borgnakke and sonntag_fundamentals of thermodynamics

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Updated June 2013

8e

SOLUTION MANUAL

CHAPTER 12

Fundamentals of

Thermodynamics

Borgnakke SonntagBorgnakke Sonntag


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Borgnakke and Sonntag

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Fundamentals of Thermodynamics 8th

Edition


CONTENT CHAPTER 12

SUBSECTION PROB NO.

In-Text concept questions a-f

Study guide problems 1-15Clapeyron equation 16-33

Property Relations, Maxwell, and those for

Enthalpy, internal Energy and Entropy 34-43Volume Expansivity and Compressibility 44-59

Equations of State 60-81

Generalized Charts 82-120

Mixtures 121-133Helmholtz EOS 134-138

Review problems 139-148


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The following table gives the values for the compressibility, enthalpy departure and the

entropy departure along the saturated liquid-vapor boundary. These are used for all the

problems using generalized charts as the figures are very difficult to read accurately(consistently) along the saturated liquid line. It is suggested that the instructor hands out

copies of this page or let the students use the computer for homework solutions.

Tr P

r Z

f Z

g d(h/RT)

f d(h/RT)

g d(s/R)

f d(s/R)

g

0.96 0.78 0.14 0.54 3.65 1.39 3.45 1.10

0.94 0.69 0.12 0.59 3.81 1.19 3.74 0.94

0.92 0.61 0.10 0.64 3.95 1.03 4.00 0.82

0.90 0.53 0.09 0.67 4.07 0.90 4.25 0.72

0.88 0.46 0.08 0.70 4.17 0.78 4.49 0.640.86 0.40 0.07 0.73 4.26 0.69 4.73 0.57

0.84 0.35 0.06 0.76 4.35 0.60 4.97 0.50

0.82 0.30 0.05 0.79 4.43 0.52 5.22 0.45

0.80 0.25 0.04 0.81 4.51 0.46 5.46 0.39

0.78 0.21 0.035 0.83 4.58 0.40 5.72 0.35

0.76 0.18 0.03 0.85 4.65 0.34 5.98 0.31

0.74 0.15 0.025 0.87 4.72 0.29 6.26 0.27

0.72 0.12 0.02 0.88 4.79 0.25 6.54 0.23

0.70 0.10 0.017 0.90 4.85 0.21 6.83 0.20

0.68 0.08 0.014 0.91 4.92 0.18 7.14 0.17

0.66 0.06 0.01 0.92 4.98 0.15 7.47 0.15

0.64 0.05 0.009 0.94 5.04 0.12 7.81 0.12

0.60 0.03 0.005 0.95 5.16 0.08 8.56 0.08

0.58 0.02 0.004 0.96 5.22 0.06 8.97 0.07

0.54 0.01 0.002 0.98 5.34 0.03 9.87 0.04

0.52 0.0007 0.0014 0.98 5.41 0.02 10.38 0.03


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In-Text Concept Questions


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12.a

Mention two uses of the Clapeyron equation.

If you have experimental information about saturation properties down to

a certain temperature Clapeyron equation will allow you to make an intelligent

curve extrapolation of the saturated pressure versus temperature function Psat(T)for lower temperatures.

From Clapeyrons equation we can calculate a heat of evaporation, heat of

sublimation or heat of fusion based on measurable properties P, T and v. The

similar changes in entropy are also obtained since

hfg= Tsfg; hif= Tsif; hig= Tsig


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12.c

If I raise the pressure in an isentropic process, does hgo up or down? Is that

independent upon the phase?

Tds = 0 = dh vdP , so h increases as P increases, for any phase. The

magnitude is proportional to v (i.e. large for vapor and small for liquid andsolid phases)


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12.d

If I raise the pressure in a solid at constant T, does sgo up or down?

In Example 12.4, it is found that change in s with P at constant T is negatively

related to volume expansivity (a positive value for a solid),

dsT= - v PdPTso raising P decreases s.


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12.e

What does it imply if the compressibility factor is larger than 1?

A compressibility factor that is greater than one comes from domination of

intermolecular forces of repulsion (short range) over forces of attraction (long

range) either high temperature or very high density. This implies that the densityis lower than what is predicted by the ideal gas law, the ideal gas law assumes the

molecules (atoms) can be pressed closer together.

12.f

What is the benefit of the generalized charts? Which properties must be known

besides the charts themselves?

The generalized charts allow for the approximate calculations of enthalpy

and entropy changes (and P,v,T behavior), for processes in cases where specific

data or equation of state are not known. They also allow for approximate phaseboundary determinations.

It is necessary to know the critical pressure and temperature, as well asideal-gas specific heat.


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Concept-Study Guide Problems


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12.1

The slope dP/dT of the vaporization line is finite as you approach the critical

point, yet hfgand vfgboth approach zero. How can that be?

The slope is

dP

dT sat= hfg

Tvfg

Recall the math problem what is the limit of f(x)/g(x) when x goestowards a point where both functions f and g goes towards zero. A finite limit for

the ratio is obtained if both first derivatives are different from zero so we have

dP/dT [dhfg/dT] /d(Tvfg)/dT as T Tc


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12.2

In view of Clapeyrons equation and Fig. 2.4, is there something special about ice

I versus the other forms of ice?

Yes. The slope of the phase boundary dP/dT is negative for ice I to liquid

whereas it is positive for all the other ice to liquid interphases. This also meansthat these other forms of ice are all heavier than liquid water. The pressure must

be more than 200 MPa 2000 atm so even the deepest ocean cannot reach thatpressure (recall about 1 atm per 10 meters down).


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12.3

If we take a derivative as (P/T)vin the two-phase region, see Figs. 2.7 and 2.8,does it matter what v is? How about T?

In the two-phase region, P is a function only of T, and not dependent on v.

The slope is the same at a given T regardless of v. The slope becomeshigher with higher T and generally is the highest near the critical point.

P

T

v

V

L

S

C.P.


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12.4

Sketch on a P-T diagram how a constant v line behaves in the compressed liquid

region, the two-phase L-V region and the superheated vapor region?

P

T

VL

Cr.P.

S

v < vc

vlarge

vmedium

v > vcsmall

P

T

v

V

L

S

C.P.

vlarge

v

v > vc

medium

v < vc

The figures are for water where liquid is denser than solid.


