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NOTES ON FLUID MECHANICS
Peter E. ClarkDepartment of Chemical Engineering
November 12, 2008
1
Contents
1 Introduction 51.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.1 Density . . . . . . . . . . . . . . . . . . . . . 61.1.2 Specific Gravity . . . . . . . . . . . . . . . . . 61.1.3 Specific Weight . . . . . . . . . . . . . . . . . 6
1.2 Newtonian Fluids . . . . . . . . . . . . . . . . . . . . 71.2.1 Shear Stress . . . . . . . . . . . . . . . . . . . 71.2.2 Shear Rate . . . . . . . . . . . . . . . . . . . . 71.2.3 Viscosity . . . . . . . . . . . . . . . . . . . . 8
1.3 Non-Newtonian Fluids . . . . . . . . . . . . . . . . . 91.3.1 Power Law Fluids . . . . . . . . . . . . . . . . 91.3.2 Bingham Plastic . . . . . . . . . . . . . . . . . 101.3.3 Herschel-Bulkley Fluids . . . . . . . . . . . . 111.3.4 Dilatant Fluids . . . . . . . . . . . . . . . . . 111.3.5 Time Dependent Fluids . . . . . . . . . . . . . 11
1.4 Kinematic Viscosity . . . . . . . . . . . . . . . . . . . 121.5 Surface Tension . . . . . . . . . . . . . . . . . . . . . 121.6 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Fluid Statics 152.1 Basic Equation of Fluid Statics . . . . . . . . . . . . . 162.2 Pressure - Depth Relationships . . . . . . . . . . . . . 16
2.2.1 Constant Density Fluids . . . . . . . . . . . . 162.2.2 Variable Density Fluids . . . . . . . . . . . . . 17
2.3 Pressure Forces . . . . . . . . . . . . . . . . . . . . . 182.4 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Pressure Measurement . . . . . . . . . . . . . . . . . 20
2.5.1 Manometers . . . . . . . . . . . . . . . . . . . 20
2
Contents
2.5.2 Bourdon Tube . . . . . . . . . . . . . . . . . . 232.5.3 Pressure Transducers . . . . . . . . . . . . . . 24
2.6 Accelerated Rigid-Body Motion . . . . . . . . . . . . 24
3 Balance Equations 273.0.1 Equation of Continuity . . . . . . . . . . . . . 28
3.1 Control Volume . . . . . . . . . . . . . . . . . . . . . 293.2 Fluid Velocity in a Confined Region . . . . . . . . . . 30
3.2.1 Flow Regimes . . . . . . . . . . . . . . . . . . 313.2.2 Plug or Creeping Flow . . . . . . . . . . . . . 323.2.3 Laminar Flow . . . . . . . . . . . . . . . . . . 333.2.4 Turbulent Flow . . . . . . . . . . . . . . . . . 33
3.3 Unsteady-State Mass Balances . . . . . . . . . . . . . 333.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . 35
4 The First Law of Thermodynamics 374.1 Energy Transfer . . . . . . . . . . . . . . . . . . . . . 374.2 Energy Balance . . . . . . . . . . . . . . . . . . . . . 38
4.2.1 Sign Conventions . . . . . . . . . . . . . . . . 394.2.2 Potential Energy . . . . . . . . . . . . . . . . 394.2.3 Kinetic Energy . . . . . . . . . . . . . . . . . 40
4.3 Internal Energy . . . . . . . . . . . . . . . . . . . . . 404.4 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . 41
5 Bernoulli Equation 435.1 Applying the Bernoulli Equation . . . . . . . . . . . . 445.2 Bernoulli Equation With Friction . . . . . . . . . . . . 475.3 Gas Flows . . . . . . . . . . . . . . . . . . . . . . . . 485.4 Non-Flow Work . . . . . . . . . . . . . . . . . . . . . 505.5 Flow Measurement . . . . . . . . . . . . . . . . . . . 50
5.5.1 Pitot Tube . . . . . . . . . . . . . . . . . . . . 515.5.2 Static Pitot Tube . . . . . . . . . . . . . . . . 525.5.3 Venturi Meter . . . . . . . . . . . . . . . . . . 535.5.4 Orifice Meter . . . . . . . . . . . . . . . . . . 545.5.5 Rotameters . . . . . . . . . . . . . . . . . . . 55
5.6 Unsteady Flows . . . . . . . . . . . . . . . . . . . . . 565.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . 60
3
Contents
6 Fluid Friction in Steady One-Dimensional Flow 61
7 Momentum Balance 637.1 Newton’s Laws . . . . . . . . . . . . . . . . . . . . . 637.2 Control Volumes . . . . . . . . . . . . . . . . . . . . 637.3 Forces on a Control Volume . . . . . . . . . . . . . . . 657.4 Steady Flow . . . . . . . . . . . . . . . . . . . . . . . 667.5 Rotational Motion and Angular Momentum . . . . . . 73
7.5.1 Review . . . . . . . . . . . . . . . . . . . . . 73
4
1 Introduction
Outline1. Review
a) Density
b) Specific gravity
c) Specific weight
2. Newtonian Fluids
a) Stress
b) Strain
c) Strain rate or shear rate
d) Viscosity
3. Non-Newtonian Fluids
a) Pseudoplastic
b) Bingham Plastic
c) Yield pseudoplastic or Herschel-Bulkly Fluid
d) Dilatant
e) Time dependent
i. Rheopeticii. Thixotropic
4. Kinematic Viscosity
5. Surface Tension
5
1. Introduction
6. Pressure
7. Computer Problems
1.1 Review
There are a few concepts that need to be reviewed to aid inunderstanding the text.
1.1.1 Density
ρ =mass
volumeUnits :
kgm3 ;
gcm3 :
lbm
f t3 (1.1)
Density is an extremely important property of matter. The density of amaterial can be considered continuous except at the molecular level.Density can also be thought of as the constant that relates mass tovolume. This makes it easy to convert between the two.
1.1.2 Specific Gravity
SG =ρ
ρre fWhere ρre f = 1000
kgm3 ;1
gcm3 ;62.4
lbm
f t3 ;8.33lbm
gal(1.2)
Specific gravity is used instead of density to tabulate data for differentmaterials. Using the specific gravity, the density in any set of unitsmay be found by picking the reference density in the desired units.Note: When the reference density is expressed in g
cm3 , the density andspecific gravity have the same numerical value.
1.1.3 Specific Weight
γ = ρggc
Units :Nm3 ;
dynecm3 ;
lb f
f t3 (1.3)
The specific weight (1.3) is a quantity that is used frequently in fluidmechanics. In the American Engineering Series (AES) of units, it isnumerically equal to the density. The units are lb f rather than lbm.
6
1.2. Newtonian Fluids
The choice of γ (gamma) for the symbol for specific weight wassomewhat unfortunate since another important variable in fluidmechanics (strain) also uses γ . To avoid this problem, some booksuses Γ for strain. This is non-standard and can cause confusion sinceΓ is rarely used for this purpose in the literature.
1.2 Newtonian Fluids
A fluid is defined as a material that can not support a stress or as amaterial that is continuously deformed by the application of a stress.
Figure 1.1: A fluid element before deformation.
Figure 1.2: Fluid element after the application of a force acting tan-gentially on the top of the element.
1.2.1 Shear Stress
The shear stress is simply the force divided by the area over which itacts. As we will see, it has the same units as pressure.
shear stress(τ or σ) =f orcearea
Units :Nm2 ;
dynecm2 ;
lb f
f t2 ;lb f
in2 (1.4)
1.2.2 Shear Rate
Strain is a quantity commonly found in mechanics. It is the ratio ofthe change in length divided by the original length. In fluids the shear
7
1. Introduction
strain is defined as the shear displacement divided by the height of thefluid element. Strain is unitless.
strain (γ) =displacementelement height
=xy
(1.5)
Since a fluid is continuously deformed by a force acting on it, straindoesn’t mean much, but the rate of strain or shear rate (γ) is important.
strain rate or shear rate (γ) =dγ
dt=
d xy
dtUnits :
1time
(1.6)
1.2.3 Viscosity
Newton first proposed that the shear stress could be related to theshear rate by
τ = Constant γ
The constant is termed the viscosity (µ). It is a constant ofproportionality between shear stress and shear rate. Viscosity isanalogous to a modulus.
