AFD1 The Mechanic Energy Equation

Applied Fluid Dynamics
AFD1 The Mechanic Energy Equation
Chemical Engineering Guy
www.ChemicalEngineeringGuy.com

AFD1 Block Overview

Textbook, Reference and Bibliography

• Chapters:
– 6  Fluid Flow and Bernoulli
Equation
– 7  Mechanical Energy Equation
Applied Fluid Mechanics. Robert
Mott 6th Edition

• Section 2: Fluid Mechanics
– CH4: Basic Equations of Fluid Flow
– CH5: Flow of Incompressible Fluids
Unit Operation of Chemical
Engineering. McCabe 7th Edition

AFD1 Block Overview
• Section 0: Review
– Density, Viscosity, Friction, Mass, Weight, Gravitational Constant. Pressure, Temperature, etc.
• Section 1: Why M.E.E
– Systems
• Section 2: Kinetic + Potential Energy
– Kinetic Energy  Velocity
– Potential Energy  Relative Position
• Section 3: Pressure Head
– Pressure  Force per unit area
• Section 4: Inlet/Outlet Work
– Work inlet (Pumps, Compressors, etc…)
– Work Outlet (Turbine, Mill, etc…)
• Section 5: Friction Loss  Trajectory
– Nature of Friction Loss
• Section 6: Application - Mechanical Energy Equation
– Bernoulli’s Law
– Torricelli’s Law
– M.E.E. Exercises

Section 0: Review

Mass
• Mass is a property of a physical body:
– determines the strength of its mutual gravitational attraction to other
bodies
– its resistance to being accelerated by a force
– theory of relativity gives the mass–energy content of a system
• The SI unit of mass is the kilogram (kg).
• The English unit of mass is the pound (lb)

Mole
• The mole is a unit of measurement for amount
of substance.
• It is defined as the amount of any chemical
substance that contains as many elementary
entities in 6.022*10^23 entities
• Entity:
– Electrons
– Atoms
– Molecules

Molecular Mass/Weight
• Is the mass of a molecule
• It is measured in mass per unit mol
• Example:
– The Molecular Mass of 1 mol of Carbon is 12 grams
– In 36 grams of Carbon, you should have 3 moles

Density
• Amount of mass per unit volumen
occupied by that mass.
• SI Units  kg/m3
• English Units  lb/ft3

Density
• Amount of mass per unit volumen
occupied by that mass.
• For Incompressible Flow:
– Density is dependant of Temperature
– The change is small
– Calculation may be avoided  Constant
Value

Specific Gravity
• Specific gravity is the ratio of
the density of a substance to the
density (mass of the same unit
volume) of a reference substance.
• The reference substance is nearly
always water at its densest (4°C) for
liquids and for gases it is air at room
temperature (21°C)

Specific Gravity
• Specific Gravity of Gasoline is 0.78 with
respect of water at 4°C…
• What is the density of Gasoline?
– Dwater = 1000 kg/m3
– Dgasoline = Dwater*S.G = 1000*0.78 = 780 kg/m3

Specific Weigth
• The specific weight is the weight per
unit volume of a material.
• The symbol of specific weight is γ (the Greek
letter Gamma).
• A commonly used value is the specific weight
of water on Earth at 5°C which is 9.807
kN/m3 or 62.43 lbf/ft3.

Viscosity
• The viscosity of a fluid is a measure of
its resistance to gradual deformation by shear
stress or tensile stress.
• For liquids, it corresponds to the informal
concept of "thickness".
• For example, honey has a much higher
viscosity than water
– Dynamics vs. Kinematic

Viscosity of Liquids @25°C

Viscosity of Liquids
• In general…
– As T increases  Viscosity decreases

Dynamic Viscosity
• AKA Absolute Viscosity
• Is the actual value of Viscosity @T
• SI Units are
– kg/m-s
– cP (CentiPoise)
– Pa-S
• English Units are:
– lbf s ft-2
– lb ft-1 s-1
• The use of the Greek letter miu (μ) for the dynamic stress
viscosity is common among mechanical and chemical
engineers, as well as physicists

Kinematic Viscosity
• The kinematic viscosity (also called
"momentum diffusivity") is the ratio of the
dynamic viscosity μ to the density of the
fluid ρ.
• It is usually denoted by the Greek letter
nu (ν).

