05 chapter 5 notes first law of thermodynamics

7/27/2019 05 Chapter 5 Notes First Law of Thermodynamics

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Adiabatic work

SystemSystem

Work Transfer

SurroundingsSurroundings

Adiabatic Boundary

4

Equivalence of adiabatic processes

First LawFirst Law-- AllAlladiabatic processesadiabatic processesbetween twobetween two

equilibrium statesequilibrium statesrequire the samerequire the sameadiabatic workadiabatic work to beto be

done on the systemdone on the systemregardless of theregardless of thepathpath

State 1State 1

State 2State 2


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First Law Statement

The adiabatic work done onThe adiabatic work done on

a closed system to connecta closed system to connect

two equilibrium statestwo equilibrium states

is independent of path, thereforeis independent of path, therefore

Identifies a change in a thermodynamicIdentifies a change in a thermodynamic

property called energy.property called energy.

6

The First Law

,1 2 2 1adbyW E E =

Work doneWork doneby (out of)by (out of)the systemthe system

isis

positive by conventionpositive by conventionminusminus

work by is work on (in)work by is work on (in)..


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For a non adiabatic process, the Heat

transfer to the system, Q, is defined by

,1 2 ,1 2

,1 2 2

,1 2

,1 2 1

H eat transfer is def ined asthe d i ffe rence be tw een the w ork byand the ad iaba t ic w ork byfor the sam e chang e in s ta te

ato b yy

o

d b

b yt

Q W W

Q W E E o r

=

= +

8

Define Internal energy, U change between two states at

at the same extrinsic state , i.e. the same velocity andposition but different intrinsic properties.

,1 2 2 1adbyW U U

Work done by the system is positive byWork done by the system is positive by

convention.convention. KEKE11--22 = 0, and= 0, andPEPE11--22= 0= 0

Which are the adiabatic work on the system atWhich are the adiabatic work on the system at

same intrinsic state but different extrinsic state.same intrinsic state but different extrinsic state.


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11

MotorinW

fromQ

inmass sys out motorsystem

m

W W

PE

=

=

mm

Example 1Example 1

For steady operation, thereFor steady operation, there

is no change of the state ofis no change of the state of

the system, and for the motor systemthe system, and for the motor system

0in out loss

W W Q =

12

Example 2 - unsteady

MechanicalMechanicalWork to RaiseWork to Raise

the Mass, mthe Mass, m

Heat Loss fromHeat Loss fromfrom Motorfrom Motor

Force = mgForce = mgBattery has storedBattery has storedenergy. It is releasedenergy. It is releasedvia electrical current.via electrical current.

MBB

mm

System BoundarySystem Boundary


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Example 3: Q5 Q1 W=Ebattery + EmotorQQ55

To correctly analyze this arrangement, all energy & work transfersTo correctly analyze this arrangement, all energy & work transfers

must be counted. Where is the primary energy source? What aremust be counted. Where is the primary energy source? What are

the possible systems for analysis?the possible systems for analysis?

B

W

Motor

QQ11

QQ22

QQ33

QQ44

Fan

Battery

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Cyclic Processes

P1

P2

SS11

SS22

21,2121 = byWQdU

1 2 1 1 2 1 1 2 1

0by

dU Q W

= =


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Initial SystemInitial System

State 1State 1

Deformed SystemDeformed System

BoundaryBoundary

State 2

xx

yy

zz P1

1

2

P3

P2

Process PathProcess Path

dW=PdAdN

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MOVING BOUNDARY WORKReversible-Moving boundary work(P dVwork): The expansion andcompression work in a piston-cylinderdevice. Observer moving with the system

The work associatedwith a moving

boundary is calledboundary work.A gas does a

differentialamount of workWbas it forcesthe piston tomove by adifferentialamount ds.

Quasi-equilibrium process:A process during which the systemremains nearly in equilibrium at alltimes.

Wbis positive for expansionWbis negative for compression


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The area under the processcurve on a P-Vdiagramrepresents the boundary work.

The boundarywork done

during a processdepends on the

path followed aswell as the end

states.

The net work doneduring a cycle is thedifference between

the work done bythe system and the

work done on thesystem.

