ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

The difference between v′R and v′o R, areillustrated with respect to TR for these equations as PR→ 0 in Figure 11.If a= b = 0 in any of the real gas equations of state, these equations a

P = (RT/v) + (1/v2) (bRT – a/Tn) (63)Therefore,

Where B(T) = b – a/RT(n+1) As P → 0, v → ∞, and Z → 1

Solving for v from Eq (64),

Figure 11: The deviation function for Berthelot, RK,and VW gases, PR→0

where n = 0, 1, 1/2, a∗ = (27/64), (27/64), 0.4275, and b∗ = 0.125, 0.125, 0.08664, respectively,for the Van der Waals, Berthelot and RK equations The difference between v′R and v′o R, areillustrated with respect to TR for these equations as PR→ 0 in Figure 11.

If a= b = 0 in any of the real gas equations of state, these equations are identical tothe ideal gas state equation

8 Three Parameter Equations of State

If v = vc (i.e., along the critical isochore), employing the Van der Waals equation,

Clausius developed a three parameter equation of state which makes use of mentally measured values of Zc to determine the three parameters, namely

Figure 12 Illustration of Pitzer factor estimation

molecule, making it uneven or polar The compressibility factors for nonsymmetric or polarfluids are found to be different from those determined using two parameter equations of state.Therefore, a third factor, called the Pitzer or acentric factor ω has been added so that the em-pirical values correspond with those obtained from experiments This factor was developed as

a measure of the structural difference between the molecule and a spherically symmetric gas(e.g., a simple fluid, such as argon) for which the force–distance relation is uniform around themolecule In case of the saturation pressure, all simple fluids exhibit universal relations for

PRsat with respect to TR (as illustrated in Figure 14) In Chapter 7 we can derive such a relationusing a two parameter equation of state For instance, when TR = 0.7, all simple fluids yield

PRsat≈ 0.1, but polar fluids do not The greater the polarity of a molecule, the larger will be itsdeviation from the behavior of simple fluids Figure 14 could also be drawn for log10 Prsat vs.1/TR as illustrated in Figure 12 The acentric factor ω is defined as

ω = –1.0 – log10 (PRsat)TR=0.7 = –1 – 0.4343 ln (PRsat)TR=0.7 (67)Table A-1 lists experimental values of ω” for various substances In case they are not listed, it

is possible to use Eq (68)

With these observations, we are able to establish the following relation, namely.(Z (ω,TR,PR)– Z(0) (PR, TR)) = ωZ(1)

Evaluation of Z(ω,TR,PR) requires a knowledge of Z(1), w and Z(0) (PR, TR)

i Saturation Pressure Correlations

The function ln(Psat) varies linearly with T–1, i.e.,

Using the condition T = Tc, P = Pc, if another boiling point Tref is known at a pressure Pref, thenthe two unknown parameters in Eq (70) can be determined Therefore, the saturation pressure

at T = 0.7Tc can be ascertained and used in Eq (69) to determine ω

ii Empirical Relations

Empirical relations are also available, e.g.,

ω = (ln PRsat – 5.92714 + 6.0964/TR,BP+1.28862 ln TR,BP – 0.16935 TR,BP)/

where PR denotes the reduced vapor pressure at normal boiling point (at P = 1 bar), and TR,NBP

the reduced normal boiling point

An alternative expression involves the critical compressibility factor, i.e.,

Another such relation has the form

9 Other Three Parameter Equations of State

Other forms of the equation of state are also available

a One Parameter Approximate Virial Equation

For values of vR > 2 (i.e., at low to moderate pressures),

where B1(TR) = B(0)(TR) + ωB(1)(TR), B(0)(TR) = (0.083 TR–1) – 0.422 TR–2.6, and B(1)(TR) =0.139 – 0.172 TR–5.2

b Redlich–Kwong–Soave (RKS) Equation

Soave modified the RK equation into the form

where a = 0.42748 R2Tc2Pc–1, b = 0.08664 RTcPc–1, and α (ω,TR) = (1 + f(ω)(1 – TR0.5))2 , which

is determined from vapor pressure correlations for pure hydrocarbons Thus, f(ω) = (0.480 +1.574 ω – 0.176 ω2

Figure 13: An illustration of the variation in the compressibility factor with

respect to the acentric factor

where a = 0.45724 RTcPc , b = 0.07780 RTcPc and α(ω,TR) = (1 + f(ω) (1 – TR )), f(ω)

= 0.37464 + 1.54226 ω – 0.26992 ω2

Equation (75) can be employed to predict the variation

of Psat with respect to T, and can be used to explicitly solve for T(P,v)

10 Generalized Equation of State

Various equations of state (e.g., VW, RK, Berthelot, SRK, PR, and Clausius II) can

be expressed in a general cubic form, namely,

In terms of reduced variables this expression assumes the form

PR = TR/(v′R– b´) – a´α (ω,TR)/(TRn (v′R + c´) (v′R + d´)), (77)where a´ = a/(Pcv′c2 Tcn), b´ = b/v′c, c´ = c/v′c, and d´ = d/v′c Tables are available for the pa-rameters a´ to d´ Using the relation Z = PRv′R/TR, we can obtain a generalized expression for

Z as a function of TR and PR, i.e.,

Z3 + Z2 ((c´ + d´ – b´) PR/TR – 1) + Z (a´α (ω,TR) PR/TR2+n –

(1 + b´PR/TR) (c´ + d´)PR/TR + c´d´ PR2/TR2) –

(a´α (ω,TR) b´PR2/TR(3+n) + (1 + PR b´/TR) (c´d´PR2/TR2)) = 0 (78)Writing this relation in terms of v′R,

′

vR3 PR + v′R2((c´ + d´ – b´)(PR/TR) –1) + v′R ((c´d´ – b´c´ –b´d´)PR – (c´ + d´)TR+ a×/TRn) – PR b´c´d´ – a´α (ω,TR)b´/TR – TRc´d´ = 0 (79)Using this equation along with the relation TR = PRv′R/Z, the compressibility factor can beobtained as a function of PR and v′R, i.e.,

