felix termodinamica quimica ch03

8/13/2019 felix termodinamica quimica ch03

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Chapter 3 Solutions Engineering and Chemical Thermodynamics

Wyatt TenhaeffMilo Koretsky

Department of Chemical EngineeringOregon State University

[email protected]


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3.1Since entropy is a state function

2,1, stepsysstepsyssys sss +=

Step 1 is a constant volume process. Therefore, no work is done. After neglecting potential ankinetic energy effects, the energy balance for a reversible process becomes

qu = qdu = qdT cv =

Using the definition of entropy,

==

final

initial

T

T

vrevstepsys

T

dT c

T

qs

2

1

1,

Step 2 is an isothermal process. For and ideal gasu is zero and the energy balance is (PE andKE neglected)

wq = wdq =

For an ideal gas, the following can be shown

dvv RT

Pdvwrev ==

Therefore,

===

1

22, ln

2

1vv

Rdvv

RT q

s final

initial

v

v

revstepsys

Combination of both steps yields

+= 1

2ln2

1vv R

T dT cs

T

T

vsys


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3.2

Equation 3.62 states

=

2

112

ln

T

T

Psys P

P RdT T

cs

Substituting the equation forPc yields

++=

2

112

2ln

T

T sys P

P RdT

T CT BT A

s

( ) ( )

++

=

122

12212

12 ln

2ln

PP

RT T C

T T BT T

Assys


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3.3Equation 3.3 states

surr sysuniv S S S +=

where

)( 12 ssmS sys =

surr

surr surr T

QS = (the temperature of the surroundings is constant)

First, we will calculate sysS . From the steam tables:

=K kg

kJ 1228.71s (300 C, 10 bar)

Since the container is rigid and mass is conserved in the process,

12 vv =

=kgm 25794.0

31v (300 C, 10 bar)

=kgm 25794.0

32v

To find the number of phases present in state 2, compare the specific volume of state 2 to thespecific volume for saturated water and saturated water vapor at 1 bar. From the steam tables,

=kgm 001043.0

3,

2sat lv (sat. H2O(l) at 1 bar)

=kgm 6940.1

3,

2sat vv (sat. H2O(v) at 1 bar)

Since sat vsat l vvv ,22,

2


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5

152.0= x

From the steam tables:

= K kgkJ 3025.1,2 sat ls (sat. H2O(l) at 1 bar)

=K kg

kJ 3593.7 ,2sat vs (sat. H2O(v) at 1 bar)

The entropy of state 2 can be calculated using the following equation:

( ) sat vsat l s xs xs ,2,

22 1 +=

=K kg

kJ 2231.22

s

Now the entropy change of the system can be calculated.

[ ]( ) =

=K kJ 0.49

K kgkJ 1228.7

K kgkJ 2231.2kg10sysS

To find the change in entropy of the surroundings, an energy balance will be useful. Since nowork is done, the energy balance is

QU = We also know

QQsurr =

To following expression is used to solve for the internal energy:

( )( 1,2,2 1 u xuu xmU sat vsat l +=

From the steam tables

=kgkJ 2.27931u (300 C, 10 bar)

=kgkJ 33.417 ,2

sat lu (sat. H2O(l) at 1 bar)


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=kgkJ 1.2506 ,2

sat vu (sat. H2O(v) at 1 bar)

The expression for internal energy yields

[ ]kJ77.20583=U

Therefore,

[ ]kJ77.20583=Q

and

[ ][ ]

===K kJ 25.70

K 293kJ77.20583

surr

surr surr T

QS

Now the change in entropy of the universe can be calculated

=+=+=K kJ 25.21

K kJ 25.70

K kJ 0.49surr sysuniv S S S

SinceS univ > 0, this process is possible.


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3.4Entropy Balance:


Considering the copper block to be the system, no work is done on the system; thus, the energy balance is

qdu = (neglecting PE and KE)qdT cv =

This can be used in the following expression for entropy

=2

1

T

T

revsys T

qs

=2

1

T

T

vsys dT T

cs

From Table A.2.3

RcP 723.2= Rcc Pv 723.2== (for liquids and solids)

Therefore,

[ ][ ]

=

== K 15.373

K 280lnK mol

J 314.8723.2ln723.2723.212

2

1T T

RdT T

Rs

T

T sys

=K mol

J 50.6syss

[ ] =

=K J 1024

K molJ 50.6

molkg

063465.0

kg10sysS

Since the temperature of the lake remains constant, the change in entropy of the surroundings c be calculated as follows

( ) ( )surr

Cu

surr

PCu

surr surr

surr surr T

T T RnT

T T cnT

QT Q

S 1212723.2 ====


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[ ] [ ] [ ]( )

[ ]

=K 280

K 15.373K 280K mol

J 314.8723.2

molkg 0.063546

kg10surr S

=K J 1184surr S

From the definition of entropy:

=+=+=K J 160

K J 1184

K J 1024surr sysuniv S S S


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3.5

(a)Since the process is adiabatic and reversible,

0= sysS

(b)The change in entropy is calculated by

( )=

=== K J 9.73

K 111molkJ2.8mol1liquid

vapor b

vaprev

sys T hn

T q

nS

The sign is negative because there is less randomness in the liquid phase.

(c)For this situation

dT cq P=

Therefore,

( )

=== K 15.373K 15.273ln

K gJ2.4g0148.18

K 15.273

K 373.15

f

i

T

T

Prevsys dT T

cT q

mS

=K J6.23sysS

The sign is negative because as the water cools, less translational energy states are occupied bythe molecules. Therefore, the randomness decreases.

(d)First, calculate the temperature at which the blocks (block A and block B) equilibrate. Theenergy balance for the process is

( ) ( ) 0K 15.473K 15.373 22 =+ T cnT cn P BP A Since the heat capacities number of moles of A and B are equal, we find that

K 15.4232 =T

Now, calculate the change in entropy:


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Bsys Asyssys S S S ,, +=

+

=

BP B

AP Asys T

T cn

T T

cnS ,12

,12 lnln

Substituting values, we obtain

=K J 337.0sysS

The sign is positive because two objects at different temperatures will spontaneously equilibratto the same temperature when placed together.


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3.6

The solution below compares problems 2.14 and 2.15, the calculation of 2.13 was erroneouslyincluded in the problem statement of the first printing and is shown at the end of this problem.

Problem 2.14Since the system is well-insulated no heat is transferred with the surroundings. Therefore, theentropy change of the surroundings is zero and

sysuniv S S =

The gas in the piston-cylinder system is ideal andPc is constant, so

=

12

12 lnln

PP

RT T

cnS Psys

From the problem statement, we know

[ ][ ] bar 1 bar 2

2

1==

P

P

The ideal gas law can be used to solve1T .

[ ]( )( )

[ ]( )[ ]K 6.240

mol1K mol

bar L 0.08314

L10 bar 2111 =

==

nRV P

T

Solving for 2T is slightly more involved. The energy balance for this system where potential andkinetic energy effects are neglected is

W U =

Conservation of mass requires

21 nn = Let

21 nnn ==

The energy balance can be rewritten as

=2

1

2

1

V

V E

T

T v dV PdT cn


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Since vc and E P are constant

( ) ( )1212 V V PT T nc E v =

2V and 1T can be rewritten using the ideal gas law

22

2 PnRT

V =

nRV P

T 111 =

Substituting these expressions into the energy balance, realizing that 2PP E = , and simplifyingthe equation gives

nR

V PPT

27

25

1122

+

=

Using the following values

[ ][ ]

[ ][ ]

=

==

==

K mol bar L 08314.0

mol0.1 L10

bar 1 bar 2

1

2

1

R

nV

P

P

results in

[ ]K 2062 =T

Since both states are constrained, the entropy can be calculated from Equation 3.63:

[ ]( ) [ ][ ][ ][ ]

==

bar 2 bar 1ln

K molJ 314.8

K 6.240K 206ln

K molJ 314.85.3mol0.1sysuniv S S

==K J 24.1sysuniv S S


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Problem 2.15Since the initial conditions in the piston-cylinder assembly are equal to the initial conditions ofProblem 2.14, 1T and 1P are known. Moreover,2P is known, so we only need to find2T inorder to calculate the entropy change. For adiabatic, reversible processes, the followingrelationship (Equation 2.89) holds:

const PV k =

This can be used to find2V .

k k V PP

V

1

121

2

=

Noting that5

7==v

P

c

ck and substituting the proper values results in

[ ]L4.162 =V

Th polytropic expression can also be manipulated to yield

.1 const TV k =

Therefore,

12

1112

=k

k

V

V T T

Substitution of the appropriate variables provides

[ ]K 4.1972 =T

Now the entropy can be calculated,

[ ]( ) [ ]

[ ][ ][ ]

== bar 2 bar 1

lnK molJ

314.8K 6.240K 97.41

lnK molJ

314.85.3mol0.1sysuniv S S

==K J 004.0sysuniv S S

The value of the entropy shown represents round-off error. Since the process is reversible andadiabatic, we know from Table 3.1 and the related discussion in Section 3.3 that the entropychanges of the system, surroundings, and universe will be zero.


