Mech 280: Frigaard

Lecture 1: Kinematics, ideal mechanical systems and Bernoulli’s equation

Should be able to:

Understand some terms used in flow visualization and

kinematics

Understand Lagrangian and Eulerian frames of reference

Understand the material (or total) derivative

Derive Bernoulli’s theorem using momentum balance along a

streamline

Solve simple problems using the Bernoulli theorem

Mech 280: Frigaard

Streamlines, Pathlines, Streaklines

A streamline is everywhere tangent to the velocity field In steady 2D flow can integrate

A pathline is the actual path travelled by a fluid particle over some time interval E.g. with a tracer particle

A streakline is locus of fluid particles that pass through a given point in the flow Common experimentally: e.g. inject

dye in water, smoke in air

uv

dxdy

Vds

wdz

vdy

udx

=

===

Mech 280: Frigaard

Natural ways to describe flows?

Lagrangian description A fluid is comprised of a large number of fluid

particles having mass, momentum, internal

energy, and other properties. Mathematical

laws can then be written for each fluid particle.

Eulerian description In the Eulerian description of fluid motion, we

consider how flow properties change at a fluid

element that is fixed in space and time (x,y,z,t),

rather than following individual fluid particles.

Mech 280: Frigaard

Eulerian description = most common

Eulerian description of fluid flow: a flow domain or control volumeis defined by which fluid flows in and out

We define field variables which are functions of space and time Pressure field: P=P(x,y,z,t) Velocity field: V=V(x,y,z,t) = u(x,y,z,t)i + v(x,y,z,t)j + w(x,y,z,t)k Acceleration field, a=a(x,y,z,t) = ax(x,y,z,t)i + ay(x,y,z,t)j + az(x,y,z,t)k These (and other) field variables define the flow field.

Suitable for formulation of initial boundary-value problems (PDE's). Leonhard Euler (1707-1783).

Particle’s velocity at a point is same as fluid velocity

Time derivative of particle position is fluid velocity

What about acceleration?

( ) ( ) ( )tzdtdty

dtdtx

dtd

particleparticleparticleparticle ++== VV

Acceleration in Eulerian coordinates?

Acceleration is the time derivative of the particle's velocity:

To take the time derivative of the velocity, chain rule must be used

Advective acceleration term: acceleration of fluid even in steady flow

Advective acceleration is nonlinear: source of many phenomena and main

challenge in solving fluid flow equations

( ) ( ) ( )tzdtd

zty

dtd

ytx

dtd

xt particleparticleparticleparticle ∂∂

+∂∂

+∂∂

+∂∂

=VVVVa

onaccelerati Advectiveonaccelerati Local

zw

yv

xu

tparticle ∂∂

+∂∂

+∂∂

+∂∂

=VVVVa

particleparticle dtd Va =

( ) ( ) ( )( )ttztytx particleparticleparticleparticle ,,,VV =

Material derivative & acceleration:

Total derivative operator is called material derivative

Other names for the material derivative include: total, particle, Lagrangian, Eulerian, and substantial derivative

Lagrangian:

Eulerian:

Consider advective acceleration tangential & normalto steady streamline:

∇⋅+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

= Vtz

wy

vx

utDt

D

∇⋅+∂∂

= VtDt

Ddtd

Mech 280: Frigaard

( )

sVV

tVa

dsdAmgdsdAW

WdAdPPPdAFma

s

ss

∂∂

+∂∂

=

==

−+−== ∑

ρρ

θsin

Bernoulli’s equation

Apply Newton’s second law along a streamline Assume no viscous effects

Assume no heat transfer

Acceleration = material derivative of velocity V along a streamline s

Steady incompressible flow, divide by m and take ds0

Along a streamline: constant=++

=+∂∂

+∂∂

gzPVdsdzg

sP

sVV

ρ

ρ

2

01

2

Bernoulli’s equation

Mech 280: Frigaard

Frictionless flow with no energy transfer

3 terms in Bernoulli equation Potential energy

Flow energy

Kinetic energy

Alternatively, dividing by g: Static head

hydrostatic head (pressure head)