Atmospheric, Gauge, Absolute,
Vaccum Pressures
• There are many types of pressures
• Lets make them clear!
• Common “Types” of Pressure
– Atmospheric
– Gauge
– Absolute
– Vaccum

Vaccum Pressures
• Atmospheric Pressure
– Pressure exerted by the atmosphere.
– It may vary with height
– 1 atm  101,325 Pascal or N/m2

Vaccum Pressures
• Gauge pressure is the additional pressure in a
system relative to atmospheric pressure.
• It is a convenient pressure measurement for
most practical applications.

Vaccum Pressures
• PSIa  Pound Force per Squarer Inch
measured in absolute pressure
• PSIg  Pound Force per Squarer Inch
measured in gauge
33+14.7 psig = 47.7 psia

Vaccum Pressures
• Aboslute Pressure
– Total Pressure exerted P = F/A
– Commonly given as 1 atm + Gauge Pressure
– We use this pressure in many applications
• Ideal Gas Law
• Mechanical Energy Equation
33+14.7 psig = 47.7 psiaAbsolute Pressure

Vaccum Pressure
• It is a “negative” Pressure
• Actually is a Pressure in the range of:
– 0 atm < P < 1 atm
– You will feel “Suction”

Temperature vs. Abs Temperature
• Temperature  any measurement of temperature
with any given reference
– i.e. 0°C for the freezing point of water
• Abs. Temperature  measurement of temperature
with the reference of the thermodynamic value of
the ABOSULTE ZERO
– In °C  -273°C
– All molecules are in ground state

Units of Force (SI vs. English)
• The SI  1 Newton = 1 kg*m/s2
• Kilogram Force  Force Exerted by 1 kg
– F = m*A = 1kg*9.8m/s2 = 9.8N

• The English System
• Pound Force  Force Exerted by 1 lb
– F = m*A = 1lb*32.17 ft/s2 = 1 lbf
Pound Mass is
not Pound
Force!

Gravitational Constant
• Since 1 lbf = 32.17 lbm*ft/s2
• The relationship between them is
• gc = 32.17
• Be sure to change all units in equations
• There must be dimensional analysis!

• In English System

Gravitational Constant
• Personal recommendation
• USE SI system to avoid problems!

SI System

Burmese System
• The traditional Burmese units of
measurement are still in everyday use
in Burma.
• According to the CIA Factbook, Burma is one
of three countries that have not adopted
the International System of Units (SI) metric
system as their official system of weights and
measures
https://en.wikipedia.org/wiki/Burmese_units_of_measurement

End of Section 0: Review

Section 1: Why M.E.E

M.E.E Systems

M.E.E Systems
Incompressible Flow

Mechanical Energy
• Mechanical Energy  Movement and Forces
• Momentum Transport/Transfer
• Change of Velocity Profiles
• Change in position of objects
• Fluid mechanics  Movement of fluids
– Fluid  liquid + gas
– Compressible vs. Incompressible
• Avoid Heat Energy, Q

Mechanical Energy  System
• System
• Point A
• Point B
• Surrounding
• Universe

M.E.E Systems
• In Fluid Dynamics there are many “energies”
involved
– Pressure head (W = P*V)
– Velocity (Kinetic Energy)
– Due to Height (Potential Energy)
– Inlet/Outlet Energy in the System
• Pumps/Turbine
– Energy Loss due to Frictions

Mechanical Energy Equation
a
b
b
a
Emptying Filling

Pump (efficiency) Inlet Turbine (Outlet)
Friction Loss
a
b
b
a
Emptying Filling

• Pa/Pb = Pressure in A/B
• Za/Zb = Heigh of Point A/B
• Va/Vb = Velocity in Point A/B
• nWp = Energy Inlet (Pump)
• Wm = Energy Outlet (Turbine)
• hf = Energy loss due to Friction from A  B

Some NOTES

NOTE #1  [J/kg] or [m2/s2]
• J/kg [=] m2/s2
• Multiply by Mass Flow to get J/s or W
– J/kg * kg/s = J/s

NOTE #2  [m] or [ft]
• Divide by gravity to get units of “m” or “ft”
– Very common in the Pump industry

NOTE #3  Setting A&B
• Set the best A/B point
– What do you need?
– How to simplify?

