿𝐹𝑝𝑛 = 𝑝 − 𝛿𝑝𝑛 𝛿𝑠𝛿𝑦 − 𝑝 + 𝛿𝑝𝑛 𝛿𝑠𝛿𝑦

= −2𝛿𝑝𝑛𝛿s𝛿𝑦 = −𝜕𝑝

𝜕𝑛𝛿s𝛿n𝛿𝑦

= −𝜕𝑝

𝜕𝑛𝛿∀

𝜌𝛿∀𝑣2

𝑅= −𝛾𝛿∀

𝑑𝑧

𝑑𝑛−𝜕𝑝

𝜕𝑛𝛿∀

𝜌𝑣2

𝑅= −

𝜕

𝜕𝑛𝑝 + 𝛾𝑧

Integrate against 𝑑𝑛 to a direction normal to the streamline

𝑣2

𝑅𝑑𝑛 +

𝑝

𝜌+ 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡 across the stream line

incompressible

𝑝 + 𝛾𝑧 decreases in the 𝑛 direction, i.e., toward the rotation center 𝑛 What happens if R ?

3-9

Bernoulli Equation (BE)

• BE is a simple and easy to use relation between the following three variables in a moving fluid

• pressure

• Velocity

• Elevation

• It can be thought of a limited version of the 1st law of thermodynamics.

• It is valid for steady, incompressible, inviscid flows.

• Warning: BE is one of the most used, but also one of the most abused equation influid mechanics. So be careful !!!

• BE can be seen as a balance of kinetic, potential and pressure energies.

• BE in "Head" form:

𝑝 +𝜌𝑣2

2+ 𝜌𝑔𝑧 = const along a stream line

Press PE

𝑝

𝜌𝑔+𝑣2

2𝑔+ 𝑧 =

KE

const along a stream line

pressure head velocity head elevation head

Total head (ℎ𝑇)

• Consider the flow of water from the syringe. The force applied to the plunger will produce a pressure greater than atmospheric pressure at point 1. The water flows from the needle (point 2) with relatively high velocity and rises up to point 3 at the top of its trajectory (Reference: Munson’s book).

3

𝑔

2

1

3-10

Applications of the BE

• Due to the frictional effects (viscous forces) the water will not go up as much as predicted by the BE.

• Such energy-loss effects arise especially at the narrowneedle exit and between the water jet and surroundingair stream.

𝐹

Point Energy type

Kinetic𝜌𝑉2/2

Potential𝜌𝑔𝑧

Pressure (gage)𝑝

1 Small Zero Large

2 Large Small Zero

3 Zero Large Zero

𝑝 +𝜌𝑉2

2+ 𝜌𝑔𝑧 = constant along a streamline

3-11

Static, Dynamic, Stagnation & Total Pressures

Static pressure

Dynamic pressure

Hydrostatic pressure

Stagnation pressure

BE says that total pressure is constant along a streamline.

Total pressure

BA

𝑉∞𝑥

𝑅

𝑦

𝑝𝐴 +𝜌𝑉𝐴

2

2= 𝑝𝐵 = 𝑝𝑠𝑡𝑎𝑔

• Between A and B along the streamline

• Pitot tube is a device used for speed measurement.

• It actually measures pressure and converts it into speed using the BE.

• It is a simple tube with a 90o bend.

• Read about the role of Pitot tube malfunctions on plane crashes.

http://www.planeandpilotmag.com/article/blocked-pitot-tubes/#.WhPvY1WWaUk

3-12

Simple Pitot Tube

Pitot tube on a F1 race car Pitot tube on an aircraft

→ 𝜌gℎ0 +𝜌𝑉𝐴

2

2= 𝜌g ℎ0 + ℎ1

3-13

Simple Pitot Tube (cont’d)

Exercise: Show that the fluid speed at point A is given by

𝑉A = 2𝑔ℎ1

• With a Pitot tube we actually measure the pressure difference between points ‘A’ and ‘𝑂’ and convert this difference to a speed difference using the BE.

