generalized compressibility chart - dr. javier ortega -...

Generalized Compressibility Chart

v►The p- -T relation for 10 common gases is

shown in the generalized compressibility chart.

1

Dr. J. Ortega - Spring 2014

Generalized Compressibility Chart

TR

pZ

v

► In this chart, the compressibility factor, Z, is plotted versus

the reduced pressure, pR, and reduced temperature TR,

where pR = p/pc TR = T/Tc

(Eq. 3.27) (Eq. 3.28) (Eq. 3.23)

The symbols pc and Tc denote the

temperature and pressure at the

critical point for the particular gas

under consideration. These values

are obtained from Tables A-1 and

A-1E.

R

8.314 kJ/kmol∙K

1.986 Btu/lbmol∙oR

1545 ft∙lbf/lbmol∙oR (Eq. 3.22)

is the universal gas constant

R

2

Generalized Compressibility Chart ► When p, pc, T, Tc, v , and are used in consistent units, Z,

pR, and TR are numerical values without units.

► Example: For air at 200 K, 132 bar, TR = 200 K/133 K = 1.5,

pR = 132 bar/37.7 bar = 3.5 where Tc and pc for air are from

Table A-1. With these TR, pR values, the generalized

compressibility chart gives Z = 0.8.

R

► With v = Mv, an alternative

form of the compressibility

factor is

Z = pv

RT (Eq. 3.24)

where

R = R

M (Eq. 3.25)

3


4


6


7


Studying the Generalized Compressibility Chart

►The solid lines labeled with TR values represent best fits to

experimental data. For the 10 different gases represented

there is little scatter in data about these lines.

►At the lowest reduced

temperature value

shown, TR = 1.0, the

compressibility factor

varies between 0.2

and 1.0. Less

variation is observed

as TR takes higher

values.

8


Studying the Generalized Compressibility Chart

►For each specified value of TR, the compressibility factor

approaches a value of 1.0 as pR approaches zero.

9


Introducing the Ideal Gas Model

►To recap, the generalized

compressibility chart shows that

at states where the pressure p

is small relative to the critical

pressure pc (where pR is small),

the compressibility factor Z is

approximately 1.

►At such states, it can be assumed with reasonable

accuracy that Z = 1. Then

pv = RT (Eq. 3.32)

10



►Three alternative forms of Eq. 3.32 can be derived

as follows:

►With v = V/m, Eq. 3.32 gives

pV = mRT (Eq. 3.33)

►With v = v/M and R = R/M, Eq. 3.32 gives

(Eq. 3.34) TRp v

►Finally, with v = V/n, Eq. 3.34 gives

pV = nRT (Eq. 3.35) 11


12



► Investigation of gas behavior at states where Eqs. 3.32-

3.35 are applicable indicates that the specific internal

energy depends primarily on temperature. Accordingly, at

such states, it can be assumed with reasonable accuracy

that it depends on temperature alone:

u = u(T) (Eq. 3.36)

►With Eqs. 3.32 and 3.36, the specific enthalpy also depends

on temperature alone at such states:

h = u + pv = u(T) + RT (Eq. 3.37)

►Collecting results, a gas modeled as an ideal gas adheres

to Eqs. 3.32-3.35 and Eqs. 3.36 and 3.37. 13


Specific Heats

► Three properties related to specific internal energy and specific

enthalpy having important applications are the specific heats cv

and cp and the specific heat ratio k.

vv

T

uc

(Eq. 3.8)

pp

T

hc

(Eq. 3.9)

vc

ck

p

(Eq. 3.10)

► In general, cv is a function of v and T (or p and T), and cp

depends on both p and T (or v and T).

► Specific heat data are provided in Fig 3.9 and Tables A-19

through A-21.

► Although cv and cp are referred to as specific heats, there is no

general relationship between them and the heat transfer term of

the energy balance denoted by Q. 14


Internal Energy and Enthalpy of Ideal Gases

►For a gas obeying the ideal gas model, specific

internal energy depends only on temperature.

Hence, the specific heat cv, defined by Eq. 3.8, is

also a function of temperature alone. That is,

(Eq. 3.38) dT

duTc )(v (ideal gas)

►On integration,

(Eq. 3.40) dTTcTuTu )(12 vT1

T2

(ideal gas)

15



►Similarly, for a gas obeying the ideal gas model,

specific enthalpy depends only on temperature.

Hence, the specific heat cp, defined by Eq. 3.9, is

also a function of temperature alone. That is,

(Eq. 3.41) dT

dhTcp )( (ideal gas)

►On integration,

(Eq. 3.43) dTTcThTh p )(12T1

T2

(ideal gas)

16



►In applications where the specific heats are

modeled as constant,

►For several common gases, evaluation of

changes in specific internal energy and enthalpy

is facilitated by use of the ideal gas tables:

Tables A-22 and A-23.

►Table A-22 applies to air modeled as an ideal

gas.