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12.5

If I raise the pressure in an isothermal process does h go up or down for a liquid

or solid? What do you need to know if it is a gas phase?

Eq. 12.25: (h

P)

T

= v T (v

T)

P

= v[1 - TP

]

Liquid or solid, Pis very small, h increases with P ;

For a gas, we need to know the equation of state.

12.6The equation of state in Example 12.3 was used as explicit in v. Is it explicit in P?

Yes, the equation can be written explicitly in P.

P =RT

v +

C

T3


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12.7

Over what range of states are the various coefficients in Section 12.5 most useful?

For solids or liquids, where the coefficients are essentially constant over a

wide range of Ps and Ts.


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12.8

For a liquid or a solid is v more sensitive to T or P? How about an ideal gas?

For a liquid or solid, v is much more sensitive to T than P.

For an ideal gas, v = RT/P , varies directly with T, inversely with P.

12.9

Most equations of state are developed to cover which range of states?

Most equations of state are developed to cover the gaseous phase, from

low to moderate densities. Many cover high-density regions as well, including

the compressed liquid region. To cover a wider region the EOS must be morecomplex and usually has many terms so it is only useful on a computer.


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12.10

Is an equation of state valid in the two-phase regions?

No. In a two-phase region, P depends only on T. There is a discontinuity

at each phase boundary. It is actually difficult to determine the phase boundary

from the EOS.

12.11

As P 0, the specific volume v . For P , does v 0?

At very low P, the substance will be essentially an ideal gas, Pv = RT, so

that v becomes very large. However at very high P, the substance eventually must

become a solid, which cannot be compressed to a volume approaching zero.


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12.12

Must an equation of state satisfy the two conditions in Eqs. 12.49 and 12.50?

It has been observed from experimental measurements that substances do

behave in that manner. If an equation of state is to be accurate in the near-critical

region, it would have to satisfy these two conditions.If the equation is simple it may be overly restrictive to impose these as it

may lead to larger inaccuracies in other regions.


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12.13

At which states are the departure terms for h and s small? What is Z there?

Departure terms for h and s are small at very low pressure or at very high

temperature. In both cases, Z is close to 1 and this is the ideal gas region.


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12.14

The departure functions for h and s as defined are always positive. What does that

imply for the real substance h and s values relative to ideal gas values?

Real-substance h and s are less than the corresponding ideal-gas values.

This is true for the range shown in the figures, Pr< 10. For higher P the isothermsdo bend down and negative values are possible.

Generally this means that there are slightly attractive forces between themolecules leading to some binding energy that shows up as a negative potential

energy (thus smaller h and u) and it also give a little less chaos (smaller s) due to

the stronger binding as compared to ideal gas. When the pressure becomes large

the molecules are so close together that the attractive forces turns into repulsiveforces and the departure terms are negative.


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12.15

What is the benefit of Kays rule versus a mixture equation of state?

Kays rule for a mixture is not nearly as accurate as an equation of state

for the mixture, but it is very simple to use and it is general. For common

mixtures new mixture EOS are becoming available in which case they arepreferable for a greater accuracy in the calculation.


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Clapeyron Equation


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12.16

An approximation for the saturation pressure can be ln Psat= A B/T, where A

and B are constants. Which phase transition is that suitable for, and what kind of

property variations are assumed?

Clapeyron Equation expressed for the three phase transitions are shown in Eqs.12.5-12.7. The last two leads to a natural log function if integrated and ideal gas

for the vapor is assumed.

dPsatdT

= Psathevap

RT2

where hevapis either hfgor hig. Separate the variables and integrate

P-1

satdPsat= hevapR-1T-2dT

ln Psat= A B/T ; B = hevapR-1

if we also assume hevapis constant and A is an integration constant. The function

then applies to the liquid-vapor and the solid-vapor interphases with different

values of A and B. As hevapis not excactly constant over a wide interval in T it

means that the equation cannot be used for the total domain.


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12.17

Verify that Clapeyrons equation is satisfied for R-410A at 10oC in Table B.4.

Clapeyron Eq.:dPsat

dT

=dPg

dT

=hfg

Tvfg

B.4: P = 1085.7 kPa, hfg= 208.57 kJ/kg, vfg= 0.02295 m3/kg

Slope around 10oC best approximated by cord from 5oC to 15oCdPgdT

=1255.4 933.9

15 5= 32.15 kPa/K,

hfg

Tvfg=

208.57

283.15 0.02295= 32.096 kPa/K

This fits very well. Use CATT3 to do from 9 to 11oC for better approximationas the saturated pressure is very non-linear in T.


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12.18

In a Carnot heat engine, the heat addition changes the working fluid from

saturated liquid to saturated vapor at T, P. The heat rejection process occurs at

lower temperature and pressure (TT), (PP). The cycle takes place in apiston cylinder arrangement where the work is boundary work. Apply both the

first and second law with simple approximations for the integral equal to work.Then show that the relation between Pand Tresults in the Clapeyron equationin the limit TdT.

qH

= Tsfg; qL= (T-T)sfg ; wnet= qH- qL= Tsfg

The boundary movement work, w = Pdv

wNET

= P(v2-v

1) +

2

3

Pdv + (P - P)(v4- v

3) +

1

4

Pdv

Approximating,

2

3

Pdv (P -P2 ) (v3- v2);

1

4

Pdv (P -P2 ) (v1- v4)

Collecting terms: wNET

P[(2

v2+v3)- (2

v1+v4)]

(the smaller the P, the better the approximation)

PT 1

2(v2+ v3)1

2(v1+ v4)

sfg

In the limit as T 0: v3 v2= vg, v4v1= vf

& limT0PT=dP

satdT =

sfg

vfg

s

P

v

P

T

T- T

1 2

34P- P

P- P P

s at Tv at Tfg fg

4 3

1 2T

T

T T


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12.19

Verify that Clapeyrons equation is satisfied for carbon dioxide at 6oC in TableB.3.