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Problem 2.13

Equation 3.3 states

surr

surr surr sysuniv

T

QssmS S S +=+= )( 12

where the temperature of the surroundings is constant. First, we will determine2s . The first lawcan be applied to constrain state 2. With potential and kinetic energy effects neglected, theenergy balance becomes

W QU +=

The value of the work will be used to obtain the final temperature. The definition of work(Equation 2.7) is

=2

1

V

V E dV PW

Since the piston expands at constant pressure, the above relationship becomes

( )12 V V PW E =


=K kg

kJ 2119.61s (10 MPa, 400 C)

=kgm 02641.0

31v (10 MPa, 400 C)

[ ]33111 m07923.0kgm 02641.0kg)3( =

== vmV

Now 2V and 2v are found as follows

[ ]36312 m4536.0Pa100.2J748740m07923.0 =

==

E PW

V V

[ ] ===

kgm 1512.0

kg3m4536.0

33

22

2 mV

v


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Since 2v and 2P are known, state 2 is constrained. From the steam tables:

=K kg

kJ 1270.72s

kg

m 0.1512 bar,203

Now U will be evaluated, which is necessary for calculatingsurr Q . From the steam tables:

=kgkJ 2.29452u

kgm 0.1512 bar,20

3

=kgkJ 4.28321u ( )C400 bar,001

( ) [ ]( ) [ ]kJ4.338kgkJ 4.2832

kgkJ 2.2945kg3 121 =

== uumU

Substituting the values ofU and W into the energy equation allows calculation of Q

W U Q = [ ]( ) [ ] surr QQ === J1009.1J748740[J]338400 6

so

[ ]( ) [ ][ ] =

=

+=

K kJ 126.1

K 15.673kJ1009.1

K kgkJ 2119.6

K kgkJ .12707kg3

)(

3

12surr

surr univ T

QssmS

Therefore, this process is irreversible.


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3.7(a)

From Equation 3.63:

=

1

2

1

2 lnlnPP R

T T cs Psys

[ ][ ]

[ ][ ]

=

=

K molJ 63.20

bar 1 bar 5.0ln

K 300K 500ln

27

K molJ 314.8syss

(b)From Equation 3.65:

+

=1

2

1

2 lnlnv

v R

T

T cs vsys

[ ][ ]

=

+

=

K molJ 85.4

molm 05.0

molm .0250

lnK 300K 500ln

25

K molJ 314.8

3

3

syss

(c)First, we can find the molar volume of state 1 using the ideal gas law.

== molm 025.03

111 P RT v

Now, we can use Equation 3.65

+

=

1

2

1

2 lnlnvv

RT T

cs vsys

[ ][ ]

=

+

=K mol

J 62.10

molm 025.0

molm .0250

lnK 300

K 500ln2

5

K mol

J 314.83

3

syss


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3.8(i). We wish to use the steam tables to calculate the entropy change of liquid water as it goesfrom its freezing point to its boiling point. The steam tables in Appendices B.1 B.5 do nothave data for subcooled water at 1 atm. However, there is data for saturated water at 0.01 C ana pressure of 0.6113 kPa. If we believe that the entropy of water is weakly affected by pressure

then we can say that the entropy of water at 0.01 C and 0.6113 kPa is approximately equal to thentropy at 0 C and 1 atm. The molar volumes of most liquids do not change much with pressuat constant temperature. Thus, the molecular configurations over space available to the watermolecules do not change, and the entropy essentially remains constant. We do not need toconsider the molecular configurations over energy since the temperature difference is so slight.So, from the steam tables:

( ) ( ) = K kgkJ 0kPa.61130,01.0atm1,0 C sC s

( ) =K kg

kJ 3068.1atm1,100 C s

Therefore,

( ) ( ) == K kgkJ 0

K kgkJ 3068.1atm1,0atm1,100 C sC ss

=K kg

kJ 3068.1s

(ii). From the steam tables:

( )=K kg

kJ 3068.1atm1,100 C s l

( )=K kg

kJ 3548.7atm1,100 C s v

Therefore,

( ) ( ) == K kgkJ 3068.1

K kgkJ 3548.7atm1,100atm1,100 C sC ss lv

=K kg

kJ 048.6s

The change in entropy for process (ii) is 4.63 times the change in entropy for process (i). Thereare many ways to reconcile this difference, but think about it from a molecular point of view. I process (i), the available molecular configurations over energy are increased as the temperatureincreases. As the temperature increases, the molar volume also increases slightly, so the


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available molecular configurations over space also increase. Now consider process (ii), wherethe molecules are being vaporized and entering the vapor phase. The molecular configurationover space contribution to entropy is drastically increased in this process. In the liquid state, thmolecules are linked to each other through intermolecular interactions and their motion islimited. In the vapor state, the molecules can move freely. Refer to Section 3.10 for a discussio

of entropy from a molecular view.


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3.9Before calculating the change in entropy, we need to determine the final state of the system. Lethe mixture of ice and water immediately after the ice has been added represent the system.Since the glass is adiabatic, no work is performed, and the potential and kinetic energies areneglected, the energy balance reduces

0= H

We can split the system into two subsystems: the ice (subscripti) and the water (subscriptw).Therefore,

0=+= wwii hmhm H and

wwii hmhm =

We can get the moles of water and ice.

[ ] [ ]kg399.0kgm 001003.0

m.00040 3

3

=

==

w

ww v

V m

( )[ ]( ) [ ]mol15.22

molkg 0180148.0

kg399.0

2

===O H

ww MW

mn

( ) [ ]mol55.5

2

==O H

i

i MW

mn

Now, lets assume that all of the ice melts in the process. (If the final answer is greater than 0C, the assumption is correct.) The following expression mathematically represents the changein internal energies (cP=c v).

( )( ) C250C100 ,,, =+ f wPw f wP fusiPi T cnC T chcn Note: Assumed the heat capacities are independent of temperature to obtain thisexpression.

From Appendix A.2.3 Rc iP 196.4, = Rc wP 069.9, =

and

=molkJ0.6 fush


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Substitution of values into the above energy balance allows calculation ofT f .

C 12.3= f T

(Our assumption that all the ice melts is correct.)

Now, we can calculate the change in entropy. From Equation 3.3

surr sysuniv sss +=

Since the glass is considered adiabatic,

0= surr s sysuniv ss =

We will again break the system into two subsystems: the ice and the water. The change inentropy of the universe can be calculated as follows

wwiiuniv snsnS +=

The definition of change in entropy is

= final

initial

rev

T q

s

Assuming the heat capacities of ice and water are independent of temperature, the expressionsfor the change in entropy of the subsystems are

+

+

=

K 15.273K 27.276ln

K 15.273K 15.263K 15.273ln ,, wP

fusiPi c

hcs

=

K 98.152K 27.276ln,wPw cs

Therefore

+

+

+

=

K 98.152K 27.276ln

K 15.273K 27.276ln

K 15.273K 15.263K 15.273ln ,,, wPwwP

fusiPiuniv cnc

hcnS

Substituting the values used before, we obtain


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=mol

J 59.6univS


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3.10

(a)The maximum amount of work is obtained in a reversible process. We also know the entropychange for the universe is zero for reversible processes. From the steam tables

=K kg

kJ 2119.61s (400 C, 100 bar)

=K kg

kJ 5434.82s (400 C, 1 bar)

Using the entropy criterion,

( ) surr univ S ssmS +== 12 0

( ) [ ]( )

== K kgkJ

2119.6K kgkJ

5434.8kg5.0 12 ssmS surr

=K kJ 1658.1surr S

Since the process is isothermal, we know that the temperature of the surroundings is constant at400 C. Therefore,

==K kJ 1658.1surr

surr

surr S T Q

[ ]kJ76.784=surr Q

To find the work obtained, perform an energy balance. An energy balance where potential andkinetic energy effects are neglected is

W QU +=

We can calculate the change in internal energy from the steam tables, and we also know thatsurr QQ = . From the steam tables:

= kgkJ 4.28321u (400 C, 100 bar)

=kgkJ 8.29672u (400 C, 1 bar)

Therefore,


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(d)Since the system is well-insulated

0= surr s

Use Equation 3.65 to calculate the change in entropy:

( )

+

=

+

==

2112

12

12

12 lnln2/5lnln

PT PT

RT T

RV V

RT T

css vsysuniv

Substituting values, we obtain,

=K mol

J 354.0syss

(e)Because the change in entropy of the universe is equal to change in entropy of the system, whicis positive in this situation, the second law is not violated.