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Polytropic, Isothermal, and Isobaric processes

Polytropic process: C, n(polytropic exponent) constants

N=1Polytropicprocess

Polytropic - for ideal gas

When n= 1(represents anisothermal process

for-ig )

Schematic andP-Vdiagram for

a polytropicprocess.

Constant pressure process


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ENERGY BALANCE FOR CLOSED SYSTEMS

Energy balance for any systemundergoing any process

Energy balancein the rate form

The total quantities are related to the quantities per unit time is

Energy balance per unit mass basiswhere represents an amount of

rather than a change of.Energy balance indifferential form

Energy balancefor a cycle

in by systemq w de =

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Equations

In solving thermodynamics problems

Define the problem

Examine data. Is the substance known? Are two intensive propertiesidentified at the each end state?

Can the work be evaluated or is it known ?

Can heat transfer be evaluated or is it known?

Write the energy equation in form appropriate to the system. What modeling approximations can be made or need to be made?

Write the conservation of mass equation, and if possible relate to thegiven data.

Are there a sufficient number of equations for the # of unknowns?

In some cases additional fundamental laws or particular equations mustbe applied- i.e. equation of state ( or equivalent use of tables) andfurther modeling assumptions made.

Verify or check validity of assumptions, i.e. iterate.


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Scientific Method- eight step format

1. Define the problem

2. Collect the data.

3. Analyze the data

4. Form an hypothesis

5. Test the hypothesis

6. Analyze the results

7. Accept or reject the hypothesis

8. If necessary, iterate

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net-in net-out system

Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies

in ,ou 2 1

in ,ou 2 1

( ) (since KE=PE=0)( )

b t

b t

Q W E

Q W U m u uQ W m u u

=

= =

= +

P

v

1

2300 kPa

75 kPa

45 A saturated water mixture at 75 kPa , 13 % quality with an initial volume of 2 m3 is contained in a

linear spring-loaded piston-cylinder device It is heated until the pressure is 300 kPa and volumeincreases to 5 m3. The heat transfer and the work done are to be determined.

Assumptions1 The cylinder is stationary and thus the kinetic and potential energy changes are

zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energystored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-

equilibrium.

Analysis We take the contents of the cylinder as the system. This is a closed

system since no mass enters or leaves. The energy balance for this stationary

closed system can be expressed as

Peq = A + B Vol


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The initial state is saturated mixture at 75 kPa. The specific volume and internal energy at this state are (Table B1.2),

kJ/kg89.658)8.2111)(13.0(36.384

/kgm28914.0)001037.02172.2)(13.0(001037.0

1

31

=+=+=

=+=+=

fgf

fgf

xuuu

xvvv

The mass of water is

kg9170.6/kgm28914.0

m23

3

1

1===

v

Vm The final specific volume is

/kgm72285.0kg9170.6

m5 33

22 ===

m

Vv

The final state is now fixed. v> v g The internal energy at this specific volume and 300 kPa pressure is(Table B1.3)

kJ/kg2.26572 =u

1 2 3,out 2 1

3

With a linear spring, the quasi-equilibrium P varies linearly with the volume, thus(75 300)kPa 1 kJ

( ) (5 2)m2 2 1 kPa m

b

P PW PdVol Vol Vol

+ + = = = =

562.5 kJ

Substituting into energy balance equation gives

kJ14,385=+=+= kJ/kg)89.658kg)(2657.29170.6(kJ5.562)( 12out,in uumWQ b

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Energy balance when sign convention is used (i.e., heat input and workoutput are positive; heat output and work input are negative).

Various forms of the first-law relationfor closed systems when signconvention is used.

For a cycle E =0, thus Q = W.

The first law cannot be proven mathematically, but no process in nature is knownto have violated the first law, and thus until a violation of the law is observed, itshould be taken as valid.

Energy EQ Using Sign General Convention

of 1st LawGeneral finite -period-

: /

FormsQ W E

Rate form Eq Q W DE Dt

NeglectingKE PE Q W Umass base q w eDifferentialmassbase q w de

=

=

=

=

=


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TANK A2 kg

1 MPa300C

TANK B3 kg

150Cx=0.5

Q

42 Two tanks initially separated by a partition contain steam at different states. Now the partition isremoved and they are allowed to mix until equilibrium is established. If the final pressure is 300 kPa ,

the temperature and quality of the steam at the final state and the amount of heat lost from the tanks

are to be determined.

