chapter 4: control volume analysis using energyjmcgrath/ln.ch4.notes1.pdf · • energy transfer...

Chapter 4: Control Volume Analysis Using Energy

Conservation of Mass and Conservation of Energy

ENGINEERING CONTEXT

The objective of this chapter is to develop and illustrate the use of the control volume forms of the conservation of mass and conservation of energy principles. Mass and energy balances for control volumes are introduced in Secs. 4.1 and 4.2, respectively. These balances are applied in Sec. 4.3 to control volumes at steady state and in Sec. 4.4 for transient applications.

Although devices such as turbines, pumps, and compressors through which mass flows can be analyzed in principle by studying a particular quantity of matter (a closed system) as it passes through the device, it is normally preferable to think of a region of space through which mass flows (a control volume). As in the case of a closed system, energy transfer across the boundary of a control volume can occur by means of work and heat. In addition, another type of energy transfer must be accounted for- the energy accompanying mass as it enters or exits.

Examples of Open Systems

How do we get from the fixed mass (closed) systems that we know, to the open systems we are interested in?

??

Q

PT

W

WMass crosses boundary-What about energy?

Consequences of mass and energy crossing boundary?

Where does mass cross boundary?

Yes

Yes

Yes

No

Does total mass in the system change over time?

Where does energy cross boundary?

Yes

Yes

??

??

What is the purpose of this device?What performance questions might we ask?Which energies are most significant?What is the relationship between these energies?Does the total mass and/or energy of system change over time?

What forms of energy cross boundary?

Mechanical Work (Out)Rotational, Torque

Chemical Energy (In)(Stored in Chemical Bonds)

Non-reacted Gas,Hot Gas (Out)

ChemicalThermal (In)

One Inlet, One Exit Control Volume Boundary

Dashed line defines control volume boundary

Inlet, iExit, e

Mass is shown in green

Developing Control Volume Mass Balance

Inlet, iExit, e

mi mcv (t)

At time t

Start with fixed mass (closed system) and allow an interaction with mass crossing a boundary. Transform to a form that accounts for flow across boundary.

Total mass is: ( )total i cvm m m t= +

Push mass mi into CV during interval ∆t


( ) ( )total i cv cv em m m t m t t m= + = + ∆ +

Inlet, iExit, e

memcv (t + ∆t)

At time t + ∆t

( ) ( )cv cv i em t t m t m m+ ∆ − = −


( ) ( )cv cv i em t t m t m mt t t

+ ∆ −= −

∆ ∆ ∆

( ) ( )cv cv i em t t m t m m+ ∆ − = −

( ) ( ) ( )lim 0 cv cv i em t t m t m mtt t t

+ ∆ −∆ → = − ∆ ∆ ∆

Mass balance

Express on rate basis over time interval ∆t, as ∆t goes to zero:

cvi e

dm m mdt

= −

Multiple Inlets and Exits Control Volume Boundary

Dashed line defines control volume boundary

Inlets, iExits, e

cvi e

i e

dm m mdt

= −∑ ∑

Conservation of Mass for Control Volume

time of time of flow time of flowmass contained within of mass across of mass acrossthe control volume inlet exit

rate of change rate ratein out

at time t i at time t e at time t

= −

i ei em m>∑ ∑

cvi e

i e

dm m mdt

• •

= −∑ ∑

Suppose

What happens to cvdmdt In = Stored + Out

Conservation of Mass for Control Volumetime of time of flow time of flowmass contained within of mass across of mass acrossthe control volume inlet exit

rate of change rate ratein out

at time t i at time t e at time t

= −

i ei em m<∑ ∑

cvi e

i e

dm m mdt

• •

= −∑ ∑

cvdmdt

What happens to if:

i ei em m=∑ ∑ Steady State0cvdm

dt=

0cvdmdt

<

Evaluating Mass Flow Rate at Ports

nA

m V dAρ= ∫

( )amount of mass crossing dA duringthe time interval t

nV t dAρ = ∆ ∆

instantaneous rateof mass flowacross dA

nV dAρ =

23

m L t L mL t

∼

23

m L mLL t t

∼

Integrate over port area:

Forms of Mass Rate Balancecv

i ei e

dm m mdt

= −∑ ∑

Alternative forms convenient for particular cases:

One-dimensional Flow

Steady State

Integral

One Dimensional Flow Model

A

m VdA AVρ ρ= =∫

V = constant

One-Dimensional Flow Model or Approximation:

Actual Two-Dimensional Flow:

V = V(r)

( ) ( )[ ]2A A

m VdA r V r rdrρ ρ π= =∫ ∫

Note simplifying 2-D to 1-D

How to handle non-uniform velocity/properties

m AVρ=

AVmv

=

1vρ

=

( ) ( )cv i i e ei i i e e e

i e i ei e

dm AV AV AV AVdt v v

ρ ρ= − = −∑ ∑ ∑ ∑

One Dimensional Flow Model

is the Volumetric Flow Rate (e.g. m3/kg)AV

Convenient form when Volume Flow Rates known

Other forms of Conservation of Mass

i ei em m

• •

=∑ ∑

Steady-State Flow:

No changes in time

All derivatives with respect to time are zero

cvi e

i e

dm m mdt

= −∑ ∑0

What goes in, goes out!

In many cases, the mass in balances the mass out

Other forms of the Conservation of Mass

( ) ( )n nV A Ai ei e

d dV V dA V dAdt

ρ ρ ρ= −∑ ∑∫ ∫ ∫

( )cvV

m t dVρ= ∫

cvi e

i e

dm m mdt

= −∑ ∑

Integral Form:

Fluid density may vary within the control volumeFluid velocity and density may vary across flow area

nA

m V dAρ= ∫

Text Examples

Feedwater Heater at Steady State

Filling a barrel with water

Energy Conservation in Open Systems

• Mass crosses the system boundary, transporting– Internal thermal energy per unit mass, (u)– Kinetic energy per unit mass– Potential energy per unit mass– Chemical energy in chemical bonds

• Perform Transformation of Control Mass to Control Volume

Conservation of Energy for Control Volume

time net at which net at whic of the energy energy is being

contained within transferred the control volume by heat transfer

rate of change rate rate

inat

time t at time t

= −

h net of energyenergy is being transfer thetransferred control volume

by work accompanying mass flow

rateinto

out

at time t

+

Approach similar to MassConservation derivation:

Closed system to open system

Mass balance to rate basis

Below:

Stored = HeatIn – WorkOut +Net “Mass-Energy” In


time net at which net at whic of the energy energy is being

contained within transferred the control volume by heat transfer

rate of change rate rate

inat

time t at time t

= −

h net of energyenergy is being transfer thetransferred control volume

by work accompanying mass flow

rateinto

out

at time t

+

2 2

2 2cv i e

i i i e e edE V VQ W m u gz m u gzdt

• • • •

= − + + + − + +

Single Inlet, Single Outlet


2 2

2 2cv i e


• • • •

= − + + + − + +

Note: Internal energy, Kinetic energy, Potential energyassociated with fluid flow at inlet & exit



2 2

2 2cv i e


• • • •

= − + + + − + +

Work term includes flow across boundary and all other forms

shaftW

(Work?)

(Work?)

Push mass mi into Control Volume

Force through a displacement is Work

Work IN, is done ON system at inlets

Concept of Flow Work

Inlet, iExit, e

mi mcv (t)Fi = PiAi

Concept of Flow Work

Inlet, iExit, e

me

mcv (t + ∆t)

Fe = PeAe

Push mass me out of Control Volume

Force through a displacement is Work

Work OUT, is done BY system at outlets

Concept of Flow Work Rate

Inlet, iExit, eme

mcv (t + ∆t)

Fe = PeAe

( )time rate of energy transferby work from the controlvolume at exit, e

e e eP A V =

Work Rate in terms of pressure & velocity

Work Rate: W dxW F FVt dt

δδ

= = =

( )W FV PA V= =

Evaluating Rate of Flow Work

( ) ( )cv e e e i i iW W P A V PA V= + −

Rate of Flow Work & Rate of All other forms of work included(rotating shafts, displacement of boundary, electrical work, etc.)