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3.12

(a)To calculatessys, we need to take the gas from state 1 to state 2 using a reversible process. The process in part a can be drawn as follows:

State 1

24.6 bar500 K

State 2

34.4 bar557 K

State I

P I T I = 557 K

rev.adiabatic rev.

isothermal

We need to find the intermediate pressure,P I where we end up at the temperature in state 2,T 2.To find it, we can use the results from pages 78-79 of the text:

( ) ( ) ( ) k k I

k k I I

k k PT PT PT /12/1/1

11 ==

Therefore,

( )

[bar]9.3511

2

1

=

=

PT T

P

k k

I (I)

If we draw the process on aPT diagram, we get:

20

40

490 560

1

2

T

P

I

Adiabatic compression Isothermalexpansion


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The change in entropy can be represented as follows:

isothermaladiabaticsys sss +=

For the reversible adiabatic process, the change in entropy is zero. Therefore,

isothermalsys ss =

For an isothermal process,

dPP

RT wq revrev ==

Therefore,

=

=== K molJ 354.0ln

22

I

P

P

final

initialsys P

P RdPP

RT qs

I

If we substitute relation (I) in the expression above, we get the same expression in the book:

( )

=

+

=

1

2

1

2

1

2

1

2 lnlnln1

lnPP

RT T

cT T

Rk

k PP

Rs Psys

(b)For this construction, we use the path diagrammed on the followingPT diagram:

20

40

490 560

1

2

T

P

Isobaric heating

Isothermalcompression

We write


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isothermalisobar sys sss +=

First, find an expression for the isobaric heating

dT cdhq Prev ==

Therefore,

==

12ln

2

1T T

cdT T c

s P

T

T

Pisobar

Now, we need an expression for the isothermal, reversible expansion

Pdvwq revrev == Therefore,

=

===

=

=

=

= 12

/

/ 21

/

/lnln

222

121

222

12PP

RPP

Rdvv

Rdv

T P

sP RT v

P RT v

P RT v

P RT visothermal

I

Combine the two steps:

= = K molJ 354.0lnln 12

1

2PP R

T T cs Psys

(c)For this construction, we use the path diagrammed on the followingPT diagram:


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20

40

490 560

1

2

T

P

Isochoric heating

Isothermalcompression

I

For isochoric heating followed by an isothermal expansion, the entropy can be expressed asfollows:

isothermalisochoricsys sss +=

For the reversible, isothermal expansion, we obtain (refer to Parts (a) and (b) to see how this isderived)

=

I isothermal P

P Rs 2ln

However, we can relate the intermediate pressure to the initial pressure through the ideal gas law

11

2 T P

T P I =

or

[bar]4.27= I P

=

=

=

12

12

1

1

222 lnlnlnln

PP

RT T

RP

T T

P R

PP

Rs I

isothermal

For the isochoric process:

dT cq vrev =

Therefore,


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3.13

(a)Since the process is reversible and adiabatic, the entropy change for the process is zero.

0= syss Furthermore, since the process is adiabatic, the changes in entropy of the surroundings and theuniverse are also zero.

(b)

Since this process is isentropic (s=0), we can apply an expression for the entropy change of anideal gas.

s =0 = cP

T T 1

T 2

dT R ln P2P1

Insert the expression for cP from Appendix A

+==

1

2263

ln10227.010506.0639.302

1PP

RdT T

T T Rs

T

T

which upon substituting

[ ][ ]

[ ] bar 06.12 bar 1

K 250

2

1

1

===

P

P

T

yields

[ ]K 4822 =T

(c)An energy balance gives

( ) ===2

1

2

1

1T

T

T

T

Pv dT cdT cuw


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3.14The problem statement states that the vessel is insulated, so we can assume that heat transfer tothe surroundings is negligible. Therefore, the expression for the entropy change of the universis

sysuniv ss =

An energy balance will help us solve for the entropy change. Neglecting potential and kineticenergy effects, the energy balance is

W QU +=

Since the vessel is insulated, the heat term is zero. Furthermore, no work is done, so the energy balance is

( ) 0 12 == uumU 21 uu =

The values for the initial pressure and temperature constrain the value of specific energy at state1. From the steam tables,

=kgkJ 2.26191u (400 C, 200 bar)

=kgkJ 2.26192u

The problem statement provides the pressure of state 2 (100 bar). Since we know the pressureand internal energy at state 2, the entropy is constrained. Information given in the problemstatement also constrains state 1. From the steam tables,

=K kg

kJ 5539.51s (400 C, 200 bar)

=K kg

kJ 7754.52s

=kgkJ 2.2619 bar,100 2u

Therefore,

===K kg

kJ 2215.0 12 ssss sysuniv

==K kJ 2215.0sysuniv S S


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3.16

(a) A schematic of the process is shown below:

1 bar, 298 K

MixingProcess

N2 N2

N2

N2

O2

O2

N2

N2

N2

N2

N2

N2 N2

N2

O2

N2 N2

N2

N2

O2

1 bar, 298 K P2 , T 2

The tank is insulated. The change in entropy of the universe can be rewritten as

22 ,, Osys N sysuniv S S S +=

since the tank is well-insulated. After the partition ruptures, the pressure and temperature willremain constant at 1 bar and 298 K, respectively. This can be shown by employing mass andenergy balances. Therefore, the partial pressures in the system are

bar 79.02,2 = N p

bar 21.02,2 =O p

Use Equation 3.62 to determine the entropies:

=

= bar 1

bar 79.0lnK mol

J314.8ln2

2 12

K 298

K 298,

N

P N sys P

P RdT

T c

s

=K mol

J96.12, N syss

=

= bar 1

bar 21.0lnK mol

J314.8ln2

2 12

K 298

K 298,

O

POsys P

P RdT

T c

s

= K molJ98.122,Osyss

Therefore,

( ) ( )

+

=

K molJ98.12mol21.0

K molJ96.1mol79.0univS


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=K J27.4univS

(b) A schematic of the process is shown below:

2 bar, 298 K

MixingProcess

N2

N2

N2

N2 N2

N2 N2

N2

O2 N2

N2

N2

N2 N2

N2 N2

N2

O2

N2

N2

N2

N2

N2

N2 N2

N2

O2

N2

N2

N2

N2 N2

N2

N2

N2

O2

1 bar, 298 K P2 , T 2

Before calculating the change in entropy, we need to find how the temperature and pressure

change during the process. The energy balance simplifies to0=U

which can be rewritten as

( ) ( ) 0K 298K 2982222 2 =+ vOv N v N O cncnT cnn

Assuming the heat capacities are equal, we can show that

K 2982 =T

We can find the pressure after the rupture by recognizing that the tank is rigid. Therefore,

tot N O V V V =+ 22

By employing the ideal gas law, we get the following equation (it has been simplified):

2,1,122

2

2

2

2

P

nn

P

n

P

n N O

O

O

N

N +=+

Substitute values and solve to obtain

bar 65.12 =P

Now, use Equation 3.62 to calculate the entropies as was done in Part (a):


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37

=

= bar 2

bar 30.1lnK mol

J314.8ln2

2 12

K 298

K 298,

N

P N sys P

P RdT

T c

s

=K mol

J 58.32, N syss

=

= bar 1

bar 347.0lnK mol

J314.8ln2

21

2K 298

K 298,

O

POsys P

P RdT

T c

s

=K mol

J8.82,Osyss

Therefore,

( ) ( )

+

=K mol

J8.8mol21.0K mol

J58.3mol79.0univ

S

=K J68.4univS


38/104


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39

3.18For this process to work, conservation of mass and the first and second laws of thermodynamicmust hold. The subscript 1 refers to the inlet stream, 2 refers to the cold outlet, and 3refers to the hot outlet. To test the conservation of mass, perform a mass balance

321 mmm &&& +=

+=s

kg 5.1s

kg 5.0s

kg 2

Clearly, the conservation of mass holds. Now test the first law of thermodynamics by writing aenergy balance. Since there is no heat transfer or work, the energy balance becomes

0 113322 =+ hmhmhm &&& (PE and KE effects neglected)

Using the conservation of mass we can rewrite the mass flow rate of stream 1 in terms of stream2 and 3:

) ) 0 133122 =+ hhmhhm && If we assume that the heat capacity of the ideal gas is constant, the equation can be written asfollows:

( ) ( ) 0 133122 =+ T T cmT T cm PP &&

Therefore,

0K)20(s

kg 5.1K)60(s

kg 5.0 =+ PP cc

This proves that the first law holds for this system. For the second law to be valid, the rate ofchange in entropy of the universe must greater than or equal to 0, i.e.,

dS dt

univ

0

Assuming the process is adiabatic, we can write the rate of entropy change of the universe forthis steady-state process using Equations 3.48-3.50:

( ) ( )133122113322 ssmssmsmsmsmdt dS

univ

+=+=

&&&&&


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40

where a mass balance was used. For constant heat capacity, we can calculate the entropydifferences using Equation 3.63:

=

=

1

2

1

2

1

2

1

212 lnln2

5lnlnPP

T T

RPP

RT T

css P

=

=

1

3

1

3

1

3

1

313 lnln2

5lnlnPP

T T

RPP

RT T

css P

Applying these relations, we get:

( )Puniv

c R MW dt

dS 057.077.21 =

Therefore, the second law holds if RcP

6.48 ; this value is far in excess of heat capacity forgases, so this process is possible.