Clapeyron Eq.:dP

satdT =

dPg

dT =h

fgTvfg

B.3: P = 4072.0 kPa, hfg= 211.59 kJ/kg, vfg= 0.00732 m3/kg

Slope around 6oC best approximated by cord from 4oC to 8oC

dPgdT

=4283.1 3868.8

8 4= 103.575 kPa/K,

hfgTvfg

= 211.59279.15 0.00732

kJ/kgK m3/kg

= 103.549 kPa/K

This fits very well.


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12.20

Use the approximation given in problem 12.16 and Table B.1 to determine A and

B for steam from properties at 25oC only. Use the equation to predict the

saturation pressure at 30oC and compare to table value.

ln Psat= A B/T dPsat

dT= Psat(-B)(-T

-2)

so we notice from Eq.12.7 and Table values from B.1.1 and A.5 that

B =hfg

R=

2442.3

0.4615

kJ/kg

kJ/kg-K= 5292 K

Now the constant A comes from the saturation pressure as

A = ln Psat+ B/T = ln 3.169 +5292

273.15 + 25= 18.9032

Use the equation to predict the saturation pressure at 30oC as

ln Psat= A B/T = 18.9032 -5292

273.15 + 30= 1.4462

Psat= 4.2469 kPa

compare this with the table value of Psat= 4.246 kPa and we have a very accurate

approximation.


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12.21

A certain refrigerant vapor enters a steady flow constant pressure condenser at

150 kPa, 70C, at a rate of 1.5 kg/s, and it exits as saturated liquid. Calculate therate of heat transfer from the condenser. It may be assumed that the vapor is an

ideal gas, and also that at saturation, vf


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12.22

Calculate the values hfgand sfgfor nitrogen at 70 K and at 110 K from the

Clapeyron equation, using the necessary pressure and specific volume valuesfrom Table B.6.1.

Clapeyron equation Eq.12.7: dPgdT

= hfgTvfg

= sfgvfg

For N2at 70 K, using values for Pgfrom Table B.6 at 75 K and 65 K, and also

vfgat 70 K,

hfgT(vg-vf) Pg

= 70(0.525 015)(76.1-17.4175-65

)= 215.7 kJ/kgsfg= hfg/T = 3.081 kJ/kg K

Table B.6.1: hfg= 207.8 kJ/kg and sfg= 2.97 kJ/kg K

Comparison not very close because Pgnot linear function of T. Using 71 K &

69 K from the software, we can then get

hfg= 70(0.525 015)(44.56-33.24

71-69) = 208.0 kJ/kg

At 110 K,

hfg110(0.014 342)(1938.8-1084.2

115-105)= 134.82 kJ/kg

sfg= hfg/T =134.82

110= 1.226 kJ/kg K

Table B.6.1: hfg= 134.17 kJ/kg and sfg= 1.22 kJ/kg K


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12.23

Find the saturation pressure for refrigerant R-410A at 80oC assuming it is higherthan the triple point temperature.

The lowest temperature in Table B.4 for R-410A is -60oC, so it must be extended

to -80oC using the Clapeyron Eq. 12.7 integrated as in example 12.1

Table B.4 at T1= -60oC = 213.15 K:

P1= 64.1 kPa, R = 0.1145 kJ/kg-K

lnP

P1

=R

hfg

T T

1

(T - T1)=

279.96

0.1145(193.15 - 213.15)

193.15 213.15 = -1.1878

P = 64.1 exp(-1.1878) = 19.54 kPa


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12.24

Ammonia at 70oC is used in a special application at a quality of 50%. Assume

the only table available is B.2 that goes down to 50oC. To size a tank to hold 0.5kg with x = 0.5, give your best estimate for the saturated pressure and the tank

volume.

To size the tank we need the volume and thus the specific volume. If we do not

have the table values for vfand vgwe must estimate those at the lower T. We

therefore use Clapeyron equation to extrapolate from 50oC to 70oC to get the

saturation pressure and thus vgassuming ideal gas for the vapor.

The values for vf and hfgdo not change significantly so we estimate

Between -50oC and 70oC: hfg= 1430 kJ/kg

and at 70oC we get: vf= 0.001375 m3/kg

The integration of Eq.12.7 is the same as in Example 12.1 so we get

lnP2

P1=

hfg

R (

T2- T1

T2T1) =

1430

0.4882

-70 + 50

203.15 223.15= -1.2923

P2= P1exp(-1.2923) = 40.9 exp(-1.2923) = 11.2 kPa

vg= RT2/P2=0.4882 203.15

11.2= 8.855 m3/kg

v2= (1-x) vf+ x vg= 0.5 0.001375 + 0.5 8.855 = 4.428 m3/kg

V2= mv2= 2.214 m3

A straight line extrapolation

will give a negative pressure.

P

T

-50-70


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12.25

Use the approximation given in problem 12.16 and Table B.4 to determine A and

B for refrigerant R-410A from properties at 0oC only. Use the equation to predict

the saturation pressure at 5oC and compare to table value.

ln Psat= A B/T dPsat

dT= Psat(-B)(-T

-2)

so we notice from Eq.12.7 and Table values from B.4.1 and A.5 that

B =hfg

R=

221.37

0.1145

kJ/kg

kJ/kg-K= 1933.4 K

Now the constant A comes from the saturation pressure as

A = ln Psat+ B/T = ln 798.7 +1933.4

273.15= 13.7611

Use the equation to predict the saturation pressure at 5

o

C as

ln Psat= A B/T = 13.7611 -1933.4

273.15 + 5= 6.8102

Psat= 907 kPa

compare this with the table value of Psat= 933.9 kPa and we have an

approximation 3% low. Notice hfgdecreases so we could have used a lower value

for the average in the interval.


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12.26

The triple point of CO2is -56.4oC. Predict the saturation pressure at that point

using Table B.3.

The lowest temperature in Table B.3 for CO2is -50oC, so it must be extended to

-56.4oC = 216.75 K using the Clapeyron Eq. 12.7 integrated as in Ex. 12.1

Table B.3: at T1= -50oC = 223.15 K, P

1= 682.3 kPa, hfg= 339.73 kJ/kg

Table A.5: R = 0.1889 kJ/kg-K

lnP

P1=

R

hfg

T T1

(T - T1)=

339.73

0.1889(216.75 - 223.15)

216.75 223.15= -0.23797

P = 682.3 exp(-0.23797) = 537.8 kPa

Notice from Table 3.2: P = 520.8 kPa so we are 3% high. As hfgbecomes larger

for lower Ts we could have estimated a more suitable value for the interval from

-50 to -56.4oC


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12.27

Helium boils at 4.22 K at atmospheric pressure, 101.3 kPa, with hfg=83.3kJ/kmol. By pumping a vacuum over liquid helium, the pressure can be lowered,

and it may then boil at a lower temperature. Estimate the necessary pressure to

produce a boiling temperature of 1 K and one of 0.5 K.