Z(3+n) (a´α(w, TR) /(v′R(2+n)PR(1+n))) (1–b´/v′R) +

Z3(1+(c´+d´–b´)/v′R–(b´/v′R2)(c´+d´–d´/vR)) – Z2(1 + (c´ + d´)/v′R–c´d´PR/v′R+

where TR in α (w, TR) expression must be replaced by PR vR´/Z Table 2 tabulates values of α,

n, a´, b´, c´, and d´ for various equations of state

Table 2: Constants for the generalized real gas equation of state

11 Empirical Equations Of State

These equations accurately predict the properties of specified fluid; however, they are not able for predicting the stability characteristics of a fluid (Chapter 10)

b Beatie – Bridgemann (BB) Equation of State

This equation is capable representing P-v-T data in the regions where VW and RKequations of state fail particularly when ρ < 0.8 ρc It has the form

Figure 14: Relation between pressure and volume for

compres-sion/expansion of air (from A Bejan, Advanced Engineering

Ther-modynamics, John Wiley and Sons., 1988, p 281).

d Lee–Kesler Equation of State

This is another modified form of the BWR equation which has 12 constants and isapplicable for any substance This relation is of the form

PR = (TR/v′R) (1+A/v′R+B/v′R2+C/v′R5+(D/v′R)(β+γ/v′R2)exp(–γ/v′R2)), (83a)

Z = PRv′R/TR = 1+A/v′R+B/v′R2+C/v′R5+(D/v′R)(β+γ/v′R2)exp(–γ/v′R2), (83b)where A = a1 – a2/TR – a3/TR2 – a4/TR3, B = b1 – b2/TR + b3/TR3, C = c1 + c2/TR, and D = d1/TR3.The constants are usually tabulated to determine Z(0) for all simple fluids and Z(ref) for a refer-ence fluid, that is usually octane (cf Table A-21) Assuming that

Z(ref) – Z(0) = ω Z(1)

Z(1) can be determined

A general procedure for specified values of PR and TR is as follows: solve for vR´ from

Eq (83a) with constants for simple fluids and use in Eq (83b) to obtain Z(0) Then repeat theprocedure for the same PR and TR with different constants for the reference fluid, obtain Z(ref),and determine Z(1) from Eq.(83c) The procedure is then repeated for different sets of PR and

TR A plot of Z(0) is contained in the Appendix and tabulated in Table A–23A The value of Z(1)

so determined is assumed to be the same as for any other fluid Tables A-23A and A-23Btabulate Z(0) and Z(1) as function of PR and TR

up to 1.5Tc

12 State Equations for Liquids/Solids

a Generalized State Equation

The volume v = v (P,T), and dv = (∂v/∂P)TdP + (∂v/∂T)PdT, i.e.,

dv = vβP dT – vβT dP, or d(ln v) = βPdT – βTdP

If β and β are constant, the general state equation for liquids and solids can be written as

ln(v/vref) = βP (T – Tref) – βT (P – Pref) (86)This relation is also referred to as the explicit form of the thermal equation of state In terms ofpressure, the relation

P = Pref + (βP/βT)(T – Tref) – ln (v/vref)/( βT vref), (87)

is an explicit, although approximate, state equation for liquids and solids Both Eqs (87) or(88) can be approximated as

(v–vref)/vref = βP (T – Tref) – βT (P – Pref) (88)Solving the relation in terms of pressure

P = Pref + (βP/βT) (T – Tref) – (v–vref)/( βT vref), (89)which is an explicit, although approximate, state equation for liquids and solids

The pressure effect is often small compared to the temperature effect Therefore, Eq.(89) can be approximated in the form

In case βP (T – Tref) « 1, then

which is another explicit, although approximate, state equation for liquids and solids

Copper has the following properties at 50ºC: v, βP, and βT are, respectively,7.002×10–3 m3 kmole–1, 11.5×10–6 K–1, and 10–9 bar–1 Therefore, heating 10 kmole of the sub-stance from 50 to 51ºC produces a volumetric change that can be determined from Eq.(87) as7.002×10–3× 10 × 11.5×10–6 = 805 cm3 If a copper bar containing 10 kmole of the substance

is vertically oriented and a weight is placed on it such that the total pressure on the massequals 2 bar, the volume of the copper will reduce by a value equal to –7.002×10–3× 10 ×0.712×10–9

= –0.05 mm3 Therefore, changing the state of the 10 kmole copper mass from50ºC and 1 bar to 51ºC and 2 bars, will result in a volumetric change that equals 805 – 0.05 =804.95 mm3

For solids βP is related to the linear expansion coefficient α The total volume V ∝ L3,and

βP = 1/V(∂V/∂T)P = 1/L3∂(L3)/∂T = (3/L) ∂L/∂T = 3α, (92)where α = (1/L) (∂L/∂T)P

δw = –vdP (for a reversible process in an open system).

∆P = 6.03×65 = 391 bar

b Murnaghan Equation of State

If we assume that the isothermal bulk modulus BT (= 1/βT) is a linear function of thepressure, then

BT(T,P) = (1/βT) = –v(∂P/∂v)T = BT(T,0) + αP (93)where α = (∂BT/∂P)T Therefore,

c Racket Equation for Saturated Liquids

The specific volume of saturated liquid follows the relation given by the Racketequation, namely,

vf v Zc c TR

= ( – 1 0 2857. )

d Relation for Densities of Saturated Liquids and Vapors.

