WATER AND STEAM

103

3 Physical Properties

Physical property data of chemical compounds are parameters frequently required for simulating chemical processes and in designing chemical plants. However, such data are not readily available outside of certain proprietary simulation design packages or large commercial databases. Physical property data can be obtained by conducting experiments to measure the properties of individual substances or of mixtures. However, because of the multitude of chemical compounds that are of interest to the chemical engineer and the vast number of thermodynamic states in which they appear in combination, it is impossible to obtain all physical property data by experiments alone. Therefore, it is necessary to establish generalizations for estimating physical properties of mixtures using only limited available experimental information.

The estimation of thermodynamic properties of vapors and liquids plays a very important role in chemical engineering computations. Numerous predictions and correlations of thermodynamic physical property data have been presented—and students, scientists, and practicing engineers alike have utilized these correlations and interpolations to obtain data of sufficient accuracy for their individual purposes.

In this chapter, physical property data and correlations for typical liquids and gases are intro- duced and reviewed. The correlations described in this chapter are limited to those that the author has deemed to have the greatest value in common practical application. The main objectives of this chapter are to provide readers with various estimation procedures and correlations for a limited set of properties of typical chemical compounds and to present MATLAB^® programs that analyze prop- erties for a range of variables and correlation constants.

of T_s(P). Region 5 represents superheated steam between 1073.15 and 2273.15 K and is covered by a fundamental equation for the specific Gibbs free energy g(P, T). The maximum pressure of regions 1, 2, and 3 is 100 MPa, and that of region 5 is 50 MPa.

3.1.2 ProPerTy equaTions

3.1.2.1 Parameters and Auxiliary Equations

The values of parameters and reference constants used in the IAPWS-IF97 are Specific gas constant: R = 0.461526 kJ/(kg · K)

Critical temperature: T_c = 647.096 K Critical pressure: P_c = 22.064 MPa Critical density: ρc = 322 kg/m³

The specific internal energy and the specific entropy of the saturated liquid at the triple point are set equal to zero. This condition is met at the temperature and pressure of the triple point T_t = 273.16 K, P_t = 611.657 Pa and at the enthalpy of the triple point H_t = 0.611783 J/kg.

The boundary between regions 2 and 3 as shown in Figure 3.1 is defined by the following qua- dratic pressure–temperature relation:

p= +n1 n2q+n3q2

where π = P/P*, θ = T/T*, P* = 1 MPa, T* = 1 K, and n₁, n₂, and n₃ are constants. This equation roughly describes an isentropic line. The IAPWS-IF97 involves many constants, and MATLAB programs presented in this book contain these constants.

3.1.2.2 Basic Equations for Region 1

The basic equation for region 1 is a fundamental equation for the specific Gibbs free energy g given by dimensionless form

g P T

RT n

i i

I_i J_i

, ,

( )

₌

( )

⁼

(

^-

) (

^-

)

å

=

g p t p t

1 34

7 1. 1 222.

P (MPa)

g(P,T) f(p,T) g(P,T)

T(P, h) T(P, h)

T(P, s) T(P, s)

g(P,T)

T (K) T_s(P)

P_s(T)

273.15 623.15 1073.15 2273.15

100 50

1 3 2

4

5 100

FIGURE 3.1 P–T diagram divided into five regions with property equations.

where π = P/P*, τ = T*/T, P* = 16.53 MPa, and ni and Ji are constants. Other thermodynamic prop- erties can be obtained from the above equation and its derivatives. For example, derivatives with respect to π and τ are used:

¶

¶ = -

(

-

) (

^-

)

^¶_¶ ^{= -}

=

-

å å

=

g

p p t g

t i tp

i i

I J

i

n I ^{i i} ni

1 34

1

1 34

7 1. 1 222. ,

(

7 1. pp

)

^IiJi

(

t^{-1 222.}

)

^Ji-1

3.1.2.3 Basic Equations for Region 2

The basic equation for region 2 is a fundamental equation for the specific Gibbs free energy g. This equation can be expressed as a combination of an ideal gas part γ^o and a residual part γ^T:

g P T RT

o T

, , , ,

( )