Solution:

Helium at 4.22 K: P1= 0.1013 MPa, h

-FG

= 83.3 kJ/kmol

dPSAT

dT=

hFG

TvFG

h

FGP

SAT

RT2 ln

P2

P1

=h

FG

R[

1

T1

1

T2

]

For T2= 1.0 K:

lnP

2

101.3=

83.3

8.3145[

1

4.22

1

1.0] => P

2= 0.048 kPa = 48 Pa

For T2= 0.5 K:

lnP

2

101.3=

83.3

8.3145[

1

4.22

1

0.5]

P2= 2.160110-6kPa = 2.1601 10-3Pa


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12.28

Using the properties of water at the triple point, develop an equation for the

saturation pressure along the fusion line as a function of temperature.

Solution:

The fusion line is shown in Fig. 2.3 as the S-L interphase. From Eq.12.5 we have

dPfusion

dT=

hif

Tvif

Assume hifand vifare constant over a range of Ts. We do not have any simple

models for these as function of T other than curve fitting. Then we can integrate

the above equation from the triple point (T1, P1) to get the pressure P(T) as

P P1=hif

vif ln

T

T1

Now take the properties at the triple point from B.1.1 and B.1.5

P1= 0.6113 kPa, T1= 273.16 K

vif= vf vi= 0.001 0.0010908 = - 9.08 105m3/kg

hif= hf hi= 0.0 (-333.4) = 333.4 kJ/kg

The function that approximates the pressure becomes

P = 0.6113 3.672 106 lnT

T1 [kPa]


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12.29

Using thermodynamic data for water from Tables B.1.1 and B.1.5, estimate the

freezing temperature of liquid water at a pressure of 30 MPa.

H2O

dT

dPif=

Tvif

hifconst

At the triple point,

vif = vf- vi= 0.001 000 - 0.001 090 8 = -0.000 090 8 m3/kg

hif= hf- hi= 0.01 - (-333.40) = 333.41 kJ/kg

dPif

dT =333.41

273.16(-0.000 090 8)= -13 442 kPa/K

at P = 30 MPa,

T 0.01 +(30 000-0.6)

(-13 442)= -2.2oC

T.P.

T

P 30 MPa


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12.30

Ice (solid water) at 3C, 100 kPa, is compressed isothermally until it becomesliquid. Find the required pressure.

Water, triple point T = 0.01oC , P = 0.6113 kPa

Table B.1.1: vf= 0.001 m3/kg, hf= 0.01 kJ/kg,

Tabel B.1.5: vi= 0.001 0908 m3/kg, hi= -333.4 kJ/kg

ClapeyrondPif

dT=

hf- hi

(vf- vi)T=

333.4

-0.0000908 273.16= -13 442 kPa/K

P dP

if

dTT = -13 442(-3 - 0.01) = 40 460 kPa

P = Ptp

+ P = 40 461 kPa


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12.31

From the phase diagram for carbon dioxide in Fig. 2.5 and 2.4 for water what can

you infer for the specific volume change during melting assuming the liquid has ahigher h than the solid phase for those two substances.

The saturated pressure versus temperature has a positive slope for carbon dioxide

and a negative slope for water.

ClapeyrondPif

dT=

hf- hi

(vf- vi)T

So if we assume hf- hi> 0 then we notice that the volume change in the melting

gives

Water: vf- vi< 0 so vf< vi

Carbon dioxide: vf- vi> 0 so vf> vi


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12.32

A container has a double wall where the wall cavity is filled with carbon dioxideat room temperature and pressure. When the container is filled with a cryogenicliquid at 100 K the carbon dioxide will freeze so that the wall cavity has a mixture

of solid and vapor carbon dioxide at the sublimation pressure. Assume that we do

not have data for CO2at 100 K, but it is known that at 90C: Psat=38.1 kPa,hIG=574.5 kJ/kg. Estimate the pressure in the wall cavity at 100 K.

Solution:

For CO2space: at T1= -90oC = 183.2 K , P1= 38.1 kPa, hIG= 574.5 kJ/kg

For T2= TcO2= 100 K: ClapeyrondPSUB

dT=

hIGTvIG

hIGPSUB

RT2

lnP2P1

=hIGR

[1

183.2

1

100] =

574.5

0.188 92[

1

183.2

1

100] = 13.81

or P2= P

11.00510-6 P

2= 3.8310-5kPa = 3.8310-2Pa


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12.33

Small solid particles formed in combustion should be investigated. We would liketo know the sublimation pressure as a function of temperature. The only

information available is T, hFG

for boiling at 101.3 kPa and T, hIF

for melting at

101.3 kPa. Develop a procedure that will allow a determination of the sublimation

pressure, Psat

(T).

TNBP

= normal boiling pt T.

TNMP

= normal melting pt T.

TTP

= triple point T.

1) TTP

TNMP

2) 0.1013 MPa

PTP

(1/PSAT

)dPSAT

TNMP

TTP

hFG

RT2dT

Since hFG

const hFG NBP

the integral over temperature becomes

lnP

TP

0.1013

hFG NBP

R[

1

TNBP

-1

TTP

] get PTP

3) hIG at TP

= hG

- hI= (h

G- h

F) + (h

F- h

I) h

FG NBP+ h

IF NMP

Assume hIG

const. again we can evaluate the integral

lnP

SUB

PTP

= PTP

PSUB

(1/PSUB

)dPSUB

TTP

T

hIG

RT2dT

hIG

R[

1

TTP

1

T]

or PSUB

= fn(T)

P

TP

TP

NMP NBP

101.3 kPa

T

Solid Liquid

Vap.

