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MECH 221 FLUID MECHANICS(Fall 06/07)
Chapter 3: FLUID IN MOTIONS
Instructor: Professor C. T. HSU
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MECH 221 – Chapter 3
3.1. Newton’s Second Law
The net force F acting on a matter of mass m leads to an acceleration a following the linear relation: F = ma
For a solid body of fixed shape, m is a constant and a is described along the trajectory of motion.
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MECH 221 – Chapter 3
3.1. Newton’s Second Law If r(t) represents the particle trajectory, the
velocity v(t) and acceleration are then given by:
v(t) = dr/dt ; a(t) = dv/dt = d2r/dt2
Therefore, F = mdv/dt
If m = m(t), then F = (mdv/dt)+(vdm/dt)=d(mv)/dt = dM/dt. where M = mv is the momentum.
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MECH 221 – Chapter 3
3.1. Newton’s Second Law
For fluids enclosed in a control volume V(t) which may deform with time along the trajectory of motion, it is only correct to use the fluid momentum M to describe the Newton’s second law
For M= , we have F = dM/dt =
)(tV
dVdt
dv
V(t)dVv
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MECH 221 – Chapter 3
3.2. Description of Fluid Flow
Fluid dynamics is the mechanics to study the evolution of fluid particles in a space domain (flow field). There are two ways to describe the flow field
Lagrangian description
Eulerian description
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MECH 221 – Chapter 3
3.2.1 Lagrangian Description
Given initially the locations of all the fluid particles, a Lagrangian description is to follow historically each particle motion by finding the particle locations and properties at every time instant.
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MECH 221 – Chapter 3
3.2.1 Lagrangian Description Therefore, if a specific fluid particle is initially
(t=t0) located at (x0, y0, z0), the Lagragian description is to determine (x(t), y(t), z(t)) and the fluid properties, such as
f(t) = f [x(t), y(t), z(t); t],v(t) = v [x(t), y(t), z(t); t] , etc.
given f0 = f (x0, y0, z0; t0),
v0 = v (x0, y0, z0; t0) etc.
Note that (x(t), y(t), z(t)) is a function of time.
(for any function f)
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MECH 221 – Chapter 3
3.2.2. Eulerian Description An Eulerian description of fluid flow is simply to
state the evolution of fluid properties at a fixed point (x, y, z) with time. Here (x, y, z) is independent of time. Hence,f(t) = f[x, y, z, t] , v = v [x, y ,z, t] etc.
Eulerian description is to observe the fluid properties of different fluid particles passing through the same fixed location at different time instant, while Lagrangian description is to observe the fluid properties at different locations following the same particle
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MECH 221 – Chapter 3
3.2.3. Relation between Lagrangian and Eulerian description
It is important to note that there is only one flow property at the same location with respect to the same time, i.e., f(x(t),y(t),z(t),t) = f(x,y,z,t). Therefore, the Lagrange differential with respective to dt,
which is equal to the total Eulerian differential:
tDt
fDf dd
dtt
fdz
z
fdy
y
fdx
x
fdf
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MECH 221 – Chapter 3
3.2.3. Relation between Lagrangian and Eulerian description
Hence, the total derivative is given by:
ft
f
t
f
z
fw
y
fv
x
fu
t
f
dt
dz
z
f
dt
dy
y
f
dt
dx
x
f
Dt
Df
v
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MECH 221 – Chapter 3
3.3. Equations of Motion for Inviscid Flow
Conservation of Mass
Conservation of Momentum
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MECH 221 – Chapter 3
3.3.1. Conservation of Mass Mass in fluid flows must conserve. The total mass in V(t) is given by:
Therefore, the conservation of mass requires thatdm/dt = 0.
where the Leibniz rule was invoked.
)(
/tV
dVdt
ddtdm
S
dt
dVdV
t)t(V
)(tV
dVm
S)t(V
ddVt
sv
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MECH 221 – Chapter 3
3.3.1. Conservation of Mass Hence:
This is the Integral Form of mass conservation equation.
0ddVt S)t(V
sv
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MECH 221 – Chapter 3
3.3.1. Conservation of Mass Integral form of mass conservation equation
By Divergence theorem:
Hence:
)()]([
tVdV
tv
= 0
dVdtVS
)()( vsv
0)(
StVddV
tsv
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MECH 221 – Chapter 3
3.3.1. Conservation of Mass As V(t)→0, the integrand is independent of V(t)
and therefore,
This is the Differential Form of mass conservation and also called as continuity equation.
0 )(
vt
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum The Newton’s second law,
is Lagrangian in a description of momentum conservation. For motion of fluid particles that have no rotation, the flow is termed irrotational. An irrotational flow does not subject to shear force, i.e., pressure force only. Because the shear force is only caused by fluid viscosity, the irrotational flow is also called as “inviscid” flow
dt
dMF
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum For fluid subjecting to earth gravitational acceleration,
the net force on fluids in the control volume V enclosed by a control surface S is:
where s is out-normal to S from V and the divergence theorem is applied for the second equality.
This force applied on the fluid body will leads to the acceleration which is described as the rate of change in momentum.
V(t)S
dVρpd gsF
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum
where the Leibniz rule was invoked.
)t(V
dVdt
ddt/d vM
S
dt
dV)(dV)(
t)t(Vvv
S)t(V
ddV)(t
svvv
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum Hence:
This is the Integral Form of momentum conservation equation.
dVpdddV)(t S )t(VS)t(V
gss vvv
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum
Integral form of momentum conservation equation
By Divergence theorem:
Hence:
dV)(d)t(VS vvvv s
dVpdV()(t)t(V )t(V
][)][
g vvv
dVpdddV)(t S )t(VS)t(V
gss vvv
dVppd)t(Vs s
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum
As V→0, the integrands are independent of V. Therefore,
This is the Differential Form of momentum conservation equation for inviscid flows.
