State Equations for Liquids/Solids

12. State Equations for Liquids/Solids

a. Generalized State Equation The volume v = v (P,T), and dv = ( ∂v/∂P) T dP + ( ∂v/∂T) P dT, i.e.,

dv = ( ∂v/∂P) T dP + ( ∂v/∂T) P dT. (85a) We define β P = (1/v)( ∂v/∂T) P ,

(85b) β T = –(1/v) ( ∂v/∂P) T ,

(85c) κ T = 1/( β T P) = (–v/P) ( ∂P/∂T) T

(85d) where β P , β T and κ T are, respectively, the isobaric expansivity, isothermal compressibility, and

isothermal exponent. The isobaric expansivity is a measure of the volumetric change with re- spect to temperature at a specified pressure. We will show in Chapter 10 that β T >0 for stable fluids. Upon substituting these parameters in Eq. (86a),

dv = v β P dT – v β T dP, or d(ln v) = β P dT – β T dP. If β P and β T are constant, the general state equation for liquids and solids can be written as dv = v β P dT – v β T dP, or d(ln v) = β P dT – β T dP. If β P and β T are constant, the general state equation for liquids and solids can be written as

pressure, the relation P=P ref +( β P / β T )(T – T ref ) – ln (v/v ref )/( β T v ref ),

(87) is an explicit, although approximate, state equation for liquids and solids. Both Eqs. (87) or

(88) can be approximated as (v–v ref )/v ref = β P (T – T ref )– β T (P – P ref ).

(88) Solving the relation in terms of pressure P=P ref +( β P / β T ) (T – T ref ) – (v–v ref )/( β T v ref ),

(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

ln(v/v ref ) ≈β P (T – T ref ). (90) In case β P (T – T ref ) « 1, then v/v ref = (1 + β P (T – T ref )).

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

Copper has the following properties at 50ºC: v, β P , and β T are, respectively, 7.002 ×10 –3 m 3 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) as

7.002 ×10 –3 × 10 × 11.5×10 –6 = 805 cm 3 . 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 mass

equals 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 mm 3 . Therefore, changing the state of the 10 kmole copper mass from 50ºC and 1 bar to 51ºC and 2 bars, will result in a volumetric change that equals 805 – 0.05 =

804.95 mm 3 . For solids β is related to the linear expansion coefficient α. The total volume V ∝ L 3 P , and

(92) where α = (1/L) (∂L/∂T) P .

β = 1/V( ∂V/∂T) = 1/L 3 P P ∂(L 3 )/ ∂T = (3/L) ∂L/∂T = 3α,

g. Example 7 Water is compressed isentropically from 0.1 bar and 30ºC to 60 bar. Determine the change in volume, and work required to compress the fluid. Treat water as a com- pressible substance, and assume that at 30ºC, β P = 2.7 ×10 –4 K –1 , β T = 44.8 ×10 –6 bar –1 ,

v = 0.00101 m 3 kg –1 , and c = 4.178 kJ kg –1 p K –1 .

Solution Since β = 44.8

×10 bar –1 and dv = – β P dP v, ln v 2 /v = – β T (P 2 –P 1 ) = – 0.00268, i.e., v 2 /v = 0.997.

Now, v 2 = 0.997 ×0.00101 = 0.001007 m 3 kg –1 , so that v – v = 0.001007 – 0.001010 = 0.000997 m 3 2 –1 kg .

δw = –vdP (for a reversible process in an open system). ∴ δw = – v (dP/dv) dv = (1/β T ) dv. Integrating this expression,

w = (1/ β T )(v 2 –v . ) = 100 kPa bar –1 ×(0.001007–0.00101)÷44.8×10 –6 = –6.76 kJ kg –1 .

h. Example 8

A defective radiator does not have a pressure relief valve and there is no drainage provision for the reservoir. The radiator water temperature increases from 25ºC to 90ºC. Assuming that the radiator is rigid, what is the final water pressure in the ra- diator?

Solution Since d ln v = β P dT – β T dP and the volume is constant,

dP/dT = β / β = 2.7 ×10 –4 K –1 /44.8 ×10 –6 bar –1

PT

= 6.03 bar K .

Assuming that β T and β P are constants, ∆P = 6.03×65 = 391 bar.

b. Murnaghan Equation of State If we assume that the isothermal bulk modulus B T (= 1/ β T ) is a linear function of the

pressure, then

B T (T,P) = (1/ β T ) = –v( ∂P/∂v) T =B T (T,0) + αP (93) where α = (∂B T / ∂P) T . Therefore, ∂P β T (1,0)/(1 + αP β T (1,0)) = –dv/v.

(94) Integrating, and using the boundary condition that as P → 0, v → v 0 , we obtain the following

relation v/v

αPβ T (1,0)) 0 (1/α) = 1/(1 + ( , i.e., (95)

P(T,v) = ((v 0 /v) α – 1) (1/ αβ T (1,0)).

c. Racket Equation for Saturated Liquids The specific volume of saturated liquid follows the relation given by the Racket equation, namely,

(– 1 T R . 0 2857 v )

f = vZ c c .

d. Relation for Densities of Saturated Liquids and Vapors. If ρ f denotes the saturated liquid density, and ρ g the saturated vapor density, then

ρ Rf = ρ

f / ρ c = 1 + (3/4)(1 – T R ) + (7/4)(1– T R ) , and (98) ρ = ρ g / ρ c R ) – (7/4)(1– T R ) Rg 1/3 = 1 + (3/4)(1 – T .

(99) These relations are based on curve fits to experimental data for Ne, Ar, Xe, O 2 , CO, and CH 4 .

It is also seen that ρ Rf – ρ Rg = (7/2)(1 – T R ) 1/3 .

At low pressures,

ρ Rf ≈ (7/2)(1 – T R ) 1/3 since ρ Rf >> ρ Rg

In thermodynamics, ρ Rf – ρ Rg is called order of parameter. If ρ Rg is known at low pressures (e.g., ideal gas law), then ρ Rf can be readily determined. Another empirical equation follows the relation

ρ R,f = 1 + 0.85(1–T R ) + (1.6916 + 0.9846 ψ)(1–T R ) 1/3 (101) where ψ ≈ ω.

e. Lyderson Charts (For Liquids) Lyderson charts can be developed based on the following relation, i.e.,

(102) The appendix contains charts for ρ R vs. P R with T R as a parameter. In case the density is known

ρ R = ρ/ρ c =v c /v.

at specified conditions, the relation can be used to determine P c ,T c and ρ c . Alternatively, if the density is not known at reference conditions, the following relation, namely,

ρ/ρ ref =v ref /v = ρ R / ρ R,ref (103) can be used.

f. Incompressible Approximation Recall that liquid molecules experience stronger attractive forces compared to gases due to the smaller intermolecular spacing. The molecules are at conditions close to the lowest potential energy where the maximum attractive forces occur. Therefore, any compression of liquids 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.