Transcript Chapter 3 from Moran and Shapiro

Thermodynamic
Properties

Property Table -from
direct measurement
 Equation of State -any equations that relates
P,v, and T of a substance
Ideal -Gas Equation of State
 Any
relation among the pressure,
temperature, and specific volume of a
substance is called an equation of state. The
simplest and best-known equation of state is
the ideal-gas equation of state, given as
where R is the gas constant. Caution should be
exercised in using this relation since an ideal
gas is a fictitious substance. Real gases exhibit
ideal-gas behavior at relatively low pressures
and high temperatures.
Universal Gas Constant
Universal gas constant is given on
Ru = 8.31434 kJ/kmol-K
= 8.31434 kPa-m3/kmol-k
= 0.0831434 bar-m3/kmol-K
= 82.05 L-atm/kmol-K
= 1.9858 Btu/lbmol-R
= 1545.35 ft-lbf/lbmol-R
= 10.73 psia-ft3/lbmol-R
Example
Determine the particular gas constant
for air and hydrogen.
kJ
8.1417
kJ
R
kmol

K
u
 0.287
Rair  M 
kg
kg

K
28.97
kmol
kJ
8.1417
kJ
kmol

K
 4.124
Rhydrogen 
kg
kg  K
2.016
kmol
Ideal Gas “Law” is a simple
Equation of State
PV  mRT
Pv  RT
PV  NRuT
P v  RuT, v  V/N
P1V1 P2V2

T1
T2
Percent error for applying ideal gas
equation of state to steam
Question …...
Under what conditions is it
appropriate to apply the
ideal gas equation of state?
Ideal Gas Law

Good approximation for P-v-T
behaviors of real gases at low
densities (low pressure and high
temperature).
 Air, nitrogen, oxygen, hydrogen,
helium, argon, neon, carbon
dioxide, …. ( < 1% error).
Compressibility Factor
 The
deviation from ideal-gas behavior
can be properly accounted for by using
the compressibility factor Z, defined as
Z represents the volume ratio
or compressibility.
Ideal
Gas
Real
Gases
Z=1
Z > 1 or
Z<1
Real Gases
 Pv
= ZRT or
 Pv = ZRuT, where v is volume
per unit mole.


Z is known as the compressibility factor.
Real gases, Z < 1 or Z > 1.
Compressibility factor

What is it really doing?
 It accounts mainly for two things
• Molecular structure
• Intermolecular attractive forces
Principle of
corresponding states
The
compressibility factor Z is
approximately the same for all
gases at the same reduced
temperature and reduced pressure.
Z = Z(PR,TR) for all gases
Reduced Pressure
and Temperature
P
T
PR 
; TR 
Pcr
Tcr
where:
PR and TR are reduced values.
Pcr and Tcr are critical properties.
Compressibility factor for ten substances
(applicable for all gases Table A-3)
Where do you find critical-point properties?
Table A-7
Mol
(kg-Mol)
R
(J/kg.K)
Tcrit
(K)
Pcrit
(MPa)
Ar
28,97
287,0
(---)
(---)
O2
32,00
259,8
154,8
5,08
H2
2,016
4124,2
33,3
1,30
H2 O
18,016
461,5
647,1
22,09
CO2
44,01
188,9
304,2
7,39
Reduced Properties

This works great if you are given a
gas, a P and a T and asked to find
the v.
 However, if you are given P and v
and asked to find T (or T and v and
asked to find P), trouble lies ahead.
 Use pseudo-reduced specific volume.
Pseudo-Reduced
Specific Volume
 When
either P or T is unknown, Z can
be determined from the compressibility
chart with the help of the pseudoreduced specific volume, defined as
not vcr !
Ideal-Gas Approximation

