Thermodynamics lecture 6

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BITS Pil i BITS Pilani Pilani Campus L t 6 G C ibilit F t L ec t ure 6 G ases, C ompress ibilit y F ac t or

Transcript of Thermodynamics lecture 6

Page 1: Thermodynamics lecture 6

BITS Pil iBITS PilaniPilani Campus

L t 6 G C ibilit F tLecture 6 – Gases, Compressibility Factor

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The Ideal Gas• PV = nȒT or Pv = ȒT

Ȓ 1 1• Ȓ is the universal gas constant 8.3145 kJ kmole-1 K-1

• n = m/M

• PV = mRT, where R is the gas constant per unit mass of a

given gas, 0.2968 kJ/kg K for N2, 0.1889 kJ/kg K for CO2 etcg g g 2 g 2

• Obeyed as a limiting law by gases at low density where

inter-molecular forces are negligibleinter-molecular forces are negligible

• What constitutes low density, and by how much does a

l d i t f id l b h i t i T d P?real gas deviate from ideal behaviour at given T and P?

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Ideal gas approximation for water vapour

Ideal gas appro imation er good hen P << P for almost

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Ideal gas approximation very good when P << Pc for almost any T, and T > ~ 2Tc for even moderate pressures up-to ~ 5Pc

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Deviations from ideality - corrections

Equations of State

• van der Waals (P + a/ṽ2)(ṽ - b) = ȒT, two parameter, cubic

equation of state

• Other equations of state – cubic and non-cubic, including

those applicable to liquids and gasesthose applicable to liquids and gases

• Virial equation of state Z = 1 + B(T)/ṽ + C(T)/ṽ2 + ....

• Generalized (compressibility factor) correlations – eg., Lee

Kesler

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Compressibility Factor Z• Z = Pv/RT• Z = 1 for ideal gas. Note below that Z → 1 as P → 0g• What do Z > 1 and Z < 1 signify?

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Compressibility Factor• Noting that Z = v/vid where vid is the specific volume of the corresponding ideal gas at the same T and p, we conclude that Z < 1 when attractive interactions dominate, and Z > 1 when repulsive ones do• Observe also that different gases display qualitatively• Observe also that different gases display qualitatively similar behaviour if compared at appropriate T and p values• Express the temperature and pressure in terms of the p p pcritical values for the chosen substance• Define the reduced variables Pr = P/Pc, Tr = T/Tc, vr = v/vc

N l t Z f ti f th d d f• Now plot Z as a function of the reduced pressure for various chosen values of the reduced temperature• The data for different gases fall close to the same line for aThe data for different gases fall close to the same line for a given Tr

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Generalized Compressibility Chart

• Suggests the principle of corresponding statescorresponding states • Two substances at the same Prand Tr are said to be in corresponding statescorresponding states

• Expressed in terms of the reduced variables all substancesreduced variables, all substances obey the same eos – approximate principle• that is, if Pr and Tr are the same,

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, r r ,then so will vr

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Compressibility Chart for N2

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Lee Kessler Compressibility Chart

• Represents average behavior• Holds well, especially for simple almost sphericalsimple, almost spherical substances, especially when away from saturation• Useful when experimentalUseful when experimental data very scarce

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Example

• Determine the specific volume for R-134a at 100 C,

3.0 MPa using (a) the table, (b) ideal gad eos, and (c)

the generalized compressibility chartthe generalized compressibility chart

(a) 0.00665 m3/kg

(b) R = 0.08149 kJ/kg K, v = 0.01014 m3/kg

(c) Tc = 374.2, Tr = 1 and Pc = 4.06 MPa , Pr = 0.74

Z = 0.67 and v = ZRT/p = 0.00679 m3/kg,

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Example• Estimate the saturation pressure for chlorine at 300 K.

From Table, Tc = 416.9 K and Pc = 7.98 MPa. So, Tr =

0.72

From D1, Pr = 0.1 + (0.15/10)x2 = 0.13

P = 0 13x7 98 = 1 04 MPaPsat = 0.13x7.98 = 1.04 MPa

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Example• A bottle with volume 0.1 m3 contains n-butane with quality

f0.75 at 298K. Estimate the mass of butane in the bottle.

Tc = 425.2 K, Pc = 3.80 MPa, R = 0.1430 kJ/kg K

Tr = 0.70, Pr = 0.1

Z = 0 9 Z = 0 022Zg = 0.9, Zf = 0.022

Z = (1 – x) Zf + x Zg = 0.68

M = PV/ZRT = 1.3 kg

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