NOTE #4  Special Cases
• Always Try to cancel Variables to simplify Maths
• Examples:
– Point A + B atmospheric Pressures
– Za = 1 m Zb = 2 … Use H = (2-1) = 1
– Point A is a static pond or lake Va = 0
– No Pump and no Turbine nWp = Wm = 0
Incompressible Flow

NOTE #4  Special Cases
• Bernoulli Equation
• Bernoulli Equation + Friction
• Bernoulli Equation + Pump
• Torricelli’s Law

Bernoulli’s Equation

Torricelli’s Law
• Torricelli Law is derived
from a special case in the
M.E.E.
Check out problems in the
“Practice Problems” Section

NOTE #5  Gc
• We will use SI system in general
• The english system must be corrected with the
factor “gc”
• This is the “gravitational” constant
• Check out that topic!

NOTE #5  Gc
• In SI system, gc = 1
• In English units gc = 32.17 lb-ft/s2
Bernoulli Equation Corrected for English System (gc)

NOTE #5  Gc
• The gc value is used in English Units, it kinda makes sense:
– How much force will 1 lb of Apple exert?
– Lets Call it 1 lbf (pound force)
– This is about 32 lb*ft/(lb*fs2)
– The thing is, there are many unit based on lbf (pound force)
• We use it as well in SI but way less extensive!
– How much force does 1 kg of Apple exert?
– Lets call it 1 kgf (which is about 9.8 N)
– We don’t use this… we prefer the Newton

NOTE #6  S.G & S.W.
• Recall that S.G.  Specific Gravity
• For S.W.  Specific Weigth
• Many times you might see this equation:
• In terms of S.W instead of density and gravity

NOTE #6  S.G & S.W.
• The Pressure head is defined as
– Pressure/S.W
• The Velocity head is defined as
– Velocity squared / 2*gravity
• The Potential Head
– Height alone

End of Section 1: Why M.E.E

Section 2: Kinetic + Potential Energy

Kinetic Energy
Ek = ½ * m* V2
Ek = ½*V2
Joule
Joule/kg

Kinetic Energy
• Energy due to the Velocity of the fluid
• If the fluid is incompressible…
– Density won’t change
– Volumetric Flow will be constant Q = m3/h
Ek = ½ * m* V2

Kinetic Energy:
alpha - correction factor
• This “correction” factor is defined as:
• Throughout the course, we will asume this is
equal to 1
• Ek = ½*a*V2/2  Ek = ½*V2/2

Kinetic Energy:
Relating Velocity and Pipe’s Diameter
• If volumetric Flow is constant…
– Pipe’s Diameter will change the Area of flow
– Less Area, more speed
– More Area, less speed

Kinetic Energy:
• Set a Volumetric Flow; 1 m3/h
• The Pipes are fixed, won’t change Area
• R1 > R2
– Suppose 2*R1 = R2
– R1 = 0.05 m
• Calculate:
– Volumetric Flow Rates in 1 and 2
– Area of Flow in 1 and 2
– Velocity of fluid in 1 and 2

Kinetic Energy:
• Recall that mass flow is constant
– Inlet = Outlet
– M1 = M2
• Density 1 = Density 2  Incompressible
• M = D*Q
– D1*Q1 = M1
– D2*Q2 = M2
– D1*Q1 = D2*Q2 and D1 = D2 so..
– Q1 = Q2

Kinetic Energy:
• Q = A*V
– Q1 = A1*V1
– Q2 = A2*V2
• For a circle, A = PI*D2/4
– A1 = PI*D12/4
– A1 = PI*D22/4
– Q1 = (PI*D12/4)*V1
– Q2 = (PI*D22/4)*V2
– (PI*D12/4)*V1 = (PI*D22/4)*V2
– D12*V1 = D22*V2
A1*V1 = c = A2*V2
D12/D22= V2/V1

Kinetic Energy:
• We are now able to Relate:
– Velocity  Volumetric Flow
– Volumetric Flow  Diameter
– Velocity  Diameter
– Kinetic Energy  Diameter, Velocity, Vol. Flow

Kinetic Energy
• Recall that alpha is equal to 1 most of the times

Potential Energy
Ep = m* g * H
Ep = g * H
Joule
Joule/kg

Potential Energy

Potential Energy
• Energy due to
– Gravitational field
– Relative Position
• Gravity S.I. = 9.8 m/s2
• Gravity E.U. = 32 ft ft/s2
• Position only in the Y-Axis
• Many times this will be so low, we may ignore it

Potential Energy
• Pump Inlet = 10.2 cm
• Pump Outlet = 34.5 cm
• Inlet V = 0.3 m/s
• Outlet V = 1.2 m/s
• P inlet = 1 atm
• P outlet = 10 atm

Potential Energy
• SI units:
• English Units:

Potential Energy  S.W.
• The Pressure head is defined as
– Pressure/S.W
• The Velocity head is defined as
– Velocity squared / 2*gravity
• The Potential Head
– Height alone