Assumptions1 The tank is stationary and thus the kinetic and potential energy changes arezero. 2 There are no work interactions.

Analysis(a) We take the contents of both tanks as the system. This is a closed system since no

mass enters or leaves. Noting that the volume of the system is constant and thus there is noboundary work, the energy balance for this stationary closed system can be expressed as

The properties of steam in both tanks at the initial state are (Tables B1.1 through B1.3)kJ/kg7.2793

/kgm25799.0

C300

kPa1000

,1

3

,1

,1

,1

=

=

=

=

A

A

A

A

uT

P v

( )[ ]

( ) kJ/kg4.15954.19270.50.66631

/kgm0.196790.0010910.392480.500.001091

kJ/kg4.1927,66.631

/kgm.392480,001091.0

50.0

C150

1,1

31,1

3

1

,1

=+=+=

=+=+=

==

==

=

=

fgfB

fgfB

fgf

gfB

uxuu

x

uux

T

vvv

vv

[ ] [ ]

in out system system


2

out 2 1 2 1

[ ) ) ]( ) ( )

(with

A B

A B

A B A B

Q W E U

m m mU mu mu muQ U U U m u u m u u

W

= =

= +

= +

= = + = +

KE=PE=0)=

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Two intensive properties are required to define final state. Assume the fina l state is an equilibrium state.The total volume and total mass of the system are

kg523

m106.1/kg)m19679.0kg)(3(/kg)m25799.0kg)(2( 333,1,1

=+=+=

=+=+=+=

BA

BBAABA

mmm

mm vvVVV

Now, the specific volume at the final state may be determined

/kgm22127.0kg5

m106.1 33

2 ===m

Vv

which fixes the final state and we can determine other properties. Table B1.2 SinceVf


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Energy balance for a constant-pressure

expansion or compression process

HWU b =+

For a constant-pressure expansionor compression process:

An example of constant-pressure, Pequ,bdy condition-not reversible-no piston-friction

General analysis for a closed systemundergoing a quasi-equilibriumconstant-pressure process. Qis tothesystem and Wis fromthe system.

Wb

Weee0.3A

Since frictional effects neg

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Net out system


e,in pw,in

e,in pw,in 2 1

(since =KE=PE=0)

( )

(

NetoutQ W E

W W PAdh U Q

W W m h h

I t

=

+ =

+ =

Volt pw,in 2 1) ( )W m h h+ =

H2OPequil.

WpwWe

v

P

1 2

----39393939 An insulated cylinder is initially filled with 5 L ofsaturated liquid water at a pressure of175 kPa. The water is heated electrically with a

current of8A for 45 min through a resistor as it is stirred by a paddle-wheel with constant pressureboundary condition. Ifone half of the

liquid is evaporated during this process and the paddle wheel work is 400 kJ, determine the voltage of the current source . Show the

process on a P-vdiagram.

AssumptionsAssumptionsAssumptionsAssumptions1111 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2222 The cylinder is well-insulated and

thus heat transfer is negligible. 3333 The thermal energy stored in the cylinder itself is negligible. 4444 The piston moves with no friction and

negligible acceleration.

AnalysisAnalysisAnalysisAnalysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance

for this stationary closed system can be expressed as

U+ Wb = HSince the initial and final P is the same a nd the endstates are-equilibrium states. The properties of water ar e (Tables

B1.1through B1.3)

( )

kg4.731/kgm0.001057

m0.005

kJ/kg1593.61.22135.001.4875.0

kPa175

/kgm0.001057

kJ/kg487.01

liquidsat.

kPa175

3

3

1

1

222

2

3kPa175@1

kPa175@11

===

=+=+=

=

=

==

==

=

v

V

vv

m

hxhhx

P

hhP

fgf

f

f

Substituting,

(400kJ) (4.731 kg)(1593.6 487.01)kJ/kg

4835 kJ

4835 kJ 1000 VA

(8 A)(45 60 s) 1 kJ/s

elect

elect

W

I t

+ =

=

= =

W = V

V 223.9 V


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Steam0.15 kg3.5 MPa Q

154 A piston-cylinder device contains 0.15 kg steam at 3.5 MPa , superheated by 5 C in a cylinder with a set of stops.

As Heat is lost from the steam, the piston moves down and hit the stops at which point the state is saturated liquid. It is

cooled further until a temperature of200 C is reached. The final pressure and quality (if mixture), the boundary work,

and the heat transfer until the piston first hits the stops and the total heat transfer are to be determined.