• Fluid flows in an open channel from left to right.

• We want to measure the speed at point A.

• Fluid fills the Pitot tube and rises inside it to alevel of ℎ1 above the free surface.

• The aim of using a Pitot tube is to create a stagnation flow at point ‘𝑂’ with zero velocity.

A O

ℎ0

ℎ1𝑝𝑎𝑡𝑚

𝑝𝐴 +𝜌𝑉𝐴

2

2= 𝑝𝑂 Why 𝑝𝐴 = 𝜌gℎ0? even in motion

3-14

Use of Pitot Tube with a Static Tube

• Static tube will give :

• Pitot tube will give:

• BE between points ‘A’ and ‘o’ will give:

• Combining all these, speed at point A is obtained as

𝑉A = 2𝑔(ℎ1 − ℎ2)

• Now the flow is inside a closed channel or a pipe.

• We need to use an additional tube called thestatic tube (piezometer).

• Static tube is used to measure the static pressure at point A.

ℎ1ℎ2

A 𝑂

ℎ0

𝑝𝑎𝑡𝑚

𝑝𝐴 = 𝑝𝑎𝑡𝑚 + 𝜌g(ℎ0 + ℎ2)

𝑝𝑂 = 𝑝𝑎𝑡𝑚 + 𝜌g(ℎ0 + ℎ1)

𝑝𝐴 + 𝜌𝑉𝐴2/2 = 𝑝𝑂 + 𝜌𝑉𝑂

2/2

3-15

Combined Pitot-Static Tube (Prandtl’s Tube)

• The required pressure difference is 𝑝0 − 𝑝𝐴 = (𝜌𝑚 − 𝜌)𝑔ℎ𝑚

• BE between ‘A’ and ‘o’ will give

𝑉A = 2(𝑝0 − 𝑝𝐴)/𝜌 →

• Instead of measuring static pressure at point Ausing a piezometer tube, a second tube is usedaround the Pitot tube.

• Static pressure holes (point 𝑎) of the outer tube are located such that they measure correct upstream static pressure, i.e. 𝑝𝑎 = 𝑝𝐴 .

• Two tubes provide the necessary pressure difference measurement using the mercury in it.

A O

𝜌𝑚

𝑎𝜌

ℎ𝑚

𝑉𝐴 = 2𝑔ℎ𝑚𝜌𝑚𝜌− 1

3-16

Errors & Another variation

What error is expected by the imperfect manufacture of tip?

𝑝1 > 𝑝 𝑝1 < 𝑝 𝑝1 = 𝑝What benefit is expected?

3-17

Combined Pitot-Static Tube (cont’d)

• Pressure variation along a

combined Pitot-static tube is as shown.

• As seen, the static holes are located carefully such that they measure the static pressure ahead of the device.

Exercise: Water flows through the pipe contraction shown. For the given 0.2 m difference in the manometer level, determine the flow rate.

Munson’s book𝑉 = ?

𝑝Stagnation

pressure at tip

Upstream and static hole pressures are equal

Static holes

Stagnation pressure on

stem

Stem

0.1 m𝑄 = ?

0.2 m

0.05 m

3-18

Bernoulli Equation Exercises: Siphon vs Cavitation

Exercise : A tube can be used to discharge water from a reservoir as shown. Determine the speed of the free jet and the minimum absolute pressure of water that occurs at the top of the bend.

3 m

This is known as siphoning. It can be used to drain gasoline from the tank of an automobile. Onceyou establish the initial flow by sucking gasolinefrom the tube, the gasoline will flow by itself.

1 m

Exercise : Cavitation?

3-19

Use of BE: 1) Free Jets - Toricelli Equation

• Consider the discharge of a liquid from a large reservoir through an orifice (hole).