τ = µ γ =⇒ µ =τ
γUnits :
F · tarea
;dyne ·s
cm2 ;N ·sm2 or Pa ·s (1.7)
The unit (dyne ·scm2 ) is called a Poise. It is more common to use
centipoise (cp) or 0.01Poise. Water has a viscosity of 1 cp whilehoney has a viscosity of about 400 cp. It is easy to confuse Poise andcentipoise when making calculations. Remember, a Poise is equal to1, dyne ·s
1cm2 . A Pa ·s is equal to 0.1 Poise. Viscosity can also be
expressed in lb f ·sf t2 . The viscosity of water in the AES system of units
is 2.1x10−5 lb f ·sf t2 . Converting from Poise to lb f ·s
f t2 is accomplished by
multiplying Poise by 1 lb f ·sf t2 /478.8 Poise. The equivalents for
viscosity are
1lb f ·s
f t2 = 47.88N ·sm2 = 478.8 Poise = 47880 centipoise
8
1.3. Non-Newtonian Fluids
1.3 Non-Newtonian Fluids
Fluids that exhibit a nonlinear relationship between stress and strainrate are termed non-Newtonian fluids. Many common fluids that wesee everyday are non-Newtonian. Paint, peanut butter, and toothpasteare good examples. High viscosity does not always implynon-Newton behavior. Honey is viscous and Newtonian while a5W30 motor oil is not very viscous, but it is non-Newtonian. Thereare several types of non-Newtonian fluids. Figure 1.3 shows severalof the more common types.
Figure 1.3: Types of non-Newtonian fluids.
1.3.1 Power Law Fluids
Power law fluids are fluids that follow the power law (Equation 1.8)over part or all of the shear rate range. These fluids are also known aspseudoplastic fluids. Where m is the consistency index (K is also usedin the literature) with units of F ·sn
area and n is the power law or flowbehavior index. For power law fluids, n ranges from 0 to 1. While it isgreater than one for dilatant fluids. The value of m or K depends uponthe system of units. Viscosity of a power law fluid is obviously afunction of shear rate and not constant as it is for Newtonian fluids.There are two different ways to describe the viscosity: the slope of thetangent line at any point on the curve and the slope of the line drawn
9
1. Introduction
from the origin to the shear rate of interest (Figure 1.4). The latter isthe preferred method and is termed the apparent viscosity and giventhe symbol (ηa). This is short for the apparent Newtonian viscositybecause it is the viscosity that a Newton fluid would have if the line isbased on a single point measurement. Because the viscosity is afunction of shear rate it is necessary to specify the shear rate at whichthe viscosity is reported (ηa(γ)).
τ = mγn (1.8)
ηa (γ) =τ
γ=
m γn
γ=
mγ1−n (1.9)
Figure 1.4: Apparent Newtonian viscosity.
1.3.2 Bingham Plastic
Bingham plastic fluids are fluids that exhibit a yield stress (Figure 1.3.This means that the fluid will support a stress up to a point before flowbegins. Good paints are Bingham plastics. The flow behavior ofBingham plastic fluids is described by
τ = τ0 + µp γ (1.10)
Where τ0 is the yield stress (the stress that must be exceeded beforeflow begins) and µp is the plastic viscosity.
10
1.3. Non-Newtonian Fluids
1.3.3 Herschel-Bulkley Fluids
The Herschel-Bulkley formulation is a generalization of the Binghamplastic equation. In the Bingham plastic model, the viscosity isconstant after the yield stress is exceeded while the Herschel-Bulkleymodel allows for power law behavior.
τ = τ0 +m γn (1.11)
Looking at the various parameters in the Herschel-Bulkley equation,we can make the following observations for
τ = τ0 +m γn
when
τ0 = 0 =⇒ τ = mγn PowerLaw
τ0 = 0 and n = 1 =⇒ τ = µγ Newtoniann = 1 =⇒ τ = τ0 + µpγ BinghamPlastic
The Herschel-Bulkley fluid model can be reduced to the three othermodels and is therefore the most general of the simple fluid models.
1.3.4 Dilatant Fluids
Dilatant fluids are shear thickening fluids. This means that theviscosity increases with shear. There are few examples of dilatantfluids. Most dilatant fluids are concentrated slurries. The power lawcan be used to describe dilatant fluid behavior (n > 1).
1.3.5 Time Dependent Fluids
Time dependent fluids are fluids that increase or decrease in viscosityover time at a constant shear rate. The construction industry uses acement slurry that thins with time at a constant pump rate because it iseasier to pump and fills forms easily. There are two types of timedependent fluids: Rheopectic and Thixotropic.
11
1. Introduction
Rheopectic Fluids
Rheopectic fluids increase in viscosity with time at constant shearrate. There are few examples of rheopectic behavior. In the Britishliterature, rheopectic behavior is called anti-thixotropic behavior.
Thixotropic Fluids
These are fluids that lose viscosity over time at a constant shear rate.Dispersing agents for cements tend to make them thixotropic.Viscosity is recovered after the cessation of shear. In some systems,the time it takes to recover is so short that a sheared sample can not bepoured out of a container before it gets too thick to flow.
1.4 Kinematic Viscosity
The kinematic viscosity, (ν), is derived from gravity driven flowmeasurements. Usually a glass capillary viscomer is used. Kinematicviscosity can be derived from the shear viscosity (µ) by dividing µ bythe fluid density.
ν =µ
ρ
Units are areatime (
f t2
s , cm2
s , etc).
1.5 Surface Tension
Surface tension, (σ), is a property of a liquid surface. It describes thestrength of the surface interactions. The units on surface tension areForcelength . Surface tension is the driving force for water beading on awaxy surface and free droplets of liquid assuming a spherical shape.Lowering of surface tension can be accomplished by adding speciesthat tend collect at the surface. This breaks up the interactionsbetween the molecules of the liquid and reduces the strength of thesurface. Pure water has a surface tension that approaches 72dyne
cm .Surface tension can be reduced by adding surface active agents(surfactants) to the liquid.
12
1.6. Pressure
1.6 Pressure
Pressure is defined as force divided by the area that the force acts overand therefore has units of F
A . It can be a result of an applied force (forexample pumping) or hydrostatic (weight of a column of fluid). Thetotal pressure is the sum of the applied and hydrostatic pressure.Pressure will be discussed in more detail in the next chapter.
P = Papplied +Phydrostatic
13
2 Fluid Statics
Outline1. Basic Equation of Fluid Statics
2. Pressure - Depth Relationships
a) Constant density fluids
b) Ideal gases
3. Pressure
4. Pressure Vessels and Piping
5. Buoyancy
6. Pressure Measurement
a) Manometers
b) Pressure gages
c) Pressure transducers
7. Static Pressure Head
8. Acceleration
15
2. Fluid Statics
2.1 Basic Equation of Fluid Statics
The pressure at a point in a static fluid is the same in all directions.What this means that the pressure on the small cube in Figure 2.1 isthe numerically the same on each face as the cube shrinks to zerovolume. At the surface, the pressure is zero gage. Since depthincreases in the downward z-direction, the sign on the specific weightin equation 2.1 is negative.
Figure 2.1: Hydrostatic pressure
A simple force balance and some algebra yields
dPdz
=−ρ g =−γ (2.1)
For angles, (θ), that deviate from vertical the equation becomes
dPdz
=−ρ g cosθ =−γ cosθ (2.2)
2.2 Pressure - Depth Relationships
2.2.1 Constant Density Fluids
To establish the relationship between pressure and depth for aconstant density fluid, the pressure - position relationship must beseparated and integrated
16
2.2. Pressure - Depth Relationships
dPdz
=−ρ g =⇒∫
dP =∫−ρ gdz
P =−ρ gz =−γ z (2.3)
Equation 2.3 is one of the most important and useful equations in fluidstatics.
Example 2.1. A 5000 f t well is filled with a drilling mud thathas a specific weight of 11.2 lb f
f t3 . What is the pressure in psig atthe bottom of the well?
h = ∆z = 0 f t−5000 f t =−5000 f t
P =−hγ
P =−h11.2 lb f
gal|7.48gal
t3 | f t2
144 in2
P =−(−5000 f t) |
11.2 lb f
gal|7.48gal
f t3 | f t2
144 in2 = 2909 psig
2.2.2 Variable Density Fluids
Most fluids are relatively incompressible, but gases are not. Thismeans that the density increases with depth so we cannot use thespecific weight as a constant in determining the pressure. If weassume an ideal gas, then the density is given by
ρ =PMRT
where M is the molecular weight, T is the absolute temperature, and Ris the gas constant in appropriate units.Replacing ρ in the differental equation (d p/dz =−ρzz) we get
17
2. Fluid Statics
dPdz
=−ρ g =⇒ dPdz
=−PMRT
g (2.4)
d pP
=−gMRT
dz =⇒∫ P2
P1
1P
dP =−gMRT
dz (2.5)
Which yields
ln(P2
P1) =−gM
RT(z2− z1) =⇒ P2 = P1 e−
gMRT ∆z (2.6)
Because the it is possible for the temperature to be non-uniform andgases are poor conductors of heat, the correct equation to use is
P2 = P1
(1− k−1
k·gM∆zRT1
) kk−1
(2.7)
T2 = T1
(1− k−1
k·gM∆z
RT
)(2.8)
Where k is the ratio of the heat capacities.
