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Unformatted text preview: 9/25/2010 Compressibility Chart and Factors
Ideal Gas Model ME 200
Thermodynamics I
Lecture 13
September 24th, 2010 Purdue University , Dr. Tim Pourpoint – timothee@purdue.edu Last Week
• Specific heat at constant volume (Cv) and specific heat at constant pressure (Cp)
Cp > Cv
Cp/Cv = k
• Incompressible substance i.e. no change in specific volume for solids and liquids
• Approximations for compressed liquid property evaluation
• Only one specific heat for solids and liquids
• Calculation of internal energy and enthalpy changes for solids and liquids
ME 200 2 1 9/25/2010 Today
• Equation of State
• Ideal Gas Model
• Compressibility Factor ME 200 3 Equation of State
• What is an Equation of State (EOS)?
– A relationship between pressure, temperature and specific volume (or density) of a given pure substance (may involve other properties). Usually in form p = f(T,v).
– Usually valid only under a limited range of conditions of pressure and temperature. • Why do we need equations of state? ME 200 4 2 9/25/2010 Equation of State
• Why do we need equations of state?
Property tables provide very accurate information; but need to be
measured with some reference state.
Most are calculated from
sophisticated equations of state.
Use of simple relations such as ideal gas EOS and van der Waals
EOS desirable for qualitative physical understanding of fluid behavior. Ref.: Simulation of High‐Pressure Metal Hydride Systems, Smith, K.C., Zheng, Y., Fisher, T.S., Mudawar, I., Pourpoint, T.L., 19th National & 8th ISHMT‐ASME Heat & Mass Transfer Conference, 2008, JNTU Hyderabad, India, Paper No: US‐53 Incorporation in numerical codes for calculation of thermodynamic
properties, system performance. ME 200 5 Ideal Gas
• Ideal gas is an imaginary substance that has certain characteristics:
– No molecular volume:
molecules are simply points is space
– No molecular forces:
molecules are on average very far apart from each other • Is this possible? Not really. However, we can get very close to such behavior at very low density
• All substances will become ideal gases at sufficiently low pressure and high temperatures
ME 200 6 3 9/25/2010 Ideal Gas Equation of State (IGEOS)
• Based on experimental data, ideal gases obey
Pv = RT
– P, v, T = pressure, specific volume, temperature
– R = specific gas constant = Ru/M
– Ru = universal gas constant
– M = molecular weight (molar mass) • Molar mass: Mass of one mole of a substance (same irrespective of system of units)
1 kmol of N2 = 28 kg or: 1 lbmol of N2 = 28 lbm ME 200 7 Summary of IGEOS Forms
On a molar basis: On a mass basis: v = V/m
N = m/M ME 200 8 4 9/25/2010 Ideal Gas Equation of State
• Ideal Gas Processes: pv RT • For a fixed mass of a given gas: p1v1 p2 v2 T1
T2 • Special cases:
– Isobaric process:
(constant pressure)
– Isothermal process:
(constant temperature) v1 T1 v2 T2
v1 p2 v2 p1 – Isochoric process at constant mass:
(constant specific volume)
ME 200 p1 T1 p2 T2 9 Notes on Using IGEOS
• The temperature, T, must be absolute (Kelvin or Rankine)
– The pressure, p, must be absolute
– The universal gas constant, Ru, is different from the specific gas constant, R
– Use the most convenient units ME 200 10 5 9/25/2010 Ideal Gas Equation of State • What are the limits for using IGEOS? Usually good for air, N2, O2, H2, He, Ar, Ne at low pressures and high temperatures (less than 1% error) Works well for water vapor below 10 kPa irrespective of temperature (less than 0.1% error) Fails close to critical point and saturated vapor line Definitely not applicable for water vapor in steam power plants owing to very high pressures ME 200 11 T (ºC) v (m3/kg)
ME 200 12 6 9/25/2010 Other EOSs
• There are several other EOSs:
–
–
–
– Virial Equation of State
Cubic Equations of State
Redlich/Kwong EOS:
Many more... • People spend scientific careers developing and calibrating EOSs a
b
p
v
T RedlichKwong constant, kJ2K0.5/kg2bar
RedlichKwong constant, m3/kg
pressure, bar
specific volume, m3/kg
temperature, K • Not emphasized in ME 200
ME 200 13 Example 13.1
• Find the mass (kg) and weight (N) of the air contained in the classroom if the temperature and pressure of the air are 22C and 0.955 atm (< 1 atm just to practice unit conversions)
• ASSUMPTIONS:
– Ideal gas law applies
– Room dimensions: 40 ft by 100 ft by 12 ft • DATA: R = 0.2870 (kPa*m3)/(kg*K) g = 9.81 m/s2 T = 22 + 237.15 = 295.15 K
ME 200 14 7 9/25/2010 Example 13.1
• CALCULATIONS:
PV = mRT
0.955 atm
0.287 1.142 20 ft kg
∙
m3 101.325 kPa
1 atm kPa∙ 3
295.15 K
kg∙K m3 777 kg 0.3048 m
0.3048 m
0.3048 m
∙ 100 ft
∙ 12 ft
1 ft
1 ft
1 ft 680 m3 777 kg ∙ 9.81 s2 7620 N ME
ME 200 15 Compressibility Factor
• What is meant by low pressure and high temperature?