Dynamic head

Unsteady, variable density version of Bernoulli

Using the Bernoulli equation

P1ρ

+V1

2

2+ g z1 =

P2ρ

+V2

2

2+ g z2 = const

P2 − P1ρ

+V2

2 − V12

2+ g z2 − z1( )= 0

( ) ( ) 021

122

12

2

2

1

2

1

=−+−++∂∂

∫∫ zzgVVdpdstV

ρ

22V

Pgz

ρ

gV

gPz

22

ρ

Bernoulli’s equation is an energy equation for an ideal mechanical system

Mech 280: Frigaard

Example 1

Air flows past a ball, as shown. It is determined that the velocity along the x-axis for x<-a is given by:

Determine the pressure variation along the x-axis and the stagnation pressure at x = -a

+= 3

3

0 1xaVV

Mech 280: Frigaard

Example 2Water flows from a garden hose. A child places his thumb to cover most of the outlet. A thin stream of water jets upwards. a) If the gage pressure in the hose just upstream of his

thumb is 400kPa what is the maximum height the jet can achieve, (assuming no energy losses at position 1)?

b) Assuming that soon above position 1 the pressure in the jet is Patm , derive an expression for the velocity VJ as a function of height z above 1

Mech 280: Frigaard

( )

R

2

,cos

Va

dndAmgdndAWWdAdPPPdAFma

n

nn

nnnnn

=

==

−+−== ∑ρρ

θ

F=ma: Normal to a streamline?

Newton’s second law normal to streamline Assume no viscous effects

Assume no heat transfer

Steady incompressible flow, divide by m and take dn 0 n = in direction of curvature

Circular streamlines at same height implies radial increase in pressure

R

21 Vdndzg

nP

=−∂∂

−ρ

Mech 280: Frigaard

Example 3:

Flows a & b have circular streamlines:

Sketch the pressure distributions, assuming p(r0) = p0

Mech 280: Frigaard

Fluids in the news:

Mech 280: Frigaard

Lecture 2: Applications and limitations of Bernoulli’s equation

Bernoulli states that total pressure is constant

Stagnation pressure is the pressure at any point where the fluid is brought to rest isentropically

This can be used with a Pitot tube and piezometer to measure fluid velocity

gzVPPtotal ρρ++=

2

2

2

2VPPstagρ

+=

Mech 280: Frigaard

Example 4:

A piezometer and a Pitot tube are tapped into a horizontal water pipe as shown. Determine the velocity at the center of the pipe?

Mech 280: Frigaard

Example 5:a) What is the pressure at the nose of a torpedo moving in salt water at

100ft/s at a depth of 30ft?b) If the pressure at a point on the side of the torpedo, at the same depth

as the nose, is 10.0psi gage, what is the fluid velocity at that point?

Mech 280: Frigaard

Free streams and jets:

Example 6: Water flows along an undulating path, as shown. What is the pressure variation: a) between points 1 & 2?b) between points 3 & 4?

Mech 280: Frigaard

Example 7: A stream of liquid flows from a hole of diameter 0.01m near the bottom of a reservoir of diameter 0.2m, with height as shown. The liquid is replenished continuously.

Find the flow rate Q that keeps the level constant

Mech 280: Frigaard

Example 8: Water flows under the sluice gate as shown.

Estimate the flow rate per unit width Q

Mech 280: Frigaard

Bernoulli: frictionless flow with no energy transfer Energy equation for an ideal

mechanical system

Track these quantities along a flow path EGL = total head

HGL = static + pressure head

Measure with series of: Pitot tubes (EGL)

Piezometric taps (HGL)

Energy & Hydraulic Grade Lines

constant 2

headDynamic

2

headPressure

headStatic

=++g

Vg

Pzρ

HGL and EGL Plot of EGL and HGL generally

decays along flow path

Variation in EGL along real

flow paths is a measure of

energy losses/gains (see later)

2

2P VEGL zg gρ

= + +

PHGL zgρ

= +

Frictionless flow: HGL indicates positive & negative pressure

Limitations on using the Bernoulli Equation

Steady flow: d/dt = 0

Frictionless flow

No shaft work: wpump=wturbine=0

Incompressible flow: ρ = constant

No heat transfer: qnet,in=0

Applied along a streamline

Mech 280: Frigaard

Fluids in the news:

Mech 280: Frigaard

What we covered

Eulerian and Lagrangian frames of reference

Converting time derivatives between the two systems

Material derivative and acceleration

Bernoulli’s equation

Derivation from momentum balance

Interpretation as an energy balance

Examples of applicationconstant=++

=+∂∂

+∂∂

gzPVdsdzg

sP

sVV

ρ

ρ

2

01

2

Mech 280: Frigaard

Lecture 3: Integral analyses of fluid flow

Systems and control volumes

Reynolds transport theorem

General principals of conservation laws for a control volume

Conservation of mass for a control volume

Systems and control volumes

Generally in mechanics we apply conservation laws to estimate quantities of interest

A (closed) system is a region in space consisting of a fixed quantity of matter No mass crosses system

boundary A control volume, is any

properly selected region in space Analogous to the Free

body diagram in Solid mechanics / dynamics

Mass can cross control surfaces

26

Mech 280: Frigaard

The fundamental conservation laws (conservation of mass, energy &momentum) apply directly to systems

However, in most fluid mechanics problems, control volume analysis is preferred over system Control volumes are chosen more relevant to the problem at hand

Therefore, we need to transform the conservation laws from a system to a control volume. This is accomplished with the Reynolds transport theorem (RTT)

At a time t, we define a system to be the mass in the control volume (CV) bounded by the control surface (CS)

Reynolds Transport Theorem (RTT)

nReynolds transport theorem: Rate of change of ‘B’ in the system is equal to the rate of change of ‘B’ in the CV plus the net flow of ‘B’ out of the CV, across the CS

Mech 280: Frigaard

Mathematical statement

What is B? Any extensive property Associated intensive property is β β is the amount of B per unit mass

∫=V

sys dvB ρβ

dmdB

=β

Mass Momentum Energy Angular momentum

B, Extensive properties m mV E mr×V

β, Intensive properties 1 V e r×V

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ

Reynolds transport theorem: Rate of change of ‘B’ in the system is equal to the rate of change of ‘B’ in the CV plus the net flow of ‘B’ out of the CV, across the CS

Mech 280: Frigaard

Moving / deforming control volume

Reynolds transport theorem

Is valid for fixed, moving and/or deforming control volumes,

The velocity Vr in the second term is the relative velocity:

Vr = V -VCS

V is the fluid velocity, (typically given in an inertial system)

VCS is velocity of the moving/deforming CS, (in the same

system as V)

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ

Mech 280: Frigaard

Conservation of Mass

Conservation of mass is one of the most fundamental principles in nature.

Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process.

For closed systems mass conservation is implicit since the mass of the system remains constant during a process.

For control volumes, mass can cross the boundaries which means that we must keep track of the amount of mass entering and leaving the control volume.

Mech 280: Frigaard

Mass Conversvation

B =m = Massn

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ

( )∫∫ ⋅+

=

CSr

CV

dAdvdtd nVρρ0

1==dmdmβ

0 =sysBdtd :conserved Mass

Mech 280: Frigaard

Mass and Volume Flow Rates

Consider a fixed CV and CS, so that Vr = V

Consider flow through a control surface, with area Ac . Flow rates obtained by integration

The amount of mass flowing through a control surface per unit time is called the mass flow rate and is denoted

The dot over a symbol is used to indicate time rate of change.

Volume flow rate through Ac

Often density doesn’t change much

c c

n cA A

m m V dAδ ρ= =∫ ∫

V

( ) cavgA

cnA

c AVdAVdAQcc

==⋅= ∫∫ nV

1

c

avg n cc A

V V dAA

= ∫

avg cm V Aρ=

Conservation of Mass Principle For fixed CV of arbitrary shape,

rate of change of mass within the CV

net mass flow rate

Therefore, general conservation of mass for a fixed CV is:

Same expression for deforming or moving CV, except net mass flow rate in/out are relative to the CS

= ∫

CVCV dv

dtdm

dtd ρ

( )∫∫ −=⋅==CS

inoutCS

net mmdAmm nVρδ

CVin out

dmm mdt

− =

Steady Flow Processes For steady flow, total amount of

mass contained in CV is constant. Total amount of mass entering

must be equal to total amount of mass leaving

For incompressible flows,in out

m m=∑ ∑

n n n nin out

V A V A=∑ ∑

Mech 280: Frigaard

Example 1:

If the average water velocity in a in a 300mm pipe is 0.5m/s, what is the

average velocity in a 75mm water jet coming from the pipe?