The compressibility chart shows
the conditions for which Z = 1 and
the gas behaves as an ideal gas:
 (a)
PR < 0.1 and TR > 1
Exercise 3-21
Steam at 600 oC & 1 MPa. Evaluate
the specific volume using the steam
table and ideal gas law
ToC
Tsat = 180oC therefore
is superheated steam 600
374
(Table A-3)
Volume = 0,4011
m3/kg
1MPa
Tcr
180
v
Solution - page 1
Part (b) ideal gas law
Rvapor = 461 J/kgK
v = RT/P , v = 461x873/106 = 0.403 m3/kg
Steam is clearly an ideal gas at this state.
Error is less than 0.5%
Check:
PR = 1/22.09 = 0.05
& TR = 873/647 = 1.35
TEAMPLAY
Find the compressibility factor to
determine the error in treating
oxygen gas at 160 K and 3 MPa as
an ideal gas.
Other Thermodynamic Properties:
Isobaric (c. pressure) Coefficient
v
P
1  v 
   0
v  T  P
 v 
 
 T  P
T
Other Thermodynamic Properties:
Isothermal (c. temp) Coefficient
v
 v 
 
 P  T
1  v 
   0
v  P  T
T
P
Other Thermodynamic Properties:
We can think of the volume as being a function
of pressure and temperature, v = v(P,T).
Hence infinitesimal differences in volume are
expressed as infinitesimal differences in P and
T, using  and  coefficients
 v 
 v 
dv    dT    dP   vdT  vdP
 T  P
 P  T
If  and  are constant, we can integrate for v:
 v 
Ln   T  T0    P  P0 
 v0 
Other Thermodynamic Properties:
Internal Energy, Enthalpy and
Entropy
u  uT, v 
h  hT, P   u  Pv
s  su, v 
Other Thermodynamic Properties:
Specific Heat at Const. Volume
u
v
 u 
Cv     0
 T  v
 u 
 
 T  v
T
Other Thermodynamic Properties:
Specific Heat at Const. Pressure
h
P
 h 
CP     0
 T  P
 h 
 
 T  P
T
Other Thermodynamic Properties:
Ratio of Specific Heat
 Cp 

  
 Cv 
Other Thermodynamic Properties:
Temperature
v
s
1

s
 
T  0
 u  v
1
T
u
Ideal Gases: u = u(T)
0
 u 
 u 
du  
 dT    dv
 T  v
 v  T
Therefore,
 u 
du  
 dT  C v ( T )dT
 T  v
We can start with du and
integrate to get the change in u:
du  C v dT, and
u  u2  u1  
T2
T1
C v (T)dT
Note that Cv does change with temperature
and cannot be automatically pulled from
the integral.
Let’s look at enthalpy
for an ideal gas:

h = u + Pv where Pv can be replaced
by RT because Pv = RT.

Therefore, h = u + RT => since u is
only a function of T, R is a constant,
then h is also only a function of T

so h = h(T)
Similarly, for a change in
enthalpy for ideal gases:
C p  C p (T )
 h 

0
 P 
&
dh  C p dT, and
h  h2  h1  
T2
T1
C p (T)dT
Summary: Ideal Gases
 For ideal
gases u, h, Cv, and Cp are
functions of temperature alone.
 For ideal gases, Cv and Cp are written
in terms of ordinary differentials as
 du 
 dh 
Cv  
; Cp  


 dT  ideal gas
 dT  ideal gas
For an ideal gas,
h
= u + Pv = u + RT
dh du

R
dT dT
Cp = Cv + R
 kJ 
 kg  K 


Ratio of specific heats
is given the symbol, 

Cp
Cv
Cp

Cp (T)
C v (T)
  (T)
R
 1
  1
Cv
Cv
Other relations with the
ratio of specific heats which
can be easily developed:
R
R
Cv 
and Cp 
-1
-1
For monatomic gases,
5
3
C p  R , Cv  R
2
2
andboth are constants.
Argon, Helium, and Neon
For all other gases,

Cp is a function of temperature and it may be
calculated from equations such as those in
Table A-5(c) in the Appendice

Cv may be calculated from Cp=Cv+R.

