Chapter5 Notes ME311 F14 PJF

8/9/2019 Chapter5 Notes ME311 F14 PJF

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MASS AND ENERGY ANALYSIS OF

CONTROL VOLUMES

Florio

Cengel

10-2014

2

Objectives• Develop the conservation of mass principle.

• Apply the conservation of mass principle to various control volumes

systems.

• Apply the first law of thermodynamics as the statement of the

conservation of energy principle to control volumes.

• Identify the energy carried by a fluid stream crossing a control surface as

the sum of internal energy, flow work Pv, kinetic energy, and potential

energy of the fluid and to relate the combination of the internal energy and

the flow work to the property enthalpy.

• Solve energy balance problems for common steady-flow devices such as

nozzles, compressors, turbines, throttling valves, mixers, heaters, and

heat exchangers.

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CONSERVATION OF MASS

3

Mass is conserved even during chemical reactions , in addition there is

conservation of the atomic species..

Conservation of mass: Mass, like energy, is a conserved property, and it cannot becreated or destroyed during a process only changed in form and/or location.

Closed m ass systems : The mass of the system remain constant during a process.

Control volumes : Mass can cross the boundaries, and thus the identity of the matterwithin the region changes; so we must keep track of the amount of mass entering andleaving the control volume.

Mass m and energy E can be converted to each other according to

where c is the speed of light in a vacuum, which is c = 2.9979 108 m/s.

For most of our problems, the mass change due to energy change is negligible.

1/2

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Mass and Volume Flow Rates

4

The volume flow rate is the volume

of fluid flowing through a cross

section per unit time.

Definition of

average velocityMass flow

rate

Volume flow rate-define average velocity

dA

avg

avg avg

m Vol

V A

Ac

Vavg is the uniform velocity necessary to

produce the same volume flow rate.

Vavg =VOLdot/Ac

PdA

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Conservation of Total Mass Principle

5

The conservation of mass principle for a control volume : The net mass transfer to or

from a control volume during a time interval t is equal to the net change (increase or

decrease) in the total mass within the control volume during t . M is the amount of mass

General conservation of mass 0 place holder for DM/Dtfmsof the fms momentarily occuping the cv at time t-

Reynolds- Transport eq.

General-exact-conservation of mass in rate form

or

Atomic species or molecular

species form is:Volume flow conservation for

steady fixed cv with

incompressible flow.

( )V N dA

∆

∆

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Mass Balance for Steady-Flow Processes

6

Conservation of mass principle for a two-

inlet –one-outlet steady-flow system.

During a steady-flow (rate of exchanges), steady state within -fixed CV process,

the total amount of mass contained within a control volume does not change with time

(mCV = constant).

Then the conservation of mass principle requires that the total rate of mass entering

a control volume must equal the total rate of mass leaving it.For steady-flow processes, we are interested in

the amount of mass flowing per unit time, that is,

the mass flow rate.

Multiple inlets

and exits

Single

streamMany engineering devices such as nozzles,

diffusers, turbines, compressors, and pumps

involve a single stream (only one inlet and one

outlet).

21

3

For a single stream, 1 in, 2 out, and using bulk properties

Gt 37

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Images

• video 1

• video 2

• Video 3

• Video 4

• Video 5

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8

/kgm001017.0 C65

MPa7 31

1

1

v

T

P 6 MPa, 450C

80 m/s

Steam7 MPa65C

5-8 Water flows through the tubes of a boiler. The velocity and volume flow rate

of the water at the inlet are to be determined. The internal diameter is 0.130 m and the water enters at 7MPa,65 C and exits at 6MPa 450 C, 80 m/s.