End of Section 2: Kinetic + Potential Energy

Section 3: Pressure Head

Pressure and Work
• Pressure changes
– P = F/A
• Change in Volume  A or H
• Change in Force
• Work = F*d
– Work = P*A*d
• Work = P*V
• Change of Pressure  Change in Work!  Energy
F = P*A
A*d = V

Pressure and Work
P = atm
P = atm

Pressure and Work
P tank = 2 atm
P outlet = 1 atm
P high < 1 atm

Pressure in “Open Systems”

Pressure and Energy
• P [=] kg*/(m*s2)
• E [=] kg m2/s2
– P  E
– kg*/(m*s2) = kg m2/s2
– C = Volume

Pressure and Energy
P = F/A*V
E = P/ρ
Joule
Joule/kg

Pressure Head Energy

End of Section 3: Pressure Head

Section 4: Inlet/Outlet Work

Section 4: Inlet/Outlet Work
• IF we want to use always positive values…
• E inlet = E outlet
• Instead of Ein - Eout = 0
• Left side “positive or inlet” Work
• Right side “negative or outlet” Work

Inlet Work
• In general  Pumps
• Any type of system that adds Work in an
external manner
– Will be affected by efficiency
• 0 % < Efficiency % < 100%
• We MULTIPLY Efficiency
• Wpump < Welectric

Outlet Work
• In general  Turbine
• Any type of system that takes out Work in an
external manner
– Will not be affected by efficiency in OUR system
• 0 % < Efficiency % < 100%
• We DIVIDE Efficiency
• Wturb = Wout

Inlet vs. Outlet Work
• If we have a 50 KW pump. How much is
actually getting the system. E = 65%
– Win = nW = 0.65*50 = 32.5 KW
• If we want to produce 6MW from the system
and the efficiency of the turbine is about 90%.
How much are we taking out?
– Wout = Wreq/n = 6MW/0.90 = 6.67 MW

• You’re Always screwed Rule
– If you need to supply eletricity to a pump system
• You will need MORE electricity than the required nWp
– If you have a turbine and want extr aelectricity
• You will have LESS electricity tan the amount being used
• For INLET WORK
– Welectricity = Wp/n
• For Outlet Work
– Welectricity = Wt*n
• NOTE: In all cases you will expend energy 

• There is no such thing as “Negative” Work
• Inlet Work  goes to the left (+)
• Outlet Work  goes to the rigth (+)
• Mathematically, you WILL have negativ
evalúes when manipulating the equation

End of Section 4: Inlet/Outlet Work

Section 5: Friction Loss  Trajectory

Friction Nature
• Friction is everywhere
• Friction may be lost in many types of forms
– Sound
– Heat
– Motion of particles
• Friction is dependent on trajectory
– More length  More Friction  More Energy loss

Friction Types
• The next equipment will make a friction loss
– Pipe Roughness
– Fittings and Valves
– Pump and Compressors

Friction Types

Friction Loss  Equation
• Since this is a loss… add it in the right SIDE
• We want positive values
• Pretty similar to Work outlet
– But we are not using it to generate energy 

Friction Loss  Calculation
• In this Section we will get the TOTAL value of
the Energy Lost due to Friction
• Check out Block 3  Friction Factor
• There we will discuss how to calculate
Friction!

Friction Loss
NOTE #1
• Many textbooks use the concept of “m” or “ft” of
friction lost
– HL
• This value is similar to Hf because it represents a
friction loss
– HL*g = Hf
• The interesting part is to compare directly how
much meters would you need to compensate
– 2 m lost due to friction  2 m higher tank
– 19.8 J/kg  2 m

Friction Loss  Exercises
• If we lost 10 J/kg of Friction (Hf) Calculate the
next
– Change in Position needed to recover
– Change in Velocity needed to recover
– Change in Pressure needed to recover
– Inlet Work needed to recover
– Outlet Work needed to recover

• Change in Position needed to recover
• g*Ha = g*Hb + Hf
• Hf = g*Ha - g*Hb
– Hf = g*(Ha - Hb)
– 10/9.8 = Ha
– Ha = 1.1 m
Assume Hb = 0

• Change in Velocity needed to recover
• Va
2/2 = Vb
2/2 + Hf
• Hf = Va
2/2- Vb
2/2
– 2Hf = Va
2 - Vb
2
– 2Hf = Va
2
– Sqrt(2*10) = Va
– 4.47 m/s= Va
Assume Vb = 0

• Change in Pressure needed to recover
• Pa/rho= Pb/rho + Hf
– 1/rho(Pa – Pb)= Hf
– (Pa – Pb)= Hf*rho
– dP = Pa-Pb= 10*1000
– dP = 10,000 Pa
Assume Water rho = 1000 kg/m3

• Pump Requirement needed to recover
• nWp = hf
– 0.72Wp = hf
– Wp = 10/0.72 = 13.88 J/kg
Assume n = 72%

• Turbine Requirement needed to recover
• It is impossible
– By definition, the turbine takes OUT energy
– You will get a negative value
• Analysis!

