generalized compressibility

7/27/2019 Generalized Compressibility

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EGR 334 Thermodynamics

Chapter 3: Section 11

Lecture 09:Generalized Compressibility Chart

Quiz Today?
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Todays 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


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Like c p and c v, todays topic is about compressible gases. This method does not work for two phase mixtures such aswater/steam. It only applies to gases .

3

Limitation:

pv Z RT

where absolute pressureabsolute temperature

molar specific volume

pT

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


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Universal Gas Constant

4

Substance Chem. Formula R (kJ/kg-K) R(Btu/lm-R)

Air --- 0.2870 0.06855

Ammonia NH 3 0.4882 0.11662Argon Ar 0.2082 0.04972

Carbon Dioxide CO 2 0.1889 0.04513

Carbon Monoxide CO 0.2968 0.07090

Helium He 2.0769 0.49613Hydrogen H 2 4.1240 0.98512

Methane CH 4 0.5183 0.12382

Nitrogen N 2 0.2968 0.07090

Oxygen O 2 0.2598 0.06206

Water H 2O 0.4614 0.11021

R can also be expresses on a per mole basis:

R R M

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


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5Sec 3.11 : Compressibility

For low pressure gases it was noted from experiment that therewas a linear behavior between volume and pressure at constanttemperature.

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

and the limit as P 0

then RT

Pv P 0lim

The ideal gas model assumeslow Pmolecules are elastic spheresno forces between molecules


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

RT Pv

Z

Define the compressibility factor Z,

Z 1 when

ideal gasnear critical pointT >> Tc or (T > 2T c)

Step 1: Thus, analyze Z by first

looking at the reduced variables

C R

C R

T

T T

P P

P Pc = Critical Pressure

Tc = Critical Pressure


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Step 2: Using the reduced pressure, p r and reduced temperature, T r 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 toa) state whether the substance behaves as an ideal gas, if Z 1

b) calculate the specific volume of the gas using

' Rc

c

vv

RT p

vv M

The figures also lets youdirectly read reducedspecific volume where

RT v Z

pwhere

R R M


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Summarize:

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

or M R R

RC

p p

p R C

T T

T ' R

c

c

vv

RT p

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

3) Calculate the missing property using

pv Z

RT pv

Z RT

where vv M

(Note: p c and T c can be found onTables A-1 and A-1E)

(Note: M for different gases can be foundon 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


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Example: (3.95) A tank contains 2 m 3 of air at -93C and a gagepressure of 1.4 MPa. Determine the mass of air, in kg. The localatmospheric pressure is 1 atm.

Sec 3.11 : Compressibility

V = 2 m 3 T = -93Cpgage = 1.4 MPapatm = 0.101 MPa


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11

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

Sec 3.11 : Compressibility

V = 2 m 3 T = -93C = 180 Kp = p gauge + p atm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar

From Table A-1 (p. 816): For Air: 16) T c = 133 K p c = 37.7 bar

150.40

37.7180

1.35133

R C

RC

p p

pT

T T

V

m p pv Z RT RT

25 315 10 2 1 1

61.1

1000 10.95 8.314 18028.97

N m

kJ kmol kmol K kg

m kJ J m kg

J N m K

Z=0.95ViewCompressibility Figure

pV pV m

R ZRT Z T M
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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 > T c

T uu and h h T u T pv u T RT


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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 a p V nb nRT

V

a attraction between particles

b volume of particles

Redlich Kwong

Peng-Robinson

m m m

RT a p

V b TV V b

2 22m m m

RT a p

V b V bV b

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

.....1 32 v

T Dv

T C vT B

Z

B Two molecule interactions C Three molecule interactions


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14

Example: (3.105) A tank contains 10 lb of air at 70F with a pressure of 30 psi. Determine the volume of the air, in ft 3. Verify that ideal gasbehavior can be assumed for air under these conditions.

m = 10 lbT = 70Fp = 30 psi


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Example: (3.105) Determine the volume of the air, in ft 3. Verify thatideal gas behavior can be assumed for air under these conditions.

Sec 3.12 : Ideal Gas

m = 10 lbT = 70F = 530Rp = 30 psi= 2.04 atm

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

2.04 0.05537.2

5302.22

239

RC

RC

p p p

T T

T

Z= 1.0 (Figure A-1)

pv pV Z 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

R

M mZ T mZRT V p p
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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 p cr = 33.5 atm, T cr = 126.2 K

At State 1, p r,1 = 200/33.5 = 5.97 and T r,1 = 252.4/126.2 = 2.

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

Following the constant v r' line until it intersects with the line atTr,2 = 189.3/126.2 = 1.5

gives

Pr,2 = 3.55.

Thus P 2 = 3.55 x 33.5 = 119 atm.

Since the chart shows Z drops down to around 0.8 at State 2, so it would notbe appropriate to treat it as an ideal gas law for this model.


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End of Slides for Lecture 09

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generalized compressibility

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