210
Introduction ENGR to Thermodynamics
Course announcements
First term exam: TOMORROW, 8 am in Main Auditorium
Closed Book, Closed Notes
Only pencils, erasers, calculator at workplace
No internet, wireless or telephony capable devices in your
workspace
Exam Details
50 minutes of work
You will need your recitation section number
You asked for a formula sheet I will provide it; you only
bring pencils, eraser, and calculator to exam
Lecture 7- Winter 201125
ENGR 210 Week 4
Recitation Problems covered this week
Chap 3: 24E, 60, 66
Xtra Problems
Chap 3: 26, 27, 29E, 42, 64
Lecture 7- Winter 201125
Ideal gas equation of state
Equation of state - equation relating P, T, v
Early results
Boyle: P v-1
Charles / Gay-Lussac: P = R (T / v) at low pressures
Ideal gas law
P v = R T
R = gas constant = Ru / M
Ru = universal gas constant = 8.31447 kJ / (kmolK);
1.98588 BTU/(lbmol-R)
M = molar mass = kg/kmol, lbm/lbmol
kmol = 6.023 x 1026 molecules
lbmol is not a specific number of things
Lecture 7- Winter 201125
Different Forms of Ideal Gas Law
PV = mRT
PV = NRuT
P = absolute pressure
usually in MPa or kPa
V = molar specific volume
n m3/kmol = V/n
T = absolute temperature
in K or R
Ru = 8.314 kJ/(kmolK)
N = number of moles
,
,
Pv = RT
Pv = RuT
Universal Gas Constant
Ru = MR
8.3145 kJ/(kmolK)
8.3145 kPam3/(kmolK)
1.986 Btu/(lbmolR)
1545.4 ftlbf/(lbmolR)
10.73 psiaft3/(lbmolR)
Lecture 7- Winter 201125
Alternative Origin of Ideal Gas Law
The ideal gas equation of state can be
derived from basic principles if one
assumes
Intermolecular forces are small
Volume occupied by the particles is small.
Lecture 7- Winter 201125
Example 1
Determine the particular gas constant for air and hydrogen.
R=
Ru
M
kJ
kmol K = 0.2870 kJ
kg
kg K
28.97
kmol
8.314
Rair =
kJ
kmol K = 4.124 kJ
kg
kg K
2.016
kmol
8.314
Rhydrogen =
Lecture 7- Winter 201125
Table A-1: Ideal Gas Information
Lecture 7- Winter 201125
Ideal Gas Rules of Thumb
The ideal gas equation of state is appropriate when:
the pressure is small compared to the critical
pressure (P< 0.25* Pcrit)
or
the temperature is larger than twice the critical
temperature and the pressure is less than 10
times the critical pressure.
Critical point data are also in Table A-1.
Lecture 7- Winter 201125
Manipulating the Ideal Gas Law
By writing the ideal gas equation twice and
simplifying, the properties of an ideal gas at
two different states 1 and 2 are related by
R = P1V1 / (m1T1) = P2V2 /(m2T2)
R = P1v1 / T1 = P2v2 /T2 .
If, in addition, we have m1 = m2, then we may write
PV1 P2V2
1
=
T1
T2
Lecture 7- Winter 201125
Example 2
An ideal gas having an initial temperature of 25C
undergoes the two processes described below. final Determine
the temperature of the gas.
Process 1-2: The volume is held constant while the pressure
doubles.
Process 2-3: The pressure is held constant while the volume
is reduced to one-third of the original volume.
T2
P
3
2
T3
T1
Ideal Gas
1
V
Lecture 7- Winter 201125
Example 2 (cont)
Process 1-3:
m1 = m3
so that
PV1 P3V3
1
=
T1
T3
but V3 = V1/3 and P3 = P2 = 2P1 , so
T3 = T1
P3 V3
P V1
1
T3 = T1
2 P V1 / 3 2
1
= T1
P V1
3
1
T3 =
2
(25 + 273) K = 198.7 K = 74.3C
3
Lecture 7- Winter 201125
Superheated H2O vapor = ideal gas?
In general NO !!!!!! Use tables for properties
All the complex interactions are included in
tabulated (empirical) values
Can water vapor EVER be treated as an ideal
gas?
YES but under certain conditions!!!
How close is the empirical value to the ideal gas value?
Lecture 7- Winter 201125
2-13
When can Steam be treated as an IG
Lecture 7- Winter 201125
What about working fluids without tables?
Need a measure of deviation from ideal gas behavior
One example is Compressibility factor: Z
Z=
P v vact ual
=
R T vide al
Also, find that Z = function of a scaled temperature
and pressure called reduced temperature and
reduced pressure
Lecture 7- Winter 201125
Reduced Pressure and Temperature
Define PR and TR to be values normalized to
critical pressure and temperature,
respectively
PR = P / Pcr
TR = T / Tcr (absolute temperature)
Universal relationship
Principle of corresponding states
Generalized compressibility chart
PR 1 (actually < 0.25) - behaves like ideal gas
TR 2 - behaves like ideal gas
Largest deviations from ideal gas behavior
near critical point
Lecture 7- Winter 201125
2-14
Generalized Compressibility: A-15
Various
gases
Lecture 7- Winter 201125
Example 3
Find specific volume, v, for superheated
water vapor at 10 MPa and 400C using
ideal gas law
generalized compressibility chart
steam tables (accurate value)
Lecture 7- Winter 201125
Example 3 (cont) Ideal Gas
T
v = Ru
MP
v=
8.314kJ /(kmol K )(400 + 273.15) K
(18kg / kmol )(10000kPa)
v = 0.0311 m3/kg
Lecture 7- Winter 201125
Example 3 (cont) Z factor
P = P / P ;T = T / T
R
crit
R
crit
P = 10MPa / 22.06MPa = 0.4533
T = 400+273.15 / 647.1 = 1.0403
R
R
From Appendix A-15, Z = 0.85
v
actual
=Z
v
ideal
v = 0.0264 m3/kg
Lecture 7- Winter 201125
Example 3 (cont) fini
Table A-6, v = 0.026436 m3/kg
Ideal Gas, v = 0.0311 m3/kg
Z, v = 0.0264 m3/kg
Ideal gas is off by about 20%
Z is pretty close
Lecture 7- Winter 201125
Next
Prepare for the Exercise for Credit
Work problems to understand concepts
Review assumptions and physical laws
invoked to solve each problem
Get a good nights sleep
Read Chapter 4
Lecture 7- Winter 201125