𝑝2 +𝜌𝑉2

2

2= 𝑝1 +

𝜌𝑉12

2+ 𝜌𝑔ℎ

Why 𝑝2 = 0?

∴ 𝑉2= 2𝑔ℎ Is not a function of 𝜌

𝑝2 = 𝑝4 = 𝑝1

What about (5)? 𝑝5 > 𝑝2? 𝑉5 > 𝑉2? If so, how much?

3-20

Use of BE: 1) Free Jets - Toricelli Equation

• Consider the discharge of a liquid from a large reservoir through an orifice (hole).

When exit (2) is small enough to neglect the spatial velocity distribution,flow discharge rate from the exit is

𝑄 = 𝐴2𝑉2= 𝐴2 2𝑔ℎ However, this is an ideal value that must be corrected in reality

3-21

Vena Contracta and Contraction Coefficient

𝐴𝑗𝐴ℎ

𝐴𝑗 < 𝐴ℎ

𝐶𝐶 = 𝐴𝑗/𝐴ℎ

Contraction Coefficient

𝑄 = 𝐴𝑗𝑉𝑗= 𝐴𝑗 2𝑔ℎ

= 𝐶𝐶𝐴ℎ 2𝑔ℎ

𝑄 = 𝐶𝑣𝐶𝐶𝐴ℎ 2𝑔ℎ

≡ 𝐶𝑑𝐴ℎ 2𝑔ℎ

Further correction

𝐶𝑑: discharge coefficient, experimentally determined; 𝐶𝑣: velocity coefficient for viscous effect

3-22

Obstruction Flow Meters

• They are used to measure flow ratesthrough pipes. General idea is

• to place an obstacle inside the pipe and force the fluid to accelerate and pass through a narrow area.

• measure the pressure differencebetween the low-velocity, high-pressure upstream and the high-velocity, low-pressure downstream.

• use the BE to relate this pressuredifference to the flow rate in thepipe.

3-23

Venturi Meter

𝑑 𝑉2𝑉1𝐷

1 2

𝐴1 =𝜋𝐷2

4𝐴2 =

𝜋𝑑2

4

• BE on a streamline along the pipe axisfrom section 1 to section 2

𝑝1 +𝜌𝑉1

2

2= 𝑝2 +

𝜌𝑉22

2

• Continuity equation between section 1to 2 for incompressible flow

𝑄 = 𝐴1𝑉1= 𝐴2𝑉2 → 𝐷2𝑉1 = 𝑑2𝑉2

• Combine two equations to eliminate 𝑉1and obtain 𝑉2 as

𝑉2 =2(𝑝1 − 𝑝2)

𝜌(1 − 𝛽4)where 𝛽 =

𝑑

𝐷

• Flow rate through the pipe is given by

𝑄 = 𝐴2𝑉2 = 𝐴22(𝑝1 − 𝑝2)

𝜌(1 − 𝛽4)

Venturi Meter (cont’d)

• This flow rate can be corrected for viscous effects using the discharge coefficient

Experimentally determined and provided by the manufacturer

(see the Slide 3-26)

• The same equation can be used for the nozzle flow meter too.

• For both the Venturi meter and the nozzle flow meter, the contraction is mostly smooth so that the contraction coefficient was set to 1 (𝐶𝐶 = 1)

𝑄 = 𝐶𝑑𝐴22(𝑝1 − 𝑝2)

𝜌(1 − 𝛽4), 𝛽 =

𝑑

𝐷

3-24

Orifice Meter

• Section o has the orifice plate with the hole diameter 𝑑.

• Section 2 is the vena contracta section. 𝑝2 is measured here.

Exercise: Derive the flow rate equation for this orifice meter.

𝑑 𝑉2𝑉1𝐷

1 2o

𝐴1 =𝜋𝐷2

4𝐴O =

𝜋𝑑2

4

𝐴2 = 𝐶𝐶𝐴O

(vena contracta)

• For the orifice meter the expansion is abrupt and 𝐶𝐶 < 1, i.e. vena contractaarea is smaller than the orifice area.