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41

3.19

(a)

T 1 = 640oC

V 1= 20 m/s Nozzle V 2=?

P1 = 4 MPa P 2 = 100 kPa

(b)Since the process is reversible and adiabatic,

=K J 0sysS

(c)

The steam tables can be used to determine the final temperature. From the interpolation of steatable data

( ) == K kgkJ 4692.7MPa4C,640 12 ss

Now two thermodynamic properties are known for state 2. From the steam tables:

C4.1212 =T

==K kg

kJ 4692.7MPa,1.0 22 sP

(d)An energy balance is required to calculate the exit velocity. The energy balance for the nozzle

( ) ( )1122 0 K K ehmehm ++= &&

Realizing that 21 mm && = allows the energy balance to be written as

( ) 212, hehe K K +=

which is equivalent to

( )[ ]2122 5.02 hV hV += rr

Substituting the following values


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42

=sm 201V

r

=kgJ 27191002h (0.1 MPa, 121.4 C)

= kgJ 37670001h (4 MPa, 640 C)

yields

=sm 14482V

r

Note: the velocity obtained is supersonic; however, the solution does not account for this type oflow.


43/104

43

3.20

A schematic of the process follows:

Propane inv1 = 600 cm3/molT 1= 350oC

Turbine

P2 = 1 atm

ws

Propaneout

Since this process is isentropic (s=0), we can apply an expression for the entropy change of anideal gas. We must be careful, however, to select an expression which does not assume a

constant heat capacity.

s =0 = cPT T 1

T 2 dT R ln P2

P1

Insert the expression for cP from Appendix A

+==

1

2263

ln10824.810785.28213.102

1PP

RdT T

T T Rs

T

T

We can also relate P1 to T1 and v1 (which are known) through the ideal gas law:

P1 = RT 1

v1

This leaves us with 1 equation and 1 unknown (T2). Integrating:

( ) + ==

1

1221226123

1

2 ln10412.4)(10785.28ln213.10 RT vPT T T T

T T s

UsingT1 = 623 K,v1 = 600 cm3/mol,P2 = 1 atm


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44

R=82.06 cm3 atm/mol K,

we get:T2 = 454 K

To solve forn

W S &

& we need a first law balance, with negligible ke and pe, the 1st law for a steady

state process becomes:

( ) S W Qhhn &&& ++= 210

If heat transfer is negligible (isentropic),

( ) +===2

1

2

1

26312 10824.810785.28213.1

T

T

T

T p

S dT T T RdT chhn

W &

&

integrating:

( ) ( ) ( )[ ]313262122312 10941.210393.14213.1 T T T T T T RnW S +=

&

&

Substituting

K 623TK 454

K molJ 314.8

1

2==

=

T

R

results in

=mol

J 19860n

W S &

&


45/104


46/104

46

Therefore,

+

==22 1

2

1

2 lnln N O

surr surr surr surr PP

PP

RT sT q

( )

+

=

22 bar 79.0

bar 1ln bar 21.0

bar 1lnK 15.293K mol

J 314.8 N O

surr q

=mol

J 4378.2surr q

and

=mol

J 4378.2q

Energy Balance:

0=++out

S out out in

inin W Qhnhn &&&&

Because the temperature of the oxygen and nitrogen doesnt change in the process, the energy balance per mole of feed becomes

qwS =

=mol

J 4378.2S w


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47

3.22

(a)Take the entire container to be the system and assume no heat or work crosses the system boundary.

Energy Balance:

0=U

After the oscillation cease, the temperature and pressure on both sides of the piston will be thesame assuming the metallic piston is very thin and the thermal conductivity coefficient is large. Now, lets rewrite the energy balance.

0Bgas,1Agas,1 =+= ununU B A

or

0BgasAgas,1

,1 =+ uun

n

B

A

41.22 1,1,

1,1,

1,1,1,

1,1,1,

,1

,1 === A B

B A

A B B

B A A

B

A

T P

T P

T V P

T V P

n

n

since

1,1,2 B A V V = so

( ) ( ) 0101.55/2K 15.7732341.2

2T

K 15.373

3-2 =++ dT T RT R

K 5852 =T

Mass Balance:

2,2,1,1, B A B A nnnn +=+

Using the ideal gas law, we get


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48

2

2

2,

2,2,

2,

2,2,

1,

1,1,

1,

1,1,

RT V P

RT

V P

RT

V P

RT

V P

RT

V P tot B

B B

A

A A

B

B B

A

A A =+=+

since the pressure and temperature of state 2 are equal on both sides. But

1,1,1, 3 A B Atot V V V V =+=

Using the volume relationships and simplifying, we get

[ ] [ ][ ]

[ ][ ]

+=

+=

K 15.373 bar 12

K 15.773 bar 10

3K 8552

3 1,1,

1,

1,22

B

B

A

A

T

P

T

PT P

[ ] bar 56.32 =P

(b)The container is well-insulated, so

sysuniv ss =

The entropy change of the system can be split into two subsystems:

Bgas,1,1

,1Agas

,1,1

,1 snn

ns

nn

ns

B A

B

B A

Asys +

++

=

Using Equation 3.63 and realizing that Rcc vP += ,

=

=

K molJ 96.1lnln5.2

41.341.2

1,

2

1,

2Agas

A A PP

RT T

Rs

=

+=

K molJ 51.1ln105.15.3

41.31

1,

23

Bgas

2

1, B

T

T PP

RdT T

T Rs

B

Therefore,

=+==K mol

J 47.3K mol

J 51.1K mol

J 96.1sysuniv ss

The process is possible.


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49

3.23

Equation 3.3 can be modified to show


Lets first calculate sysS . Before this problem is solved, a few words must be said about thenotation used. The system was initially broken up into two parts: the constant volume containeand the constant pressure piston-cylinder assembly. The subscript 1 refers to the constantvolume container; 2 refers the piston-cylinder assembly. i" denotes the initial state before thvalve is opened, and f denotes the final state.

First, the mass of water present in each part of the system will be calculated. The mass will beconserved during the expansion process. Since the water in the rigid tank is saturated and is inequilibrium with the constant temperature surroundings (200 C), the waters entropy isconstrained. From the steam tables,

[ ] kPa8.1553kg

kJ 4322.6

kgkJ 3308.2

kgkJ 12736.0

kgkJ 001156.0

,1

,1

,1

,1

=

=

=

=

=

sat

v

i

li

vi

li

P

s

s

v

v

(Sat. water at 200 C)

Knowledge of the quality of the water and the overall volume of the rigid container can be usedto calculate the mass present in the container.

( ( v il i vmvmV ,11,111 95.005.0 +=

Using the values from the steam table and 31 m5.0=V provides

[ ]kg13.41 =m

Using the water quality specification,


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50

[ ][ ]kg207.005.0kg92.395.0

11

11

==

==

mm

mml

v

For the piston-cylinder assembly, bothP andT are known. From the steam tables

=

=

K kgkJ 9665.6

kgm 35202.0

,2

3,2

i

i

s

v

(600 kPa, 200 C)

Now enough information is available to calculate the mass of water in the piston assembly.

[ ]kg284.02

22 ==

v

V m

Now the final state of the system must be determined. It helps to consider what physicallyhappens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. Whenthe valve is opened, the water will rush out of the rigid tank and into the cylinder untilequilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and thesurroundings represent a large temperature bath at 200 C, the final temperature and pressure othe entire system will match the surroundings. In other words,

==kgkJ 9665.6 ,2 i f ss (600 kPa, 200 C)

Thus, the change in entropy is given by

li

li

vi

vii f

li

visys smsmsmsmmmS ,1,1,1,1,22,1,12 )( ++=

Substituting the appropriate values reveals

=K kJ 05.3sysS

Now we calculate the change in entropy of the surroundings. Since the temperature of thesurroundings is constant,

surr surr

surr surr T

QT Q

S ==

After neglecting potential and kinetic energy effects, the energy balance becomes


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51

W QU +=

The change in internal energy and work will be calculated in order to solve forQ. The followingequation shows how the change in internal energy can be calculated.

li

li

vi

vii f

li

vi umumumummmU ,1,1,1,1,22,1,12 )( ++=


=

=

kgkJ 3.2595

kgkJ 64.850

,1

,1

vi

li

u

u

(sat. H2O at 200 C)

==kgkJ 9.2638 ,2 i f uu (600 kPa, 200 C)

Using these values and the values of mass calculated above,

[ ]kJ0.541=U

Calculating the work is relatively easy since the gas is expanding against a constant pressure of600 kPa (weight of the piston was assumed negligible). From Equation 2.7,

)( i f E V

V E V V PdV PW

f

i

==

where

[ ][ ]

[ ] [ ] [ ]3333

,2,1,12

m6.0m5.0m0.1

m55.1)(

Pa600000

=+=

=++==

i

il

iv

i f

E

V

vmmmV

P

Note: iv ,2 was used to calculate f V because the temperature and pressure are the samefor the final state of the entire system and the initial state of the piston-cylinder assembly

The value ofW can now be evaluated.