Assumpt ions Flow through the boiler is steady., and cv is fixed in size and shape

Propert ies The specific volumes of water at the inlet and exit are (Tables A-6 and A-7)

/kgm05217.0 C450

MPa6 32

2

2

v

T

P

Analysis The cross-sectional area of the tube is

222

m01327.04

m)13.0(

4

D Ac

The mass flow rate through the tube is same at the inlet and exit. It may be determined from exit data to be

kg/s20.35/kgm0.05217

m/s)80)(m01327.0(3

2

2

2 v

V Am c

The average water velocity at the inlet is then3

11 2

(20.35 kg/s)(0.001017 m /kg)/

0.01327 mc

mV Vol A

A

v 1.560 m/ s

The volumetric flow rate at the inlet is

2

1 1(0.01327 m )(1.560 m/s)

c A V 30.0207 m / sVol

Rtn 6

Qdot

10/14/2014 Dr. Florio

http://localhost/var/www/apps/conversion/tmp/scratch_4/FIG04_03.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_15.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_17.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_19.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_17.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_15.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG4_1.MOVhttp://localhost/var/www/apps/conversion/tmp/scratch_4/FIG04_03.MOV


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9

2 3

room

3 3

room

define air vol change rate = [(2.7 m)(200 m ) 540 m /room

(540 m )(0.35/h) 189 m / h 189,0

/

00 L/

]

h

room time

3150 L / min

Vol

Vol Vol

House200 m2

0.35 ACH

Ex3- The minimum fresh air requirements of a residential building is specified

to be 0.35 air volume changes per hour (volume changes), or 35% of the air in the room is to be replaced every

hour. The room is assumed to be 2.7m high and 200 m2 floor area.. The size of the fan capacity in L/min that

needs to be installed and the diameter of the duct are to be determined if the average air speed is 6 m/s.

Analysis Assume air pressure change is small so air density constant ∆P IS SMALL relative to P Steady flow

volume flow in is equal to out. The volume of the building and the required minimum volume flow rate of fresh air

are

The volume flow rate of fresh air can be expressed as

2 2( / 4) ; VD = const so Vmax, Dmin and vise versaVA V D Vol

Solving for the diameter D and substituting,

34 4(189 / 3600 m /s)

(6 m/s) D

V 0.106 m

Vol

Therefore, the diameter of the fresh air duct should be at least 10.6 cm if the velocity of air

is not to exceed 6 m/s .

Return to

7

outflow

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Control volume-Control Mass-simple- in time dt a small element of mass enters the

resulting balance equations

• Fixed mass form to the control volume form

• dM)cv = dm)in The total workby is dWother,SHAFT,SHEAR,ELECT - P in [v//v]*dvol in +P out [v/v]*dvol out ; Tointroduce the mass into the CV an effort must be expended

• [δQ- δ Wother+ (P*dvolin-Pvdmout) + e*dmin - e*dmiout)=dE)cv] - in time dt• Where Pdvol is work done by normal forces at flow port in

• dmin = dVol/v)in = [A*Vrel,n, /v]*dt

10

M,E

e*dmin through

an imaginary

mesh

Wflowon =P*d(volin)

dQdot through the

entire surface,

including flow portsurface(s)

dWdot

through the

entire surface

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FLOW WORK for A CONTINUOUSLY FLOWING FLUID

11Schematic for flow work.

Flow work-- P* voldot, : The work (or energy transfer)

required to push the mass into (on) or out (by) of the

control volume or the work done by normal stresses atthe flow ports. This work is necessary for maintaining a

continuous flow through a control volume.

In the absence of acceleration, the force

applied on a fluid by a piston is equal to the

force applied on the piston by the fluid.10/14/2014 Dr. Florio

Total Energy of a Flowing Fluid= stored +flow work

12

The total energy consists of three parts for a non-flowing fluid and for a flowing fluid energy transfer

associated with the mass consists four parts flow work and the 3 of e.

h = u + Pv

The transfer of energyat a flow port-flow work-is automatically taken

care of by enthalpy. The

product of properties

Pv represents the flow

work per unit of mass

only at the flow port

otherwise it is a

product of properties.

dPv pdV vdP

The quantity is called the methalpy , the total enthalpy

or transfer energy per unit of mass

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Energy Transport by Mass-bulk motion

iim

13

At any instant the product

is the energy transported into control

volume by mass per unit time.