• Friction if f(V)
• In this section we will either
– Get the Hf value
– Get a function of Hf = f(V)
• Many times is a parabolic function
• AFD3 Block gets a deeper understanding of
the friction loss! Check it out!

Need More Problems?
Check out the COURSE
• Courses
A p p l i e d F l u i d D y n a m i c s
P a r t 1 : I n c o m p r e s s i b l e F l o w
You’ll get SOLVED problems, Quizzes, Slides, and
much more!

End of Section 5: Friction Loss  Trajectory

Section 6: Applications of M.E.E

Applications of M.E.E
• Exercises!
• Theory
– Loss of a specie  Gain of a specie
– Imagine the cases!

• Special cases
– Pa and Pb cancel each other
– Va and Vb cancel each other
– Ha and Hb cancel each other

• Pa and Pb cancel each other
– Same Pressure in tanks (A and B)
– Ambient Conditions
• Va and Vb cancel each other
– Same Pipe Diameter
• Ha and Hb cancel each other
– Point A and B are in the same height

• Pa decrease, Vb increase
• Pa increase, hf increase
• Va decrease, Vb increase, no change in hf
• Pa increase, Va increase, Pb decrease

Bernoulli’s Law
• No Pump (Inlet Work)
• No Turbine (Outlet Work)
• No Friction

Bernoulli’s Law
Exercise 4.5
Data:
ρ = 998 kg/m3
Di = 50 mm
Vi = 1.0 m/s
Pi = 100 kN/m2
Same Elevation
Do = 20 mm
Hf = neglegible
A) Velocity at exit
B) Pressure at exit (outlet)

Bernoulli’s Law
Exercise 4.5

Torricelli’s Law
• No Pump (Inlet Work)
• No Turbine (Outlet Work)
• No Friction
• No Pressure Head

Torricelli
Exercise 4.4
Apply M.E.E to the next case!
a) At what velocity will the fluid exit?

Torricelli
Exercise 4.4

M.E.E Application
• What will happen if we actually use a Pump?
• We can’t apply Bernoulli’s Equation!
• We need M.E.E!!

M.E.E
Exercise 4.6
• For the given system
A) Calculate Power Requirement for the Pump
B) Calculate the Pressure Ratio of the Pump (Pdis/Psuc)

M.E.E
Exercise 4.6

M.E.E
Exercise 4.6
b) Find the Pressure Ratio of the Pump
That is, the ΔP between Suction and Discharge

End of Section 6: Applications of M.E.E

Brief Intro…
• This was just a brief introduction of the M.E.E
• We will learn much things to add up our
analysis of the M.E.E
– Pumping
– Friction Loss
– Piping and metering of fluids
– Parallel v.s Series System
• Check out Block 6
– Incompressible Flow Application

End of AFD1
• By now you should know:
– What type of problems we are going to solve
– Incompressible Flow  M.E.E
– Kinetic energy  Velocity and Diameters
– Potential  location of object in Y-Axis
– Pressure  Pressure head means energy
– Win  Pumps and efficiency calculations
– Wout  Turbine and efficiency calculations
– Friction Loss  What it is, and its basic
– Solve Problems in Fluid Mechanics  Incompressible flow

Questions and Problems
• Check out the SOLVED & EXPLAINED problems
and exercises!
– Don’t let this for later…
• All problems and exercises are solved in the
next webpage
– www.ChemicalEngineeringGuy.com
• Courses
– Momentum Transfer Operations

Contact Information!
• Get extra information here!
– Directly on the WebPage:
• www.ChemicalEngineeringGuy.com/courses
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• www.facebook.com/Chemical.Engineering.Guy
– My Twitter:
• www.twitter.com/ChemEngGuy
– Contact me by e-mail:
• Contact@ChemicalEngineeringGuy.com

AFD1 The Mechanic Energy Equation

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