₌^{g p t}

( )

^{=g p t}

( )

^{+g p t}

( )

g p t g p t g

p g

p t

o

i

io J T

i

i I J o o

i

n ^io n ^{i i}

= + =

(

-

)

^{= =}

= = =

å å

ln , . , ,

1 9

1 43

1

0 5 1

å

^{99 n Jio io Jtio-}1

where π = P/P*, τ = T*/T, P* = 1 MPa, T* = 540 K, and n_i, I_i, and J_i are constants. Derivatives of the ideal gas and residual parts with respect to π and τ are used in the calculation of physical properties:

¶

¶ =

(

-

)

^¶_¶ ⁼

(

^-

= -

å å

=

g

p p t g

t p t

t p

T

i

i i I J T

i i I

n I ^{i i} n ⁱJi

1 43

1

1 43

0 5. , 0 5.

))

^Ji-1

3.1.2.4 Basic Equations for Region 3

The basic equation for region 3 is a fundamental equation for the specific Helmholtz free energy f.

This equation can be expressed in dimensionless form as f T

RT n n

i

i Ii Ji

r, d t s d t

( )

₌

( )

, ^{= +}

å

=

F 1

2 40

ln

where δ = ρ/ρ*, τ = T*/T, ρ* = ρc, and T* = Tc. 3.1.2.5 Basic Equations for Region 4

The saturation line can be expressed as a dimensionless quadratic equation, which can be solved in terms of saturation pressure P_s and saturation temperature T_s:

b J^{2 2} 1b J2 b bJ bJ b J J

2 2

3 2

4 5 6 2

7 8 0

+n +n +n +n +n +n +n +n =

where β = (Ps/P)^1/4, ϑ = (Ts/T) + (n9/((Ts/T*) − n10)), P* = 1 MPa, T* = 1 K. The solution of the above equation for the saturation pressure yields

P P

C

B B AC

s

* =

- + -

é

ëê ù

ûú 2

2 4

4

where

A=J²+n1J+n2 B n= 3J2+nJ+n C=nJ +nJ+n

4 5 6 2

7 8

, ,

Similarly, the solution of the quadratic equation for the saturation temperature yields T

T

n D n D n n D

s

*= ¹⁰+ -

(

¹⁰+

)

^2-4

(

^{9+ 10}

)

2 where

D G

F F EG E n n F n n n G n n n

=- - 2 - = + + = + + = + +

2 4

2 3 6 1 2

4 7 2 2

5 8

, b b , b b , b b

3.1.2.6 Basic Equations for Region 5

The basic equation for high-temperature region 5 is a fundamental equation for the specific Gibbs free energy:

g P T RT

o T

, , , ,

( )

₌^{g p t}

( )

^{=g p t}

( )

^{+g p t}

( )

g^o p t g p t

i

io J T

i

i I J

n ^io n ^{i i}

= + =

= =

å å

ln ,

1 6

1 6

where π = P/P*, τ = T*/T, P* = 1 MPa, T* = 1000 K, and n_i, I_i, and J_i are constants.

The MATLAB function H2OPT is used to compute properties of water and steam. This function calculates thermodynamic properties of H₂O at the given temperature T (°C) and pressure P (kPa) for regions 1, 2, and 4. This function can be called as

wprop = H2OPT(P,T)

The returned parameter (wprop) is the name of the structure that stores the calculation results. wprop contains several fields: P is pressure (kPa), T is temperature (°C), v is specific volume (m³/kg), u is internal energy (kJ/kg), h is enthalpy (kJ/kg), and s is entropy (kJ/(kg · K)).

function [wprop] = H2OPT(P,T)

% Compute thermodynamic properties of H2O at the given temperature(C)

% and pressure(kPa) using the IAPWS formula(revised release on the

% IAPWS-97)(region 1, 2, 4)

% inputs

% P: kPa, T: C

% output:

% wprop: structure containing following fields:

% P(pressure, kPa), T(temperature, C), v(specific volume, m^3/kg),

% u(internal energy, kJ/kg), h(enthalpy, kJ/kg), s(entropy, kJ/kg/K)