Mech 280: Frigaard

Example 2:An airplane moves forward at speed 971km/hr. The frontal intake area of the jet

engine is 0.8m2 and the entering air density is 0.736kg/m3. A stationary observer

determines that relative to the Earth, the jet engine exhaust gases move away

from the engine with a speed 1050km/hr. The engine exhaust area is 0.558m2,

and the exhaust gas density is 0.515kg/m3.

Estimate the mass flowrate of fuel into the engine in kg/h

Mech 280: Frigaard

Example 3:

1m

2m

For an observer standing at the c.v. inlet (point 1)V1 = VJ – Uc = 7 – 2 = 5 m/s

= ρ1 V1 A1 = 998 kg/m3*5 m/s*π*.042/4 = 6.271 kg/sNote: The inlet velocity used to specify the mass flow rate is again measured relative to the inlet boundary, but now is given by VJ – Uc . Exit:

= 6.271 kg/s, Again, since ρ and A are cons., V2 = 5 m/sAgain, the exit flow is most easily specified by conservation of mass concepts.

VJ

A jet of water leaves the nozzle at mean speed 7 m/sAnd strikes the turning vane as shown. a) Compute the mean velocity exiting the CV if the

vane is stationaryb) Repeat the calculation if the vane moves to the

right at a steady velocity of 2 m/s.

Mech 280: Frigaard

Example 4:

A room contains dust at a uniform concentration, C=ρdust/ρ. The room is to be cleaned by flushing air through the room. Find an expression for time rate of change of dust mass in the room.

Vout

Mech 280: Frigaard

What we covered Derived the general Reynolds Transport Theorem (RTT)

Velocity is relative to moving / deforming control volumes

Used the RTT to derive conservation of mass for a CV

Mass flow rates:

Mass conservation can be written as: CVin out

dmm mdt

− =

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ

( )∫∫ ⋅+

=

CSr

CV

dAdvdtd nVρρ0

( )∫ ∫∫ =⋅==CS CS

nCS

net dAVdAmm ρρδ nV

Mech 280: Frigaard

Lecture 4: Conservation of momentum

Should be able to:

Use Reynolds’ transport theorem to derive the conservation of

linear momentum relation for a CV

Examples for fixed & moving CV

Examples coupled with mass conservation CV

Last lecture we derived Reynolds’ transport theorem

Applied this to give conservation of mass relation in a CV

In steady flow conservation of mass is relation between flow rates

What about computing forces?

Mech 280: Frigaard

Newton’s Laws

Newton’s laws are relations between motions of bodies

and the forces acting on them

First law: a body at rest remains at rest, and a body in motion

remains in motion at the same velocity in a straight path when

the net force acting on it is zero

Second law: the acceleration of a body is proportional to the net

force acting on it and is inversely proportional to its mass

( )VVaF mdtd

dtdmm ===

Mech 280: Frigaard

Momentum Conservation for a CV

n

B =mV = Momentumn

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ

( )∫∫∑ ⋅+

=

CSr

CVCVdAdv

dtd nVVVF ρρ

( ) VV == mdmdβ

∑= FsysBdtd :conserved Momentum

Mech 280: Frigaard

Orientation of CS’s and the Momentum Flux Correction Factor

Momentum is a vector

quantity

Flux term in momentum

balance is more complex

than for mass conservation

Nearly always we orient CS’s

to be normal to the flow

direction Does not affect mass balance

Simplifies flux term

Note that at inlet/outlet, Vavg

often only has component in

normal direction:

In an incompressible flow,

we have

β>1 always and for turbulent

flow β in range 1.01 – 1.05( ) ∫∫ ==⋅cc A

avgcnA

c mdAVdA VVnVV βρρ

∫

∫

∫

∫==

c

c

c

c

Ac

Ac

avg

Acn

Acn

avg dA

dA

dAV

dAV VV

VV

ρ

ρβ

∫

=

cAc

avgn

n

c

dAVV

A

2

,

1β

Choosing a Control Volume

CV is chosen by fluid dynamicist and there is a lot of freedom. However, selection of CV can either simplify or complicate analysis Clearly define all boundaries. Analysis is often simplified

if CS is normal to flow direction Clearly identify all fluxes crossing the CS Clearly identify forces and torques of interest acting on

the CV and CS Fixed, moving, and deforming control volumes:

For moving CV, use relative velocity,Vr = V - VCS = V -VCV

For deforming CV, use velocity relative to each deforming control surfaces

Vr = V –VCS Note that for momentum balance, as given, Newton’s

laws require an inertial frame of reference For non-inertial frames, add fictitious forces – addressed

later in course

Forces Acting on a CV

Forces acting on CV consist of body forces that act throughout the

entire body of the CV (such as gravity, electric, and magnetic forces) and

surface forces that act on the control surface (such as pressure and

viscous forces, and reaction forces at points of contact).

• Body forces act on each volumetric portion dv of the CV.

• Surface forces act on each portion dA of the CS.

n

dFdFsurface

45

Body Forces

The most common body force is

gravity, which exerts a downward

force on every differential element

of the CV

dFbody = dFgravity = ρgdV

Summing over all differential

volume elements, gives total body

force acting on CVi

dFbody = dFgravity = ρgdVk

j

g

ggF CVCV

body mdV == ∫∑ ρ

Surface Forces

Surface forces include both normal and

tangential components

Diagonal components σxx, σyy, σzz are called

normal stresses and are due to pressure and viscous stresses

Off-diagonal components σxy, σxz, etc., are

called shear stresses and are due solely to

viscous stresses

Force/area acting on CS with normal n:

Total surface force acting on CS

∫∑ ⋅=CS

surface dSσnF

( ) ∑=

=⋅321 ,,i

iijj nσσn

Body and Surface Forces

Surface integrals are cumbersome.

Careful selection of CV allows

expression of total force in terms of

more readily available quantities like

weight, pressure, and reaction forces.

Goal is to choose CV to expose only

the forces to be determined and a

minimum number of other forces.

Surface

otherviscouspressure

Body

gravity ∑∑∑∑∑ +++= FFFFF

∑∑∑ −=In

avgOut

avg mm VVF ββ

Simplification for steady flows:

Mech 280: Frigaard

Example 1: For the pipe-flow reducing section shown, D1 = 8cm, D2 = 5cm and all fluids are at 20°C. If V1 = 5m/s and the manometer reading is h = 58cm, estimate the total horizontal force on the flange bolts.

Mech 280: Frigaard

Example 2: Force on an elbow

Water flows steadily through a 90° elbow, exiting to atmosphere at position 2. If D = 5cm & P1 = 3.43 kPa, what anchoring force is needed to keep the elbow in place?

Mech 280: Frigaard

Example 3: Moving control volumeA water jet 4 cm in diameter with a velocity 7 m/s is directed towards a turning vane (θ = 40˚) that is moving with velocity, Uc = 2 m/s. Determine the force F necessary to hold the vane in steady motion.

iieeb VmVmF −=−

V1 = VJ – Uc = 7 – 2 = 5 m/s and V2 = 5 m/s inclined 40˚ .

-Fb = 6.271 kg/s * 5 m/s *cos 40˚ -6.271 kg/s * 5 m/s

and -Fb = - 7.34 kg m/s2 or Fb = 7.34 N ← ans.

Mech 280: Frigaard

Example 4: Non-uniform pressureA sluice gate across a channel of width b operates in both open and close positions, as shown. Is the anchoring force required to hold the gate in place larger when the gate is open or closed?

Mech 280: Frigaard

Fluids in the news

Mech 280: Frigaard

What we covered

Derived linear momentum equation

Momentum flux correction factor

Simplified forms for momentum balance

Sum of forces include: Pressure, body, shear and mechanical

Computed examples of: Stationary control volume

Moving, control volume in an inertial frame

( )∫∫∑ ⋅+

=

CSr

CVCVdAdv

dtd nVVVF ρρ

Mech 280: Frigaard

Lecture 5: Momentum II

Examples using the momentum balance to generate thrust

Derive the conservation of momentum for a control volume

moving in an accelerating frame of reference.