Next figure shows the temperature behavior
…. specific heats go up with temperature.
Specific Heats for Some Gases

Cp = Cp(T)
a function of
temperature
Three Ways to Calculate
Δu and Δh

Δu = u2 - u1 (table)

2

Δu =

Δu = Cv,av ΔT
1
Cv (T) dT

Δh = h2 - h1 (table)

2

Δh =

Δh = Cp,av ΔT
1
C p (T) dT
Isothermal Process
 Ideal
gas: PV = mRT
For ideal gas, PV = mRT
We substitute into the
integral
2
Wb   PdV 
1

2
1
mRT
dV
V
Collecting terms and integrating yields:
Wb  mRT 
2
1
 V2 
dV
 mRTn 
V
 V1 
Polytropic Process

PVn = C
Ideal Gas Adiabatic Process and
Reversible Work


1.
2.
3.
4.
5.
What is the path for process with expand or
contract without heat flux? How P,v and T
behavior when Q = 0?
To develop an expression to the adiabatic
process is necessary employ:
Reversible work mode: dW = PdV
Adiabatic hypothesis: dQ =0
Ideal Gas Law: Pv=RT
Specific Heat Relationships
First Law Thermodynamics: dQ-dW=dU
Ideal Gas Adiabatic Process and
Reversible Work (cont)
First Law:
Using P = MRT/V
dQ

  dW
0
PdV
dU

MC vdT
CV  dT 
 dV 


 
R  T
 V 

1  1
Integrating from
(1) to (2)
 T2   V2 
  
 T1   V1 
1  
Ideal Gas Adiabatic Process and
Reversible Work (cont)
Using the gas law :
Pv=RT, other
relationship amid
T, V and P are
 T2   V1 
  
 T1   V2 
 P2   V1 
  
 P1   V2 
developed
accordingly:
 T2   P2 
  
 T1   P1 
  1

  1 
Ideal Gas Adiabatic Process and
Reversible Work (cont)
An expression for
work is developed
using PV =
constant.
i and f represent
the initial and final
states
 

W   PdV  PV i 
dV
V


PV  i 1  

1  
V  V 
W
   1 f

PV f  PV i 

   1
i
Ideal Gas Adiabatic Process and
Reversible Work (cont)

The path representation
are lines where Pv =
constant.
 For most of the gases, 
1.4
 The adiabatic lines are
always at the righ of the
isothermal lines.
 The former is Pv =
constant (the exponent is
unity)
P
f
f
i
v
Polytropic Process
A frequently encountered process for gases
is the polytropic process:
PVn = c = constant
Since this expression relates P & V, we
can calculate the work for this path.
Wb  
V2
V1
PdV
Polytropic Process
Process

Constant pressure
 Constant volume
 Isothermal & ideal gas
 Adiabatic & ideal gas
Exponent n
0

1
k=Cp/Cv
Boundary work for a gas which
obeys the polytropic equation
Wb 

2
1
PdV  c 
2
1
v2
dV
n
V
V 
V V
 c


1 n
 1-n  v1
1-n
1 n
2
1 n
1
n  1
We can simplify it
further
n
The constant c = P1V1 =
n
2
n
P2V2
)  P1V 1n V 11 n 
1-n
1 n
2
P
2 V (V
Wb 
P2V2  P1V1