% Initialization crit = 1e-4;

Pg = 0; Tg = 0; vg = 0; ug = 0; sg = 0; hg = 0; % vapor phase Pf = 0; Tf = 0; vf = 0; uf = 0; sf = 0; hf = 0; % liquid phase stateH2O = ''; % state alarm message

warnmsg = ''; % alarm message

% Conversion of temperature and pressure units

T = T+273.15; % conversion of temperature unit: C -> K

P = P/1000; % conversion of pressure unit: kPa -> MPa

% identification of region(rg) if T <= 623.15 % region 4 Ts = rg4eqn(P);

if (Ts-T) > crit

rg = 1; % subcooled liquid elseif (T-Ts) > crit

rg = 2; % superheated steam else

rg = 3; % saturation state end

end

% auxiliary equation at the boundary between region 2 and 3 if T > 623.15 % boundary between region 2 and 3

rg = 2; % saturated steam or outside of the region

n1 = 348.05185628969; n2 = -1.1671859879975; n3 = 0.0010192970039326;

Ps23 = n1 + n2*T + n3*T^2; % auxiliary equation at the boundary if P > Ps23

rg = 5;

end end

if P < 1e-4 || T < 273.15 rg = 4;

end

% region 1: subcooled liquid

if rg == 1 % T < Ts: subcooled liquid stateH2O = 'subcooled liquid';

if T > 623.15 || P >100

warnmsg = 'Region not defined: P > 100 MPa'; rg = 6;

else

[Pf,Tf,vf,uf,sf,hf] = propPT(P,T,1); % region 1 end

end

% region 2: superheated steam

if rg == 2 % T > Ts: superheated steam stateH2O = 'superheated steam';

if T > 1073.15 || P > 100

warnmsg = 'Region not defined: T > 800 C or P > 100 000 kPa';

rg = 6;

else

[Pg,Tg,vg,ug,sg,hg] = propPT(P,T,2); % region 2 end

end

% region 3: saturated vapor + saturated liquid if rg == 3 % saturated state

stateH2O = 'saturated state';

wa rnmsg = '2 independent variables are required to determine the state';

[Pg,Tg,vg,ug,sg,hg] = propPT(P,Ts,2); % vapor is in region 2 [P f,Tf,vf,uf,sf,hf] = propPT(P,Ts,1);

% subcooled liquid is in region 1 end

% region 4 if rg == 4

stateH2O = 'Incorrect data'; warnmsg = 'P or T < 0';

end

% region 5

if rg == 5 % outside region 3

stateH2O = 'Calculation in region 3 by IAPWS-97 is not available';

warnmsg = 'Check whether the pressure is greater than 100 MPa';

end

% restore units of temperature and pressure(K->C, MPa->kPa) Tg = Tg - 273.15; Tf = Tf – 273.15; Pg = Pg*1000; Pf = Pf*1000;

% Assign values to corresponding fields of the structure to be returned.

if rg == 1 % region 1

wprop.Pf = Pf; wprop.Tf = Tf; wprop.vf = vf;

wprop.uf = uf; wprop.hf = hf; wprop.sf = sf;

elseif rg == 2 % region 2

wprop.Pg = Pg; wprop.Tg = Tg; wprop.vg = vg;

wprop.ug = ug; wprop.hg = hg; wprop.sg = sg;

elseif rg == 3 % region 3

wprop.Pf = Pf; wprop.Tf = Tf; wprop.vf = vf;

wprop.uf = uf; wprop.hf = hf; wprop.sf = sf;

wprop.Pg = Pg; wprop.Tg = Tg; wprop.vg = vg;

wprop.ug = ug; wprop.hg = hg; wprop.sg = sg;

end

wprop.stateH2O = stateH2O;

wprop.warnmsg = warnmsg;

% Print results.

fprintf('Temperature: %g C, Pressure: %g kPa\n', T-273.15, P*1000);

if rg == 1

fp rintf('liquid pressure = %g kPa\n', Pf); fprintf('liquid temp.