When the kinetic and potential energies of a

fluid stream are negligible the energy carried

in

Exact form when the properties of the

mass at each inlet or exit change with

time as well as over the cross section

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Steady State

• 1. If the states within do not change with time, the system is said tobe in a steady state.-[Black Box]

• 2. Steady flow implies all the rate exchange quantities, mdot ,Wdot,Q dot do not vary with time

• 3. For a control volume whose boundary does not change and is ina steady state, is also acting under steady flow conditions. Thus theaccumulation of all extensive properties are zero.

• 4. It is sometimes convenient to define specific work and heatquantities using the reference mass flow rate

• 5.

14

/ and w=W/mref ref

q Q m

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ENERGY ANALYSIS OF STEADY-FLOW-FIXED- SYSTEMS- any

extensive within cv is constant

15

Under steady-flow, fixed conditions, the mass

and energy contents of a control volume remain

constant.

Under steady-flow, fixed,

conditions, the fluid properties at

an inlet or exit remain constant

(do not change with time).

10/14/2014 Dr. Florio

Mass and Energy balances for a steady-

flow process

16

A water heater

in steady

operation.

Mass

balance

Energy

balance

opensystem

OR

,fixed)

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17

Under steady operation,shaft work, shear and

electrical work are the only

forms of work a simple

compressible system may

involve.

Energy balance relations with sign conventions (i.e.,

heat input and work output are positive)

when kinetic and potential energy

changes are negligible

Caution on energy units

Eq below, path must be known

out

rev shaft

in

w vdP ke pe for a fms

rev rev

rev rev

q h vdP

while

q u Pdv

Most handy form

10/14/2014 Dr. Florio

Relative importance- Steady Flow-SF

• The relative importance of various terms differ indifferent energy conversion applications.

• In some applications some terms are usually smallor negligible (but not all the times). Level of ke

• Energy units must be consistent

• Best in applications to first write the full steadyflow equations and then eliminate the negligibleterms.

• Customize in light of the circumstances andobjectives of the particular application.

18

2 2 / 2 [45 / 2]/1000 =1 k / ] If V J kg

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Series of -Analysis –applications

Not internal design use to Synthesized

• The following is a series of applications involving commoncomponents that will be packaged to form components of a system.

• Energy conversion devices, M=mech E type;i.e. work to pressure,ke,or pe

• M-M, M+T-M, M+T- [M+T]

• Rev compressible, Mechanical coupled to thermal

• Irreversible for M converted to thermal

• Synthesized

1910/14/2014 Dr. Florio

20

Nozzles and Diffusers

Shapes for subsonic flows-

converging-Nozzles and diverging-

diffusers are shaped(5-7 degrees)

so that they cause large changes in

fluid velocities and thus kinetic

energies.

Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even

garden hoses P/ρ to V2 /2 or vise-versa.

Devices to alter kinetic energies by means of

geometry changes recall ke can be

completely converted to mech work

A nozzle is a device that increases the kinetic

energy of a fluid at by the transformation of

the pressure E i.e. pressure energy. For a

compressible substance, there is a coupling

between T and P.

The cross-sectional area of a nozzle

decreases in the flow direction for subsonic

flows and increases for supersonic flows. The

reverse is true for diffusers.

A diffuser is a device that increases the

pressure of a fluid by the transformation of the

ke.Energy balance for an

approximate adiabatic

nozzle or diffuser :

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Alternate formulation- Nozzle or diffuser• Convert mechanical and thermal to kinetic, or

vise-versa in the acceleration or decelerationof the flow with no shaft or boundary work:

• For nozzles or diffusers with bulk state1 in

and bulk state2 out is

• :Using Conservation of mass and EE,assuming steady and flows

• q+ h1+V12 / 2J=h2+V2

2 /2J E

• Using the usual assumptions o f steady flow,small potential energy change and adiabatic.J is an energy conversion factor, mech E tothermal measures