2.3 Pressure Forces
Pressure acting on a surface exerts a force. This force can becalculated from
dF = PdA =⇒∫
dF =∫
PdA =⇒ F =∫
PdA
For constant pressureF = PA (2.9)
Example 2.2. Using the result of the previous example(P = 2909 psi), what is the force acting on the bottom of thehole if the hole is eight inches in diameter?
F = PA
F = 2909lb f
in2π
4(8 in)2 = 146222 lb f
18
2.4. Buoyancy
Example 2.3. You are the proud possessor of a 200 f t tall watertower. At the base of the tower (1 f oot above the ground) is aseven foot tall by two foot wide access panel that is held in placeby bolts. What is the force on the plate?
The pressure at the bottom of the plate is given by
P = 199 f t γw
P = 12418 lb f
The pressure at the top of the plate is given by
P = 192 f t γw
P = 11981 lb f
Since the panel is rectangular, the average pressure can be used.
P =−(−192 f t +(−199)) f t2
γw = 12199lb f
f t2
F =12199 lb f
f t2 |7 f t |2 f t= 170786 lb f
2.4 Buoyancy
Remembering that in a liquid at rest the pressure at a point is the samein all directions, a body submerged in the liquid will experience aforce on the top and bottom of the body. Because the body is finite inlength, the force on the bottom will be greater than the force on thetop. The easy way to think about the magnitude of the buoyant forceis that it is equal to the weight of the liquid displaced. It is probablybetter to consider it as a force balance. To see how this works we willuse a length of pipe suspended in a well.
19
2. Fluid Statics
Example 2.4.A 5000 f t well is filled with a drilling mud that has a specificweight of 11.2 lb f
gal . A 1000 f t piece of pipe (od = 8 in, id = 7 in)is submerged at the surface with only the top of the pipe exposed.What is the buoyant force on the pipe?
h = ∆z = 0 f t−1000 f t
Pb =−−1000 f t |12 inf t|11.2 lb f
gal|7.48gal
f t3 | f t3
(12 in)3 = 582lb f
in2
FB = Pbπ
4((8 in)2− (7 in)2)= 5940 lb f
What happens if the bottom of the pipe is at 3500 f t.
Pb =−−3500 f t |12 inf t|11.2 lb f
gal|7.48gal
f t3 | f t3
(12 in)3 = 2036lb f
in2
Pt =−−2500 f t |12 inf t|11.2 lb f
gal|7.48gal
f t3 | f t3
(12 in)3 = 1454lb f
in2
FB = (Pb−Pt)π
4((8 in)2− (7 in)2)= 5940 lb f
2.5 Pressure Measurement
2.5.1 Manometers
There are numerous ways that pressure can be measured. Twoexamples are shown in Figures 2.2 and 2.3.
20
2.5. Pressure Measurement
Figure 2.2: Open-end manometer.
Figure 2.3: Differential manometer.
Most books, your’s included, present a manometer equation, whichfor simple manometers is easy to use. For a manometer having morethan one fluid and a complex topography, the following three rules arefar easier to use.
Starting at one end of the manometer
1. Work through the system considering only one fluid at a time
2. Add pressure differences as you proceed down through a fluidfrom the starting point (or subtract when working upwards).
3. Move horizontally through a fluid without change in thepressure
21
2. Fluid Statics
Set the resulting equation equal to the pressure at the other end of themanometer.
Applying these rules is simple. In Figure 2.4, the tank and tube form asimple manometer. To find the pressure at point A, we start by writingthe pressure at point A as PA. Appling rule 3, we move from point Ato point B. Since both points are at the same level, the pressures areequal.
Figure 2.4: A tank with a standpipe (simple manometer).
Rule 2 says to subtract going up so hγ is subtracted from PA and sincethere is no flow PA−hγ must be equal to the pressure at the top of thetube (Patm). This produces the equation
PA−hγ = Patm
For more complicated manometers containing more fluids and morelegs, just add terms to the equation to describe each one.
22
2.5. Pressure Measurement
Example 2.5. Looking at the manometer in Figure 2.3 we can,by applying the rules, write an equation for the manometer.Begin by designating the pressure at point A as PA.Applying rule 3, move horizontally to point F where the pressureis equal to PA.Applying rule 2, add the contribution to the pressure from thecolumn of fluid (F → E), (b+h)γw.
PA +(b+h)γw
Applying rule 3, move horizontally from E to DApplying rule 2, subtract the contribution from the mercury col-umn, −hγHg
PA +(b+h)γw−hγHg
Applying rule 2, subtract the contribution from the water columnC→ B, −(b+a)γw
PA +(b+h)γw−hγHg− (b+a)γw
Since there is no flow, set the equation = to the pressure at B,PB.
PA +(b+h)γw−hγHg− (b+a)γw = PB
This becomes the manometer equation. Some algebra allows usto determine the pressure difference between points A and B, ∆P
2.5.2 Bourdon Tube
Bourdon tube pressure gages (Figure 2.5) consist of a closed-endmetallic tube that is curved in to almost a circle. Application ofpressure causes the tube to ’straighten’ a small amount and the motionis converted into a rotation of the needle on a dial. These gages aresimple and rugged, so they are used in a wide range of applications ina plant environment.
23
2. Fluid Statics
Figure 2.5: Bourdon tube pressure gage
2.5.3 Pressure Transducers
A transducer is a device that turns a mechanical signal into anelectrical signal or an electrical signal into a mechanical response.There are a number of ways to accomplish this.
1. Strain gage
2. Capacatance
3. Variable reluctance
4. Optical
2.6 Accelerated Rigid-Body Motion
Changes in motion bring about acceleration or deceleration. Agasoline truck accelerating or decelerating will cause the fluid in thetanks to move to one end of the tank or the other. The major change inthe analysis of pressure is that the force balance becomes equal to themass times accleration rather than zero.
dPdz
=−ρ
(g+
d2 zdt2
)(2.10)
This equation can be integrated for constant density fluids to yield
P2−P1 =−ρ
(g+
d2zdt2
)(z2− z1) (2.11)
24
2.6. Accelerated Rigid-Body Motion
For gauge pressure
P =−ρ h(
g+d2zdt2
)=−hγ +
(ρ
d2zdt2
)(2.12)
Equation (2.12) has the static pressure term (−hγ) and anacceleration term,
(ρ
d2zdt2
). Evaluating the static pressure term is
relatively easy, the acceleration term can be a little more difficult.
25
2. Fluid Statics
26
3 Balance Equations
Almost all of engineering comes down to balances – mass,momentum, and/or energy. We have already seen in developing themanometer equation that it is nothing more than a pressure balance.In the beginning chemical engineering course, the general balanceequation was presented. Evaluating the various components of theequation is the challenge in various aspects of chemical engineering.
Flowin−Flowout +Generation−Consumption = Rateo f Accumulation
Simple mass balances without chemical reactions leave out thegeneration and consumption terms.
min− mout = Rateo f Accumulation
So, if the rate of accumulation is zero (steady-state), we have
min = mout
Based on balance arguments, the following important relationshipscan be derived.
Q = vA (3.1)
In Equation 3.1, v is the velocity and A is the cross-sectional area. Forconstant density fluids
Qin = Qout
sovin Ain = vout Aout (3.2)
27
3. Balance Equations
3.0.1 Equation of Continuity
Because mass must be conserved
m1 = m2
and
m = ρ vA (3.3)
the continuity equation becomes
ρin vin Ain = ρout vout Aout (3.4)
or∑ρ vA = 0 (3.5)
The equation of continuity can also be written in terms of the specificweight.
Example 3.1. Water flows into a cylindrical tank throughpipe 1 at the rate of 20 ft/s and leaves through pipes 2 and3 at the rate of 8 ft/s and 10 ft/s, respectively. At 4 wehave an open air vent. The following are the inside diame-ters of the pipes: D1 = 3 in, D2 = 2 in, D3 = 2.5 in, and D4 = 2 in
Solution1. Calculate the mass flow rates in using m = ρ vA
m1− m2− m3 = mtank
ρ v1 A1−ρ v2 A2−ρ v3 A3 = ρ vtank Atank
It is clear that the value of ρ is the same on both sides of the equationso it can be eliminated. Since area = π
4 d2 and the area term is on bothsides of the equation, the π
4 term cancels.