Examples of accelerating frames of reference

Mech 280: Frigaard

Example 1:A liquid jet of velocity Vj and area Aj strikes a single 180˚ bucket on a turbine wheel rotating at angular velocity Ω. a) Find an expression for the power deliveredb) At what Ω is the power a maximum

Mech 280: Frigaard

Example 2:A static thrust stand as sketched is to bedesigned for testing a jet engine. The following conditions are known for a typical test: • Intake air velocity: 200m/s • Exhaust gas velocity: 500m/s • Intake cross-sectional area: 1m2

• Intake static gage pressure: -22.5kPa• Intake static temperature: 268K• Exhaust static gage pressure: 0kPaEstimate the nominal anchoring force for design

Mech 280: Frigaard

Non-inertial frames of reference

Absolute Acceleration = Non-inertial Acceleration +

Relative Acceleration

Consider fluid particle, with position:

ri =

Velocity:

Vi =

V = velocity in non-inertial frame

How about acceleration?

X

Y

Z

y

x

z

Mech 280: Frigaard

Relative acceleration

X

Y

Z

y

x

z

reli dtd aVa +=

Mech 280: Frigaard

Non-inertial frames of reference

imaF = Newton’s 2nd Law

dtdmm

dtdmm relreli

VaFFaVa =−⇒=

+= ∑∑

∫∫ ∫∑ ∫ −+∂∂

=−ie A

icv A

ecvcv

cv mdmdVdt

md VVVaF ρ

Effect on control volume analysis is to add fictitious forces to the net force balance

In general, situations can be very complex. In many mechanical systems we restrict motion to specific degrees of freedom.

Mech 280: Frigaard

Example 1: Accelerating control volume

A water jet 4cm in diameter with a velocity of 7m/s is directed to a turning vane moving with velocity Uc(t), and with angle θ = 40˚ as shown. Determine the acceleration of the vane as a function of time, assuming initially stationary.

∫∫ ∫∑ ∫ −+∂∂

=−ie A

icv A

ecvcv

cv mdmdVdt

md VVVaF ρ

Mech 280: Frigaard

Example 2: Rocket acceleration

Calculate the rocket’s angular velocity, given the rocket’s initial mass, the propellant exit velocity and the exit mass flow rate (both assumed constant)?

atmeee ppVm =,,

ω

∫∫ ∫∑ ∫ −+∂∂

=−ie A

icv A

ecvcv

cv mdmdVdt

md VVVaF ρ

Mech 280: Frigaard

What we covered:

Examples of using the linear momentum balance to

calculate thrust, in inertial frames

We derived the fictional forces that we experience

in a non-inertial frame of reference.

Note that the fictional force (acceleration) is negative.

Examples

∫∫ ∫∑ ∫ −+∂∂

=−ie A

icv A

ecvcv

cv mdmdVdt

md VVVaF ρ

https://www.youtube.com/watch?v=X_UDguQ420IDon’t do this at home

Mech 280: Frigaard

Lectures 7 & 8: Energy balance for CV’s

Should be able to: Use the Reynolds Transport Theorem (RTT) to derive the

conservation of energy for a control volume

Understand forms of energy and how work is done on

system

Understand efficiency

Relationship with Bernoulli equation for single stream

systems

Application examples

Mech 280: Frigaard

General Energy Equation

One of the most fundamental laws in nature is the 1st law of

thermodynamics, which is also known as the conservation of

energy principle.

It states that energy can be neither created nor destroyed during a

process; it can only change forms

• Falling rock, picks up speed as PE is converted to KE.

• If air resistance is neglected, PE + KE = constant

• Enthalpy?

General Energy Equation

Energy in a system is in 3 forms Kinetic energy: 0.5mV2

Potential energy: mgz Enthalpy: H

The energy content of a closed system can be changed by two mechanisms: heat transfer Q and work transfer W

Conservation of energy for a closed system can be expressed in rate form as

Net heat transfer to the system:

Net power input to the system:

In open system (CV) work comes from Pressure work Viscous work Shaft work (pump / turbine)

Mech 280: Frigaard

Energy Conservation for a CV (from RTT)

B = Esys

β = e = u+0.5V2+gz = (internal + kinetic + potential) specific energies

General Energy Equation:

( )∫∫ ⋅+

=

CSr

CVsys dAdv

dtdB

dtd nVρβρβ n

𝑄𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑛𝑛 + 𝑊𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑛𝑛 =𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑

=𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

( )∫∫ ⋅+

==+

CSr

CV

sysinnetinnet dAedve

dtd

dtdE

WQ nVρρ,,

Pressure work

Where does expression for pressure work come from? When piston moves down ds under the influence of F=PA,

the work done on the system is dWboundary=PAds. If we divide both sides by dt, we have

For generalized control volumes:

Note sign conventions: n is outward pointing normal vector Negative sign ensures that work done is positive when is

done on the system.