= %g C\n', Tf);

fprintf('volume = %g m^3/kg\n', vf);

fprintf('internal energy = %g kJ/kg\n', uf);

fp rintf('enthalpy = %g kJ/kg\n', hf); fprintf('entropy = %g kJ/

kg*K\n', sf);

fprintf('state = %s\n', stateH2O);

elseif rg == 2

fp rintf('vapor pressure = %g kPa\n', Pg); fprintf('vapor temp.

= %g C\n', Tg);

fprintf('volume = %g m^3/kg\n', vg);

fprintf('internal energy = %g kJ/kg\n',ug);

fp rintf('enthalpy = %g kJ/kg\n', hg); fprintf('entropy = %g kJ/

kg*K\n', sg);

fprintf('state = %s\n', stateH2O);

elseif rg == 3

fp rintf('liquid pressure = %g kPa\n', Pf); fprintf('liquid temp.

= %g C\n', Tf);

fp rintf('vapor pressure = %g kPa\n', Pg); fprintf('vapor temp.

= %g C\n', Tg);

fprintf('liquid volume = %g m^3/kg\n', vf);

fprintf('liquid internal energy = %g kJ/kg\n', uf);

fprintf('liquid enthalpy = %g kJ/kg\n', hf);

fprintf('liquid entropy = %g kJ/kg*K\n', sf);

fprintf('vapor volume = %g m^3/kg\n', vg);

fprintf('vapor internal energy = %g kJ/kg\n', ug);

fprintf('vapor enthalpy = %g kJ/kg\n', hg);

fprintf('vapor entropy = %g kJ/kg*K\n', sg);

fprintf('state = %s\n', stateH2O);

end

Example 3.1: Calculation of H2O Properties

Calculate the specific volume, the internal energy, the enthalpy, and the entropy of H₂O at the specified temperatures and pressures given below. Compare the results with those recorded in steam tables that can be found in standard thermodynamic texts.

1. 750 kPa, 167.76°C 2. 2100 kPa, 214.85°C Solution

Results of calculations by the function H2OPT are shown below. These results match the values in steam tables:

>> wp = H2OPT(750,167.76);

Temperature: 167.76 C, Pressure: 750 kPa vapor pressure = 750 kPa

vapor temp. = 167.76 C volume = 0.255506 m^3/kg internal energy = 2574.02 kJ/kg enthalpy = 2765.65 kJ/kg entropy = 6.68356 kJ/kg*K

>> wp = H2OPT(2100,214.85);

Temperature: 214.85 C, Pressure: 2100 kPa liquid pressure = 2100 kPa

liquid temp. = 214.85 C volume = 0.001181 m^3/kg internal energy = 917.44 kJ/kg enthalpy = 919.92 kJ/kg entropy = 2.46999 kJ/kg*K

The function H2OPT calls another function propPT, which in turn calls other functions. Each function performs a specific task as shown in Table 3.1. The functions shown in Table 3.1 can be found in Appendix A.

TABLE 3.1

The MATLAB^® Functions Used in the Calculations of H₂O Properties at Given Temperature and Pressure

Name Description

H2OPT The main function: assigns the results to the structure variable propPT Calculates properties at regions 1, 2, and 4

rg1IJn Defines 34 coefficients of the basic equation for region 1

rg2IJn Defines coefficients (γ^o: 9, γ^T: 43) of the basic equation for region 2 rg1eqn Calculates the basic equation for region 1

rg2eqn Calculates the basic equation for region 2 rg4eqn Calculates the basic equation for region 4 rg1dgdp Calculates (∂γ/∂π)|_τ

rg1dgdT Calculates (∂γ/∂τ)|_π

3.1.3 ProPerTiesof saTuraTeD sTeam

In this section, we will focus on the thermodynamic properties of saturated steam valid from 273.15 to 647.15 K. The following description is based on the revised IAPWS Formulation 1995 for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use.²

At a given temperature T (K), the saturated vapor pressure can be approximated as

p p T

T a a a a a a

c c

= ^expé_ëê

(

^1t+ ^{2t1 5.} + ^3t3+ ^{4t3 5.} + ^5t4+ ^{6t7 5.}

)