T TT

P


42/218






Property Relations


43/218






12.34

Use Gibbs relation du = Tds Pdv and one of Maxwells relations to find an

expression for (u/P)Tthat only has properties P, v and T involved. What is thevalue of that partial derivative if you have an ideal gas?

du = Tds Pdv divide this by dP so we get

u

P T= T

s

P T P

v

P T= T

v

T P P

v

P T

where we have used Maxwell Eq.12.19. Now for an ideal gas we get

Ideal gas: Pv = RT v =RT

P

then the derivatives are

v

T P=R

P and

v

P T= RTP2

and the derivative of u is

u

P T= T

v

T P P

v

P T= TR

P P( RTP2) = 0

This confirms that u is not sensitive to P and only a function of T.


44/218






12.35

The Joule-Thomson coefficient Jis a measure of the direction and magnitude of

the temperature change with pressure in a throttling process. For any three

properties x,y,z use the mathematical relation

x

y z

y

z x

z

x y= 1

to show the following relations for the Joule-Thomson coefficient:

J=

T

P h=

T

v

T P v

CP=

RT2

PCP

Z

T P

Let x = T, y = P and z = h and substitute into the relations as:

T

P h

P

h T

h

T P= 1

Then we have the definition of specific heat as CP=

h

T P so solve for the first

term

J=

T

P h=

1

CP/

P

h T=

1

CP

h

P T

The last derivative is substituted with Eq.12.25 so we get

J= TP h=

T

v

T P v

CP

If we use the compressibility factor then we get

Pv = ZRT

v

T P=

ZR

P+

RT

P

Z

T P=

v

T+

RT

P

Z

T P

so then

T

v

T P v = v +

RT2

P

Z

T P v =

RT2

P

Z

T P

and we have shown the last expression also.

J=

T

P h=

T vT P v

CP=

RT2

PCP

Z

T P


45/218






12.36

Find the Joule-Thomson coefficient for an ideal gas from the expression given in

Problem 12.35

J= T

P h=

T

v

T P v

CP= RT

2

PCP

ZT P

For an ideal gas: v = RT/P so then the partial derivative

v

T P=

R

P T

v

T P v =

RT

P v = v v = 0

For an ideal gas Z = 1 so the very last derivative of Z is also zero.


46/218






12.37

Start from Gibbs relation dh = Tds + vdP and use one of Maxwells equation to

get (h/v)Tin terms of properties P, v and T. Then use Eq.12.24 to also find anexpression for (h/T)v.

Find (hv)T and (hT)vdh = Tds + vdP and use Eq.12.18

(hv)T= T (sv)T+ v(

Pv)T = T (

PT)v+ v(

Pv)T

Also for the second first derivative use Eq.12.24

(hT)v= T(sT)v+ v(

PT)v= Cv+ v(

PT)v


47/218






12.38

From Eqs. 12.23 and 12.24 and the knowledge that Cp> Cvwhat can you

conclude about the slopes of constant v and constant P curves in a T-s diagram?

Notice that we are looking at functions T(s, P or v given).

Solution:

The functions and their slopes are:

Constant v: T(s) at that v with slope

T

s v

Constant P: T(s) at that P with slope

T

s P

Slopes of these functions are now evaluated using Eq.12.23 and Eq.12.24 as

T

s P=

s

T P

-1

=T

Cp

T

s v=

s

T v

-1

=T

Cv

Since we know Cp> Cvthen it follows that T/Cv > T/Cpand therefore

T

s v >

T

s P

which means that constant v-lines are steeper than constant P lines in a T-sdiagram.


48/218






12.39

Derive expressions for (T/v)uand for (h/s)vthat do not contain the properties

h, u, or s. Use Eq. 12.30 with du = 0.

(Tv)u= -(uv)T/(uT)v= CvP-T(

P

T)v (see Eqs. 12.33 and 12.34)

As dh = Tds + vdP => (hs)v= T+v(

Ps)v= T-v(

Tv)s (Eq.12.20)

But (Tv)s= -(

sv)T/(

sT)v= - Cv

T(PT)v

(Eq.12.22)

(hs)v= T +

vT

Cv(

PT)v


49/218






12.40

Evaluate the isothermal changes in the internal energy, the enthalpy and theentropy for an ideal gas. Confirm the results in Chapters 3 and 6.

We need to evaluate duT, dhT and dsTfor an ideal gas: P = RT/v.

From Eq.12.31 we get

duT= [ T (PT)v P ] dvT= [ T (

R

v) P ] dvT= [ P P] dvT= 0

From Eq.12.27 we get using v = RT/P

dhT= [ v T (vT)P] dPT= [ v T (

R

P) ] dPT= [ v v ] dPT= 0

These two equations confirms the statements in chapter 5 that u and h arefunctions of T only for an ideal gas.

From Eq.12.32 or Eq.12.34 we get

dsT= (vT)PdPT = (

PT)vdvT

= R

PdPT =

R

vdvT

so the change in s can be integrated to find

s2 s1= R lnP2

P1= R ln

v2

v1 when T2= T1


50/218






12.41

Develop an expression for the variation in temperature with pressure in a constant

entropy process, (T/P)s, that only includes the properties PvTand the specific

heat, Cp.Follow the development for Eq.12.32.

(TP)s= -

(sP)T

(sT)P

= -

-(vT)P

(CP/T)

=T

CP

(vT)P

{(sP)T= -(

vT)P, Maxwell relation Eq. 12.23 and the other is Eq.12.27}


51/218






12.42

Use Eq. 12.34 to get an expression for the derivative (T/v)s. What is the generalshape of a constant s process curve in a T-v diagram? For an ideal gas can you say

a little more about the shape?

Equation 12.34 says

ds = CvdT

T + (

PT)vdv

so then in a constant s process we have ds = 0 and we find

(Tv)s=

T

Cv(

PT) v

As T is higher the slope is steeper (but negative) unless the last term (P/T)vcounteracts. If we have an ideal gas this last term can be determined

P = RT/v (PT)v=

R

v

(Tv)s= TCv

Rv

= PCv and we see the slope is steeper for higher P and a little lower for higher T as Cvis

an increasing function of T.