Assumptions1 The kinetic and potential energy changes are negligible,. 2 The friction between the piston and the cylinder is negligible.

Analysis(a) We take the steam in the cy linder to be the system. This is a closed system since no masscrosses the system boundary. The energy balance for this stationary closed system can be ex pressed as

net-in net-out system system


b,in out (since KE PE 0)

Q W E U

W Q U

= =

= = =

Denoting when piston first hits the stops as state (2) and the final state as (3), the energy balance relations may be written as

30

)u-(

)u-(

133-out,1inb,

122-out,1inb,

umQW

umQW

=

=

The properties of steam at various states are (Tables B1.1 through B 1.3)[email protected] MPa

1

242.56 C

242.56 5 247.56 Csat

T

T T dSH

=

= + = + =

kJ/kg3.2617

/kgm05821.0

C56.247

MPa5.3

1

31

1

1

=

=

=

=

uT

P v

kJ/kg4.1045

/kgm001235.0

0

MPa5.3

2

32

2

12

=

=

=

==

ux

PP v

kJ/kg55.851C200

/kgm001235.0

3

3

3

3

323

=

=

=

=

==

u

P

x

TkPa1555

0.00062vv

(b) Noting that the pressure is constant until the piston hits the stops during which the boundary work is done, it can be determined from

its definition as

kJ29.91=== 3211inb, 0.001235)m21kPa)(0.058kg)(350015.0()( vvmPW

(c) Substituting into energy balance relations, yields Q out 1-2 = m (h1h2 ) or

kJ265.7== kJ/kg)3.26174.1045(kg)15.0(kJ91.292-out,1Q

(d) kJ294.8== kJ/kg)3.261755.851(kg)15.0(kJ91.293-out,1Q

Specific volume comparison

out,1-2 2 1 b,in,1-3

out,1-3 3 1 b,in,1-3 out,1-2 out,2-3

( -u )+( -u )+

Q m u W

Q m u W Q Q=

= = +


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SPECIFIC HEATS- properties of a substance

Specific heat at constant volume, cv: Informal definition-The energyrequired to raise the temperature of the unit mass of a substance byone degree as the volume is maintained constant.

Specific heat at constant pressure, cp: Informal definition-The energyrequired to raise the temperature of the unit mass of a substance byone degree as the pressure is maintained constant.

Specific heat is a propertythat can be qrev/dT as the

energy required to raise thetemperature of a unit massof a substance by onedegree in a specified way.

Constant-

volume andconstant-

pressure specificheats cvand cp(values are for

helium gas).

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The equations in the figure are valid for anysubstance undergoing anyprocess.

cvand cpare properties.

cvis related to the changes in internal energyand cpto the changes inenthalpy.

A common unit for specific heats is kJ/kg C or k J/kg K. Are these unitsidentical?

The specific heat of a substancechanges with temperature.

cpis always greater orequal to cv.

Formal definitions of cvand cp.


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If the heat transfer results in changes of the translational ke of the molecule, thespecific heat will remain constant; but if some of the input energy is diverted to the

rotational or viberational modes than the specific heat will increase, pjf

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v1

v2

U(T,v)

T

T

u

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INTERNAL ENERGY, ENTHALPY,AND SPECIFIC HEATS OF IDEAL GASES

For ideal gases, u,h, cv, and cpvarywith temperatureonly(Tables--.

Internal energy andenthalpy change of

an ideal gas

General internal energy relation for apure , simple compressible substance

( , )

/ [ / ]v v

u T v

du u T dT T P T P dv

=

= +

Molecular-kinetic + potential


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Specific heats

Quantities are defined that are related to the variables applicable toreversible work.

The specific heat at constant volume for a simple compressible sub isdefined to be cv = u/T)v at constant displacement (volume).

For a reversible constant volume process this is equivalent to c v = qrev/dT)v

Another specific heat is defined in terms of the Thermodynamic Force

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Specific heat at const force

The specific heat at constant pressure for a scs is defined as cP = h/T)P ,at const

Thermo force. [h=u +p v>u]

For a reversible process,

this is equivalent to cP = qrev /T)P .