If ρf denotes the saturated liquid density, and ρg the saturated vapor density, then

ρRf = ρf/ρc = 1 + (3/4)(1 – TR) + (7/4)(1– TR)1/3, and (98)

ρRg = ρg/ρc = 1 + (3/4)(1 – TR) – (7/4)(1– TR)1/3 (99)These relations are based on curve fits to experimental data for Ne, Ar, Xe, O2, CO, and CH4

It is also seen that

At low pressures,

ρRf ≈ (7/2)(1 – TR)1/3 since ρRf >> ρRg

In thermodynamics, ρRf – ρRg is called order of parameter If ρRg is known at low pressures(e.g., ideal gas law), then ρRf can be readily determined Another empirical equation followsthe relation

ρR,f = 1 + 0.85(1–TR) + (1.6916 + 0.9846ψ)(1–TR)1/3 (101)where ψ ≈ ω

e Lyderson Charts (For Liquids)

Lyderson charts can be developed based on the following relation, i.e.,

The appendix contains charts for ρR vs PR with TR as a parameter In case the density is known

at specified conditions, the relation can be used to determine Pc, Tc and ρc Alternatively, if thedensity is not known at reference conditions, the following relation, namely,

can be used

Recall that liquid molecules experience stronger attractive forces compared to gasesdue to the smaller intermolecular spacing The molecules are at conditions close to the lowestpotential energy where the maximum attractive forces occur Therefore, any compression ofliquids results in strong repulsive forces that produce an almost constant intermolecular dis-tance This allows us to use the incompressible approximation, i.e., v = constant

D SUMMARY

This chapter describes how some properties can be determined for liquids, vapors,and gases at specified conditions, e.g., the volume at a given pressure and temperature Com-pressibility charts can be constructed using the provided information and fluid characteristics,such as the Boyle temperature, can be determined The relations can be used to determine thework done as the state of a gas is changed Various methods to improve the predictive accu-racy are discussed, e.g., by introducing the Pitzer factor State equations for liquids and solidsare also discussed

a Case I: γ > 0

i Case Ia: α > 0

There is one real root for this case, i.e

Z = Z = (α + γ 0.5)1/3 + ( α – γ 0.5)1/3 + 1/3

ii Case Ib: α < 0

Again, only one real root exists If tan ϕ = (–p)1.5/q, tan θ = (tan(ϕ/2))1/3 if ϕ > 0, and–(tan(–ϕ/2))1/3 if ϕ<0, then

For the third case, a× = 0.3204, b× =0.05693, a1 =.2602, a0 = -0.01824, and

α = 0.02437, β = 0.002789, and γ = -6.78×10-6

Consider the RK equation of state Determine Z at TR = 1.5 and PR = 1.2, and at TR =

Case II applies, and there are three roots to the equation.

The spreadsheet software uses this methodology for solving the cubic equation

2 Another Explanation for the Attractive Force

The net force acting on the molecules on a wall is proportional to the number of rounding molecules that exert an attraction force The net force on each molecule near thewall equals the force exerted on the wall by collision minus the attraction force Therefore, for

sur-n molecules osur-n the wall, (sur-n × the force exerted on the wall by collisions per molecule) – (n ×attractive force per molecule) = (n × net force on the wall per molecule) Since the attractionforce per molecule ∝ n of surrounding the system, then (n × force exerted on the wall by colli-sion per molecule) – (n × n × constant) = (n × net force) Therefore, the net pressure equals thepressure that would have been exerted in the absence of attraction forces minus the term (n2×constant), i.e.,

P =(RT/(V – b´) – attraction force (which is ∝ n2)

= RT/(V – b´) – attraction force ∝ N2/V2 = RT/(V – b´) – a´/V2

If we compare the attractive force component to the LJ force function (cf Chapter 1),the attractive force ∝ 1/l 6

, i.e., the attractive force being proportional to the n2 seems to be thereason that the exponent is 6 in the attractive force relation (In the context of the gravitationallaw F = G mEm´/r2, where G = 6.67×10-14

kN m2 kg–2, since g = 9.81 m s–2 at r = rE, F =

GmE/rE2 If the radius of the earth is known, then its mass can be determined This derivationalso enables a simplistic relation for the pressure due to inter-planetary forces between planets

in the universe.)

3 Critical Temperature and Attraction Force Constant

Consider an l×l cross section of a wall containing a single molecule M Othermolecules that collide with M impart a momentum to it due to their velocity V Themomentum transfer rate to M is mV2/3l The molecule M also experiences attraction forces.The attraction force between a molecular pair is 4(ε/σ)σ7

/r7 according to the LJ model

Now consider a semicircular segment characterized by the dimensions dr and dθlocated at a radial distance r from M There are πrn´rdθdr molecules within that shell pulling

M away from the wall in the radial direction The net force on the molecules in that direction is(r2 dr dθ cosθ πn´ 24(ε/σ) σ7

)/r7) Assuming the force field to be continuous and integratingthis expression over r = σ to ∞ and θ = 0 to π/2, the net force on M equals 3πn´(εσ2) We mustsubtract this force from the momentum transfer rate Dividing by the area l2, the pressureequals mV2/3l3 – πn´(3εσ2)/l2 Since N´= l–3, the pressure

n´mV2/3 – πn´2(3εσ2) l ∝ RT/(v – b) – a/v2

Therefore, a = NAvag3πεσl is a weak function of the intermolecular spacing In case l ≈ σ, a =

NAvag2 3πεσ3 A more rigorous derivation based on the potential gives the relation a = 2.667

NAvag2

proper-B IDEAL GAS PROPERTIES

The molecules of ideal gases can be considered to be point masses that are enced by intermolecular attractive forces, and follow the state relationship

The molecular energy of an ideal gas uo can be determined if the molecular structure and locity are known (The subscript o is taken to denote ideal gas properties, which can be inter-preted as the condition P → 0) The value of uo depends only upon the temperature Using therelation ho (T) = uo(T) + (Pv)o = uo (T) + RT, the internal energy may be expressed as:

C JAMES CLARK MAXWELL (1831–1879) RELATIONS

Maxwell provided relations for several nonmeasurable properties in terms of able properties (e.g., T, v and P) The basis for the derivation of relations is as follows If

measur-dz = (M(x,y) dx + N(x,y) dy)

is an exact differential, it must then satisfy the exactness criterion, i.e.,

(∂M/∂y) x = (∂N/∂x)y

The variable M (x,y) is called the conjugate of x and N(x,y) is the corresponding conjugate of

y If the exactness criterion is satisfied, the sum (M(x,y) dx + N(x,y) dy) = dZ, the integration

of which yields a point (or state) function Z(x,y) that is a property (Chapter 1) Inversely if Z is

a property, then dZ is exact and, since dZ equals the aforementioned sum, the criterion for anexact differential is satisfied

1 First Maxwell Relation

The First law for a process occurring in a closed system can be expressed in the form

For a process occurring along an internally reversible path

δQrev = TdS and δWrev = PdV,

so that Eq (7) can be written as

Assume that you are to visit a planet on which only s and v can be measured, but forsome reason not T and P Equation (9) can be written in the form

The slopes of u at specified s and v are

(∂u/∂s)v = us = T(s,v), and (∂u/∂v)s = uv = – P(s,v) (11)The temperature T is the conjugate of s and (–P) the conjugate of v It is noted that(∂u/∂s) v→ 0 as T → 0 and hence u = u(v) as T → 0 Obtaining total differential ofT(s,v) and using Eq (10),

dT = (∂T/∂s)v ds + (∂T/∂v)s dv, = uss ds + usv dv, (12)where uss = ∂2u/∂s2, usv = ∂2u/∂s∂v Similarly,

These relations are useful in stability analyses (cf Chapter 10)

At constant volume, i.e., along an isometric curve Eq (9) yields the expression

Tdsv = duv, or T(∂s/∂T)v = (∂u/∂T)v = cv (14)Using the first of these two relations, the area under the resulting curve on a T–s dia-gram represents the internal energy change for the isometric process Rewriting Eq.(10) in the form, we obtain the fundamental relation in entropy form s = s(u,v), i.e.,

ds = (1/T(s,u)) du + (P(s,u)/T(s,u)) dv

Since du is an exact differential, Eq (10) must satisfy the corresponding criterion,namely,

which is known as the First relation Table 1 summarizes the relations

2 Second Maxwell Relation

Adding the term d(Pv) to Eq (9) and simplifying, we obtain the expression

At the critical point, ∂T/∂s = 0 (cf Chapter 3), and cp→ ∞.

The second relation has the form

Conjugate Maxwell Relation

use the result from Chapter 3 that c ∝ T3 at low temperatures Thereafter, assuming c

= cv, cv/T ≈ T2 Therefore, for all three cases, cv→ 0 as T → 0

Similarly, using the relation (∂s/∂T)P = cP/T, we can show that cP→ 0 as T → 0

3 Third Maxwell Relation

The Helmholtz function is defined as

Figure 1: Illustration of Maxwells Relations in terms of s and v

urable.) relations are illustrated in Figure 1 and Figure 2 Since a = u – Ts and s =–(∂a/∂T)v, a/T = u/T + (∂a/∂T)v Therefore,

∂((a/T)/∂T) = (1/T) ∂u/∂T – u/T2 – (∂s/∂T) v = cv/T – u/T2 – (∂s/∂T)v

From the fundamental relation in entropy form, T(∂s/∂T)v = (∂u/∂T)v = cv, so that

Furthermore, since

daT = P dvT,

the area under an isotherm on a P–v diagram represents the Helmholtz function The worktransfer during an isothermal process results in a change in the Helmholtz function Recallthat “a” is a measure of the availability in a closed system Knowing P=P (v,T), one canobtain Helmholtz function “a”

The Massieu function j is defined as

j = –a/T = s – u/T, i.e.,

dj = –da/T + a/T2dT = (1/T2)(PT dv – u dT) = j(T,v)

b Example 2

The fundamental relation for the entropy of an electron gas can be approximated as

B = 23/2π4/3kBm1/2Navag1/6/(31/3hP) (B)

0 2000 4000 6000 8000 10000

0.8 1

Here, kB denotes the Boltzmann constant that has a value of R/NAvag = 1.3804×10

kJ K–1, hP is the Planck constant that has a value of 6.62517×10–37

kJ s, m denotes theelectron mass of 9.1086×10–31 kg, N the number of kmoles of the gas, V its volume in

m3, and U its energy in kJ Determine s, T, and P when u = 4000 kJ k mole–1, and v

P (T, v)= (B/6) T/v (J)

If this state equation (in terms of P, T and v) is known, it does not imply that s, u,

h, a, and g can be subsequently determined This is illustrated by considering thetemperature and pressure relations

One can use Eq (J) in (K) These expressions indicate that Eqs (K) are differentialequations in terms of s and, in order to integrate and obtains =s(T, v), an integra-tion constant is required which is unknown Therefore, a fundamental relation is thatrelation from which all other properties at equilibrium (e.g., T, P, v, s, u, h, a, g,

cp, and cv) can be directly obtained by differentiation alone While the Eq (A) sents a fundamental relation, we can see that the relation Eq (J) does not

Remarks

Alternatively, one can use Eq (23) and get u shown in Eq (D) directly

The Gibbs energy is a measure of the chemical potential, and

g = h – Ts = (5/12) B2T2 v2/3+ (1/2) B2T2v2/3 = (11/12) B2T2 v2/3

The above relation suggests that the value of the chemical potential becomes largerwith an increase in the temperature A temperature gradient results in a gradient in-volving the chemical potential of electrons The state equation, P = (1/6) B2T2/v1/3

indicates that v increases (or the electron concentration decreases) as T increases atfixed P Hence, the warmer portion can have a lower electron concentration

where B = 5.21442 kg1/2 K–1 kmole1/6 s Determine the functional relations for

prop-Example 3 illustrates that the relation a = a (T,v) is a fundamental equation thatcontains all the relevant information to construct a table of properties for P,u, h, g, s, etc., (e.g.Tables A-4 for H2O, A-5 for R134a, etc.) One can plot the variation in h, g, and s with respect

to temperature as illustrated in Figure 3

Note that the attractive force constant a is different from “a” Helmholtz function.