( ) ( )cv e e e i i iW W m Pv m Pv= + −

m A Vρ= A Vmv

=

v is specific volume

or

( ) ( )PA V m Pv=

Out:

Other forms of Conservation of Energy

( ) ( )Flow Work =cv cv e e e i i iW W W m Pv m Pv• • • • •

= + + −

h u P v= + ⋅Recall:

2 2

2 2cv i e


• • • •

= − + + + − + +

2 2

2 2cv i e

cv cv i i i e e edE V VQ W m h gz m h gzdt

• • • • = − + + + − + +

Energy Rate Balance:


Multiple Ports

2 2

2 2cv i e

cv cv i i i e e ei e

dE V VQ W m h gz m h gzdt

• • • • = − + + + − + +

∑ ∑

Energy Rate Balance:

Integral Form:2 2

2 2cv cv n nV A Ai ei e

d V VedV Q W h gz V dA h gz V dAdt

ρ ρ ρ• •

= − + + + − + +

∑ ∑∫ ∫ ∫

Variations within CV and at boundaries:

( )2

2cvV V

VE t edV u gz dVρ ρ

= = + +

∫ ∫

Analyzing Control Volumes at Steady State

Transient start up and shutdown periods not included here

Steady-State Forms of Mass and Energy Rate Balances

Control volume at Steady-State (SS):

• Conditions (states/properties) of mass within CV and at boundaries do not vary with time

• No change of mass (+ or -) within CV• Mass flow rates are constant• Energy transfer rates by heat & work are

constant



• Conditions (states/properties) of mass within CV & at boundaries do not vary with time

time

time

ρ (x, y, z)

T (x, y, z)

time

time

P1

h2



• No change of mass (+ or -) within CV

time

time

mcv

mcv

Yes:

No:



• Mass flow rates are constant

time

time

im

emYes:

No:



• Energy transfer rates by heat & work are constant

inQ

outQ

netW

time

time

Steady State

Steady State Flow:2 2

02 2i e


V VQ W m h gz m h gz• • • •

= − + + + − + +

∑ ∑

Special case when no changes occur in time

2 2

2 2i e

cv i i i cv e e ei e

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

i ei em m

• •

=∑ ∑cvi e

i e

dm m mdt

= −∑ ∑0

Net Energy Flow In = Net Energy Flow Out

One Inlet, One Outlet Steady State

i em m m= =

( ) ( ) ( )2 2

02

i ecv cv i e i e

V VQ W m h h g z z

− = − + − + + −

Special case when no changes occur in time

( ) ( ) ( )2 2

02

i ecv cvi e i e

V VQ W h h g z zm m

−= − + − + + −

Note energy DIFFERENCES- Reference cancels out

Modeling Control Volumes at Steady State

• Outer surface is well insulated• Outer surface too small for significant heat transfer• Temperature difference between surface and surroundings

so small that heat transfer can be ignored• Fluid passes through CV so quickly that not enough time

for significant heat transfer to occur

Common modeling assumptions- The Art of Engineering

Heat Transfer- can be ignored (adiabatic) when:

Work Transfer- can be ignored when:

• No rotating shafts, displacements of boundary, electrical effects or other work mechanisms

Modeling Control Volumes at Steady State

• Are small compared to other energy forms

Common modeling assumptions- The Art of Engineering

Kinetic and/or Potential Energy- can be ignored when:

Steady state may be OK if:

• Properties only fluctuate slightly about average values• Periodic time variations are observed: time-average

P

t

P

t

Examples

Nozzles and Diffusers

m AVρ=

1 1 1 1 2 2 2 2m AV m AVρ ρ= = =m A Vρ=

State & Process Information Needed to Determine Pressures, etc

Steady Flow Devices


Photo courtesy of NASA Ames

AME Windtunnels

Low-Speed Closed Return Wind Tunnel Facility

AME Windtunnels

Nozzle & Diffuser Steady Flow Analysis

Nozzles and Diffusers 1 2cvdm m mdt

= −

1 2m m m= =

0cvdmdt

=

2 2

1 202 2

cv i ecv cv i i e e


= = − + + + − + +

( )2 2

1 21 20

2cvQ V Vh hm

−= + − +

cvQm Often negligible

1 2 3


( )2 2

1 21 20

2V Vh h −

= − +

Common Form of 1st Law:

Text Example

Calculate Exit Area of a Steam Nozzle

Turbines

Photo courtesy of U.S. Military Academy

Photo courtesy of M.I.T. Microturbine lab

1100kW Helicopter Engine 50 watt Microturbine

Turbine schematics courtesy ofwww.howstuffworks.com

( ) ( )2 2

1 21 2 1 22

W V Vh h g z zm

•

•

−= − + + −


Single Inlet, Single OutletHeat transfer is often small and neglected. ∆KE & ∆PE often small and neglectedWork produced is equal to enthalpy difference

