EGR 334 ThermodynamicsChapter 3: Section 11

Lecture 09: Generalized Compressibility Chart Quiz Today?

Today’s main concepts:• Universal Gas Constant, R• Compressibility Factor, Z.• Be able to use the Generalized Compressibility to solve

problems• Be able to use Z to determine if a gas can be considered

to be an ideal gas.• Be able to explain Equation of State

Reading Assignment:

Homework Assignment:

• Read Chap 3: Sections 12-14

From Chap 3: 92, 93, 96, 99

3

Like cp and cv, today’s topic is about compressible gases….This method does not work for two phase mixtures such as water/steam. It only applies to gases.

Limitation:

pvZ

RT

where absolute pressure

absolute temperature

molar specific volume

p

T

v

mol

f mol

8.314 kJ/kmol K

1.986 Btu/lb

1545 ft lb /lb

o

o

R R

R

and

Compressibility Factor, Z

4

Universal Gas Constant

Substance Chem. Formula

R (kJ/kg-K) R(Btu/lm-R)

Air --- 0.2870 0.06855

Ammonia NH3 0.4882 0.11662

Argon Ar 0.2082 0.04972

Carbon Dioxide CO2 0.1889 0.04513

Carbon Monoxide

CO 0.2968 0.07090

Helium He 2.0769 0.49613

Hydrogen H2 4.1240 0.98512

Methane CH4 0.5183 0.12382

Nitrogen N2 0.2968 0.07090

Oxygen O2 0.2598 0.06206

Water H2O 0.4614 0.11021

R can also be expresses on a per mole basis:

RR

M

where M is the molecular weight (see Tables A-1 and A-1E)

5Sec 3.11 : Compressibility

For low pressure gases it was noted from experiment that there was a linear behavior between volume and pressure at constant temperature.

The constant R is called the Universal Gas Constant.Where does this constant come from?

and the limit as P0

then RT

PvP

0

lim

The ideal gas model assumes low P molecules are elastic spheres no forces between molecules

6Sec 3.11 : Compressibility

To compensate for non-ideal behavior we can use other equations of state (EOS) or use compressibility

RT

PvZ

Define the compressibility factor Z,

Z1 whenideal gasnear critical pointT >> Tc or (T > 2Tc)

Step 1: Thus, analyze Z by first looking at the reduced variables

CR

CR

T

TT

P

PP

Pc = Critical Pressure

Tc = Critical Pressure

Fig03_12

Step 2: Using the reduced pressure, pr and reduced temperature, Tr determine Z from the Generalized compressibility charts. (see Figures A-1, A-2, and A-3 in appendix).

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Step 3: Use Z to a) state whether the substance behaves as an ideal gas, if Z ≈ 1 b) calculate the specific volume of the gas using

'Rc

c

vv

RTp

vv

M

The figures also let’s you directly read reduced specific volume where

RTv Z

p

where

RR

M

9Sec 3.11 : Compressibility

Summarize:

1) from given information, calculate any two of these:

or M

RR

RC

pp

p R

C

TT

T 'R

c

c

vv

RTp

2) Using Figures A-1, A-2, and A-3, read the value of Z

3) Calculate the missing property using

pvZ

RT pv

ZRT

where v

vM

(Note: pc and Tc can be found on

Tables A-1 and A-1E)

(Note: M for different gases can be found on Table 3.1 on page 123.)

mol

f mol

8.314 kJ/kmol K

1.986 Btu/lb

1545 ft lb /lb

o

o

R R

R

10

Example: (3.95) A tank contains 2 m3 of air at -93°C and a gage pressure of 1.4 MPa. Determine the mass of air, in kg. The local atmospheric pressure is 1 atm.

Sec 3.11 : Compressibility

V = 2 m3 T = -93°C pgage = 1.4 MPa

patm = 0.101 MPa

11

Example: (3.95) Determine the mass of air, in kg

Sec 3.11 : Compressibility

V = 2 m3 T = -93°C = 180 Kp = pgauge + patm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar

From Table A-1 (p. 816): For Air: 16) Tc = 133 K pc = 37.7 bar 15

0.4037.7

1801.35

133

RC

RC

pp

p

TT

T

Vmppv

ZRT RT

2

5 315 10 2 1 161.1

1000 10.95 8.314 18028.97

Nm

kJ kmolkmol K kg

m kJ Jm kg

J N mK

Z=0.95 View

Compressibility Figure

pV pV

mRZRT Z TM

12Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model

Ideal Gas pv RT

Equations of State: Relate the state variables T, p, V

Alternate Expressions pV mRT

pv mRT

When the gas follows the ideal gas law,Z = 1p << pc and / or T >> Tc

Tuu and h h T u T pv u T RT

13Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model

Ideal Gas pv RT

Equations of State: Relate the state variables T, P, V

Van der Waals 2

2

n ap V nb nRT

V

a attraction between particles

b volume of particles

Redlich–Kwong

Peng-Robinson

m m m

RT apV b TV V b

2 22m m m

RT apV b V bV b

virial 2 31 .....Z B T p C T p D T p

.....1

32

v

TD

v

TC

v

TBZ

B Two molecule interactionsC Three molecule interactions

14

Example: (3.105) A tank contains 10 lb of air at 70°F with a pressure of 30 psi. Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions.

m = 10 lbT = 70°F p = 30 psi

15

Example: (3.105) Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions.

Sec 3.12 : Ideal Gas

m = 10 lbT = 70°F = 530°R p = 30 psi= 2.04 atm

For Air, (Table A-1E, p 864)Tc = 239 °R and pc = 37.2 atm

2.040.055

37.2

5302.22

239

RC

RC

pp

p

TT

T

Z= 1.0 (Figure A-1)

pv pVZ

RT mRT

2

2 2

1

3

144

1545 28.97(10 )(1.0) / 53065.4

(30 )

f mol

mol m

f

ft lb

lb R

lbm lb

lb inin ft

lb RV ft

ViewCompressibility

Figure

RM

mZ TmZRTV

p p

16

Example 3: Nitrogen gas is originally at p = 200 atm, T = 252.4 K. It is cooled at constant volume to T = 189.3 K. What is the pressure at the lower temperature?

SOLUTION:From Table A-1 for Nitrogen pcr = 33.5 atm, Tcr = 126.2 K

At State 1, pr,1 = 200/33.5 = 5.97 and Tr,1 = 252.4/126.2 = 2.

According to compressibility factor chart , Z = 0.95 vr' = 0.34.

Following the constant vr' line until it intersects with the line at Tr,2 = 189.3/126.2 = 1.5 gives Pr,2 = 3.55.

Thus P2 = 3.55 x 33.5 = 119 atm.

Since the chart shows Z drops down to around 0.8 at State 2, so it would not be appropriate to treat it as an ideal gas law for this model.

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End of Slides for Lecture 09