[ ]kJ570=W


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52

The energy balance can be used to obtainQ.

[ ] [ ]( ) [ ]kJ1111kJ570kJ0.541 === W U Q

Therefore,[ ][ ]

==K kJ 35.2

K 15.473kJ1111

surr S

and

==+=K kJ 70.0

K kJ 35.2

K kJ 05.3surr sysuniv S S S


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53

3.24Let the subscript 1 represent the state of the system before the partition is removed. It has twcomponents: componenta and componentb. Subscript 2 represents the system after the partition is removed.

Mass balance:2,1,1 nnn ba =+

Energy balance for the adiabatic process:

W U sys = ) ) )babbvbaava V V V PT T cnT T cn ,1,12,12,,1,12,,1 +=+

The external pressure,P, for the above energy balance is equal toP 1,a . To find the final

temperature, first find the volumes using the ideal gas law.

[ ]( ) ( )

[ ]( ) [ ]( )[ ]

[ ]32

2-,1m05.0

m098.0sm8.9kg1000

K 300K mol

J 314.8mol2=

=aV

[ ]( )

[ ]( ) [ ]( )[ ][ ]324

22-

22 m103256.3

m098.0sm8.9kg1000

K molJ 314.8mol4

T

T

V =

=

Substitute the volumes into the energy balance and solve for2T :

( ) ( ) ( ) ( ) ( )[ ]324522 m15.0103256.3Pa101K 30023mol2K 300

23mol2 =

+

T T RT R

K 4.3602 =T

To calculate the change in entropy, we can use the following relationship

bPb

aPasys P

P R

T T

cnPP

RT T

cnS

+

=

12

12

12

12 lnlnlnln

The only unknown in the above equation isP 1,b , so we can calculate it with the ideal gas law:


54/104


55/104

55

3.25First perform an energy balance on the process:

W U sys =

The change in internal energy can be written:

unununU i A f B f B f A f A ,,,,, +=

A mass balance gives:

f B f Ai A nnn ,,, +=

These two expressions can be substituted to give:

) ) W T T cnT T cnU i f Bv f Bi f Av f A =+= ,,,,

Using the ideal gas law and Rcv 25= , we obtain

W V PV PV P i Ai A f A f A f B f B =+ ,,,,,, 25

25

25

Now, find an expression for the work

=

f

i

V

V E

dV PW

The pressure is not constant; its value is given by

Amg

A

V V k P

Amg

Akx

PP i B Batmatm E ++=++= 2

,

[ ]Pa10284.81006.1 55 B E V P +=

Integrating10142.41006.1 2

,

5

,

5 f B f B

V V W =

Substituting this expression into the energy balance gives:

2,

5,

5,,,,,, 10142.41006.12

525

25

f B f Bi Ai A f A f A f B f B V V V PV PV P =+


56/104


57/104


58/104

58

=kgkJ 9.28272h

Energy balance:

( ) sW Qhhm &&& += 12

CalculateW s:

( ) ( ) ( ) [ ]kW86.69kgkJ1.3422

kgkJ9.2827kg/s25.1 12

== QhhmW s &&&

[ ]kW673=sW &

(c)The isentropic efficiency is defined as follows:

( )( )reversibleW

W

S

actualS &

&=

For this situation,

( ) [ ]( ) [ ]kW5.447kW673665.0 ==actualS W &

(d)The real temperature should be higher since not as much energy is converted into work.

(e)Use the energy balance:

) sW Qhhm &&& += 12 =++=++=

kgkJ 2.3008

kgkJ1.3422

kg/s25.1kW-447.5kW86.69 12 hm

W Qh s

&

&&

At 1 MPa, water has this value of enthalpy when

( ) C2.2802 =actualT


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59

3.27Since the compressor is adiabatic, the energy balance after neglecting potential and kineticenergy becomes

( ) S W hhn && = 12

Using the ideal gas law and Appendix A.2, the above equation becomes

( ) ++++= 2

1

322111

T

T S dT ET DT CT BT AT

V PW

&&

Substituting the following values

[ ] [ ]Pa101 bar 1 51 ==P

=s

m 1 31V &

[ ]K 15.2931 =T [ ]K 15.4732 =T

;355.3= A ;10575.0 3= B ;0=C ;10016.0 5= D 0= E (Table A.2.2)

gives

[ ]kW8.218=S W &

This is the work of our real turbine (with 80% isentropic efficiency). We can use the isentropicefficiency to calculate the work of an equivalent 100% efficient process.

( )( )compressor S

reversibleS compressor W

W &

&=

( ) ( ) ( ) [ ]( )kW8.2188.0== compressor compressor S reversibleS W W && ( ) [ ]kW75.41=reversibleS W &

Since the process is adiabatic, we can use the following equation again to calculate what the fintemperature would be in a reversible process.

( )2

1

322

111 ++++=

T

T S dT ET DT CT BT AT

V PW

&&


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60

[ ] ( ) ++++= 2

1

W175400 322111

T

T

dT ET DT CT BT AT V P &

Substituting the values from above and solving for2T , we obtain

K8.164,2 =revT

To obtain the pressure the final pressure for the real compressor, we can calculate the final pressure for the reversible process because the final pressure is the same in both cases. For theisentropic expansion

==

12ln

T0

2

1PP

RdT c

sT

T

p

++++==

12

322ln0

2

1PP

RdT T

ET DT CT BT A Rs

T

T

( )[ ]

+==

=

=

bar 1ln160010575.0355.30 2

K 438

K 15.293

232

1

PdT

T T T

RsT

T

Solve for P2:

[ ] bar 16.42 =P


61/104

61

3.28Isentropic efficiency for a turbine is defined as

( )( )revs

actuals

w

w=

If the rate of heat transfer is assumed negligible, the energy balance for this process is

sW hhm && += 21 0

For a reversible, adiabatic process,

0= syss ( )MPa10C,500 1,2 ss rev =


==K kg

kJ 5965.6 1,2 ss rev

Since sat vsat l sss ,22,

2 , a mixture of liquid and vapor exists. The quality of the water can becalculated as follows

( ) sat vsat lrev s xs xs ,2,2,2 1 +=

( )

+

=

K kgkJ .35937

K kgkJ .302511

K kgkJ 5965.6 x x

874.0= x

Therefore,

( ) sat vsat lrev h xh xh ,2,2,2 1 +=

( )

+

=kgkJ 675.52874.0

kgkJ 17.444874.01 ,2 revh

=kgkJ 0.2391 ,2 revh

Also, from the steam tables,

=kgkJ 6.33731h (500 C, 10 MPa)


62/104


63/104


64/104

64

Substitute these expressions into the entropy balance and solve forP 1:

bar 85.11 =P


65/104


66/104

66

=kgkJ 12.31561u (540 C, 60 bar)

Hence,

[ ]( ) [ ]kJ1373kgkJ 12.3156

kgkJ 15.2881kg5 =

=W

(d)The specific volume can be found in the steam tables

=kgm 14173.0

32v

Therefore,

[ ]( ) [ ]332 m709.0kgm 14173.0kg5 =

=V


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67

3.31A schematic of the process is shown below:

well-insulated

very tinyleak hole

P1 = 60 bar

m = 5 kg

Pure H2O

T 1 = 540oCP2 = 20 bar

Pure H2O

T ' 2 = ?oC

T surr = 25oCPsurr = 1 bar

In order to leave the system, the gas must do flow work on the surroundings. The initial state ithe same as for Problem 3.30 and the final pressures are the same. Since the water only expandagainst 1 bar, the work is lower than that for the differential process described in Problem 3.30.Thus, this adiabatic process looses less energy, leading to a higher final temperature.

Another way to view this argument is to look at this process as a closed system. This depictionis the expansion analog of the compression process depicted for Example 2.5 in Figure E2.5B(page 57). We can represent this process in terms of two latches, one that keeps the process in initial state at 60 bar and one that stops the expansion after the pressure has reduced to 20 bar.The process is initiated by removal of the first latch and ends when the piston comes to restagainst the second latch. Such a process is depicted as Problem 3.31 below. Thecorresponding reversible process of Problem 3.30 is shown next to it for comparison. Clearly th process on the left does less work, resulting in a greater final temperature.

Problem 3.31 Problem 3.30

P1 = 60 bar

m = 5 kg

Pure H2O

T 1 = 540oC

P2 = 20 bar

m = 5 kg

Pure H2O


irreversible

latches

T ' 2 = ?oC

P1 = 60 bar

m = 5 kg

Pure H2O

T 1 = 540oC

P2 = 20 bar

m = 5 kg

Pure H2O

T 2 = ?oC


well-insulated

reversible


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68

3.32

(a)Consider the tank as the system.