3-25

Typical 𝑪𝒅 Graphs for Obstruction Flow Meters

𝛽 = 0.20.58

104 105 106 107 108

0.66

0.64

𝐶𝑑 orifice0.62

0.60

Dd

V

𝑅𝑒=𝜌𝑉𝐷

𝜇

𝛽 =𝑑𝐷 = 0.7

0.94104 105 106 107 108

𝑅𝑒

1.00

0.98

𝐶𝑑nozzle

0.96𝐷 𝑑V

𝛽 = 0.2

𝛽 = 0.8

Range of values depending on specific

Venturi geometry

0.94104 105 𝑅𝑒 107 108

1.00

0.98

𝐶𝑑Venturi

0.96

3-26

3-27

Obstruction Flow Meter Exercises

Exercise : (Munson’s book) a) Determine the flow rate through the Venturi meter shown. b) Flow rate is increased until cavitation is first observed. At that instant𝑝1 is read as 276 kPa gage. At what flow rate will this occur? Vapor pressure of theflowing fluid is 3.6 kPa and the atmospheric pressure is 100 kPa.

31 mm 19 mm

𝑝2 = 550 kPa𝑝1 = 735 kPa

Q

𝛾= 9.1 kN/m3

𝑄 = 𝐴2𝑉2 = 𝐴22(𝑝1 − 𝑝2)

𝜌(1 − 𝛽4)

Note that p1 is a gage pressure and vapor pressure is in absolutepressure unit. Recall at which total pressure cavitation begins.

3-28

Obstruction Flow Meters (cont’d)

• Comparison of obstruction type flow meters

• Other types flow meters

• Rotameter

• Thermal flow measurement

• Vortex type flow meter

• Ultrasonic flow meter

• Coriolis flow measurement

• Turbine flow meter

• Weirs (for open channels)

( youtube.com/watch?v=2dfIWNYJwZM )

( youtube.com/watch?v=YfQSf2NBGqc )

( youtube.com/watch?v=GmTmDM7jHzA )

( youtube.com/watch?v=Bx2RnrfLkQg )

( youtube.com/watch?v=XIIViaNITIw )

Cost Ease of Installation Pressure Loss

Orifice meter Cheap Difficult High

Nozzle flow meter Medium Difficult Medium

Venturi meter High Difficult Low

Be Careful in Using the Bernoulli Equation

• The simplest and the most commonly used BE that we studied in the previous slides may lead to unphysical results for problems similar to the following ones.

Çengel’s book

1

1

1

1

1

2

2

2

2

2

A sudden expansion

A longnarrowtube

Flow through a

valve

Flow through an energy delivering

or extracting device

Flow in highlyviscous regions

A boundary layer

A wake

Flow with excessive heat

transfer

3-29

3-30

Extended Bernoulli Equation (EBE)

• It is a modified version of the BE to include effects such as viscous energy-loss effect and shaft works.

• Remember the energy conservation equation for a single inlet (1), single exit (2)CV with uniform properties.

𝑝

𝜌+𝑉2

2+ 𝑔𝑧

1

=𝑝

𝜌+𝑉2

2+ 𝑔𝑧

2

+𝑤𝑓 +𝑤𝑠

where 𝑤𝑠 is a sharp work out per unit mass, for example, 𝑤𝑠 is negative for pump work in, but positive for turbine work out; 𝑤𝑓 is an energy loss by friction.