• For a general reversible process-1D-steady,

• h2-h1= ∫v dP ****+ qrev

• For incompressible, rev, this is ∆h = v ∆ P+qrev

• Bernouli’s- internally reversible

• ∫v dP +V22 /2 -V1

2 /2=0

• Problem

21

Eq 1

Eq 2

10/14/2014 Dr. Florio

22

kJ/kg3196.7

/kgm0.057838

C004

MPa5

1

31

1

1

hT

P v

Steam1 2

120 kJ/sHeat loss

Ex 3 Steam at 5 Mpa, 400 C is accelerated in a nozzle from a velocity of 80 m/s

and leaves at 2 Mpa, 300 C. The nozzle inlet area is 50 cm2 and heat is lost at

a rate of 120 kW The mass flow rate, the exit velocity, and the exit area of

the nozzle are to be determined.

Assumpt ions 1 This is a steady-flow process since there is no change with time.

2 Potential energy changes are negligible. 3 There are no work interactions.

Propert ies From the steam tables (Table A-6) speed of soumd –steam=SQRT(kRT)= =24.7*SQRT(T) m/s

and

kJ/kg3024.2

/kgm0.12551

C003

MPa2

2

32

2

2

hT

P v

Analysis (a) There is only one inlet and one exit, and thus

m m m1 2

. The mass flow rate of steam is

kg/s6.92 )m1050)(m/s80(/kgm0.057838

11 24311

1

AV mv

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23

(b ) Recall m2 /s2 is a J/kg We take nozzle as the system, which is a control volume sincemass crosses the boundary.

The energy balance for this steady-flow system can be expressed in the rate form as

0 (steady+fixed)

in out system

Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies

in out

/ 0 E E dE dt

E E

2 2

2 12 1

(since W pe 0)

2in

V V Q m h h

J

Substituting, the exit velocity of the steam is determined to be

22

222

/sm1000

kJ/kg1

2

m/s)(803196.73024.2kg/s6.916kJ/s120

V

It yields V 2 = 562.7 m/s; speed of sound =591 m/s

(c ) The exit area of the nozzle is determined from

32 2

2 2 2

2 2

6.916 kg/s 0.12551 m /kg1

562.7 m/s

mm V A A V

4 2

15.42 10 mv

v

10/14/2014 Dr. Florio

• Polytropic nozzle

24

(1 )/

1

Assume we have a reversible polytropic nozzle with

Pv constant and / ( 1)

If ideal gas then P const as well asTv

n

n n

n

vdP n pv n

T const

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EX1-Nozzle problem-

• Using the preceding NOZZLE cv. Fluid is water

• Given: Mdot = 0.1 kg/s, state 1; P=1000 kPa, T= 400 C,• state 2, P=500 kPa, T= 350 C• Find Vexit, Exit Area.

• Assuming SS and fixed, adiabatic and change in pe negligiblethen

• Inlet state 1 V1 =small ;0.0, SH Tables h1= 3263.88 kJ/kg• Exit state 2; h2= 3167.65 kJ/kg, v2 =0.57012 m

3 /kg• EE eq : V2

2 /(2J) = h1-h2 + V12/2J , J=1000J/kJ

• V2= sqrt(2*J*(h1-h2)) =sqrt( 2*1000*(3263.88-3167.65))

• V2= 438.7 m/s (property relation dh= v dP0 )• A=area = mdot*v/V = 0.1*0.57012/438.7=1.3E-04=1.3cm

2

25

2 1m m

2 1m m

10/14/2014 Dr. Florio

Work in devices• The cv is a fluid cv.

• Assuming steady state, fixedcontrol surface. If heatinformation is unknown,safe to assume small.Applying the 1st law, theequation for the adiabaticwork done by the impelleron the fluid is given by

• Wdotin= =mdot{(h2+ke2+pe2)-(h1+ke1+pe1)}

• Usually flow passage isdesigned so as densityincreases, the flow area

decreases, thus shape andstatement

• Usually ke and pe changesby design are small- but maynot be true for hydraulicsystems

• problem

26

compressor

Pumps-liquids- low dP high flow

Force exerted by impeller, rotor, on the fluid

and moves in direction of the force

2

1

Fby fluid

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Aside: cv is the fluid*****

• Defining specific exchange quantities : q to be Qdot/Mdot, and w tobe Wdot/mdot: where Mdot is usually the net inflow rate. For work

devices and nozzles Qdot is small and Mdot large so q small relative tow .