Mass balance

inout in mmmdt dm

&&& ==

Separating variables and integrating:

=t

in

m

m

dt mdm0

2

1

&

or=t

in dt mmm0

12 &

Energy balanceSince the potential and kinetic energy effects can be neglected, the open system, unsteady stateenergy balance is

++=

out s

inininout out

sysW Qhmhm

dt dU &&&&

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet streamand not outlet stream. Therefore, the energy balance simplifies to

ininsys

hmdt

dU &=

The following math is performed

in

t

inin

t

inin

U

U

hmumU

dt mhdt hmdU

2222

000

2

1

==== =

&&

where the results of the mass balance were used. Thus,

inhu 2 =


69/104

69

From the steam tables,

=kgkJ 5.3456inh (3 MPa, 773 K)

Now the water in the tank is constrained. From the steam tables:

=

=

kgm14749.0

K kgkJ 743.7

32

2

v

s

=kgkJ 5.3456 MPa,3 2u

Compute the change in entropy. An entropy balance gives:

ininsysuniv

smdt dS

dt dS

&

=

Integrating withsin constant

( )int

ininuniv ssmdt mssmS 220

22 == &

From the steam tables:

=K kg

kJ 2337.7ins (3 MPa, 773 K)

Therefore,

=

=K kJ 173.0

K kgkJ 2337.7

K kgkJ 743.7

kgm

14749.0

m05.03

3univS

(b)If it tank sits in storage for a long time and equilibrates to a final temperature of 20 C, some orall of the vapor will condense and exchange heat with the surroundings. Let the subscript 3designate the final state of the water when it has reached a temperature of 20 C. The change inentropy between state 2 and state 3 is given by the following equation


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70

( )T

QssmS surr univ += 232

Let the subscript 3 designate the final state of the water when it has reached a temperature of

20 C. First, find quality of the water. From the steam tables:

=

=

kgm79.57

kgm001002.0

3,3

3,3

sat v

sat l

v

v

(sat. water at 20 C)

Calculate the quality as follows:

( ) sat vsat l v xv xvv ,3,323 1 +==

( )

+

=

kgm79.57

kgm 001002.01

kgm14749.0

333

x x

0025.0= x

Now calculate the entropy of state 3

( ) sat vsat l s xs xs ,3,33 1 +=

Substitute values from the steam tables:

( )

+

=K kg

kJ6671.80025.0K kg

kJ 2966.00025.013s

=K kg

kJ .317503s

Now calculate the amount of heat transferred to the surroundings:

( )232 uumQQ surr ==

Calculate the internal energy of state 3:

( ) sat vsat

l u xu xu ,3,33 1 +=

Substituting values from the steam tables:


71/104

71

=kgkJ 74.893u

Therefore,

[ ]kJ1141kgkJ 5.3456

kgkJ 89.74

kgm 14749.0

m05.03

3=

=surr Q

Now calculate the change in entropy of the universe

[ ]K 293kJ1411

K kgkJ 743.7

K kgkJ .31750

kgm 14749.0

m05.03

3

+

= univS

=K kJ 1.38univS

The entropy change for both processes can be fount by adding together the entropy change fromPart (a) and Part (b):

( ) =K kJ1.55(b)and(a)univS


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72

3.33A schematic is given below

valve maintainspressure in system

constant

T 1 = 200 oCx 1 = 0.4V = 0.01 m 3

v

l

Mass balance

out out in mmmdt dm

&&& ==

Separating variables and integrating:

=t

out

m

m

dt mdm0

2

1

&

or=t

out dt mmm0

12 &

Entropy Balance:

surr out out

syssurr

surr out out

sysuniv T

Qsm

dt dS

T Q

smdt dS

S &

&&

&& +

=++

=

Integrating withsout constant

( )surr

out univ T Q

smmsmsmS = 121122 (1)

From steam tables:


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73

=+=+=K kg

kJ 9715.34323.64.03309.26.0)1(1 g f xuu xs

=K kg

kJ 4323.62s is the same as 2K kg

kJ 4323.6 ssout ==

Thus Equation 1 simplifies to

surr univ T

QssmS = )( 121 (2)

Energy Balance to findQ :

Qhmdt dU

out out sys

&& +=

Integrating

[ ] +=+=t t t

out out out out

um

um

dt Qdt mhdt QhmdU 0 00

22

11

&&&&

Substituting in the mass balance and solving forQ

( ) out hmmumumQ

121122 = To find the mass in each state:

=+=+=kgm 0.0511274.04.0001.6.0)1(

31 g f xvv xv

=kgm 1274.0

32v

[ ]kg196.0kgm 0.051

m01.03

3

11

1 === vV

m and [ ]kg0785.0kgm 0.1274

m01.03

3

22

2 === vV

m

From the seam tables:


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74

=+=+=kgkJ 54915.25974.064.8506.0)1(1 g f xuu xu

=kg

kJ 3.25952u and =kg

kJ 2.2793out h

Plugging in:

( ) [ ]kJ228 121122 == out hmmumumQ

Back into Equation 2

( ) 0

s

K 473

skJ 228

K kgkJ 9715.34323.6

skg 196.0)( 121 =

==

surr univ T

QssmS

Note: Vaporization is reversible ifT surr = T sys.


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75

3.34The process can be represented as:

State 1 State 2

V 1 = 1 m3 V 2 = 10 m3P

1 = 10 bar

T 1 = 1000 K Work out

Process

Q = 0

P2

= ?T 2 = ?

Solving for the number of moles:

n = P1V 1 RT 1

=120.3 mol

The maximum work is given by a reversible process. Since it is also adiabatic, the entropychange of the system is zero:

s =0=cP ln T 2T 1

R ln P2P1

SincecP =52

R

12

2

1

22/5

1

2

PV nRT

PP

T T ==

Solving forT 2 gives

( ) K 2153/2

12

2/51

2 == PV T nR

T

An energy balance on this closed system gives:

U = Q +W =W =n c vdT = n cP R( )dT T 1

T 2

T 1T 2

Solving for work, we get

W = n 32

R T 2 T 1( )= 1.18106 J


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76

3.35 The maximum efficiency is obtained from a Carnot cycle. From Equation 3.32

H

C

T T

n =1

where temperature is in Kelvin. Hence,

614.0K 15.773K 15.2981 ==n


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77

3.36The labeling described in Figure 3.8 will be used for this solution. First, a summary of knownvariables is provided.

State P (bar) T (C)

1 30 5002 0.134

From our knowledge of ideal Rankine cycles, the table can be expanded as follows

State ThermodynamicState

P (bar) T (C)

1 Superheated Steam 30 5002 Sat. Liq. and Vapor 0.1

3 Sat. Liquid 0.14 Subcooled Liquid 30

The saturation condition constrains state 3. First, start with the turbine. Since the rate of heattransfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, thefollowing is known

21 ss = and

( )12 hhmW s = && From the steam tables:

=K kg

kJ 2337.71s (500 C, 30 bar)

=

=

K kgkJ 1501.8

K kgkJ 6492.0

2

2

v

l

s

s

(sat. H2O at 0.1 bar)

= kgkJ 5.34561h (500 C, 30 bar)

=

=

kgkJ 6.2584

kgkJ 81.191

2

2

v

l

h

h

(sat. H2O at 0.1 bar)


78/104


79/104

79

Now we find the rate of heat transfer for the evaporator. For the entire Rankine cycle, Equation3.84 gives

0=+++=+ C s H C net net W W QQW Q &&&&&&

We have

C C s H QW W Q &&&& =

Using the values calculated above:

[ ]MW2.326= H Q&

Equation 3.84:

C S net W W W &&&

= [ ]MW08.116=net W &

The efficiency can be calculated with Equation 3.82

[ ][ ]

356.0MW02.326MW08.116 ===

H

net

Q

W &

&


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80

3.37To make the Rankine cycle more efficient, we need to increase the area that represents the network in Figure 3.8. This can be done in a variety of ways:

1. Increase the degree of superheating of steam in the boiler. This process is sketched in the

upper left handTs diagram below. This change reduces the moisture content of steamleaving the turbine. This effect is desirable since it will prolong the life of the turbine;however, if the steam is heated too high, materials limitations of the turbine may need to beconsidered.

2. Lower the condenser pressure. Lowering the pressure in the condenser will lower thecorresponding saturation temperature. This change will enlarge the area on theTs diagram,as shown on the upper right below. Thus, we may want to consider lowering the pressure below atmospheric pressure. We can achieve this since the fluid operates in a closed loop.However, we are limited by the low temperature off the heat sink that is available.

s

1

2

3

4

wnet

T

s

1

2new3new

4new

wnet

T

lower the condenser pressure

s

1new

2new3

4

wnet

T

increase the superheating of steam in the boiler

s

1

23

4wnet

T

increase the boiler pressureTurbine 1

Turbine 2

5

6

2 stage turbine with reheat


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81

3. Increase the boiler pressure. This will increase the boiler temperature which will increase tharea as shown on the bottom leftTs diagram.