20 f ts
(3 in | f t
12 in
)2
− 8 f ts
(2 in | f t
12 in
)2
− 10 f ts
(2.5 in | f t
12 in
)2
= vtank (12 f t)2
28
3.1. Control Volume
Figure 3.1: Example: Flow into and out of a tank.
Solving for vtank yields
vtank = 0.148f ts
Later it will be necessary to think of the vtank term as the differential(dhtank
dt
). Using Equation 3.2 and assuming that the air in the tank is
incompressiblevtank Atank = v4 A4
0.148 f ts|π4|(
2 f t)2
= v4|π
4|(
2in | f t12 in
)2
Solving for v4 gives
v4 = 21.38f ts
3.1 Control Volume
The control volume is an imaginary sub-volume in a flow systemthrough which flows occur. It can be denoted with a dotted or dashedline enclosing the volume (Figure 3.2).This may seem like an unnecessary complication, but with anothersimple addition the control volume concept becomes quite useful. Ifthe area through which the flow occurs is declared to be a vector
29
3. Balance Equations
Figure 3.2: Control volume
quantity directed outward from the control volume(~A = A~n
), it can
be combined with the velocity vector using the dot product to producethe proper sign for the flow when Equation 3.5 is used
(∑ρ v ·~A = 0
).
The dot product produces v ·~A =⇒ vA cos(θ), where θ is the anglebetween the area and velocity vector. Since the cosine of zero is oneand the cosine of 180 is (−1), the direction of flow is set relative tothe area vector. It is important to draw the control volume such thatthe velocity vector is perpendicular to the control volume surface(Figure 3.3).
Figure 3.3: Control volume with area and velocity vectors
The previous example would become
ρ v1 A1 cos180+ρ v2 A2 cos0+ρ v3 v3 cos0+ρ v4 A4 cosθ = 0
Where the sign of the final answer would determine the direction offlow. Plugging in numbers we find that A4 cosθ is positive so the flowis outward.
3.2 Fluid Velocity in a Confined Region
There are two ways to consider the velocity of a fluid in a flowchannel. The average velocity is probably the most straightforward. Itis simply the volumetric flow rate divided by the area of the conduit.
vave =QA
(3.6)
30
3.2. Fluid Velocity in a Confined Region
In reality, unless the fluid is flowing very fast or very slow, there is asignificant velocity distribution that for flow of a Newtonian fluid in apipe is parabolic (Figure 3.4). One of the basic assumptions of fluidmechanics is that the fluid velocity at the wall is zero i.e. no slip.With the velocity equal to zero at the wall, the fluid velocity is at amaximum at or near the centerline of the flow (depends upon thegeometry of the flow channel).
Figure 3.4: Velocity distribution for a Newtonian fluid in laminar flow.
3.2.1 Flow Regimes
Depending upon the fluid velocity, the fluid can flow in different flowregimes. At low flow rates, the fluid particles move in a uniformfashion along the channel with only a small difference in the velocitybetween the wall and the centerline. As the velocity increases, theflow remains uniform, but the difference between the velocity at thecenterline and the wall becomes large. This flow regime is termed thelaminar region or simply laminar flow. At some point the velocityincreases enough that the flow becomes chaotic and the uniformity inthe flow disappears (termed turbulent flow). The velocity profilebecomes flat and the flow is strongly mixed. Flow regimes can beseparated by the value of the Reynolds number. This is a
31
3. Balance Equations
dimensionless group that represents the ratio of the inertial forces tothe viscous forces and is given by
NRe =ρ d v
µ
where d is the characteristic diameter of the flow channel, ρ is thefluid density, v is the velocity, and µ is the fluid viscosity.
3.2.2 Plug or Creeping Flow
At extremely low flow rates, the fluid is normally in plug or creepingflow. This flow regime has a flat velocity profile with much smallerdifference between the velocity of the fluid flowing near the wall andat the centerline (Figure 3.5). Experimentally, it is a difficult flowregime to maintain. This flow regime is mathematically simpler thanthe other two flow regimes so it has been the subject of a great deal ofstudy. Reynolds numbers are typically less than one for plug orcreeping flow.
Figure 3.5: Velocity distribution for a Newtonian fluid in plug flow andlaminar flow.
32
3.3. Unsteady-State Mass Balances
3.2.3 Laminar Flow
As velocity increases above the plug flow limit, a velocity profilesimilar to that labeled as laminar in Figure 3.5. The Reynolds numberrange for laminar flow is one to 2,100 (it may be somewhat higher innon-Newtonian fluids). In steady (time derivative equal to zero),uniform flow a fluid particle will flow along the same streamline. Thismeans that there is no mixing of fluid across streamlines in laminarflow. In extreme cases where vibrations have been excluded, thelaminar flow regime can extend to a Reynolds number of 60,000 or so.
3.2.4 Turbulent Flow
Turbulent flow profiles exhibit a flat velocity profile with a thinlaminar layer near the wall. The motion of fluid particles in turbulentflow is chaotic and the flow is strongly mixing. If you have everwatched smoke rise from a cigarette, you should have noticed that itrises in a narrow channel (laminar flow) and then breaks into a plumewhen the velocity becomes high enough to reach the turbulent flowregime. Almost all flow in water pipes, air ducts, and pipelines isturbulent.
3.3 Unsteady-State Mass Balances
There are a number of real world problems that are not steady-stateproblems. What this means is that the rate of accumulation term in theequation below is not zero but it is equal to dm
dt .
min− mout = Rateo f Accumulation
So the equation becomes
min− mout =dmdt
Example 3.7 in de Nevers needs some explanation. In this example, atank that is initially full of air is being pumped down with a vacuumpump. At a constant volumetric flow rate (Q = 1 f t3
min ), how long does it
33
3. Balance Equations
take to reduce the pressure to 0.0001 atm? As with any balanceproblem, the best place to start is by writing the balance equation
min− mout =dmdt
In this system, there is no flow in so the min term is zero and theequation reduces to
dmdt
=−mout
At this point, use the definition of density and rearrange the equationto find the mass of gas in the system using the density and systemvolume
msystem = ρsystem V system
Taking the time derivative of both sides of the equation yields(dmdt
)system
= V systemdρsystem
dt+ρsystem
dV system
dt
but the system volume is constant so the last term is equal to zero andthe equation becomes(
dmdt
)system
= V systemdρsystem
dt
Assuming that the gas is not compressible, m = ρ V , however, we areinterested in the mass flow rate m so Q is used instead of V , yielding
mout = Qout ρout
Since Q is constant ρout = ρsystem which leads to
mout = Qout ρsystem
We can now write the differential equation for the system
V systemdρsystem
dt=−Qout ρsystem
34
3.4. Summary
Separating the variables yields
dρsystem
ρsystem=− Qout
V systemdt
Integrating the equation yields
ln(
ρsystem f
ρsystemi
)=
Qout
V system∆t
Remembering that
ρgas =PMRT
this equation can be solved for ∆t.
3.4 Summary
Mass balances are extremely important in fluid flow. Missing massusually means that there is a leak. Using the equation of continuityseems difficult at first and dealing with the dot product of velocity andarea
(v ·~A)
appears to add an unnecessary complication, learninghow to set up problems using this formulation will help to minimizemistakes in the long run. In Chemical Engineering, unsteady statemass balances are important. It is worth spending some time to try tounderstand them.
35
4 The First Law of Thermodynamics
In the last chapter, we covered mass balances. As a prelude to theBernoulli equation in the next chapter, we need to discuss energybalances. The energy balance, like the mass balance, does not havecreation or destruction terms and so the general balance equationreduces to
f lowin− f lowout = Accumulation
There are several forms of energy that we must deal with in fluidmechanics.
1. Kinetic energy – energy due to motion
2. Potential energy – energy due to position
3. Internal energy – energy of the atoms and molecules that makeup a system
4. Surface energy – energy due to surface interactions
The other forms of energy are generally not useful in fluid mechanics,but may find specialized uses in areas such as magentohydrodynamics.
4.1 Energy Transfer
Energy may be transfered through heat flow or work. Heat flows fromhot bodies to colder bodies. This form of transfer forms the basis forthermodynamics. Work is defined as
Work =∫
f orce ·d(distance) =∫
F dx
Work can also be done by rotating shafts.