• For the conditions of ss ,fixed cv , one dimensional and reversibleflow; Using the 2nd Law, h2-h1= 1 2 v*dP ****+ qrev

• the inclusion into the energy equation reduces the energy to

• a. –w)byrev = (ke/J +pe/J+ 1 2 v*dP)• Equation a.(Is 1st +2nd) and equation b.1st

• b. qrev – wbyrev = h + ke/J + pe/J for the steady , fixedconditions.

• These two equations , a and b , are independent.

2710/14/2014 Dr. Florio

EX4 Reversible partial compressor problem cv fluid- Given a non- adiabatic compressor

• NH3 is the flowing fluid. Mdot is 4 kg/s. Inlet state 1 : T= - 20 C, P= 150kPa: outlet 2: T2=80 C, P2=400 kPa.

• Assume the process is a reversible polytropic process.

• Using Pvn =const and assuming ke, pe are small• Wdot rev by = - [ 1 2 v dP] mdot = - ∆h integrating by parts yields• wby)rev = - n(P2*v2-P1*v1) /(n-1)********

• State 1 is defined going to NH3 tables you would find v1=0.79774 m3/kg;

h1=1422.9 kJ/kg, state 2 is defined : v2=0.4216, h2=1636.7 kJ/kg.

• n is required : Using the path eq : Pvn = c

• or n= ln(P2/P1)/ln (v1/v2):

• Natural log yields:n=0.5189, w=-53.1 kJ/kg, ∆h =213.8; qrev=160.7 kJ/kg

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29

Turbines and

Compressors Turbine drives the electric generator In steam,gas, or hydroelectric power plants.

As the fluid passes through the turbine, work is

done on the blades, which are attached to theshaft. As a result, the shaft rotates, and the

turbine produces work.

Compressors, as well as pumps and fans,

are devices used to increase the pressure of a

fluid. Work is supplied to these devices from an

external source through a rotating shaft.

A fan increases the pressure of a gas slightly

and is mainly used to move a gas Pout./Pin


16/16

31

SOME STEADY-FLOW ENGINEERING DEVICES

Many engineering devices operate essentially under the same conditions

for long periods of time. The components of a steam power plant (turbines,

compressors, heat exchangers, and pumps), for example, operate nonstop formonths before the system is shut down for maintenance. Therefore, these devices can

be conveniently analyzed as steady-flow devices.

At very high velocities,

even small changes invelocities can cause

significant changes in the

kinetic energy of the fluid.

10/14/2014 Dr. Florio

32

31 1

1 1

10 MPa 0.029782 m /kg

450 C 3242.4 kJ/kg

P SH

T h

v

STEAMm = 12

kg/s

·

P 1 = 10 MPa

T 1 = 450C

V 1 = 80 m/s

P 2 = 10 kPa

x 2 = 0.92

V 2 = 50 m/s

W ·

Ex-5 Steam expands in a turbine. IF THE MASS FLOW RATE INTO THE TURBINE IS 12 KG/S, The

change in kinetic energy, the power output, and the turbine inlet area are to be determined. See Fig.Assumpt ions 1 This is a steady-flow process since there is no change with time.

2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Propert ies From the steam tables (Tables A-4 through 6)

and

kJ/kg2392.52392.10.92191.8192.0

kPa1022

2

2

fg f h xhh

x

P

Analysis (a) The change in kinetic energy is determined from

2 22 2

2 1

2 2

50 m/s (80 m/s) 1 kJ/kg

2 2 1000 m /s

V V ke

J

1.95 kJ/ kg

m m m1 2

10/14/2014 Dr. Florio

Chapter5 Notes ME311 F14 PJF

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