Streamline

1

2

𝑤𝑠

𝑤𝑓

• "Head form" of the BE

𝑝

𝜌𝑔+𝑉2

2𝑔+ 𝑧

1

=𝑝

𝜌𝑔+𝑉2

2𝑔+ 𝑧

2

+ ℎ𝑓 + ℎ𝑠

Pressurehead

Velocityhead

Elevationhead

Frictionhead

Pump head in < 0, turbine head out > 0

Total head at 2(ℎ𝑇2)

Pump and Turbine Head (h𝒔)Pump head ℎ𝑠 is related to the power delivered to the fluid by thepump (𝒫𝑓 ) as follows

• Power delivered to the fluid (𝒫𝑓) is related to the power consumed by the pump (𝒫pum p ) through thepump efficiency

𝜂 =pump

𝒫𝑓

𝒫pump

• For a turbine, power extracted from the fluid is calculated in a similar way.

𝒫𝑓 = 𝜌𝑔𝑄ℎ𝑠

• Power produced by the turbine (𝒫turbine) is smaller than the extracted fluid power

𝜂turbine =𝒫turbine

𝒫𝑓

1

2

𝑄

𝑄

Centrifugal pump

3-31

𝑃𝑓 = 𝑚𝑤𝑠 = 𝜌𝑄𝑤𝑠 → 𝑃𝑓= 𝜌𝑔𝑄ℎ𝑠

3-32

Extended BE Exercises

Exercise : The pump shown below pumps water steadily at a volumetric rate of0.005 m3/s through a constant diameter pipe. At the end of the pipe there is a nozzle with an exit area that is equal to half of the pipe area. Neglecting frictional losses, determine the power that must be supplied to the pump, if it is working with 70 % efficiency.

Pump

25 m

7 m

15 m

9 m

3-33

Extended BE Exercises (cont’d)

Exercise : A pump is used to transport water between two large reservoirs. Desired volumetric flow rate through the pipes is 0.016 m3/s. Cross-sectional area of the pipes are 0.004 m2. Total frictional head losses between two reservoirs is estimated to be 2 m. Efficiency of the pump is 75 %. Determine

a) the required pump head.

b) the power delivered to the water by the pump.

c) the power required to drive the pump.

Pump

6 m

Suction pipe

Discharge pipe

Suctionreservoir

Dischargereservoir

3-34

Hydraulic Grade Line vs Energy Line in Head form of BE

ℎ𝑡𝑜𝑡 = 𝐻 = 𝑧2 +𝑝2𝛾+𝑉22

2𝑔= 𝑧3 +

𝑝3𝛾+𝑉32

2𝑔EL 𝑧1 +𝑝1𝛾> 𝑧2 +

𝑝2𝛾> 𝑧3 +

𝑝3𝛾

HGL

3-35

Hydraulic Grade Line vs Energy Line in Head form of BE

3-36

Hydraulic Grade Line vs Energy Line in Head form of BE

3-37

Restrictions on Use of BE

1) Compressibility Effect

• For a simple compressible gas flow where temperature remains constant along a streamline,the ideal gas law can be used to consider the density variation along the streamline.

𝑑𝑝

𝜌+1

2𝑣2 + 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡 → 𝑅𝑇

𝑑𝑝

𝑝+1

2𝑣2 + 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡

The constant of integration is readily obtained if the state of gas flow is known in terms of 𝑧1, 𝑝1, and 𝑉1 at any location on the streamline. The result is

𝑉12

2𝑔+ 𝑧1 +

𝑅𝑇

𝑔𝑙𝑛

𝑝1𝑝2

=𝑉22

2𝑔+ 𝑧2

Valid for inviscid, isothermal, compressible, stead-state, along the gas streamline

• For a more common condition, say, isentropic expansion or contraction process which means there is no friction or heat transfer

𝑝

𝜌𝑘= 𝐶 →

𝑑𝑝

𝜌+1

2𝑣2 + 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡 → 𝐶1/𝑘

𝑑𝑝

𝑝1/𝑘+1

2𝑣2 + 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡

𝑘

𝑘 − 1

𝑝1𝜌1

+𝑉12

2+ 𝑔𝑧1 =

𝑘

𝑘 − 1

𝑝2𝜌2

+𝑉22

2+ 𝑔𝑧2

Valid for inviscid, isentropic, compressible, stead-state, along the gas streamline

3-38

Restrictions on Use of BE

1) Compressibility Effect

𝑘

𝑘 − 1

𝑝1𝜌1

+𝑉12

2+ 𝑔𝑧1 =

𝑘

𝑘 − 1

𝑝2𝜌2

+𝑉22

2+ 𝑔𝑧2 Difference b/w incompressible BE

21

𝑉1

𝑅𝑝1

𝑀𝑎1

𝑇1

• For a stagnation flow, let us learn the difference between compressible and incompressible.