4. We can be more creative about how we use the energy available in the boiler. One way is todivide the turbine into two stages, Turbine 1 and Turbine 2, and reheat the water in the boile

between the two stages. This process is illustrated below.

turbine1

coolingwater

FuelAir

Boiler

compressor

condenser

1

4

56

Tur-bine

2

2

3

Rankine cycle with reheat

W s W s

W c

Q H

QC

The pressure in Turbine 1 will be higher than the pressure in turbine 2. This process isschematically shown on the bottom rightTs diagram. This process leads to less moisturecontent at the turbine exit (desirable) and limits the temperature of the superheat (desirable)


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82

3.38The labeling described in Figure 3.8 will be used for this solution. First, a summary of knownvariables is provided.

State P (bar) T (C)

1 100 5002 134



P (bar) T (C)

1 Superheated Steam 100 500

2 Sat. Liq. and Vapor 13 Sat. Liquid 14 Subcooled Liquid 100

The saturation condition constrains state 3. First start with the turbine (state 2). Since the rate heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, thefollowing is known

21 ss =

From the steam tables,

=K kg

kJ 5965.61s (500 C, 100 bar)

=

=

K kgkJ 3593.7

K kgkJ 3025.1

2

2

v

l

s

s

(sat. H2O at 1 bar)

The quality of the water can be calculated as follows

vl s xs xs 221 )1( +=

By substituting the steam table data, we find

874.0= x


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83

States 1, 2, and 3 are constrained; therefore, the enthalpies can be determined using the steamtables. The enthalpy of state 4 can be calculated from Equation 3.80:

( )3434 PPvhh l +=

where ==kgm 001043.0

33vvl since specific volume of liquids are insensitive to pressure

changes. From the steam tables:

State 1 2 3 4

h kgkJ 240.83

kgkJ 391.02

kgkJ 17.444

kgkJ 8.724

Note: vl h xh xh 222 )1( +=

Equation 3.84:

C H net QQW &&& =

Therefore,

23412341 hhhhmhhmhhmW net +=+= &&&&

and

( ) ( )[ ][ ]

+

=+

=

kgkJ 0.2391

kgkJ 27.84

kgkJ 5.417

kgkJ 240.83

kW10100

3

2341 hhhh

W m net

&&

=s

kg 3.116m&

Now consider the non-ideal turbine and compressor. Use the definition of isentropicefficiencies:

( )( )reversibles

actualsturbine W

W &

&=

( ) ( ) ( ) reversibleturbinereversibleS turbineactualS hhmW W 12 == &&&


84/104

84

( )( )actualC

reversibleC compressor W

W &

&=

( ) ( ) ( )compressor

reversible

compressor

reversibleC actualC

hhmW W

34 ==

&&&

Equation 3.84 is used to find the mass flow rate

C snet W W W &&& =

( ) +=compressor

reversiblereversibleturbinenet

hhhhmW

3412

&&

Substituting the enthalpy values from the table shown above and the efficiency values gives

=s

kg 147m&

A higher flow rate is needed as compared to the reversible process.


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85

3.39The labeling shown in Figure 3.8 will be used for this solution. First, a summary of knownvariables is provided.

State Thermodynamic

State

P (MPa)

1 Sat. Vapor 1.72 0.734



P (MPa)

1 Sat. Vapor 1.7

2 Sat. Liq. And Vapor 0.73 Sat. Liquid 0.74 Subcooled Liquid 1.7

The enthalpy of states 1 and 3 can be found using the NIST website. For states 2 and 4, use thefollowing expressions to find the enthalpies

vl h xh xh 222 )1( += ( ) ( )34333434 PPvhPPvhh l +=+=

To find x, use the fact that the turbine is isentropic.

21 ss = vl s xs xs 221 )1( +=

From the NIST website:

=K mol

J 38.1461s (sat. vapor at 1.7 MPa)

=

=

K molJ 07.150

K molJ

45.90

2

2

v

l

s

s (sat. mixture at 0.7 MPa)

By substituting the above values into the entropy relationship, we find


86/104


87/104

87

3.40

(a)Using the information in the problem statement and our knowledge of ideal vapor compressioncycles, the following table can be created


P (MPa)

1 Sat. Mixture 0.122 Sat. Vapor 0.123 Superheated Vapor 0.74 Sat. Liquid 0.7

Note: Refer to Figure 3.9 to review labeling convention.


( )12 hhnQC = &&

which upon combination with Equation 3.89 gives

( )42 hhnQC = &&


=molkJ 295.392h (sat. vapor at 0.12 MPa)

=molkJ 181.244h (sat. liquid at 0.7 MPa)

Therefore,

[ ]kW557.7molkJ181.24

molkJ 295.39

smol 5.0 =

=C Q&

(b) The power input to the compressor can be calculated with Equation 3.86:

( )23 hhnW C = &&


32 ss =


88/104

88


( )==K mol

J 89.177MPa0.12atvaporsat.23 ss

Now, state 3 is constrained.

=molkJ 031.433h

=

K molJ 89.177MPa,7.0 3s

Therefore,

[ ]kW87.1molkJ 295.39

molkJ 031.43

smol 5.0 =

=C W &

(c)Equation 3.90 states:

05.4

molkJ 295.39

molkJ 031.43

molkJ 181.24

molkJ 295.39

2342

2312 =

=

=

=

hhhh

hhhh

COP


89/104

89

3.41Using the information in the problem statement and our knowledge of ideal vapor compressioncycles, the following table can be created

State Thermodynamic

State

P (MPa)

1 Sat. Mixture 0.122 Sat. Vapor 0.123 Superheated Vapor 0.74 Sat. Liquid 0.7

Note: Refer to Figure 3.9 to review labeling convention.

From Equation 3.90

23

12

hhhh

W Q

COPC

C

== &

&

The enthalpy of state 2 can be found directly from the NIST website, but the enthalpies of state1 and 3 require the use of additional information.

=molkJ 295.392h (sat. vapor at 0.12 MPa)

For the process between state 2 and state 3,

32 ss =


==K mol

J 89.17723 ss (sat. vapor at 0.12 MPa)

Now, state 3 is constrained.

=molkJ 031.433h

=

K molJ 89.177MPa,7.0 3s

The process between state 4 and state 1 is also isentropic.

==K mol

J 09.11541 ss (sat. liquid at 0.7 MPa)

The quality of the R134a can be calculated as follows


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90

( ) vl xss xs 111 1 +=

where

=K mol

J 649.901ls (sat. R134a(l) at 0.12 MPa)

=K mol

J 89.1771vs (sat. R134a(v) at 0.12 MPa)

Therefore,

( )

+

=

K molJ 89.177

K molJ 649.901

K molJ 09.115 x x

280.0= x Using the quality of R134a, the enthalpy of state 1 can be calculated as follows

( ) vl xhh xh 111 1 +=


=molkJ 412.171lh (sat. R134a(l) at 0.12 MPa)

=molkJ 295.391vh (sat. R134a(v) at 0.12 MPa)

Therefore,

( )

+

=

molkJ 295.3928.0

molkJ 412.1728.011h

=molkJ 74.201h

Now, everything needed to calculate COP is available. Using Equation 3.90,

97.4

molkJ 295.39

molkJ 031.43

molkJ 74.20

molkJ 295.39

2312 =

=

=

hhhh

COP


91/104

91

Is this modification practical? No. An isentropic turbine adds significant level of complexity to the cycle. Turbines areexpensive and wear over time. Furthermore, the real turbine added to the cycle will not be 100efficient, so the COP will not increase as much. The cost of the turbine is not justified by theincrease in COP.


92/104


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93

(b)Performing an energy balance we find

( )56 hhnQ cC = &&

We can also use the following relationship

==mol

J 2109985 hh


=mol

J392956h (sat. R134 vapor at 0.12 MPa)

Therefore,

[ ]kW48.7mol

J 21099mol

J39295s

mol 411.0 =

=C Q&

(c)The power input is calculated as follows:

hC cC totalC W W W ,,, &&& +=

where

( )67, hhnW ccC = && ( )23, hhnW hhC = &&

We have all of the required enthalpies excepth3. State 3 is constrained because

==K mol

J95.17523 ss (sat. R134a vapor at 0.35 MPa)

MPa7.03 =P


=mol

J424283h (R134a vapor at 0.7 MPa with= K molJ95.175s )

Now compute the power for each unit:


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94

[ ] [ ]

+

=

molJ 40967

molJ 42428mol/s5.0

molJ 39295

molJ 41510mol/s411.0,totalC W &

[ ]kW64.1, =totalC W &

(d)The coefficient of performance is calculated using the following equation:

56.4kW64.1kW48.7

,===

totalC

C W

QCOP &

&

(e)The COP for the cascade is 4.56, while the COP is 4.05 in Problem 3.40. The cascade systemCOP is 12.6% greater.


95/104

95

3.43One design follows; your design may differ:

In order to cool a system to -5 C, the temperature of the fluid must be colder than -5 C so thatheat transfer will occur. We arbitrarily specify that the working fluid evaporates at -10 C.