37
4. The First Law of Thermodynamics
4.2 Energy Balance
Returning to the general balance equation and remembering thatenergy cannot be created or destroyed we get
f lowin− f lowout = Accumulation
Your book uses lower case symbols for specific values as opposed tothe carrot or hat above the symbol. For instance
u≡ U
It also used upper case symbols for the non specific values as shownbelow
U = mu
Energy may enter a system in several ways. It may be carried in acrossthe flow boundary by mater entering or leaving the system. Heattransfer to or from the system is also a conduit for energy transfer.Work may done on or by the system can also serve as a transfermechanism. Figure 4.1 shows an example of each of the mechanisms.
Figure 4.1: Energy input from flow, shaft work, and heat transfer
So the overall energy balance for the system becomes
d[m(u+ pe+ke)]sys = [(u+ pe+ke)indmin+dQin+dWin−[(u+ pe+ke)outdmout +dQout +dWout ]
38
4.2. Energy Balance
This equation can be simplified by defining
dQ = dQin−dQout
anddW = dWin−dWout
yielding
d[m(u+ pe+ke)]sys = [(u+ pe+ke)indmin−[(u+ pe+ke)outdmout +dQ+dW(4.1)
4.2.1 Sign Conventions
Sign conventions have always been a bit confusing. It seems difficultto imagine that one can just arbitrarily pick a sign for flow into or outof the system and get the correct answer. But, it works as long as youpick a sign and stick with it. The sign convention in your book is
• Work done on the system (flowing in) is positive
• Heat transfered to the system (flowing in) is positive
This results indQ+dW = 0
4.2.2 Potential Energy
The specific form of the potential energy equation simplifies to
pe = gz
If one looks at a flowing system, the specific form of the potentialenergy equation becomes
∆pe = pein− peout = g(zin− zout)
To get the total potential energy change we need to multiply thespecific quantity by the mass of the system.
∆PE = m∆pe
39
4. The First Law of Thermodynamics
4.2.3 Kinetic Energy
The kinetic energy term reduces to
ke =v2
2
or
KE = mke = mv2
2
For systems with velocity change
∆KE = m(
v2in2− v2
out2
)
4.3 Internal Energy
For a closed system of constant mass with no kinetic or potentialenergy, the internal energy can be written as
mdu = dQ+dW
Integration yields
U = mu = Q+W + constant
The constant is based on some reference point. Typically, it is water atthe triple point where u is set equal to zero.
4.4 Work
Work on a compressible system is given by
dW = Fdx = PAdx =−PdV
Since the volume decreases, the sign is negative. Injection work is thework required to inject a mass across system boundaries. Rather than
40
4.5. Summary
deal with the injection work and the internal energy, the enthalpy(h = u+PV ) is used. So that the energy balance equation becomes
d[
m(
h+gz+v2
2
)]sys
=(h+gz+
v2
2
)in
dmin
−(
h+gz+v2
2
)out
dmout
+ dQ + dW
(4.2)
4.5 Summary
The first law can be used to analyze energy flow into and out of asystem. Equation 4.2 provides the basis for the Bernoulli equationdiscussed in the next chapter.
41
5 Bernoulli Equation
One of the most useful equations in fluid mechanics is the Bernoulliequation (BE). Your book presents it in the differential form
∆
(Pρ
+gz+v2
2
)= 0
where ∆P is Pout−Pin. Looking at the units we find
Fl2 |
l3
mass+
lt2 |
l+
l2
t2 |2In order to straighten this mess out, it is necessary to divide ρ bygc orto express force in terms of mass×acceleration. Either way yieldsunits of l2
t2 .Fl2 |
l3
mass|mass l
t2 F+
lt2 |
l+
l2
t2 |2There is a better way to think about the Bernoulli equation. Normally,the BE is written between two points in the flow.
P1
ρ+gz1 +
v21
2=
P2
ρ+gz2 +
v22
2(5.1)
Neglecting friction and work and dividing both sides by g theBernoulli equation may also be written as
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g(5.2)
This form has the advantage of having units of length or head.Otherwise, ρ in Equation 5.1 has to be divided by gc to make the unitswork out.
43
5. Bernoulli Equation
5.1 Applying the Bernoulli Equation
In applying the BE, it is best to pick the two points (1 and 2) to zeroout as many terms as possible. For instance picking the two pointsalong the same horizontal line eliminates z1 and z2. As an example,consider a tank full of water that is draining from an opening near thebottom of the tank (Figure 5.1)By selecting points 1 and 2 (point 2 is just at the exit) as shown in
Figure 5.1: Bernoulli equation tank draining example
Figure 5.1, z1 and z2 are equal and cancel each other out. V1 is smallrelative to V2 and can be ignored and P2 is equal to Patm so for gaugecalculations it is equal to zero. We are left with P1 and v2 to evaluate.The position of the points can be selected in a slightly differentmanner to make the problem even easier as shown in Figure 5.2.In this case, P1, z1, V1, and P2 are all equal to zero. This leaves z2 and
Figure 5.2: Bernoulli equation tank draining example
v2 to be evaluated. The quantity, z2, is just the distance from the
44
5.1. Applying the Bernoulli Equation
surface of the liquid to the midpoint of the flow and is equal to (−h).Evaluating the velocity, v2
0+0+0 = 0+(−h)+v2
22g
Solving for v2
v2 =√
2gh (5.3)
Which is known as Torricelli’s equation.
Example 5.1. Example: In the venturi meter shown below, theD = 40cm and d = 10cm. What is the water discharge (flow rate)in the system, if the pressure at A is 100kPa, Patm = 101.325kPa,the water temperature is 10◦C, and cavitation has just started atpoint 2?
Figure 5.3: Venturi tube.
Solution:
• The pressure at point 1 is given byP1 = 100,000Pa+101,325Pa
• The pressure at point 2 is set by the vapor pressure of water at10◦C. P2 = 1230Pa
• Since the points are along the same horizontal line, z1 = z2.
45
5. Bernoulli Equation
Writing the Bernoulli equation between points 1 and 2
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g
canceling the elevation terms and rearranging the equation we get(P1
γ− P2
γ
)=
v22
2g−
v21
2gNow we must express v2 in terms of v1.
v1A1 = v2A2 =⇒ v2 =v1A1
A2
and solve the BE equation for V1
(P1
γ− P2
γ
)=
(v1
A1A2
)2
2g−
v21
2g
v21
2g=
(P1γ− P2
γ
)(
A1A2
)2
V1 =
√√√√√√2g
(P1γ− P2
γ
)(
A1A2
)2
substituting in the numeric values yields
v1 = 1.25
Since we know Q = vA
Q = 0.157m3
s
It should be noted that cavitation is just boiling. When the pressureequals the vapor pressure of the liquid at the specified temperature,boiling occurs. So when a problem specifies cavitation, the pressureat the point of cavitation is fixed by the vapor pressure of the liquid atthe specified temperature.
46
5.2. Bernoulli Equation With Friction
5.2 Bernoulli Equation With Friction
Up to now we have neglected frictional forces. These forces increasepressure drop and in some cases cause heating. The Bernoulliequation can be modified to include friction by adding a friction term(F ) to the right side of the equation. Note: In deNevers the frictionterm has a negative sign because he has rearranged the equation toyield the ∆ values to be final state minus initial state i.e. P2−P1 sodon’t be confused. The Bernoulli equation becomes
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g+
F
g(5.4)
The friction term F takes the form of a constant times(
v2
2g
).
Frictional forces arise from simple flow, flow through valves, elbows,orifices, etc. Values for the constant can be obtained from tables orfrom the friction factor charts.
Example 5.2. Fluid flowing through the pipe and elbow shownbelow exhibits frictional losses both in the pipe and the elbow.The pipe friction will be treated in Chapter 6 and will be ne-glected in this example.
Figure 5.4: Flow through an elbow.
The Bernoulli equation for this flow system becomes
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g+
F
g
47
5. Bernoulli Equation
If the elbow is in plane perpendicular to the page, z1 = z2. Since thereis no change in velocity, v1 = v2. So the Bernoulli equation reduces to
P1
γ=
P2
γ+
F
g
which simplifies toP1−P2
γg = Fb
Measuring ∆P then allows us to calculate the friction associated withthe elbow. If the flow rate is known then the constant, Kb in theequation below can be evaluated.
F = Kv2
2
5.3 Gas Flows
Gasses at low velocities can be treated as incompressible fluids. Thepressure drop must also be low or the change in volumetric flow rateand thus, the fluid velocity will cause problems. This means that theBE can be used in many situations that are of interest to ChemicalEngineers.