52/218






12.43

Show that the P-v-T relation as P(v b) = RT satisfies the mathematical relation

in Problem 12.35.

x

y z

y

z x

z

x y= 1

Let (x, y, z ) be (P, v, T) so we have

P

v T

v

T P

T

P v= 1

The first derivative becomes, P = RT/(v b)

P

v T= RT (v b)-2= P/(v b)

The second derivative, v = b + RT/P

v

T P= R/P

The third derivative, T = (P/R)(v b)

T

P v= (v b)/R

Substitute all three derivatives into the relation

P

v T

v

T P

T

P v=

RT

(v b)2

R

P

v b

R=

RT

(v b)

1

P= 1

with the last one recognized as a rewrite of the original EOS.


53/218






Volume Expansivity and Compressibility


54/218






12.44

What are the volume expansivity p, the isothermal compressibility T,and theadiabatic compressibility sfor an ideal gas?

The volume expansivity from Eq.12.37 and ideal gas v = RT/P gives

p=1v(vT)P=

1v(

RP

) =1T

The isothermal compressibility from Eq.12.38 and ideal gas gives

T=

1

v(

vP)T=

1

v( RT P2) =

1

P

The adiabatic compressibility sfrom Eq.12.40 and ideal gas

s= 1

v(

vP)s

From Eq.12.32 we get for constant s (ds = 0)

(

T

P)s =T

Cp(

v

T)P =T

Cp

R

P =

v

Cpand from Eq.12.34 we get

(vT)s =

Cv

T(

PT)v=

Cv

Tv

R =

Cv

P

Finally we can form the desired derivative

(vP)s= (

vT)s(

TP)s=

Cv

P

v

Cp=

v

kP

s= 1

v(

vP)s = (

1

v)(

v

kP)=

1

kP=

1

k

T


55/218






12.45

Assume a substance has uniform properties in all directions with V = LxLyLzand

show that volume expansivity p= 3T. Hint: differentiate with respect to T anddivide by V.

V = LxLyLzFrom Eq.12.37

p=1

V(VT)P =

1

LxLyLz( T

LxLyLz) P

=LyLz

LxLyLz(

LxT )P+ LxLyLz

LxLz( T

Ly)P+

LxLy

LxLyLz(

LzT )P

=1

Lx( T

Lx)P+

1

Ly( T

Ly)P+

1

Lz( T

Lz) P

= 3 TThis of course assumes isotropic properties (the same in all directions).


56/218






12.46

Determine the volume expansivity, P, and the isothermal compressibility,

T, for

water at 20C, 5 MPa and at 300C, and 15 MPa using the steam tables.

Water at 20oC, 5 MPa (compressed liquid)

P=

1v(vT)P

1v(vT)P ; T= -

1v(vP)T-

1v(vP) T

Estimate by finite difference using values at 0oC, 20oC and 40oC,

P

1

0.000 99950.001 0056 - 0.000 9977

40 - 0 = 0.000 1976 oC-1

Using values at saturation, 5 MPa and 10 MPa,

T

-1

0.000 99950.000 9972-0.001 0022

10 - 0.0023=0.00050 MPa-1


P

1

0.001 3770.001 4724 - 0.001 3084

320 - 280 = 0.002 977 oC-1

T-

1

0.001 3770.001 3596-0.001 3972

20 - 10=0.002731 MPa-1


57/218






12.47

Use the CATT3 software to solve the previous problem.

The benefit of the software to solve for the partial derivatives is that we can

narrow the interval over which we determine the slope.


P=

1

v(

vT)P

1

v(

vT)P ; T= -

1

v(

vP)T-

1

v(

vP) T

Estimate by finite difference using values at 19oC, 20oC and 21oC,

P

1

0.000 99950.000 9997 - 0.000 9993

21 - 19 = 0.000 40 oC-1

Using values at saturation, 4.5 MPa and 5.5 MPa,

T-

1

0.000 99950.000 9993-0.000 9997

5.5 - 4.5=0.000 40 MPa-1


P

1

0.001 3770.001 385 - 0.001 369

302 - 298 = 0.011 619 oC-1

T-

1

0.001 3770.001 373-0.001 381

16 - 14=0.002 905 MPa-1


58/218






12.48

A cylinder fitted with a piston contains liquid methanol at 20C, 100 kPa andvolume 10 L. The piston is moved, compressing the methanol to 20 MPa at

constant temperature. Calculate the work required for this process. The isothermal

compressibility of liquid methanol at 20C is 1.22 10-9m2/N.

1w2= 1

2

Pdv=

P(vP)TdPT= -1

2

vT

PdPT

For v constant & Tconstant the integral can be evaluated

1w2= -v

T

2(P22- P21)

For liquid methanol, from Table A.4: = 787 m3/kgV

1= 10 L, m = 0.01 787 = 7.87 kg

1W2=0.011220

2[(20)2- (0.1)2] = 2440 J = 2.44 kJ


59/218






12.49

For commercial copper at 25oC (see table A.3) the speed of sound is about 4800

m/s. What is the adiabatic compressibility s?

From Eq.12.41 and Eq.12.40

c2= (P)s= v2(Pv)s =

1

-1

v(vP)s

=1

s

Then we get using density from Table A.3

s=1

c2=1

480028300 m2kgs2m3

=1000

4800283001

kPa

= 5.23 109 kPa 1

Cu


60/218






12.50

Use Eq. 12.32 to solve for (T/P)sin terms of T, v, Cpand p. How large a

temperature change does 25oC water (p= 2.1 10-4K-1) have, when

compressed from 100 kPa to 1000 kPa in an isentropic process?

From Eq.12.32 we get for constant s (ds = 0) and Eq.12.37

(TP)s=T

Cp(

vT)P =

T

Cppv

Assuming the derivative is constant for the isentropic compression we estimatewith heat capacity from Table A.3 and v from B.1.1

Ts= (TP)sPs=

T

Cppv Ps

= 273.15 + 254.18

2.1 10-40.001003 (1000 100)

= 0.013 K barely measurable.