For an ideal gas u and h are functions only of temperature. Thus, so to are the specific

heats.

For an ideal gas, cp = cv + R. (important)

For a gas, at any T the energy is distributed between the translational ke, rotational

ke and vibrational energy and electronic. Only the first is directly proportional to

the temperature,T.

For an ideal gas undergoing a polytropic process the equivalent specific heat is

Cn = Cv + 1/(1-N)

For a given substance rotation of the molecule begins at a critical velocity of impact . As

more molecules collide at or above the critical speed the more will be put into rotation.

When the velocities become high enough to cause the molecules to vibration and result in

an increase in the specific heat. Molecules shake , rotate while they move. Rotational

energy mostly transform to translational energy on cooling, while vibration converts to

thermal radiation.


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Diatomic molecule lower level Energy

Translation, rotation,vibration

At higher T the heat in part goes to increase the translational ke, resulting in T increase,part to rotation, then part to vibration; thus for a given amount of Q the more that goes

into other than translation results in a smaller increase in T, the ratio of dQ/dT increases due to smaller increase in T.

Translation - T

rotation not proportional to T

vibrationnot propto T

Q (sameamount)

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Specific heats for ideal gases. ~(f+2)/2

(Ideal gas tables)

tri

di

mon

rotation

vibration

translation


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Real gas Pressure effect on cp, low P ~ig as P increase deviation from ig

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Ideal-gas constant-

pressure specificheats for somegases (see TableA6 for cpequations).

At low pressures, all real gases approachideal-gas behavior, and therefore theirspecific heats depend on temperature only.

The specific heats of real gases at lowpressures are called ideal-gas specificheats, or zero-pressure specific heats, andare often denoted cp0 and cv0.

uand hdata for a number ofgases have been tabulated, A-7.

These tables are obtained bychoosing an arbitrary referencepoint and performing theintegrations by treating state 1as the reference state.

In the preparation of ideal-gastables, 0 K is chosen as thereference temperature.


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(kJ/kg)

For small temperature intervals, thespecific heats may be assumed to vary

linearly with temperature, for such acase the average value is at the

average T.

Internal energy and enthalpy change and

use the mean value theorem for specificheat average value

The relation u = cvTis valid for anykind ofprocess, constant-volume or not.

Area=h2h1

42

1. By using the tabulated uand h, A7 -8data. This is the easiest and mostaccurate way when tables arereadily available.

2. By using the cvor cprelations (TableA-6) as a function of temperatureand performing the integrations. Thisis very inconvenient for hand

calculations but quite desirable forcomputerized calculations. Theresults obtained are accurate.

3. By using average specific heats.This is very simple and certainly veryconvenient when property tables arenot available. The results obtainedare reasonably accurate if thetemperature interval is not verylarge. A.5 at 25 C

Three ways of calculating u andh

Three ways of calculating uWith varying accuracy.

For a linear variation, cp,avg =(cp(T2) + cp(T1))/2

cp


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Specific Heat Relations of Ideal Gases

The cpof an ideal gas can bedetermined from a knowledge ofcvand R.

On a molar basis

The relationship between cp, cvand R

Specificheat ratio

The specific ratio varies withtemperature, but this variation isvery mild.

For monatomic gases (helium,argon, etc.), its value is essentiallyconstant at 1.667.

Many diatomic gases, including air,have a specific heat ratio of about1.4 at room temperature.

dh= cpdT and du= cvdT

44

0peke

QO2

T1 = 25CT2 = 300C

QO2

T1 = 25CT2 = 300C

71 1-kg of Oxygen is heated from 25 to 300 C to experience a specified temperature change. The heat transfer

is to be determined for two cases., a) a constant volume process and b) a constant pressure process for both

cases assume the initial volume is 1 unit ofvolume.

Assumptions1 Oxygen is an ideal gas s ince it is at a high temperature and low pressure relative to its critical

point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible,

. 3 a. Constant specific heats can be used for oxygen, b. Variable specific heats..

PropertiesThe specific heats of oxygen at the average temperature of (25+300)/2=162.5C=436 K are cp=

0.952 kJ/kgK (Table A-6).