Compressibility factor quantifies the deviation of a pure substance from ideal gas behavior at a given temperature and pressure Z = 1 (ideal gas)
For a real gas, Z > 1 or Z < 1
For Z very close to unity, we can assume ideal gas behavior, most of the time Compressibility Factor Plot for Hydrogen
ME 200 16 8 9/25/2010 Continue Compressibility Factor
• Z is a measure of the deviation from ideal gas behavior
• Can plot Z as a function of p for constant T for each individual fluid.
• However, when these charts are studied, they are found to be “qualitatively” similar • Thus, more useful is to plot Z as a function of the reduced pressure pR
for constant reduced temperature TR Principle of corresponding states:
At the same reduced pressure and reduced temperature, gases behave similarly
PR = reduced pressure
TR = reduced temperature
Pcr = critical pressure
Tcr = critical temperature
ME 200 ME 200 17 Generalized Compressibility Chart 18 9 9/25/2010 Generalized Compressibility Chart
• Calculate two reduced properties: PR, TR and vR PR = P/Pcr TR = T/Tcr vR ≠ v/vcr; vR = v/(RTcr/Pcr) (pseudoreduced specific volume) • Determine Z
• When can ideal be assumed? If PR << 1, then ideal gas at all temperatures If TR >> 2, then ideal gas at all pressures (except when PR >> 1) Near critical point and saturated vapor, then real gas behavior
ME 200 19 Continue Compressibility Factor
• Notes: If pR < 0.1 Z = 1.0 If TR > 2.0 Z = 1.0 • For example, air at 27oC and 1 bar:
pR TR ME 200 p
1bar 0.027 pR 0.1 !
pcr 37.7bars
T 300 K 2.3
Tcr 133K TR 2.0 ! 20 10 9/25/2010 Example 13.2
• Determine the value of the specific volume, in m3/kg, of ammonia (NH3) vapor at 100 bar and 420 K using:
a) The compressibility chart
b) The ideal gas model Which of the above values is more correct? Explain ME 200 21 Solutions for Example 13.2
a) With compressibility chart
Using data from Table A.1:
and: TR = T/Tc = 420/406 = 1.034
PR = P/Pc = 100/112.8 = 0.886
From Figure A.1 (page 911 in the book):
v’R 0.6 (and NOT 0.74 as I somehow had on my notes)
and: Z 0.53 c) The chart value is more correct. It takes into account the real gas behavior of ammonia at this high pressure and temperature. The ideal gas model over‐predicts the specific volume by about 50%
ME 200 22 11 9/25/2010 Another Example
• Determine the compressibility factor of CO2 (Tcr = 304 K; pcr = 7.39 MPa) at 5 MPa and 300 K using the generalized compressibility chart.
• Given:
– T = 37oC ; p = 1.54 Mpa • Find:
– Z = ? • Basic Equations:
p
T
pR TR pcr
Tcr • Solution: • Assumptions:
– None p
5MPa 0.68
pcr 7.39MPa Z 0.7
T 300 K 0.97
TR Tcr 304 K pR ME 200 23 Continue Compressibility Factor
• Percent deviation in Z‐factor for two fluids: Nitrogen
0
25 225
103 0.13 63.15
91.15 0.29 1.44
7.66 24.84 3.63 3390 kPa 0.71 40.57 164.9 100 22.22
26.54
32.37
53.51 6000 kPa
4000 kPa 22.15
1500 kPa
1500 kPa 2.12 500 kPa
101.3 kPa 102 101
3 v [m /kg] ME 200 38.08 T [°C] T [°C] 200 0.61 3.40 10.95 100 175 1.56 5.53 75 150 R134a 200 0.05 0.23 47.92 2.79 50 125 0.63 1.23 100 0
104 103 102 101 3 v [m /kg] 24 12 9/25/2010 Summary and Conclusions
• In this lecture, we:
– Discussed the concept of ideal gas
– Talked about applicability of ideal gas equation of state • Next Lectures
– More Ideal Gas Properties
– Polytropic Processes TRAIN YOURSELF:
Under what conditions, can a pure substance be modeled as an ideal gas?
What is the utility of compressibility factor?
Is it appropriate to assume ideal gas behavior near critical point? Why or Why not? ME 200 25 13 ...
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This note was uploaded on 12/11/2011 for the course ME 200 taught by Professor Gal during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 GAL

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