From Chap. 11 (compressible)

𝑝2 − 𝑝1𝑝1

= 1 +𝑘 − 1

2𝑀𝑎1

2

𝑘/(𝑘−1)

− 1

When 𝑀𝑎1 → 0, consider

← 1 + 휀 𝑛= 1 + 𝑛휀 + n n − 1휀2

2+ ⋯

𝑝2 − 𝑝1𝑝1

=𝑘𝑀𝑎1

2

21 +

1

4𝑀𝑎1

2 +2 − 𝑘

24𝑀𝑎1

4 +⋯𝑀𝑎1→0 𝑘𝑀𝑎1

2

2(compressible)

← 𝑀𝑎1=𝑉1𝑐1

=𝑉1

𝑘𝑅𝑇1𝑝1 +

𝜌𝑉12

2= 𝑝2 →

𝑝2 − 𝑝1𝑝1

=𝑉12

2𝑅𝑇1=𝑘𝑀𝑎1

2

2

• For an incompressible stagnation flow

3-39

Restrictions on Use of BE

2) Unsteady Effect

𝑎 =𝜕𝑉

𝜕𝑡+ 𝑉

𝜕𝑉

𝜕𝑠 𝑒𝑡 +

𝑣2

𝑅 𝑒𝑛 ↔ 𝑎 = 𝑉

𝜕𝑉

𝜕𝑠 𝑒𝑡 +

𝑉2

𝑅 𝑒𝑛(unsteady) (steady)

① a change in velocity of a particle resulting from a change in its positionalong a streamline

② a change in velocity of a particle at a fixed position

𝜕𝑉

𝜕𝑡𝑑𝑠 +

𝑑𝑝

𝜌+ 𝑑

𝑉2

2+ 𝑔𝑑𝑧 = 0

𝑝1 +𝜌𝑉1

2

2+ 𝛾𝑧1 = 𝜌

𝑆1

𝑆2𝜕𝑉

𝜕𝑡𝑑𝑠 + 𝑝2 +

𝜌𝑉22

2+ 𝛾𝑧2

Valid for unsteady, incompressible, inviscidalong a streamline

Uneasy to calculate this integral, but simplified for irrotational flow (Chap. 6)

3-40

Restrictions on Use of BE

3) Rotational Effect: e.g., for incompressible, uniform flow w/ V

𝑝1 +𝜌𝑉1

2

2+ 𝛾𝑧1 = 𝑝2 +

𝜌𝑉22

2+ 𝛾𝑧2 = 𝐶12

𝑝1 = 𝑝2 = 𝑝0, 𝑧1 = 𝑧2 = 0, 𝑉1 = 𝑉2 = V

𝐶12 = 𝑝0 +𝜌𝑉2

2

𝑝3 +𝜌𝑉3

2

2+ 𝛾𝑧3 = 𝑝4 +

𝜌𝑉42

2+ 𝛾𝑧4 = 𝐶34

𝑝3 = 𝑝4 = 𝑝0 − 𝛾ℎ,𝑧3 = 𝑧4 = ℎ, 𝑉3 = 𝑉4 = V

𝐶34 = 𝑝0 − 𝛾ℎ +𝜌𝑉2

2+ 𝛾h = 𝐶12

For a uniform flow that is "irrotational flow", there is no restriction of "along the streamline"

BE w/ a same constant can be applied for (1-2), (3-4), (2-4), but not for (4-5)