From the third relation Eq (22) and Eq (A),

(∂s/∂v)T = (∂P/∂T)v= R/(v–b) + (1/2) a/(T3/2v(v+b)) (B)Integrating Eq (B),

s2(T,v2) –s1(T,v1) =

Rln((v–b)/(v–b)) +(1/2)(a/(T3/2b)) ln(v(v+b)/(v (v+b))) (C)

Figure 3: Illustration of the variation in some properties,

e.g., h, s and g, with temperature

Obtain an expression for the entropy change in an RK gas when the gas is mally compressed Determine the entropy change when superheated R–12 is isother-mally compressed at 60ºC from 0.0194 m3 kg–1 (state 1) to 0.0126 m3 kg–1 (state 2)

isother-From Table 1 for R–12, Tc = 385 K, and Pc = 41.2 bar Therefore a =208.59 bar (mkmole–1) 2 K1/2, and b = 0.06731 m3 kmole–1 The molecular weight M = 120.92 kgkmole–1, and

a = a/M2 = 208.59 bar (m3 kmole–1) 2K1/2÷120.92 2

(kg kmole–1)2 = 1.427 k Pa (m3 kg–1) 2 K1/2, and

b = b/M = 0.557×10–3 m3 kg–1

Since, R = 8.314 ÷ 120.92 = 0.06876 kJ kg–1 K–1,

s2 – s1 = 0.06876 ln[(0.0126 – 0.000557) ÷ (0.0194 – 0.000557)]

+ (1/2){172.5 ÷ (3331.5 0.000557)} ln [0.0126 (0.0194 + 0.000557) ÷ {0.0194 × (0.0126+ 0.000557)}]

= – 0.06876 × 0.448 – 0.211 × 0.01495 = –0.03396 kJ kg–1 K–1

4 Fourth Maxwell Relation

By subtracting d(Ts) from both sides of Eq (16) and using the relation g = h – Ts, weobtain the state equation

where g represents the Gibbs function (named after Josiah Willard Gibbs, 1839–1903) tion (24) is another form of the fundamental equation The intensive form g (= g(T,P)) is alsoknown as the chemical potential µ

Note the discontinuities regarding the entropy and Helmholtz function during thephase change, while g is continuous.

At constant temperature, dgT = vdP Therefore, the area under the v–P curve for anisotherm represents Gibbs function

The Planck function is represented by the relation

r = r(T,P) = –g/T = s – h/T, i.e.,

dr = – dg/T + g/T2 dT = (1/T2)(- vT dP + h dT)

The relations for du, dh, da and dg can be easily memorized by using the phrase

“Great Physicists Have Studied Under Very Articulate Teachers” (G, P, H, S, U, V,

A, T) by considering the mnemonic diagram (cf Figure 4) In that figure a square isconstructed by representing the four corners by the properties P, S, V, T, and by rep-resenting the property H by the space between the corners represented by P and S, theproperty U by the space between S and V, etc., as illustrated in Figure 4 Diagonalsare then drawn pointing away from the two bottom corners Such a diagram is alsoknown as a thermodynamic mnemonic diagram If an expression for dG is desired(that is located at the middle of the line connecting the points T and P), we first formthe differentials dT and dP, and then link the two with their conjugates as illustratedbelow

dG = – S(the conjugate of T with the minus sign due to the diagonal pointing towardsT) × dT + V (that is conjugate of P with the plus sign due to the diagonal pointingaway from P) × dP

dia-For a point function, say, P = P(T,v), it can be proven that

(∂P/∂T)v (∂T/∂v)P (∂v/∂P)T = –1, i.e., (∂v/∂T)P = – (∂P/∂T)v/(∂P/∂v) T (29)Equation (29) is useful to obtain the derivative (∂v/∂T) P if state equations are available for thepressure (e.g., in the form of the VW equation of state) Likewise, if u = u(s,v),

of sound in a two-phase mixture Assume ideal gas behavior for the vapor phase andShow that both the isothermal expansivity βP = (1/v)(∂v/∂T)P and the isobaric com-

Assuming ideal gas behavior for the vapor phase, and if duf = cdT, then

Similarly,

Considering constant entropy in Eqs (A) and (B), using Eqs (C) and (D), dividing by

dT, we obtain the relation

(dP/dT)s = (x cp,o + (1 – x) cf)/(v – (1 – x)vf), i.e., (E)

Dividing Eq (E) by Eq (F), we obtain the relation

–(dP/dv)s = (x cp,o + (1 – x)cf)(P/(v – (1 – x)vf))/(x cvo + (1 – x)cf) (G)Using the definition of the sound speed,

Using the values for P = 690 kPa, R = 0.08149 kJ kg–1 K–1, cvo = 0.7697 kJ kg–1 K–1,

vf = 0.000835 m3 kg–1, and γ = 1.1 For the conditions x = 1, T = 298 K,

c = 163.4 m s–1

A plot for c with respect to x is illustrated in Figure 5 The expression for the speed ofsound in a solid–vapor mixture is similar except that cf must be replaced by cS, spe-cific heat of solid

D GENERALIZED RELATIONS

We will now derive generalized thermodynamic relations that express the urable properties of any substance, such as s, u, and h, in terms of its measurable properties,e.g., P, T, and v We will first obtain generalized relations for the differential quantities (e.g.,

nonmeas-ds, du, and dh,) and, then, we will integrate these relations after applying the state equations