Axial flow steam of gas turbine

( )1 2Ww h hm

•

•= = −

Steam and Gas Turbines

Hydraulic Turbine in a Dam

( ) ( )2 2

1 21 2 1 22

W V Vh h g z zm

•

•

−= − + + −


Heat transfer is small and neglected.Work produced is associated primarily with change in PE

Text Example

Calculate Heat Transfer from a Steam Turbine

Compressors and Pumps

Compressors are devices where work is done on a GAS to raise pressure, change state

Pumps are devices where work is done on a LIQUIDto raise pressure, change state

Compressors and Pumps

Rotating compressors

Reciprocating compressor


( ) ( )2 2

1 21 2 1 22

W V Vh h g z zm

•

•

−= − + + −

[ W < 0 ]

Text Example

Calculate Compressor Power

Power Washer

Heat Exchangers

Heat exchangers are devices that exchange thermal energy between fluids at different temperatures by heat transfer, Q

Work, changes in KE and PE are not usually important

Changes in enthalpy are important

Balance is between Q and changes in enthalpy

Heat Exchangers

Direct Contact

Tube-within-a-tube parallel flow

Tube-within-a-tube counterflow

Cross-flow


2 2

2 2i e

i i i e e ei e

V Vm h gz m h gz• •

+ + = + +

∑ ∑

Text Example

Power plant condenser

Cooling computer components

Throttling Devices

These are devices designed to reduce the pressure of a fluid (gas or liquid)

2 2

2 2i e

i i i e e ei e

V Vm h gz m h gz• •

+ + = + +

∑ ∑

Throttling Devices

Common Form of 1st Law: i eh h=

Work = 0

Adiabatic assumption often reasonable (why?)

Changes in KE and PE are not usually important (why not?)

Single inlet, single outlet implies enthalpy balance

Text Example

Measuring steam quality

System Integration

Engineers creatively combine/integrate components into systems to achieve overall objective, subject to constraints such as cost, weight, pollution

Example of simple power plant given:(Turbine/generator, condenser, pump, boiler)

Text Example

Waste heat recovery system

Transient (Unsteady) Analysis

• Many devices undergo periods of transient operation in which state changes with time.

• Examples include startup and shutdown ofturbines, compressors, and motors.

• Filling and emptying vessels are other examples

• Steady state assumptions not valid• Special care in applying mass and energy analyses

Transient Mass Balance

0 0 0

t t tcv

i ei e

dm dt m dt m dtdt

= − ∑ ∑∫ ∫ ∫

( ) ( ) ( ) ( )0 00

t t

cv cv i ei e

m t m m dt m dt− = −∑ ∑∫ ∫

Recall that: cvi e

i e

dm m mdt

= −∑ ∑

Integrate mass balance from t = 0 to final time, t:

Transient Mass Balance

0 0 0

t t tcv

i ei e

dm m dt m dtdt

= − ∑ ∑∫ ∫ ∫

0

t

i im m dt= ∫

0

t

e em m dt= ∫

amount of massentering the controlvolume through inlet i,from time 0 to t

amount of massexiting the controlvolume through exit e,from time 0 to t

( ) ( )0cv cv i ei e

m t m m m− = −∑ ∑

( ) ( ) ( ) ( )0 00

t t

cv cv i ei e

m t m m dt m dt− = −∑ ∑∫ ∫

Change of mass in CV = Total massgoing IN minus Total mass going OUT

Transient Energy Balance

( ) ( )0 0

0t t

cv cv cv cv i i e ei i

U t U Q W m hdt m h dt − = − + −

∑ ∑∫ ∫

( ) ( )0cv cv cv cv i i e ei e

U t U Q W m h m h− = − + −∑ ∑

2 2

2 2cv i e



• • • • = − + + + − + +

∑ ∑

Ignore KE and PE and Integrate this wrt time:

0 0

t t

i i i i i im h dt h m dt hm= =∫ ∫0 0

t t

e e e e e em h dt h m dt h m= =∫ ∫

For special case where states at inlet & outlets are constant with time:

Another special case where intensive properties within CV are uniform with position at each instant:

( ) ( )( )

cvcv

V tm t v t=

( ) ( ) ( )cv cvU t m t u t=

Text Examples

Withdrawing steam from tank at constant pressure

Using Steam for Emergency Power Generation (new)

Storing compressed air in a tank

Temperature variation in a well-stirred tank

chapter 4: control volume analysis using energyjmcgrath/ln.ch4.notes1.pdf · • energy transfer...

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