^ù_ûú

where τ = 1 − (T/T_c), p_c and T_c are critical pressure and critical temperature, respectively, and a_i(i = 1, 2, …, 6) are constants. Differentiation of the vapor pressure with respect to the saturation temperature gives

dp dT

p

T a a a a a a

= -æ èç ö

ø÷ 7 5 6 6 5+4 +3 5 +3 +1 5 +

5 3

4 2 5 3 2

2 0 5

. t^. t . t^. t . t^. 1++ æ

èç ö ø÷ é

ëê ù

ûú ln p

pc

Alternatively, the saturation temperature can be calculated at a given pressure by converting the pressure equation into a nonlinear equation. By taking logarithm on the pressure equation, we have

ln p ^{. . .}

p T

T a a a a a a

a

c

æ c

èç ö

ø÷ =

(

+ + + +

)

= -

1 2 1 5

3 3 4 3 5

5 4 6 7 5

1

1 1

t t t t t t

t

(

t ++^{a2t1 5.} +^a3t3+^{a4t3 5.a5t4}+^{a6t7 5.}

)

Rearrangement of the above equation gives

f a a a a a a p

pc

t t t t t t t t

( )

^{= + + + + - -}

( )

^æ

èç ö ø÷ =

1 2 1 5

3 3 4 3 5

5 4

6 7 5 1 0

. . . ln

At a given temperature T (K), the densities of the saturated liquid and steam can be represented respectively as

rL=rcéë1+b1t1 3+bt +bt +bt +bt +bt ùû

2 2 3 3 5 3

4 16 3

5 43 3

6 110 3

/ / / / / /

rG=rc^exp

(

c¹t^{1 3/} +c²t^{2 3/} +c³t^{4 3/} +c⁴t³+c⁵t^{37 6/} +c⁶t^{71 6/}

)

where ci(i = 1, 2, …, 6) are constants.

The specific enthalpy of the saturated liquid and steam at a given temperature T (K) can be obtained, respectively, from the following relations:

h T dp

dT h T dp

L dT

L

G

G

= +a = +

r a

, r

a a= ⁰éëd_a+d¹q^-19+d²q+d³q^{4 5.} +d⁴q⁵+d⁵q^{54 5.} ùû where di(i = 1, 2, …, 5) are constants.

The specific entropy of the saturated liquid and steam at a given temperature T (K) can be obtained, respectively, from the following equations:

s dp

dT s dp

L dT

L

G

G

= +f = +

r f

r

1 , 1

f f= é f+ q + q+ q + q + q

ëê

ù û

0 1 -20

2 3 3 5

4 4

5 53 5

19 20

9 7

5 4

109

d d d ln d ^. d 107d ^. úú

The MATLAB function satsteam calculates the properties of the saturated steam. This function calculates the properties of the saturated steam at the given temperature Ts (°C), and the results pro- duced by this function are stored in the structure stpr. The calling format of this function is

stpr = satsteam(Ts)

stpr contains several fields: stpr.T = temperature of the saturated steam (K); stpr.P = pressure (Pa); stpr.

vL = liquid volume (m³/kg); stpr.vV = steam volume (m³/kg); stpr.hL = liquid enthalpy (J/kg); stpr.

hV = steam enthalpy (J/kg); stpr.sL = liquid entropy (J/(kg · K)); stpr.sV = steam entropy (J/(kg · K)).

function stpr = satsteam(Ts)

% Computes properties of the saturated steam at a given temperature Ts(C)

% stpr: a structure containing results of calculations

% Critical properties

Tc = 647.096; % critical temperature(K) Pc = 22064000; % critical pressure(Pa) rhoc = 322; % critical density(kg/m^3)

% Definition of parameters and constants Ts = Ts + 273.15;

alpha0 = 1000; % alpha_0(J/kg) phi0 = 1000/647.096;

a = [-7.85951783 1.84408259 -11.7866497 22.6807411...

-15.9618719 1.80122502];

b = [1.99274064 1.09965342 -0.510839303 -1.75493479...

-45.5170352 -674694.45];

c = [-2.03150240 -2.68302940 -5.38626492 -17.2991605...