1 Entropy ds Relation

Any thermodynamic property for a simple compressible substance can be expressed

as a function of two other independent thermodynamic properties Therefore, for instance,

where cv = f(T,v) Using the third relation (cf Eq 22) the above expression can be expressed

in terms of measurable properties, namely,

Tv

Both Eqs (32) and (33) provide relations for ds in terms of the measurable properties P, v, and

T Equation (32) is suitable with a state equation of the form P = P(T,v) (e.g., the Van derWaals equation), while Eq (33) is more conveniently used when the state equation is ex-pressed in the form v = v(T,P)

a Remarks

Since (∂s/∂T)v = cv/T (and cv > 0 when T > 0), the gradient (∂s/∂T)v (taken along anisometric curve) has a finite and positive slope for all T > 0 Similarly, the gradient(∂s/∂T)P = cP/T has positive values (along isobars), since cP > 0 when T > 0

Along an isotherm, dsT = – (∂v/∂T)P dPT (cf Eq 33) Since in the vicinity of T = 0 K,(∂v/∂T)P = 0, this implies that s is independent of pressure at a temperature of abso-lute zero

Incompressible solids and liquids undergo no volumetric change, i.e., dv = 0 Since,

Eq (32) states that ds = cv dT/T, and cv = cp = c = constant for this case,

s = c ln T + constant

Consider Eq (33),

ds = cp dT/T – (∂v/∂T)P dP = cp dT/T – βP v dP, (34)where βP = (1/v)(∂v/∂T)P (and, likewise, βT = –(1/v)(∂v/∂P)T) (In Chapter 6 we havelearned that for liquids and solids the following relation relates the volume to the twocompressibilities, namely, v/vref = exp(βP (T–Tref) – βT (P–Pref)).)

For isentropic processes, ds = 0 and Eq (34) yields the relation

dT/T = (βp v dP)/cp, so that

(∂T/∂P)s = T (∂v/∂T)P/cP = T v βP/cP = T βP/cp´, (35)where cp´ = cp/v, kJ/K m3 If cP, βP and v are all approximately constant, then by inte-grating Eq (34), we can obtain isentropic relations for solids and liquids, i.e.,

If βP≈ 0, the temperature remains constant as the pressure changes In case βP > 0(i.e., the liquid or solid expands upon heating), the temperature increases during com-pression and decreases during expansion Likewise, if βP < 0 (e.g., for rubber, or wa-ter between 0 and 4ºC at 1bar), the temperature decreases with an increase in pres-sure

Alternately, using Eq (32),

The term βT is related to the stress–strain relation in solids From the above relation

we find that when dvs < 0 (i.e., during isentropic compression) dTs > 0 if βP > 0 (i.e.,the material expands upon heating) which implies that the temperature rises On the

other hand, when a material is isentropically stretched (dvs > 0), dTs < 0 (This issimilar to the phenomenon of first compressing a gas and then releasing it.).

We now examine the phenomenon of the bending of a beam that is illustrated inFigure 6 During the bending process, the upper layers of the beam are compressedand dTs > 0 The bottom layers are stretched where dTs < 0 Therefore, temperaturegradients develop within the beam The increase in the energy in the upper layers isstored in the form of a vibrational energy increase in the atoms (that results in thehigher temperature) and partly as intermolecular potential energy If the beamchanges periodically between the expanded and compressed states, the oscillationcauses the material to stretch, thereby increasing the intermolecular potential energy,but decreasing the thermal energy If the material has a negligible thermal conductiv-ity, there is no heat transfer between the various isothermal layers within the beam,and each layer will act as an adiabatic reversible system Thus, if the beam is placed

in a vacuum, the material will keep vibrating However, material with a finite thermalconductivity will behave differently, since there is heat transfer within the variouslayers This heat transfer is a result of the transfer of vibrational energy between thevarious layers in the beam, which results in a lower stretching for successive portions

of the cycle The amount of energy available for stretching is less resulting inthermo–elastic damping

Consider a liquid at its saturation temperature at a specified pressure As heat is plied to it isobarically, a portion of the liquid vaporizes, but the temperature is un-

sup-changed Since dT = dP = 0, Eq (33) seems to suggest that ds = 0, which is an

incor-rect interpretation In this case, cp (= (∂h/∂T)P) and ∂v/∂T both tend to infinity (due tothe fact that hfg and vfg are finite, while the temperature remains constant during theboiling process) However, we can use other relations, such as T ds + v dP = dh toobtain the relation ds = dh/T during isobaric vaporization which on integration yields

sfg = hfg/T

g Example 7

its original state once the load is removed) determine:

The change in the solid temperature

The internal energy change

The temperature after the load is removed

pressible substance

Solution

Since the process is adiabatic

and reversible we will use

Figure 6: The bending of a beam

Assume that To = 250 K, βP = 48×10–6 K–1, βT = 7.62×10–7 bar–1, v = 1.11×10–4 m3

kg–1, cp = 0.372 kJ kg–1 K–1, and cv = 0.364 kJ kg–1 K–1 Treat Cu as simple Weights are gradually placed on an insulated copper bar in order to compress it to

com-1000 bars If the compression is adiabatic and reversible (i.e., the material reverts to

Applying the First law to an adiabatic reversible process

dus = – Pdvs

Recall from Eq (37) that

dvs = –dTsβTcv/(T βP), i.e.,

dus = {P βTcv/(T βP)}dTs

Integrating and assuming that P is not a function of temperature and remaining at an

average value of 500 bar.