When gases flow through a venturi meter, the velocity increases as thepassage gets smaller and decreases as passage returns to the originalsize. Figure 5.5 is an idealized venturi tube with sharp changes in theshape of the tube. Normally, the changes are more gradual, withoutabrupt changes (like those shown in the figure). This means that thepressure will decrease as the velocity increases on the left side of therestriction and recover on the right side of the restriction. This can beseen in Figure 5.6. Pressure recovery on the downstream side of arestriction is a common occurrence. To understand this, we need tolook at a cross-section of the flow in the venturi tube at several points
48
5.3. Gas Flows
Figure 5.5: Idealized venturi tube.
Figure 5.6: Pressure profile in a venturi tube.
Figure 5.7: Flow in a venturi tube.
49
5. Bernoulli Equation
At the first increment, ∆x, the pressure, P1, must be greater than P2for flow to occur. The increment near 2 in Figure 5.7 is characterizedby an increase in velocity that necessitates an increase in the pressuredifferential across the increment. In the third increment, the velocityis decreasing and P3 must be greater than P2 to provide the forcenecessary to decelerate the flow.
5.4 Non-Flow Work
In the previous chapter, it was shown that the flow work could beburied in the enthalpy. This left all of the non-flow work to beaccounted for. Non-flow work is often called shaft work, since in fluidflow it is usually the result of a shaft turning. The Bernoulli equationcan be extended to include the work term by adding it to the left sideof the equation. Again be careful when comparing Equation 5.5 withthe equation in your book.
P1
γ+ z1 +
v21
2g+
dWn f
g=
P2
γ+ z2 +
v22
2g+
F
g(5.5)
The work term consists of all of the non-flow work. It includes workdone on the system by stirrers, mixers, pumps, etc. or work done bythe system turning turbines, shafts, or other rotating machinery. Wewill return to non-flow work in Chapter 6.
5.5 Flow Measurement
The Bernoulli equation can be exploited to measure fluid flow rates.Before we explore the different ways that the BE equation can beused, we need to talk about the nature of flow around bodies. For afixed body in a flow, the fluid will approach the body and flow aroundit as shown in Figure 5.8. The dark point in the center of the bodyrepresents the stagnation point. At this point, the flow has a velocityof zero i.e. the flow is stagnant. Since the fluid velocity at thestagnation point is zero, it gives us a point in the system to anchor theBE equation.
50
5.5. Flow Measurement
Figure 5.8: Stagnation point.
5.5.1 Pitot Tube
A pitot tube is a simple device used to measure flow rates. It worksfor both liquid and gas flows. The device, in its simplest form,consists of a small diameter hollow tube bent into the shape of a L.Usually the upstream opening is smaller than the diameter of the tube.The height, above the center-line, of the fluid in the in the vertical legof the tube is related to the velocity of the fluid in the flow. Anexample is shown in Figure 5.9.
Figure 5.9: Pitot tube
51
5. Bernoulli Equation
Writing the Bernoulli equation between points 1 and 2
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g
and zeroing out z1, z2, and v2 we get
P1
γ+
v2
2g=
P2
γ
Assigning values to the various parameters
P1 = Patm +bγ
P2 = Patm +(b+h) γ
Substituting into the Bernoulli equation, neglecting friction, andsolving for v1 we obtain
v1 =√
2gh (5.6)
5.5.2 Static Pitot Tube
Some books refer to the first example of a pitot tube as a stagnationtube and the static pitot tube (pitot static tube) as a pitot tube. We willuse deNevers nomenclature. The difference between the pitot tubeand the static pitot tube is thesmall opening on the side of the submerged part of the tube. Unlikethe stagnation tube (Figure 5.9), a direct measurement of ∆P ispossible. Using the Bernoulli equation and neglecting friction,
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g=⇒ P1
γ+
v21
2g=
P2
γ
solving for v1
v1 =
√2g
∆Pγ
=
√2
∆Pρ
(5.7)
52
5.5. Flow Measurement
Figure 5.10: Pitot static tube.
Figure 5.11: Venturi meter
5.5.3 Venturi Meter
We have previously discussed the venturi tube, but not as a method offlow measurement. It has the advantage of no moving parts and if it isdesigned properly friction can be ignored.The Bernoulli equation can be written between points 1 and 2. Forhorizontally mounted tubes, the elevation terms can be zeroed out.
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2gIf the diameter of the tube at 1 is d1 and the diameter at 2 is d2, thevelocity at point 1 can be expressed in terms of the velocity at point 2by solving v1 A1 = v2 A2 for v1. Furthermore, the area ratio can beexpressed as the ratio of the squares of the diameters. This leads tothe expression of the Bernoulli equation
53
5. Bernoulli Equation
P1
γ+
(v2
d22
d21
)2
2g=
P2
γ+
v22
2g
This equation can be solved for v2
v2 =
√√√√ 2 (P1−P2)
ρ
(1− d2
2d2
1
) (5.8)
Application of the Bernoulli equation to the venturi tube in thismanner ignores friction and other important factors. This results in asmall error in the measured flow rates. To correct for this error, themeters are calibrated and a calibration constant provided. This resultsin the corrected venturi tube equation 5.9.
v2 = Cv
√√√√ 2 (P1−P2)
ρ
(1− d2
2d2
1
) (5.9)
5.5.4 Orifice Meter
Orifice meters are, in some respect, an extreme example of a venturimeter. The orifice meter has an abrupt change in diameter rather thanthe more gradual change in the venturi meter. An orifice meter has aplate, usually removable, with a small hole machined in the center.The upstream side of the hole is flat, while the downstream side iscone-shaped. One of the common mistakes in installing an orificemeter is reversing the plate. This results in erroneous measurements.Gas measurements made with orifice meters are analyzed using acomplex formula (Figure 5.12) that goes far beyond the Bernoulliequation. The full orifice meter equation even includes the localgravitational acceleration!
Needless to say the full equation is seldom used, but it is easy to seewhy a great deal of accuracy is needed. Consider the case of a naturalgas well flowing into a pipeline. If the well is flowing 5MMc f d (5million cubic feet/day) and there is a 0.1 percent error in the meter
54
5.5. Flow Measurement
Figure 5.12: Full orifice meter equation
reading, the gain or loss in gas sales is about thirty dollars per day oreleven thousand dollars per year at the current price of natural gas. Fora field that produces a billion cubic feet per day (there are fields thatproduce this much) the difference is in excess of two million dollars!
5.5.5 Rotameters
A rotameter is a simple device that is used when a high degree ofaccuracy is not needed. The gas or liquid flow is used to levitate a ballof known diameter and density. Usually the rotameter is calibrated sothat a measured position of the ball corresponds to a flow rate. Caremust be exercised when using rotameters. They are most accurate
55
5. Bernoulli Equation
when the ball is located in about the middle two-thirds of the meter.Measurement error increases at the ends.
5.6 Unsteady Flows
There are a number of situations where unsteady flows areencountered. The simplest is the draining of a tank where the flowrate varies with the height of the fluid in the tank. The Torricelliequation (5.3) really only gives you a snapshot of the flow rate intime. If the tank diameter is large the flow rate changes slowly, but ifit is small the flow rate change can be significant.
For a simple tank (Figure 5.13) Since Torrecelli’s equation gives the
Figure 5.13: Unsteady flow example.
instantaneous flow rate we can write the velocity at 2 as
v2 =√
2gh
and if we solve v1 A1 = v2 A2 for v2, we now have
v1A1
A2=√
2gh
56
5.6. Unsteady Flows
v1 is only the rate at which the height is dropping(dh
dt
). Rearrange
and solve for(dh
dt
)yields the differential equation
− dhdt
=A2
A1
√2g,h (5.10)
This equation can be solved by separating the variables and integratingor by using the ordinary differential equation solver in MatLab.∫ h2
h1
1√h
dh =A2
A1
√2g∫ t2
t1dt (5.11)
When the tank has parallel sides this problem is relatively easy,however, if the tank is cone shape the area of the surface becomes afunction of height. Finding the functional form of the diameter is firstproblem and integrating it is the second. For the cone-shaped tankshown in Figure 5.14 as the level in the tank drops, the cross-sectionalarea A1 decreases.
Figure 5.14: Cone shaped tank.
The area of a cone as a function of height can be derived using the twodiameters and the height. Since we are dealing with a truncated cone,we have to break the problem down into two parts (Figure 5.15). Firstthe area between the two horizontal lines is constant and equal to π
4 d22 .
Next we have to determine angle,θ . We know the value of hinitial andwe can derive the length of the opposite side of the triangle (o) by
o =d1−d2
2
57
5. Bernoulli Equation
Using h and o, tanθ can be calculated
tanθ =oh
Figure 5.15: Cone shaped tank.