(Eq. 3.50)

(Eq. 3.51)

u(T2) – u(T1) = cv[T2 – T1]

h(T2) – h(T1) = cp[T2 – T1]

17


18


TABLE A-22

Ideal Gas Properties of Air

T(K), h and u(kJ/kg), so

(kJ/kg∙K)

when s = 0 when s = 0

T h u so pr vr T h u s

o pr vr

250 250.05 178.28 1.51917 0.7329 979. 1400 1515.42 1113.52 3.36200 450.5 8.919

260 260.09 185.45 1.55848 0.8405 887.8 1420 1539.44 1131.77 3.37901 478.0 8.526

270 270.11 192.60 1.59634 0.9590 808.0 1440 1563.51 1150.13 3.39586 506.9 8.153

280 280.13 199.75 1.63279 1.0889 738.0 1460 1587.63 1168.49 3.41247 537.1 7.801

285 285.14 203.33 1.65055 1.1584 706.1 1480 1611.79 1186.95 3.42892 568.8 7.468

290 290.16 206.91 1.66802 1.2311 676.1 1500 1635.97 1205.41 3.44516 601.9 7.152

295 295.17 210.49 1.68515 1.3068 647.9 1520 1660.23 1223.87 3.46120 636.5 6.854

300 300.19 214.07 1.70203 1.3860 621.2 1540 1684.51 1242.43 3.47712 672.8 6.569

305 305.22 217.67 1.71865 1.4686 596.0 1560 1708.82 1260.99 3.49276 710.5 6.301

310 310.24 221.25 1.73498 1.5546 572.3 1580 1733.17 1279.65 3.50829 750.0 6.046


► Example: Using Table A-22, determine the change in specific

enthalpy, in kJ/kg, for a process of air from an initial state where

T1 = 300 K, p1 = 1 bar to a final state where T2 = 1500 K, p2 = 10 bar.

► Solution: h1 = 300.19 kJ/kg; h2 = 1635.97 kJ/kg

h2 – h1 = 1335.78 kJ/kg

19

Property Data Use in the

Closed System Energy Balance

Example: A closed, rigid tank consists of 1 kg of air

at 300 K. The air is heated until its temperature

becomes 1500 K. Neglecting changes in kinetic

energy and potential energy and modeling air as an

ideal gas, determine the heat transfer, in kJ, during

the process of the air.

1 kg of air

Q

T1 = 300 K T2 = 1500 K

State 1 State 2 2 ●

●

T

v

T1 = 300 K 1

T2 = 1500 K

p2

p1

20


Property Data Use in the

Closed System Energy Balance Solution: An energy balance for the closed system is

KE + PE +U = Q – W 0 0 0

where the kinetic and potential energy changes are neglected

and W = 0 because there is no work mode.

Thus Q = m(u2 – u1)

TABLE A-22

Ideal Gas Properties of Air

T(K), h and u(kJ/kg), so

(kJ/kg∙K)

when s = 0 when s = 0

T h u so pr vr T h u s

o pr vr

250 250.05 178.28 1.51917 0.7329 979. 1400 1515.42 1113.52 3.36200 450.5 8.919

260 260.09 185.45 1.55848 0.8405 887.8 1420 1539.44 1131.77 3.37901 478.0 8.526

270 270.11 192.60 1.59634 0.9590 808.0 1440 1563.51 1150.13 3.39586 506.9 8.153

280 280.13 199.75 1.63279 1.0889 738.0 1460 1587.63 1168.49 3.41247 537.1 7.801

285 285.14 203.33 1.65055 1.1584 706.1 1480 1611.79 1186.95 3.42892 568.8 7.468

290 290.16 206.91 1.66802 1.2311 676.1 1500 1635.97 1205.41 3.44516 601.9 7.152

295 295.17 210.49 1.68515 1.3068 647.9 1520 1660.23 1223.87 3.46120 636.5 6.854

300 300.19 214.07 1.70203 1.3860 621.2 1540 1684.51 1242.43 3.47712 672.8 6.569

305 305.22 217.67 1.71865 1.4686 596.0 1560 1708.82 1260.99 3.49276 710.5 6.301

310 310.24 221.25 1.73498 1.5546 572.3 1580 1733.17 1279.65 3.50829 750.0 6.046

Q = (1 kg)(1205.41 – 214.07) kJ/kg = 991.34 kJ

Substituting values for specific

internal energy from Table A-22

21

22


23


24


25


(Eq. 3.52)

Polytropic Process

►A polytropic process is a quasiequilibrium

process described by

pV n = constant

26


27

►The exponent, n, may take on any value from

–∞ to +∞ depending on the particular process.

►For any gas (or liquid), when n = 0, the process is a

constant-pressure (isobaric) process.

►For a gas modeled as an ideal gas, when n = 1, the

process is a constant-temperature (isothermal)

process.

(3.54)

(3.55)


28

►When the ideal gas equation of state is introduced

into the previous equations, the following expressions

are obtained respectively.

►For a gas modeled as an ideal gas, when n = 1, the

process is a constant-temperature (isothermal)

process.

29


30


generalized compressibility chart - dr. javier ortega -...

Documents