61/218






12.51

Sound waves propagate through a media as pressure waves that cause the mediato go through isentropic compression and expansion processes. The speed of

sound cis defined by c2=(P/)sand it can be related to the adiabatic

compressibility, which for liquid ethanol at 20C is 9.4 10-10

m2/N. Find the

speed of sound at this temperature.

c2= (P)s= v2(Pv)s =

1

-1

v(vP)s

=1

s

From Table A.4 for ethanol, = 783 kg/m3

c = ( 194010-12783

)1/2= 1166 m/s


62/218






12.52

Use Table B.3 to find the speed of sound for carbon dioxide at 2500 kPa near

100oC. Approximate the partial derivative numerically.

c2

= (P

)s= v2

(P

v)sWe will use the 2000 kPa and 3000 kPa table entries. We need to find the change

in v between two states with the same s at those two pressures.

At 100oC, 2500 kPa: s = (1.6843 + 1.5954)/2 = 1.63985 kJ/kg-K,

v = (0.03359 + 0.02182)/2 = 0.027705 m3/kg

2000 kPa, s = 1.63985 kJ/kg-K: v = 0.031822 m3/kg

3000 kPa, s = 1.63985 kJ/kg-K: v = 0.0230556 m3/kg

c2v2(Pv)s= 0.0277052

3000 - 2000

0.0230556-0.031822kJ

kg= 87 557.8

J

kg

c = 87 557.8 = 295.9 m/s


63/218






12.53


At 100oC, 2500 kPa: s = 1.636 kJ/kg-K, v = 0.02653 m3/kg

101oC, s = 1.636 kJ/kg-K: v = 0.02627 m3/kg, P = 2.531 MPa

99oC, s = 1.636 kJ/kg-K: v = 0.02679 m3/kg, P = 2.469 MPa

c2v2(Pv)s= 0.026532

2531 - 2469

0.02627 - 0.02679kJ

kg= 83 919.5

J

kg

c = 83 919.5 = 289.7 m/s


64/218






12.54

Consider the speed of sound as defined in Eq. 12.41. Calculate the speed of sound

for liquid water at 20C, 2.5 MPa, and for water vapor at 200C, 300 kPa, usingthe steam tables.

From Eq. 12.41: c2= (P)s= -v2(Pv) sLiquid water at 20oC, 2.5 MPa, assume

(Pv)s(Pv) T

Using saturated liquid at 20oC and compressed liquid at 20oC, 5 MPa,

c2= -(0.001 002+0.000 99952

)2( 5-0.00230.000 9995-0.001 002

)MJkg

= 2.002106J

kg => c = 1415 m/s

Superheated vapor water at 200oC, 300 kPa

v = 0.7163 m3/kg, s = 7.3115 kJ/kg K

At P = 200 kPa & s = 7.3115 kJ/kg K: T = 157oC, v = 0.9766 m3/kg

At P = 400 kPa & s = 7.3115 kJ/kg K: T = 233.8oC, v = 0.5754 m3/kg

c2= -(0.7163)2( 0.400-0.2000.5754-0.9766

) MJkg

= 0.2558 106m2/s2

=> c = 506 m/s


65/218






12.55


From Eq. 12.41: c2= (P

)s= -v2(P

v) s

Liquid water at 20oC, 2.5 MPa, assume (Pv)s(Pv)s

and CATT3: v = 0.001001 m3/kg, s = 0.2961 kJ/kg-K

Using liquid at 3 MPa and 2 MPa at the same s = 0.2961 kJ/kg-K,

c2= - 0.0010012( 3 - 20.001 - 0.001 001

)MJkg

= 1.002106J

kg

=> c = 1001 m/sSuperheated vapor water at 200oC, 300 kPa

CATT3: v = 0.7163 m3/kg, s = 7.311 kJ/kg K



c2= -(0.7163)2( 0.310 - 0.2900.6986 - 0.7351

) MJkg

= 0.28114 106m2/s2

=> c = 530 m/s


66/218


67/218






12.57

Liquid methanol at 25oC has an adiabatic compressibility of 1.05 10-9m2/N.What is the speed of sound? If it is compressed from 100 kPa to 10 MPa in aninsulated piston/cylinder, what is the specific work?

From Eq.12.41 and Eq.12.40 and the density from table A.4

c2= (P)s= v2(Pv)s =

1

s=

1

1.05 10-9787

= 1.210 106 m2/s2c = 1100 m/s

The specific work becomes

w = P dv = P (-sv ) dP = sv P dP = sv 1

2

P dP

= sv 0.5 (P22 P

21)

= 1.05 10-9m2/N 0.5

787m3/kg (10 0002 1002) 10002Pa2

= 66.7 J/kg


68/218






12.58

Use Eq. 12.32 to solve for (T/P)sin terms of T, v, Cpand p. How much higherdoes the temperature become for the compression of the methanol in Problem

12.57? Use p= 2.4 10-4K-1for methanol at 25oC.

From Eq.12.32 we get for constant s (ds = 0) and Eq.12.37

(TP)s=T

Cp(

vT)P =

T

Cppv

Assuming the derivative is constant for the isentropic compression we estimate

with heat capacity and density (v = 1/) from Table A.4

Ts= (TP)sPs=

T

Cppv Ps

=298.152.55

K kg K

kJ2.4 10-4K-1 1

787kgm3

(10 000 100) kPa

= 0.353 K


69/218






12.59

Find the speed of sound for air at 20C, 100 kPa using the definition in Eq. 12.41and relations for polytropic processes in ideal gases.

From problem 12.41 : c2= (P

)s= -v2(P

v) s

For ideal gas and isentropic process, Pvk= constant

P = Cv-k Pv= -kCv

-k-1= -kPv-1

c2= -v2(-kPv-1) = kPv = kRT

c = kRT = 1.40.287293.151000 = 343.2 m/s

For every 3seconds after thelightning the

sound travels

about 1 km.


70/218






Equations of State


71/218






12.60

Use Table B3 and find the compressibility of carbon dioxide at the critical point.

Pv = Z RT

At the critical point from B.3:P = 7377.3 kPa, T = 31C = 304.15 K, v = 0.002139 m3/kg

from A.5: R = 0.1889 kJ/kg-K

Z =Pv

RT=

7377.3 0.002139

0.1889 304.15= 0.27


72/218






12.61

Use the equation of state as shown in Example 12.3 where changes in enthalpy

and entropy were found. Find the isothermal change in internal energy in a similarfashion; do not compute it from enthalpy.