Combining the numeric value for tanθ with any value for the heightyields the length of o at the selected height. The diameter, d1 at anyheight between h0 and h can now be expressed by
d1(h) = 2(tanθ)(h)+d2
Integrating Equation 5.11 becomes a bit more difficult because A1 isnow a function of height. The A1 terms now looks like
A1 =π
4(2(tanθ)(h)+d2)
2
and the integral is
−∫ h2
h1
(2(tanθ)(h)+d2)2
√h
dh =A2π
4
√2g∫ t2
t1dt (5.12)
This integral is somewhat more difficult to integrate than the one inEquation 5.11, but there are a number of calculators or computerprograms (MatLab, Maple, or Mathematica) that can do the integral.
58
5.6. Unsteady Flows
Example 5.3. Example: In Figure 5.16, the diameter of thetop of the tank is 2 f t and the diameter of the bottom is 0.5 f t.The tank is 15 f t high. Derive an equation to calculate the timeneeded to empty the tank given
−dh√h
=A2
A1
√2gdt
Figure 5.16: Conical tank example.
Expressing A1 as
A1 =π
4(2(tanθ)(h)+d2)
2
The integral that must be solved is
−∫ 15
0
A1√h
dh = A2√
2g∫ t2
t1dt
Assigning values to the various parameters
A2 =π
4(0.5 f t)2 = 0.196 f t2
tanθ =0h
=d2−d1
2h
=2 f t−0.5 f t
215 f t
= 0.05
A1 =π
4(2(0.05)(h)+0.5 f t)2 = (0.1h+0.5)2
59
5. Bernoulli Equation
The integral is now
−∫ 15
0
(0.1h+0.5)2√
hdh = 0.196 f t2
√2(
32.2f ts2
) ∫ t2
t1dt
∆t =−
∫ 150
(0.1h+0.5)2√
hdh
0.196 f t2
√2(
32.2 f ts2
)Analytical solution of this equation is difficult, but it is easy toevaluate numerically. The answer is
∆t = 4.6s
5.7 Summary
The Bernoulli equation describes the relationship between velocityand pressure. It is useful in a wide variety of flow problems. Caremust be taken in choosing the start and end points. In general, thepoints should be selected to minimize the number of variables thatmust be evaluated. In the next chapter we will see how the Bernoulliequation can be extended to more complex flow systems wherefriction is important and work is done on or by the system.
60
6 Fluid Friction in SteadyOne-Dimensional Flow
Unlike the systems that were covered in the last chapter, real fluids donot flow without frictional losses. Flow through pipes, valves,expansions, contractions, bends or into or out of tanks are just a fewexamples of sources of friction. One of the results of friction ispressure drop.
61
7 Momentum Balance
7.1 Newton’s Laws
1. A body at rest remains at rest and a body in motion remains inmotion at the same velocity in a straight path when the net forceacting on it is zero.
2. The acceleration of a body is proportional to the net force actingon it and is inversely proportional to its mass.
3. When a body exerts a force on a second body, the second bodyexerts an equal and opposite force on the first. This is the socalled reaction force.
Expressing the second law in equation form
F = ma = mdvdt
=d (mv)
dt(7.1)
The mv term is called the linear momentum or simply the momentum.Looking at the second law equation (7.1), it can also be thought of asthe rate of change of the momentum of a body is equal to the net forceacting on the body. For our purposes, this is probably a better way ofthinking about the second law. It is important to remember that theforce, acceleration, velocity, and momentum are vectors, i.e. theyhave both a magnitude and direction.
7.2 Control Volumes
Most of the work in working momentum problems is in thebookkeeping. Properly selecting a control volume and control surface
63
7. Momentum Balance
makes the process easier. If the subject of the analysis is moving, thenit is usually better to let the control volume move. For a fixed flowsystem, the control volume should be fixed. For example, the systemshown in 7.1 should have a control volume that encompasses thenozzle and the plate. For a jet exiting a moving airplane, the velocity
Figure 7.1: Force on a plate exerted by a jet of fluid impinging on theplate at a 90◦ angle.
of interest is the relative velocity. An example is shown in Figure 7.2.In this case, the velocity is given by
vx = v jet− vCV
Where the jet is moving in the negative x-direction while the control
Figure 7.2: Moving airplane
volume is moving in the positive x-direction. This yields a negativeoverall velocity
(v =−v jet− vCV
). An example of a deformable
control volume is the control volume inside a piston chamber in amotor (Figure 7.3). The control volume expands and contracts as thepiston moves up and down.
64
7.3. Forces on a Control Volume
Figure 7.3: Deformable control volume.
7.3 Forces on a Control Volume
There are two types of forces that act on a control volume: Body andSurface. The body forces act throughout the whole body while thesurface forces act on the control surface. Body forces can be a gravity,electric, or magnetic field. Surface forces of interest in fluid flow arepressure and viscous forces. Normally, we can ignore all of the bodyforces except gravity.
∑F = ∑Fbody +∑Fsur f ace (7.2)
Gravitational force on a fluid element
dFgravity = ρ gd V (7.3)
Total force acting on a control volume
∑Fbody =∫
CVρ gd V = mCV g (7.4)
Surface force acting on a small surface element
d~Fsur f ace = σ ·~ndA (7.5)
Where~n is an outwardly pointing vector normal to the surface dA andσ is the stress on the surface, dA (Figure 7.4).The total force on a body is
∑~F = ∑~Fbody +∑~Fsur f ace =∫
CVρ gd V +
∫CV
σ~ndA (7.6)
65
7. Momentum Balance
Figure 7.4: Surface forces acting on a small area.
7.4 Steady Flow
In steady flow, the equation reduces to
∑~F = ∑out
m~v−∑in
m~v
Which reduces to∑F = m(~vout−~vin) (7.7)
Example: Liquid Jet Striking a FlatPlateWater strikes a flat plate at a rate of 10kg/s with a velocity of 20m/sand exits in all directions in the plane of the plate. What is the forcenecessary to hold the plate in place?Solution: All of the entering velocity is in the positive x-direction andnone of the fluid exits in the x-direction. This means that v2x is equalto zero.
∑F = m(~v2x−~v1x)
reduces to−Fx = 0− mv1
Fx = mv1
Fx =10kg
s|20m
s| N ·s2
kg ·m︸ ︷︷ ︸gc
= 200 N
Example: Liquid Jet Striking aCurved Plate
66
7.4. Steady Flow
Figure 7.5: Force on a curved vane
A 3-inch diameter jet with a velocity of 50 ft/s is deflected through anangle of 70◦ when it hits a stationary vane, determine the horizontaland vertical components of the force of the water on the vane.Solution: Neglecting the weight of the fluid, height changes in thefluid, the weight of vane and assuming no frictional losses, thevelocity of the exiting liquid jet must be the same as the entering jet.This does not preclude changes in direction or cross-sectional shape.Vector analysis must be used to determine the x and y velocities.
∑F = ∑out
m~v−∑in
m~v
The x-component of the velocity is given by
v2,x = v2 cosθ =50 f t
s|cos
(70◦ | 2π
360◦
)= 17.1
f ts
The y-component of the velocity is given by
v2,y = v2 sinθ =50 f t
s|sin
(70◦ | 2π
360◦
)= 46.98
f ts
Calculating the mass flow rate
m = ρ q where q = v1 A
q =50 f t
s|0.0491 f t2
= 2.455f t3
s
m =62.4 lbm
f t3 |2.455 f t3
s= 153.19
lbm
s
67
7. Momentum Balance
The reaction force in the x-direction is given by
Fx = m(v2,x− v1,x)
Fx =153.19 lbm
s| 1gc|(
17.1f ts−50
f ts
)= -156.5 lb f
The reaction force in the y-direction is given by
Fy = m(v2,y− v1,y
)Fy =
153.19 lbm
s| 1gc|(
46.98 f ts
−0f ts
)= 223.5 lb f
If we want the force of the fluids acting on the plate, the signs arereversed.
Now we will complicate the problemby having the vane move with a velocity in the x-direction with avelocity of 15 f t/s.
Example: Liquid Jet Striking aMoving Curved Plate
Figure 7.6: Force on a moving curved vane
Using the relative velocity in place of v1
v2,x = v2 cosθ =35 f t
s|cos
(70◦ | 2π
360◦
)= 11.97
f ts
v2,y = v2 sinθ =35 f t
s|sin
(70◦ | 2π
360◦
)= 32.89
f ts
68
7.4. Steady Flow
Fx =153.19 lbm
s| 1gc|(
11.97f ts
35f ts
)= -76.7 lb f
Fy =153.19 lbm
s| 1gc|(
32.89 f ts
−0f ts
)= 109.5 lb f
Since the vane is moving away from the jet, the forces will decrease asthe jet lengthens (the distance between the jet and the vane increases).