∫ ⋅−=cs r

p dAPdt

dW)( nV

n

V

𝛿𝛿𝑊𝑝𝑝𝑝𝑝𝑛𝑛𝑠𝑠𝑠𝑠𝑝𝑝𝑝𝑝𝑛𝑛 = 𝛿𝛿𝑊𝑏𝑏𝑏𝑏𝑝𝑝𝑛𝑛𝑏𝑏𝑏𝑏𝑝𝑝𝑠𝑠 = 𝑃𝑃𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑃𝑃𝑃𝑃𝑉𝑉𝑝𝑝𝑖𝑖𝑠𝑠𝑛𝑛𝑏𝑏𝑛𝑛

𝛿𝛿𝑊𝑏𝑏𝑏𝑏𝑝𝑝𝑛𝑛𝑏𝑏𝑏𝑏𝑝𝑝𝑠𝑠 = −𝑃𝑃𝑑𝑑𝑃𝑃𝑉𝑉𝑛𝑛 = −𝑃𝑃𝑑𝑑𝑃𝑃(𝑽𝑽𝒓𝒓.n)

Mech 280: Frigaard

Viscous work

Viscous work is the work done on the control volume surface due

to shear force

Can be a real problem because we don't know velocity

distribution, therefore don't know shear stress.

Solution! Draw control volume carefully, e.g.

Perpendicular to flow

Where the velocity is zero, e.g., at a solid boundary

Or where the shear stress is zero, (or small, or perpendicular to the velocity)

∫ ⋅⋅−=cs ij dA

dtdW nV τν

Mech 280: Frigaard

Energy Analysis of Steady Flows

Simplifications: Kinetic energy correction factor set to 1 (see later)

Rate of change of energy content = 0

Well-chosen CV means rate of viscous work is zero

Pressure work rate is included in flux terms

• Note, could use enthalpy: h=u+P/ρ, but confusing later when h= “head”.

Single inflow/outflow: specific quantities

+++−

+++=+ ∑∑ gzVuPmgzVuPmWQ

inoutinnetshaftinnet 22

22

,,, ρρ

Fixed control volume

In

Out

2

1

innetshaftinnet WQ ,,, +

+++ 1

21

11

1

2gzVuPm

ρ

+++ 2

22

22

2

2gzVuPm

ρ

[ ]

lossmechturbinepump

e

innetinnetshaft

innetshaftinnetinnetshaftinnet

ewgzVPgzVPw

quugzVPgzVPw

gzVuPm

Wm

Qwq

lossmech

,2

22

2

21

21

1

1

,122

22

2

21

21

1

1,,

2

1

2,,,

,,,

22

22

2

,

++++=+++

−−+++=+++

+++=+=+

ρρ

ρρ

ρ

Mech 280: Frigaard

Example 1:A pump delivers water at a steady rate of 300 gal/min as shown. Upstream [sect. (1)]

pipe diameter is 3.5 in. & the pressure is 18 psi. Downstream [sect. (2)] the pipe diameter

is 1 in. & the pressure is 60 psi. The change in elevation is zero. The rise in internal

energy of water, u2-u1, associated with a temperature rise across the pump is 93 ft lb/lbm.

Q. Determine the power that the pump requires.

Mech 280: Frigaard

Example 2:A steam turbine generator unit is used to produce electricity. The steam enters the turbine with

a velocity of 30 m/s and enthalpy, h1, of 3348 kJ/kg. The steam leaves the turbine as a mixture

of vapor and liquid having a velocity of 60 m/s and an enthalpy h2 of 2550 kJ/kg. The flow

through the turbine is adiabatic, and changes in elevation are negligible.

Q. Determine the work output involved, per unit mass of steam through-flow

Mech 280: Frigaard

Head form of energy equation (single stream CV)

Divide by g to express each

term in units of length

Magnitude of each term is

expressed as an equivalent

height of fluid, i.e. a head

Note, with no shaft work or

irreversible head losses,

similar to Bernoulli’s equation

2 21 1 2 2

1 21 22 2pump turbine LP V P Vz h z h h

g g g gρ ρ+ + + = + + + +

Example 3:

Given the initial exit velocity what is the pump

power & the time taken to empty the tank?