AnalysisWe take the oxygen as the system. This is a closed systemsince no mass crosses the boundaries of

the system. The energy balance for a constant-volume process can be expressed as

net-in net-out system 2 1


in 2 1

( )

( )v

Q W E m u u

Q U m u mc T T

= =

= = =

The energy balance during a constant-pressure process (such as in a piston-cylinder device) can be ex pressed as

in out system

Net energy transfer Change in internal, kinetic,by heat, work, potential, etc. energies

in ,out

in ,out

in , 2 1

( )( )

b

b

p avg

Q W E

Q W U

Q W P Vol U Q H m h mc T T

=

=

= = +

= = =

since U+ Wb = Hduring a constant p ressure quasi-equilibrium process. Substituting for both cases,


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45

in, const , 2 1( ) (1 kg)(0.952 kJ/kg K)(300 25)K

P p avgQ mc T T

== = = 261.8 kJ

in, const 2 1( ) (1 kg)(0.692 kJ/kg K)(300 25)KQ mc T T = = = = 190.3 kJV v,avg

46

INTERNAL ENERGY, ENTHALPY, ANDSPECIFIC HEATS OF SOLIDS AND LIQUIDS

The specific volumes ofincompressible substancesremain constant during aprocess.

The cvand cpvalues ofincompressible substances areidentical and are denoted by c.

Incompressible substance: A substance whose specific volume(or density) is constant. Solids and liquids are approximatelyincompressible substances ; (T). dhp = du(T)


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Internal Energy Changes for an incompressiblesubstance u(T,v) ; du=

Enthalpy Changes; cv=cp=c

The enthalpy of acompressed liquid

Which is more accurate relation than

/ ) / )v T

u T dT u v dv +

cv=cp=cdhp = dup or cp = cv =c

48

in out system in

Net energy transfer Change in internal, kinetic,by heat, work, and mass potential, etc. energies

waterE E E Q U = =

Food

Reactionchamber

Water

T = 3.2C

141 A 2 g sample of a food is burned in a bomb calorimeter, that contains 3 kg of water. The

water temperature rises by 3.2C when equilibrium is established in a reaction chamber that

contains and 100 g of air.. The energy content of the food is to be determined in kJ/kg by

neglecting the energy supplied to the mixer and the thermal storage in the reaction chamber.

Assumptions1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constantspecific heats. 3 The energy stored in the reaction chamber is negligible relative to the energy stored in water. 4 Theenergy supplied by the mixer is negligible.PropertiesThe specific heat of water at room temperature is c= 4.18 kJ/kgC (Table A-4). The constant volume specificheat of air at room temperature is c

v= 0.718 kJ/kgC (Table A-5).

AnalysisThe chemical energy released during the combustion of the sample is transferred to the water as heat.Therefore, disregarding the change in the sensible energy of the reaction chamber, the energy content of the food issimply the heat transferred to the water. Taking the water as our system, the energy balance can be written as

( )( ) ( )

food

2 1water water

[outofreaction ChemicalE in water

in Water

Q E Q

Q U mc T T

= =

= = Substituting,

Qin = (3 kg)(4.18 kJ/kgC)(3.2C) = 40.13 kJfor a 2-g sample. Then the energy content of the food per unit

mass is

40.13 kJ 1000 g

2 g 1 kg

=

20,060 kJ/kg- food

To make a rough estimate of the error inv olved in neglecting the thermal energy stored in the contents of the reactionchamber, we treat the entire mass within the chamber as air and determine the change in sensible internal energy:

( ) ( )[ ] ( )( ( kJ0.23C3.2CkJ/kg0.718kg0.102chamber12chamber

===

TTmcU v

which is less than 1% of the internal energy change of water. Therefore, it isreasonable to disregard the change in the sensible energy content of the reactionchamber in the analysis.

&

Watersys

For reaction chamber


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Summary

Moving boundary work

Wbfor an isothermal process

Wbfor a constant-pressure process

Wbfor a polytropic process

Energy balance for closed systems

Energy balance for a constant-pressure expansion orcompression process

Specific heats

Constant-pressure specific heat, cp Constant-volume specific heat, cv

Internal energy, enthalpy, and specific heats of ideal gases Specific heat relations of ideal gases

Internal energy, enthalpy, and specific heats ofincompressible substances (solids and liquids)

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05 chapter 5 notes first law of thermodynamics

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