Similarly, in order to eject heat to the 20 C reservoir, the fluid must condense at a temperaturegreater than 20 C. Arbitrarily, we choose 25 C. One possible refrigeration cycle is presented below.

evaporator

compressor

condenser

1

4

3

2

T

s

3

21

4

valve

refrigeratedunit at -5 C

20 Creservoir

W c

Q H

Q H

QC

QC

A number of fluids will work sufficiently for this system, but the design process will beillustrated using R134a. For states 3 and 4, the pressure is constant, and for states 1 and 2, the pressure is constant at a different value. From the NIST website, we find

MPa201.021 == PP (T sat = -10 C)MPa665.043 == PP (T sat = 25 C)

(Note: The temperatures of each state are not constant. The listed saturation temperatureare the temperatures at which the fluid evaporates and condenses.)

Now that the pressures are known, we can compute the required flow rate required in order to provide 20 kW of cooling.

( )12 hhnQC = &&

We can geth2 from the thermodynamic tables for saturated R134a. In order to findh1 , we canuse the following relationship:

41 hh =


=mol

J 400642h (sat. R134 vapor at 0.201 MPa)


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96

=mol

J 239314h (sat. R134 liquid at 0.665 MPa)

Now, calculate the required flow rate of R134a.

( )[ ]

=

=

molJ 23931

molJ 40064

J/s000,2012 hh

Qn C

&&

=s

mol24.1n&


97/104

97

3.44The schematic of the process and the correspondingTs is presented below

evaporator

compressor

condenser

1

63

2T

s

2

1

4

valve

high Treservoir

valve

Fridge

Freezer evaporator

54

T = -15 oC

T = 5 oC

3

5

6

5 oC

-15 oC

QC,R

QC,F Q H

Q H

QC,F

QC,R

W c

W c

A number of refrigerants will work for this system, but the design process will be illustrated forR134a only. To define each state, we need to thermodynamically constrain each state. Youshould note that the problem doesnt state what temperature the fluid condenses at. Therefore,we can assume the temperature is 25 C. By using information in the Ts diagram and from the NIST website, we find

MPa34966.021 == PP (T 1=T 2=T sat = 5 C)MPa16394.043 == PP (T 3=T 4=T sat = -15 C)MPa66538.065 == PP (T 6 = 25 C)

Before solving for additional pressures and temperatures, we will list the known temperatures

and pressures.State Temperature (C) Pressure (MPa)

1 5 0.243342 5 0.243343 -15 0.163944 -15 0.163945 25 0.665386 25 0.66538

States 4, 5, and 6 are completely constrained as confirmed by Gibbs phase rule. Now, we needto find the liquid and vapor compositions of states 1, 2, and 3 to completely constrain the statesSince the heat duties are equal for the refrigerator and the freezer, we have the followingrelationship:

RC F C QQ ,, && =


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98

which upon performing an energy balance becomes

1234 hhhh =

Energy balances around the valves provide:

61 hh = 32 hh =

Substituting these relationships into the expression that equates the heat loads results in

264

32hh

hh +==


=mol

J 397554h

=mol

J 239316h

Now constrain states 1, 2, and 3 by determining the enthalpies:

==mol

J 2393161 hh

=+

==mol

J 318432

molJ 23931

molJ 39755

32 hh

Technically, the states are now all constrained, but we would like also like to know the vapor anliquid compositions of each state. The compositions of states 4, 5, and 6 are already known; wcan calculate the compositions of states 1, 2, and 3 as follows:

( ) sat vsat

l h xh xh ,11,111 1 +=

( ) sat

vsat

l h xh xh ,22,222 1 += ( ) sat v

sat l h xh xh ,33,333 1 +=

From the NIST website,


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99

=mol

J 21095,1sat

lh = molJ 40965,1

sat vh

=mol

J 21095,2sat

lh = molJ 40965,2

sat vh

= molJ 18380,3sat lh = molJ 39755,3

sat vh

Now, we can solve the composition of vapor for each state:

143.01 = x 541.02 = x 630.03 = x

The following table presents a summary of our results:

State Temperature(C)Pressure

(MPa) PhasesLiquid

CompositionVapor

Composition

1 5 0.24334 Saturated Liquidand Vapor 0.857 0.143

2 5 0.24334 Saturated Liquidand Vapor 0.459 0.541

3 -15 0.16394 Saturated Liquidand Vapor 0.37 0.630

4 -15 0.16394 Saturated Vapor 0 1

5 25 0.66538 SuperheatedVapor 0 1

6 25 0.66538 Saturated Liquid 1 0


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100

3.45In this case, the working material is a solid. The four states of the magnetic material are shownof thesT diagram below. Note the axis are shifted from the usual manner.

1

2

3

4

QC Q H

(a) The heat expelled by the cold reservoir can be approximated by:

( )12 ssT q C C

whereT is the average temperature between states 1 and 2, which is approximately 1 K. Takinvalues ofS/R from thesT diagram, we get:

( )= K molJ0.53.19.1 RqC

(b) Similarly, the heat absorbed by the hot reservoir can be approximated by:

( ) ( )==K mol

J519.12.18.844 RssT q H H

(b) The coefficient of performance is given by:

11.0,

=

==C H

C

totalC

C

qqq

wq

COP

(d) The value of COP is much lower than a typical refrigeration process (COP= 4-6); in generarefrigeration processes at these low temperatures are much less efficient.

F. Work is supplied to magnetize the material and to spin the wheel.


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101

3.46 When the polymer is unstretched it is in a more entangled state. When stretched the polymerchains tend to align. The alignment decreases the spatial configurations the polymer can have,and therefore, reduces that component of entropy. If the process is adiabatic, the entropy of thesystem cannot decrease. Consequently, its thermal entropy must increase. The only way this c

be accomplished is by increased temperature.


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102

3.47Assume the temperature is 298 K. The following data was taken from Table A.3.2:

Species f h (kJ/mol) f g (kJ/mol)Cu2O (s) -170.71 -147.88O2 (g) 0 0

CuO (s) -156.06 -128.29

The data listed above were used to create the next table using

T

ghs f f f

=

Species f s (kJ/mol K)

Cu2O (s) -0.0766O2 (g) 0CuO (s) -0.0932

Now the change in entropy of the reaction can be calculated in a method analogous to Equation2.72

( )

+

== K mol

kJ 0.0932-4K mol

kJ 0.0766-2i f irxn

svs

=mol

kJ 22.0rxns

Does this violate the second law of thermodynamics? This problem shows that the entropychange of the system is negative, but nothing has been said about the entropy change of theuniverse. We must look at the change in entropy of the surroundings to determine if the secondlaw is violated. By looking at the enthalpies, we see that the reaction is exothermic, whichmeans that heat is transferred from the system to the surroundings. Therefore, the entropy of thsurroundings will increase during this reaction.


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103

3.48The temperature and pressure terms in the equation for entropy do not contribute anything to thentropy change because they are constant. Therefore, the only remaining entropy contribution,the randomness of the atomic arrangements, must be considered. The randomness does notincrease when CdTe forms. In pure crystals of Cd and Te, the location of each atom is known

because the crystal lattice constrains the atomic locations. In CdTe, the crystal lattice stilldefines the location of each atom, so the randomness has not increased. Therefore, the change entropy is zero.

3.49This argument is not scientifically sound. Morris is arguing that since evolution results in moreorder, the second law of thermodynamics is violated, so evolution must be impossible. Howevthe flaw in this argument is caused by ignoring the entropy change of the entire universe. Thesecond law states that the entropyof the universe will remain constant or increase for any process. Morris argument was based on the entropy of the system undergoing evolution notthe entropy of the entire universe. A system can decrease in entropy if the entropy of the

surroundings increases by at least that much.


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3.50Entropy is related to the number of configurations that a state can have. The greater the numbeof configurations, the more probable the state is and the greater the entropy. We canqualitatively relate this concept to the possible hands in a game of poker, but it is moreinteresting to quantify the results using some basic concepts of probability.

We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards.. Therare a finite number of permutations in which we can arrange a 52 card deck in 5 cards. For thefirst card in the hand, we select from 52 cards, the second card can be any other card so we selefrom 51 cards, the third card has 50 cards, and so on. Thus the number of permutations of 5cards is:

P =5251504948 =311,875,200

However, we do not care the order in which the cards are dealt, so we must divide this number by the number of ways we can come up with the same hand. We do this math in a similar way.

For a given hand there are five cards we can pick first, four we can pick second, . So thenumber of ways we can make the same hand is:

N =54321=120

The number of unique configurations can be found by dividingP by N . Thus, the cards candisplay

C = P N

=2,598,960

or 2,598,960 unique configurations. To find the entropy of a given hand, we need to find ouhow many of these unique configurations belong to the hand

Consider a four of a kind. There are 13 different possible ranks of four of a kind, one for eachnumber A, 2, 3, 4, 5, 6, 7,8, 9, 10, J, Q, K. The fifth card in the hand could be any of the other48 cards. Therefore, the number of combinations of four of a kind is: 13 x 48 = 624

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