Using the criterion for an exact differential,

exp ((s2´)o/R – (s1´)o/R) = (exp (s2/R)/T2)/(exp(s1/R)/T1) = (v2 – b)/(v1 – b) (E)Upon defining vr = exp (so/R)/T, Eq (E) can be written in the form

Values of vr are usually tabulated Once the volume ratio v2/v1 is specified, T2 can bedetermined from Eq (F) Using the VW equation of state, we can then determine P2.Since, v1 =0.006 m3 kg–1 at T1 = 200 K, the VW equation yields

The tabulated values indicate that at vr2 = 263, T2 = 423 K

Finally, using the VW equation of state

k Example 11

measurable properties of a simple compressible substance

Show that cp/cv = k = βT/βs

Determine a relation for the sound speed for an ideal gas

Determine a relation for the sound speed for a VW gas

In the case of liquids and solids, a very large pressure is required to cause a smallchange in the volume so that (∂P/∂v)T→ ∞ and, consequently, c → ∞ Therefore,sound travels at faster speeds in liquids and solids.

Applying Eq (L) to the case of an ideal gas, co = k(T) RT, and dividing Eq.(I) by co

we obtain the expression,

(c2/co2) = – (k(TR, vR´)/(ko(TR) TR)) vR2 (∂PR/∂vR´)TR

2 Internal Energy (du) Relation

Combining the First and Second laws

Equation (38) is called the “Calorific Equation of State” and indicates that the internalenergy of a simple compressible substance (the solid, liquid, or gas phases) is a func-tion of temperature and volume (cf Figure 7) In it, the first term, cv dT, representsthe thermal portion of the energy (which varies due to changes in the translational, vi-brational, and rotational energies, with each degree of freedom contributing anamount equal to 2kBT per molecule, as discussed in Chapter 1)

Applying the First law to a closed system undergoing a reversible process, in thecontext of Eq (38)

δQrev = dU + PdV = mcvdT + (T∂P/∂T – P)dV + PdV (38’)Eq.(38’) states that energy transfer δQrev is used to raise the thermal energy by the in-cremental amount mcvdT (e.g., te, re, ve), to overcome the intermolecular potentialenergy barrier (T∂P/∂T – P)dV (i.e., ipe required to move the molecules against at-tractive forces) and to perform boundary PdV work For an ideal gas, the intermo-lecular potential energy equals zero so that heat transfer is used for PdV work if theprocess is isothermal

In an adiabatic system the internal energy change cvdT + (T∂P/∂T – P)dv equals thework performed on the system Note that if V and T are held fixed, δQrev =0 How-ever, this does not imply that Q = Q(V,T) since the functional form for the relationchanges, depending upon the process path between the initial and final states (i.e., Q

is not a property)

Consider Example 6 in Chapter 2 As we place a large weight on the piston, the gas isadiabatically compressed and we perform Pdv work The potential energy decrease ofthe weight is converted into internal energy increase of the gas The intermolecularpotential energy decreases during compression (T∂P/∂T – P)dv < 0, since the secondterm in Eq (38) decreases as the intermolecular spacing becomes closer Eq (38) im-plies that the term cvdT must increase If gas is ideal, then change in ipe =0; this im-plies that temperature changes to a greater extent in a real gas than in an ideal gas).For incompressible substances, dv = 0 and, consequently from Eq (38), du = cv dT.Attractive forces are very strong in such substances so that the intermolecular poten-tial energy remains virtually constant

In the case of solids and liquids it is useful to express Eq (38) in terms of the isobaricexpansivity and isothermal compressibility Since (∂P/∂v)T(∂v/∂T)P(∂T/∂P)v = –1, and(∂P/∂T)v = βP/βT, Eq (38) can be written in the form

→∞, i.e., βT→ ∞ Similarly, (∂v/∂T)P→ ∞ and hence βp → ∞ for a two-phase ture Thus as we approach the critical point, βT → ∞ and βP→ ∞

mix-Since the relation for du in Eq (38) is an exact differential, applying the appropriatecriterion,

Figure 7: Illustration of the variation of the internal

en-ergy with respect to temperature

Applying the state relations for liquids and solids (cf Chapter 6) it can be shown thatT(∂2P/∂T2) v≈ 0, i.e.,

cv = cv(T) alone Integrating Eq (39) between the limits v → ∞ (i.e., as P → 0) and v

= v, we obtain an expression for the deviation of the constant volume specific heatfrom its ideal gas value, i.e.,

The intermolecular potential energy of the solid

The state of a copper bar is initially at a pressure of 1 bar and temperature of 250 K It

is compressed so that the exerted pressure is 1000 bar Assume that the compression

is adiabatic and reversible (i.e., the material reverts to its original state once the load

is removed), and that βP = 48×10–6 K–1, βT = 7.62×10–7 bar–1, v = 1.11×10–4 m3 kg–1,

cp = 0.372 kJ kg–1 K–1, and cv = 0.364 kJ kg–1 K–1 Determine the change in the

inter-∆(ipe) = (TβP/βT – P) dvwhere the second term is same as on the RHS of Eq (C) Therefore

∆(ipe) = (250 K × 48×10–6

K–1 ÷ 7.62x10–7

bar–1) × (-8.45x10-8 m3kg-1)×100 kPa bar-1 + 0.00423, or

The net change in the internal energy of the solid is

du = 0.1329 - 0.1288 = 0.0041 kJ kg–1.Which is approximately the same as the answer in Eq ( C) In this example, the tem-perature increases by 0.365 K, increasing the thermal portion of the internal energy by0.1329 kJ, but the IPE decreases by 0.1288 kJ (Eq.(F)) At the minimum intermo-lecular potential energy (Chapter 1) , compression should cause the "ipe" to increase.Since the ipe decreases with compression, this indicates that the solid is not at thatminimum value

Eq (E), (dT/dv) < 0 Thus, upon compression (dv < 0), and dT > 0 Those

sub-A substance undergoes an adiabatic and reversible process Obtain an expression for(∂T/∂v)s in terms of cv, βP, βT and T What is the value of (∂T/∂v)s for copper, giventhat βP = 5×10–5

stances which contract upon heating (with a negative coefficient of thermal sion, e.g., rubber, water below 4ºC), βP < 0 Eq (E) yields, dT > 0, upon expansion,since dv > 0.