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MECH 221 – Chapter 3
3.3.2. Conservation of Momentum By invoking the continuity equation,
The momentum equation can take the following alternative form:
which is commonly referred to as Euler’s equation of motion.
0 )
v(
t
g
p)()(t
vvv
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MECH 221 – Chapter 3
3.4. Bernoulli Equation for Steady Flows Bernoulli equation is a special form of the Euler’s
equation along a streamline. For a first look, we restrict our discussion to steady flow so that the Euler’s equation becomes:
Assuming that g is in the negative z direction, i.e., g =- and using the following vector identity,
the Euler’s equation for steady flows becomes
g p)( vv
zg
)()(2
1)( vvvvvv
)(gv(2
1pvv
2 z )
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MECH 221 – Chapter 3
3.4. Bernoulli Equation for Steady Flows We now take the scalar product to the above equation
by the position increment vector dr along a streamline and observe that
Thus, the result leads to
The above equation now can be integrated to give
0gdv(ddp
2 z /2)
constantz 2
g2
vdp
(along streamline)
dfdf r 0d r]vv )([; (for any function f)
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MECH 221 – Chapter 3
3.4. Bernoulli Equation for Steady Flows For incompressible fluids where ρ = constant, we
have
For irrotational flows, everywhere in the flow domain and
constantz 2
g2
vp
(along streamline)
0 v
0gv(2
1p
2 z )
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MECH 221 – Chapter 3
3.4. Bernoulli Equation for Steady Flows Since, for dr in any direction, we have:
For anywhere of irrotational fluids
For anywhere of incompressible fluids
dfdf r
constantz 2
g2
vp
constantz 2
g2
vdp
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MECH 221 – Chapter 3
3.5. Static, Dynamic, Stagnation and Total Pressure
Consider the Bernoulli equation,
The static pressure ps is defined as the pressure associated with the gravitational force when the fluid is not in motion. If the atmospheric pressure is used as the reference for a gage pressure at z=0.
constantz 2
g2
vp
(for incompressible fluid)
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MECH 221 – Chapter 3
3.5. Static, Dynamic, Stagnation and Total Pressure
Then we have as also from chapter 2.
The dynamic pressure pd is then the pressure deviates from the static pressure, i.e., p = pd+ps.
The substitution of p = pd+ps. into the Bernoulli equation gives
zgps
constant 2
2
vpd
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MECH 221 – Chapter 3
3.5. Static, Dynamic, Stagnation and Total Pressure
The maximum dynamic pressure occurs at the stagnation point where v=0 and this maximum pressure is called as the stagnation pressure p0. Hence,
The total pressure pT is then the sum of the stagnation pressure and the static pressure, i.e., pT= p0 - ρgz. For z = -h, the static pressure is ρgh and the total pressure is p0 + ρgh.
od p2
vp
2
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MECH 221 – Chapter 3
3.6. Energy Line and Hydraulic Grade Line
In fact the Bernoulli equation also states that the energy density (per unit volume) possessed by the fluid particle is constant not only along a streamline but also at everywhere in fluid domain for irrotational flow. The energy consists of pressure energy (p), kinetic energy (ρv2/2) and gravitational potential energy (ρgz).
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MECH 221 – Chapter 3
3.6. Energy Line and Hydraulic Grade Line
It becomes simpler if this total energy is interpreted into a total head H (height from a datum) by dividing the Bernoulli equation with ρg such that
where p/ρg is the pressure head, v2/2g is the velocity head and z is the elevation head.
Hgg
p 2
z 2v
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MECH 221 – Chapter 3
3.6. Energy Line and Hydraulic Grade Line
A piezometric head is then defined as that consists of only the pressure and elevation heads, i.e.,
The variations of H and Hp along the path of fluid flow can be plotted into lines and are termed as “energy line” and “hydraulic grade line”, respectively.
It is noted that H is always higher than Hp and that a negative pressure (below atmospheric pressure) occurs when Hp is below the fluid streamline
zg
pH p
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MECH 221 – Chapter 3
3.6. Energy Line and Hydraulic Grade Line
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MECH 221 – Chapter 3
3.7. Applications of Bernoulli Equation
Pitot-Static Tube
Free Jets
Flow Rate Meter
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MECH 221 – Chapter 3
3.7.1. Pitot-Static Tube Pitot-static tube is a device that measures the
difference between h1 and h2 so that the velocity of the fluid flow at the measurement location can be determined from
)hh(g2 12 v
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MECH 221 – Chapter 3
3.7.1. Pitot-Static Tube
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MECH 221 – Chapter 3
3.7.2. Free Jets Free jets are the flow from an orifice of an
apparatus that converts the total elevation head h into velocity head, i.e.,
gh2 v
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MECH 221 – Chapter 3
3.7.3. Flow Rate Meter The commonly used flow rate meter is the
Venturi meter that determines the flow rate Q through pipes by measuring the difference of piezometric heads at locations of different cross-sectional areas along the pipe.
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MECH 221 – Chapter 3
3.7.3. Flow Rate Meter If hp1 and hp2 represent the piezometric heads at
section 1 and 2 with cross-sectional areas A1 and A2 respectively, we have:
2211vv AAQ
gh
gh
pp 2
v
2
v2
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1
22
and
2
12
21
2 )/(1
)(2 v
AA
hhgpp
2
12
21
2 )/(1
)(2
AA
hhgAQ pp
Then, Therefore,