The equation of state isPv

RT= 1 C

P

T4

and to integrate for changes in u from Eq.12.31 we make it explicit in P as

P = T4(v

RT3+ C )1

Now perform the partial derivative of P

(PT)v= 4 T

3(v

RT3+ C )1T4(

v

RT3+ C )23

v

RT2

= 4P

TT4P2

3v

RT2= 4P

T3P

TPv

RT=P

T[ 4 3Pv

RT]

Substitute into Eq.12.31

duT= [ T (PT)v P ] dvT= [ P( 4 3

Pv

RT) P ] dvT

= 3 P ( 1 Pv

RT) dvT= 3 P C

P

T4 dvT

The P must be eliminated in terms of v or the opposite, we do the latter as fromthe equation of state

v =

RT

P C R

1

T3 dvT= RT

P2 dPT

so now

duT= 3 CP2

T4dvT= 3 C R

1

T3 dPT

and the integration becomes

u2 u1= 3 C R T3(P2 P1)


73/218






12.62

Use Table B.4 to find the compressibility of R-410A at 60oC and a) saturatedliquid b) saturated vapor and c) 3000 kPa.

Table A.2: R = 8.31451 / 72.585 = 0.1145 kJ/kg-K

a) Table B.4.1: P = 3836.9 kPa, v = 0.001227 m3/kg

Z =Pv

RT=

3836.9 0.001227

0.1145 333.15= 0.1234

b) Table B.4.1: P = 3836.9 kPa, v = 0.00497 m3/kg

Z =Pv

RT=

3836.9 0.00497

0.1145 333.15= 0.5

c) Table B.4.2: P = 3000 kPa, v = 0.00858 m3/kg

Z =Pv

RT=

3000 0.00858

0.1145 333.15= 0.675

The R-410A is not an ideal gas at any of these states.


74/218






12.63

Use a truncated virial EOS that includes the term with B for carbon dioxide at 20o

C, 1 MPa for which B = 0.128 m3/kmol and T(dB/dT) = 0.266 m3/kmol. Findthe difference between the ideal-gas value and the real-gas value of the internal

energy.

virial eq.: P =RT

v+

BRT

v2 ; (

PT) v=

R

v+

BR

v2+

RT

v2(

dB

dT)

u-u EEA

v[ART2

Ev2 A

(A

dB

dTA

)]dvEA = ARTvE

A[T ( AdBdTE

A

)]*= -

v[(PT) v- P]dv =

ESolution of virial equation (quadratic formula):

AvE

A= A1

2E

AEAARAT

E

PE

A[1 + EA 1 + 4BP/ARATEA] where: EAARATE

PE

A= A8.3145293.15

1000E

A= 2.43737

AvE

A= A1

2EA2.43737 [1 +

A 1 + 4(-0.128)/2.43737EA]= 2.3018 mA3

E

A/kmol

Using the minus-sign root of the quadratic formula results in a compressibility

factor < 0.5, which is not consistent with such a truncated equation of state.

u uA

*E

A

=A

-8.3145 293.152.3018E

A

[0.266]/ 44.01 = 6.4 kJ/kg


75/218






12.64

Solve the previous Problem with Table B.3 values and find the compressibility of

the carbon dioxide at that state.

B.3: v = 0.05236 mA3EA/kg, u = 327.27 kJ/kg, A5: R = 0.1889 kJ/kg-K

Z = APv

RTEA= A

1000 0.05236

0.1889 293.15EA= 0.9455 close to ideal gas

To get u A*EAlet us look at the lowest pressure 400 kPa, 20 AoEAC: v = 0.13551 mA3EA/kgand u = 331.57 kJ/kg.

Z = Pv/RT = 400 0.13551/(0.1889 293.15) = 0.97883

It is not very close to ideal gas but this is the lowest P in the printed table.

u uA*E

A

= 327.27 331.57 = 4.3 kJ/kg


76/218


77/218






12.66

Extend problem 12.63 to find the difference between the ideal-gas value and the

real-gas value of the entropy and compare to table B.3.

Calculate the difference in entropy of the ideal-gas value and the real-gas value

for carbon dioxide at the state 20C, 1 MPa, as determined using the virialequation of state. Use numerical values given in Problem 12.63.

COA2

E

Aat T = 20AoEAC, P = 1 MPa

sA*

P*E

A-sAP

E

A=A

v(P)

E

RT/P*

(PT)vdvEA; ID Gas: sA*

P*E

A-sAP

E

A=A

v(P)

E

RT/P*

R

vdv EA= RlnA

P

P*E

Therefore, at P: sA*

PE

A- sAP

E

A= -RlnAP

P*EA+A

v(P)

E

RT/P*

(PT)vdv EA

virial: P =A

RTvE

A

+A

BRTv2E

A

and (APTEA)Av EA= ARv E A+ ABRv2E A+ ARTv2E

A(AdBdTEA)Integrating,

sA

*

PE

A

- sA

PE

A

= -RlnA

P

P*EA

+ RlnA

RT

P*vEA

+ R[B + T(A

dB

dTEA

)](A

1

vEA

-A

P*

ERT EA

)

= R[lnA

RT

Pv EA

+ (B + T(A

dB

dTEA

))A

1

vEA

]

Using values for COA

2E

A

from solution 12.63 and R = 0.1889 kJ/kgK

sA*

PE

A- sAPEA= 0.1889[lnA

2.437 37

2.3018E

A+(-0.128 + 0.266)A

1

2.3018E

A

]= 0.02214 kJ/kg K

From Table B.3 take the ideal as the lowest P = 400 kPa:

sA*

PE

A- sAP

E

A= 1.7904 1.6025 + 0.1889 ln(400/1000) = 0.0148 kJ/kgK

The lowest P = 400 kPa in B3 is not exactly ideal gas ( Z = Pv/RT = 0.9788)


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12.67

Two uninsulated tanks of equal volume are connected by a valve. One tank

contains a gas at a moderate pressure PA1

E

A, and the other tank is evacuated. The

valve is opened and remains open for a long time. Is the final pressure PA

2E

A

greater

than, equal to, or less than PA

1EA

/2? Hint: Recall Fig. 12.5.

Assume the temperature stays constant then for an ideal gas the pressure will bereduced to half the original pressure. For the real gas the compressibility factor

maybe different from 1 and then changes

borgnakke and sonntag_fundamentals of thermodynamics

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