2 2

2 2in in out out

in out f p tV P V Pz z h h h

g g g gρ ρ+ + = + + + − +

Mech 280: Frigaard

Example 4:The pump shown in the figure adds 10 horsepower to the water as it pumps water

from the lower lake to the upper lake. The elevation difference between the lake

surfaces is 30 ft and the head loss is 15 ft.

Q. Determine the flow rate and power loss associated with this flow

Mech 280: Frigaard

Efficiency

Transfer of emech is usually accomplished by a rotating shaft: shaft work

Pump, fan, propulsion: receives shaft work (e.g., from an electric motor) and transfers it to the fluid as mechanical energy

Turbine: converts emech of a fluid to shaft work. In absence of irreversible losses (e.g., friction), mechanical

efficiency of a device or process can be defined as

If ηmech < 100%, losses have occurred during conversion.

, ,

, ,

1mech out mech lossmech

mech in mech in

E EE E

η = = −

Pump and Turbine Efficiencies

In fluid systems, we are usually interested in increasing the pressure, velocity, and/or elevation of a fluid.

In these cases, efficiency is better defined as the ratio of (supplied or extracted work) vs. rate of increase in mechanical energy

Overall efficiency must include motor or generator efficiency.

Mech 280: Frigaard

Example 5:The pump of a water distribution system is powered by a 15-kW electric motor (efficiency 90%). The water flow rate is 50l/s. The inlet/outlet pipe diameters are the same and there is negligible increase in elevation across the pump. Determine a) the mechanical efficiency of the pump; b) the temperature rise of the water as it flows through the pump, due to inefficiency.

Mech 280: Frigaard

Example 6:In a hydroelectric power plant, 100m3/s of water flows from an elevation of 120m to a turbine. The total irreversible head loss is determined to be 35m. If the overall efficiency of the turbine generator is 80%, estimate the electric power output.

Mech 280: Frigaard

Example 7:An axial-flow ventilating fan driven by a motor thatdelivers 0.4 kW of power to the fan blades produces a 0.6-m diameter axial stream of air having a speed of 12 m/s. The flow upstream of the fan involves negligible speed.Q. Determine how much of the work to the air actually produces useful effects, that is, fluid motion and a rise in available energy. Estimate the mechanical efficiency of this fan.

Mech 280: Frigaard

Up to this point, the velocity used in the kinetic energy term of the energy equation has been the mass average velocity obtained from the definition of flow rate:𝑚 = 𝜌𝜌𝑃𝑃 𝑉𝑉

Kinetic Energy Correction Factor

3 3av

1 1u d A V A2 2

ρ ρ β=∫

3

av

1 u=A V

d Aβ

∫

However, since V varies over the flow area and the kinetic energy term varies with the square of the velocity, using this definition of Vmay not result in an accurate evaluation of the kinetic energy term for the flow. This problem can be corrected through the use of the kinetic energy correction factor, defined for incompressible flow from

Solving for 𝛽𝛽 we obtain

For fully developed laminar flow, β= 2 and for turbulent pipe flow, a value from 1.04 to 1.11 is common

Mech 280: Frigaard

Example 8:The small fan shown moves air at a mass flowrate of 0.1 kg/min. Upstream the pipe diameter is 60 mm, the flow is laminar and the kinetic energy coefficient, is 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent and the kinetic energy coefficient, is 1.08. The rise in static pressure across the fan is 0.1 kPa, and the fan motor draws 0.14 W.Q. Compare the value of losses calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions.

Mech 280: Frigaard

What we covered

Derived conservation of energy for a CV

Converted to ‘head’ form for steady, single stream CV

Note: Bernoulli’s equation as a frictionless flow with no

energy in / out.

Efficiency of pumps & turbines

Examples using energy equation in different formats

2 2

2 2in in out out

in out f p tV P V Pz z h h h

g g g gρ ρ+ + = + + + − +

P1ρ

+V1

2

2+ g z1 =

P2ρ

+V2

2

2+ g z2 = const

Mech 280: Frigaard

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