There is another way to think aboutthe momentum equation. If we write the equation as
∑F = ∑ρ vv ·A (7.8)
This really is not that different. Remember that m = ρ vA and bytaking the dot product between the velocity and the area vector (italways points outward from the control surface) we can keep the signsstraight. Reworking the first part of the last example we get thefollowing.
Example: Force on a Curved Vane
∑F = ∑ρ vv ·A
The x-component of the velocity is given by
v2,x = v2 cosθ =50 f t
s|cos
(70◦ | 2π
360◦
)= 17.1
f ts
The y-component of the velocity is given by
v2,y = v2 sinθ =50 f t
s|sin
(70◦ | 2π
360◦
)= 46.98
f ts
The reactive force in the x-direction is given by
Fx =ρ
gcv1,xv1 ·A+
ρ
gcv2,xv1 ·A
Fx =62.4 lbm
f t3 |lb f s2
32.2lbm f t|50 f t
s|50 f t
s|cos(180) |0.0491 f t2
+62.4 lbm
f t3 |32.2lb f s2
lbm f t|17.1 f t
s|50 f t
scos(0) |0.0491 f t2
= -156.5 lb f
69
7. Momentum Balance
The reactive force in the y-direction is given by
Fy =ρ
gcv1,yv1 ·A+
ρ
gcv2,yv1 ·A
Fy =62.4 lbm
f t3 |lb f s2
32.2lbm f t|0 f t
s|50 f t
s|cos(180) |0.0491 f t2
+62.4 lbm
f t3 |32.2lb f s2
lbm f t|46.98 f t
s|50 f t
scos(0) |0.0491 f t2
= 223.5 lb f
Let us look at a slightly different problem involving flow of a fluidaround a bend. The flow is not open to the atmosphere as it was in theprevious examples so pressure forces must be accounted for.
Example: Force on a Pipe BendWater flows through a 180◦ vertical reducing bend. The discharge is0.25m3/s and the pressure at the center of the inlet of the bend is150kPa. If the bend volume is 0.1m3 and it is assumed thatBernoulli’s equation is valid, what is the reaction force required tohold the bend in place. Assume the metal in the bend weighs 500N.
Figure 7.7: Force on a bend.
Solution: Starting with the basic momentum equation, evaluate all ofthe variables that can be easily evaluated.
∑F = ∑ρ vv ·A
A1 =π
4d2
1 = 0.071m3 A2 =π
4d2
2 = 0.0177m3
70
7.4. Steady Flow
v1 =q
A1= 3.54
ms
v2 =q
A2= 14.15
ms
Using the Bernoulli equation to evaluate the missing pressure.
P1
γ+ z1 +
v21
2g=
P2
γ+ z2 +
v22
2g
P1 = 150kPa z1 = 0m z2 =−(
d1
2+0.1m+
d2
2
)=−0.325m
Solving for P2P2 = 59.37kPa
Returning to the momentum equation we can write
Fx +P1A1 +P2A2 =ρ
gcv1 v1 ·A1 +
ρ
gc(−v2) v2 ·A2
Inserting values for the variables and solving for Fx we get
Fx =−10.60kN−1.049kN−0.884kN−3.537kN = −16.07kN
The force in the y-direction is simply the sum of the weight of thefluid in the bend and the weight of the bend.
Fy = WB + V ρggc
= 500N +0.1m3
|1000kgm3 |9.81m
s2 | N s2
1kgm
Fy = 1.48kN
Note: If the bend in the previous example is turned around, the PAforces have a negative sign in front of them and v1 is negative ratherthan v2. This results in a change in sign, but not magnitude, of Fx.The analysis of a bend of less than 180◦ is similar, but the angles haveto be taken into account when the velocities are calculated.
Example: Flow in a 30◦ Bend.A pipe that is 1 m in diameter has a 30◦ horizontal bend in it, as shown,and carries crude oil (SG = 0.94) at a rate of 2 m3
s . If the pressure inthe bend is assumed to be essentially uniform at 75kPagage, if the
71
7. Momentum Balance
Figure 7.8: Forces on a 30◦ bend.
volume of the bend is 1.2m3, and if the metal in the bend weighs 4kN,what forces must be applied to the bend to hold it in place?Solution: Starting with the momentum equation
∑F = ∑vρ v ·A
A1 = A2 =π
4d2 = 0.785m2 v1 =
qA1
= 2.55ms
The velocity at the exit, v2,x, is not equal to v1, but is equal to
v2,x = v1 cos(
302π
360
)= 2.205
ms
So the momentum equation becomes
Fx+P1A1−P2A2cos(
302π
360
)=
ρoil
gcv1,x v1 A1 cos(π)+
ρoil
gcv2,x v2 A2 cos(0)
Solving for Fx
Fx =−59.05kN +51.01kN−4.79kN +4.15kN = 8.64kN
There is no pressure force acting in the y−direction on the entranceside of the bend nor is there a y− component of the velocity so
Fy +P2A2 sin(
302π
360
)=
ρoil
gcv2,y v2 A2 cos(0)
Fy = 29.45kN +2.39kN = 31.84kN
72
7.5. Rotational Motion and Angular Momentum
The force in the z−direction results from the weight of the bend andthe weight of the fluid enclosed within it.
Fz = WB +Wf = 4kN +1.2m |(0.94)1000kg
m3 |9.81ms2 | N s2
1kgm
Fz = 15.1kN
7.5 Rotational Motion and Angular Momentum
7.5.1 Review
Angular velocity The angular distance traveled per unit time.
ω =dθ
dt=
d (l/r)dt
=1r
dldt
=vr
α =dω
dt=
d2θ
dt2 =1r
dvdt
=at
rv = rω
at = rα
Where at is the acceleration in the tangential direction and v is thelinear velocityFor constant rotational motion there must be a force actingtangentially to maintain the angular acceleration. The moment ortorque (M) is proportional to the length of the arm that connects thepoint of action and the center of rotation.
M = rFi = rmat = mr2α
In integral form
M =∫
r2αdm =
(∫r2dm
)︸ ︷︷ ︸
I
α = Iα
73
7. Momentum Balance
Figure 7.9: Angular velocity
I is the moment of inertia. Remembering that mv is the momentum ina linear system, the moment of momentum or angular momentum fora rotating point mass is given by
H = rmv = r2mω
For a rotating rigid body, the total angular momentum is given by
H =∫
r2ωdm =
(∫r2dm
)︸ ︷︷ ︸
I
ω = Iω
Angular velocity ω = 2π n60 yields rad/s.
In vector form, the momentum force equation becomes
~M =~r×~F
Where the magnitude of the moment of force is given by
M = Fr sinθ
The moment of momentum in vector form becomes
~H =~r×m~v
For a differential mass dm
74
7.5. Rotational Motion and Angular Momentum
dm = (~r×~v)ρ dV
Integration yields
~Hsys =∫
sys(~r×~v)ρ dV
For steady flow the net torque can be written as
∑ ~M = ∑out
~r× m~v−∑in
~r× m~v
Example: A large lawn sprinklerwith four identical arms is to be converted into a turbine to generateelectrical power by attaching a generator. Water enters the sprinklerfrom the base along the axis of rotation at a rate of 20 liters/s andleaves the nozzles in the tangential direction. The sprinkler rotates ata rate of 300 rpm in a horizontal plane. The diameter of each jet is1cm, and the normal distance between the axis of ration and the centerof each nozzle is 0.6m. What is the torque on the shaft available togenerate electrical power?Solution: We want to solve the equation
∑M = ∑out
rmv−∑in
rmv
The last term is equal to zero since rin = 0. Calculate m
m = ρq =1000kg
m3 |0.02m3
s= 20
kgs
Calculate the nozzle velocity
vn =0.02m3
s|14| 4π (0.1m)
= 63.66ms
Calculate the angular velocity
ω =(2π)300rev
min|min60s
= 31.42rad
sCalculate the tangential velocity
75
7. Momentum Balance
vnozzle = rω = (0.6m)(
42s
)= 18.85
ms
Calculate the relative velocity
vrel = vn− vr = 63.66−18.85 = 44.81ms
Calculate the torque
Tsha f t = rmtotalvr =0.6m |20kg
s|44.81m
s|N ·s2
kg ·m= 537.7N ·m
Calculating the power
Po = ωTsha f t =31.42
s|537.7N ·m | kW s
1000N ·m= 16.9kW
76