expan-If cv, v and βP are constants, then Eq (E) can be integrated to yield

ln (T2/T1) = (v2 – v1) βP/(βT cv) (F)which is the same as Eq.(B) in Example 12

3 Enthalpy (dh) Relation

Obtaining the total differential of the enthalpy h = h (T, P)

dh = (∂h/∂T)P dT + (∂h/∂P)T dP = cP dT + (∂h/∂P)T dP (40)Since dh = Tds + v dP, dividing the expression by dP, the following relation applies at constantT

Obtain an expression for dh and du for a liquid in terms of cp, βP, βT, cv, dT and dP

For incompressible liquids βP = βT = 0, and Eqs (A) can be expressed in the form

Further,

For an incompressible fluid Eq (C) assumes the form

Upon comparing Eqs (E) and (F),

Note that du/dT = (∂u/∂T)v, since v is constant Therefore,

(∂h/∂T)p = du/dT = cp = cv = c only for incompressible liquids (H)

4 Relation for (c p –c v )

We will develop relations for (cp–cv) in terms of measurable properties and show that

cp > cv, which implies that k > 1 Differentiating the relation for enthalpy h = u + Pv,

(∂h/∂T)P = cp = (∂u/∂T)P +P(∂v/∂T)p

The first term on the RHS (∂u/∂T)P can be simplified by dividing Eq (38) by dT in order toobtain , (du/dT)P = cv + (T(∂P/∂T)v – P) (dv/dT)P ; then the above equation becomes,

cp = cv + (T∂P/∂T – P)(∂v/∂T)p + P(∂v/∂T)p, or (cp – cv) = (T(∂P/∂T)(∂v/∂T)p).Using the cyclic relation (∂P/∂T)(∂T/∂v)(∂v/∂P) = –1, (∂v/∂T)P = –(∂P/∂T)/(∂P/∂v)), we obatinthe expression

We can rewrite Eq (45b) in the form (cp – cv) = Acp T, where the relation A =

vβP /(βTcP) is called the Nernst–Lindmann equation For liquids and solids, v, βP, βT,and cP are approximately constant and, consequently, A is a constant

For incompressible liquids cp = cv However for water at a temperature of 80ºC, cp =4.19 kJ kg–1 K–1 and cv = 3.86 kJ kg–1 K–1

The Gruneisen constant γg = ∂ ∂( P/ T v c)v / v Its value is constant over a wide perature range for many solid metals Typical values of γg lie between 1 and 3 Usingthe RHS of the Eq (45a), the difference in specific heats equals (–T (a finite value) ÷0) which tends to infinity at the critical point

tem-If (∂P/∂v) >0 (recall from Chapter 6 that P decreases with v for real gases at peratures lower than Tc for a certain volumetric range), then (cp – cv) < 0, i.e., cp < cv.You will learn in Chapter 10 that these states are unstable.

tem-Using the relation βs = –1/v(∂v/∂P)s, we note that βT/βs = (∂v/∂P)T/(∂v/∂P)s

Since the properties of ideal gases are generally known, we will evaluate the constant f(T) interms of ideal gas properties For an ideal gas b → 0 (since the molecules are point masses)

and a → 0 (as there are no intermolecular attraction forces) Therefore,

The latter procedure is appropriate for Martin–Hou and Kesler type state relations that do not

contain specific terms (such as a) related to the intermolecular attraction forces and body

vol-ume b For RK fluid

a(T,v) – ao(T,v) = RT ln {v/(v–b)} + {a/(bT1/2)} ln {v/(v+b)}, where (47)

ao(T,v) = uo(T) – T so(T,v)

Note that as v → ∞, the RHS of Eq (47) approaches 0 indicating that a → ao We can describe

a residual Helmholtz function aRes, which is a correction to ideal gas behavior, as

Thereafter, dividing Eq (47) by RTc, and using the equalities a = 0.4275 R2Tc2.5/Pc, b =0.08664 RTc/Pc, v = vR´vc´ (where vc´ = RTc/Pc), that equation can be expressed in dimension-less form, i.e.,

(a(T,vR´) – ao(T,vR´))/(RTc) =

–TR ln(1 – (0.08664/vR´)) – (4.934/TR2) ln (1 + (0.08664/vR)), or (49)(ao(T,vR´) – a(T,vR´))/(RTc) = TR ln (vR´/(vR´ –0.08664))

– (4.934/TR2) ln ((vR´+ 0.08664)/vR´)

p Example 16

0.0136 m3 kg–1 Determine:

The residue {a(T,v) – ao(T,v)} and the pressure using the RK state equation

The pressure using ideal gas state equation

A mass of water exists at a temperature of 600ºC Its specific volume is given as

The ideal gas volume vo at 600ºC at which the pressure equals that predicted by the

Likewise, the pressure, using the ideal gas state equation,

From Example (16) it is apparent that (a(T,v) – ao(T,v)) ≠ (a(T,P) – ao(T,P)), sincealthough the temperature and volume are unchanged, the ideal gas pressure differsfrom the real gas pressure If the specific volume of the ideal gas state is changed so

as to obtain the same pressure P as that in the real gas state, then

(a(T,v)–ao (T,vo) = a(T,P)–ao(T,P)) = (a(T,v)–ao(T,v)) – (ao(T,vo)–a(T,v)) (50)The first term in parentheses on the RHS can be evaluated using Eq (50), while thesecond term in parentheses can be evaluated as outlined below

ao(T,vo)–ao(T,v) = (uo(T)–Tso(T,vo)) – (uo(T)–Tso(T,v)) = T(so(T,v) – so(T,vo)).Recall that dso = cvo dT/T + Rdv/v, at a specified temperature

aRes (T,v) = a(T,v) – a(T,v) = – 1236.6 kJ kmole–1

A mass of water is maintained at a temperature of 600ºC and a pressure of 250 bar