-44.7586581 -63.9201063];

d = [-5.65134998e-8 2690.66631 127.287297 -135.003439 0.981825814];

alphad = -1135.905627715;

phid = 2319.5246;

theta = Ts/Tc;

tau = 1 - theta;

% saturated steam pressure

tw = (Tc/Ts)*(a(1)*tau + a(2)*tau^1.5 + a(3)*tau^3+...

a(4)*tau^3.5 + a(5)*tau^4 + a(6)*tau^7.5);

Ps = Pc*exp(tw);

% density of saturated liquid

rhoL =rhoc*(1 + b(1)*tau^(1/3) + b(2)*tau^(2/3) + b(3)*tau^(5/3) +...

b(4)*tau^(16/3) + b(5)*tau^(43/3) + b(6)*tau^(110/3));

% density of saturated steam

vw = c(1)*tau^(1/3) + c(2)*tau^(2/3) + c(3)*tau^(4/3) +...

c(4)*tau^3 + c(5)*tau^(37/6) + c(6)*tau^(71/6);

rhoV = rhoc*exp(vw);

% specific volume

vL = 1/rhoL; % saturated liquid vV = 1/rhoV; % saturated steam

% alpha

alpha = alpha0*(alphad + d(1)*theta^(-19) + d(2)*theta +...

d(3)*theta^4.5 + d(4)*theta^5 + d(5)*theta^54.5);

% phi

phi = phi0*(phid + (19/20)*d(1)*theta^(-20) + d(2)*log(theta) +...

(9/7)*d(3)*theta^3.5 + (5/4)*d(4)*theta^4 +...

(109/117)*d(5)*theta^53.5);

% dp/dT

tv = 7.5*a(6)*tau^6.5 + 4*a(5)*tau^3 + 3.5*a(4)*tau^2.5 +...

3*a(3)*tau^2 + 1.5*a(2)*tau^0.5 + a(1) + log(Ps/Pc);

dpdT = (-Ps/Ts)*tv;

% enthalpy

hL = alpha + (Ts/rhoL)*dpdT; % saturated liquid hV = alpha + (Ts/rhoV)*dpdT; % saturated steam

% entropy

sL = phi + (1/rhoL)*dpdT; % saturated liquid sV = phi + (1/rhoV)*dpdT; % saturated steam

% result structure stpr.T = Ts;

stpr.P = Ps;

stpr.vL = vL;

stpr.vV = vV;

stpr.hL = hL;

stpr.hV = hV;

stpr.sL = sL;

stpr.sV = sV;

end

Example 3.2: Physical Properties of the Saturated Steam

Calculate the specific volume, the enthalpy, and the entropy of the saturated steam at 150°C.

Compare the results with those recorded in steam tables that be found in standard thermody- namic texts.

Solution

Next is an example of calling the function satsteam and storing the results in a structure called stpr. Outputs of this function are displayed by using the built-in function fprintf and are summa- rized in Table 3.2.

>> stpr = satsteam(150);

>> fprintf('Temperature(saturated steam): %g C\n', stpr.T-273.15);

>> fprintf('Pressure = %g kPa\n', stpr.P/1000);

>> fprintf('Liquid volume = %g cm^3/g\n', stpr.vL*1000);

>> fprintf('Steam volume = %g cm^3/g\n', stpr.vV*1000);

>> fprintf('Liquid enthalpy = %g kJ/kg\n', stpr.hL/1000);

>> fprintf('Steam enthalpy = %g kJ/kg\n', stpr.hV/1000);

>> fprintf('Liquid entropy = %g kJ/kg*K\n', stpr.sL/1000);

>> fprintf('Steam entropy = %g kJ/kg*K\n', stpr.sV/1000);

Temperature(saturated steam): 150 C Pressure = 476.159 kPa

Liquid volume = 1.0905 cm^3/g Steam volume = 392.48 cm^3/g Liquid enthalpy = 632.147 kJ/kg Steam enthalpy = 2746.05 kJ/kg Liquid entropy = 1.84172 kJ/kg*K Steam entropy = 6.83735 kJ/kg*K

3.2 HUMIDITY