,
1 n
n 1
Polytropic Process
Wb  
2
1
PdV  
2
1
c
dV
n
V
P2V2  P1V1

, n 1
1 n
V 2 
 , n  1
 PVn
 V1 
Exercise 3-11
A bucket containing 2 liters of water at 100oC is
heated by an electric resistance.
a) Identify the energy interactions if the system
boundary is i) the water, ii) the electric coil
b) If heat is supplied at 1 KW, then how much
time is needed to boil off all the water to steam?
(latent heat of vap at 1 atm is 2258 kJ/kg)
c) If the water is at 25oC, how long will take to
boil off all the water (Cp = 4.18 J/kgºC)
Solution - page 1
Part a)
If the water is the system then Q > 0
and W = 0. There is a temperature
difference between the electric coil and
the water.
If the electric coil is the system then Q
< 0 and W < 0. It converts 100% of the
electric work to heat!
Solution - page 2
Part b)
The mass of water is of 2 kg. It comes from 2liters
times the specific volume of 0.001 m3/kg
To boil off all the water is necessary to supply all
the vaporization energy:
Evap = 2285*2=4570 KJ
The power is the energy rate. The time necessary to
supply 4570 KJ at 1KJ/sec is therefore:
Time = 4570/1 = 4570 seconds or 1.27 hour
Solution
page
3
Part c)
The heat to boil off all the water initially at 25oC is
the sum of : (1) the sensible heat to increase the
temperature from 25oC to 100oC, (2) the vap. heat
of part (b)
The sensible heat to increase from 25oC to 100oC is
determined using the specific heat (CP = 4.18
kJ/kgoC)
Heat = 4.18*2*(100-25)=627 KJ
The time necessary to supply (4570+627)KJ at
1KJ/sec is :
Time = 5197/1 = 5197 seconds or 1.44 hour
Exercise 3-12
A bucket containing 2 liters of R-12 is left outside
in the atmosphere (0.1 MPa)
a) What is the R-12 temperature assuming it is in
the saturated state.
b) the surrounding transfer heat at the rate of
1KW to the liquid. How long will take for all R12 vaporize?
Solution - page 1
Part a)
From table A-2, at the saturation
pressure of 0.1 MPa one finds:
• Tsaturation = - 30oC
• Vliq = 0.000672 m3/kg
• vvap = 0.159375 m3/kg
• hlv = 165KJ/kg (vaporization heat)
Solution - page 2
Part b)
The mass of R-12 is m = Volume/vL,
m=0.002/0.000672 = 2.98 kg
The vaporization energy:
Evap = vap energy * mass = 165*2.98 =
492 KJ
Time = Heat/Power = 492 sec or 8.2 min
Exercise 3-17
A rigid tank contains saturated steam
(x = 1) at 0.1 MPa. Heat is added to
the steam to increase the pressure to
0.3 MPa. What is the final
temperature?
Solution - page 1
P
Tcr = 374oC
superheated
0.3MPa
133.55oC
0.1MPa
99.63oC
v
Process at constant volume, search at the
superheated steam table for a temperature
corresponding to the 0.3 MPa and v = 1.690
m3/kg -> nearly 820oC.
Exercise 3-30
a)
b)
c)
d)
Air is compressed reversibly and adiabatically
from a pressure of 0.1 MPa and a temperature
of 20oC to a pressure of 1.0 MPa.
Find the air temperature after the compression
What is the density ratio (after to before
compression)
How much work is done in compressing 2 kg of
air?
How much power is required to compress 2kg
per second of air?
Solution - page 1

In a reversible and
adiabatic process P, T and
v follows:
 T2   V1 
  
 T1   V2 
 P2   V1 
  
 P1   V2 
 T2   P2 
  
 T1   P1 
  1
P
f
f

  1 
i
v
Solution - page 2
Part a)
The temperature after compression is
 P2 
T2  T1   
 P1 
  1 
0.4 1.4
1
 
 T2  293   
 566K ( 293oC)
 0.1 
Part b)
The density ratio is
 V2   1   P2 
   
 V1   2   P1 
1 
 2   1 1 1.4
  
 5.179
 1   0.1 
Solution - page 3
Part c)
The reversible work:
WREV

PV 2  PV 1  M  RT2  T1 




  1
2  287  566  293
Part d)
The power is:
0.4
  1
 391KJ
 R T2  T1 
dW
M
P

 391KW
  1
dt
Exercícios – Capítulo 3
Propriedades das Substâncias
Puras
Exercícios Propostos: 3.6 / 3.9 / 3.12 /
3.16 / 3.21 / 3.22 / 3.26 / 3.30 / 3.32 / 3.34
Team Play: 3.1 / 3.2 / 3.4
Ex. 3.1) Utilizando a Tabela A-1.1 ou A-1.2, determine se os estados
da água são de líquido comprimido, líquido-vapor, vapor
superaquecido ou se estão nas linhas de líquido saturado ou vapor
saturado.
 a) Vapor superaquecido
 a) P=1,0 MPa; T=207 ºC
 b) Líquido comprimido
 b) P=1,0 MPa; T=107,5 ºC
 c) P=1,0 MPa; T=179,91 ºC; x=0,0  c) Líquido saturado
 d) P=1,0 MPa; T=179,91 ºC; x=0,45  d) Líquido-Vapor
 e) Líquido comprimido
 e) T=340 ºC; P=21,0 MPa
 f) Vapor superaquecido
 f) T=340 ºC; P=2,1 MPa
 g) T=340 ºC; P=14,586 MPa; x=1,0  g) Vapor saturado
 h) T=500 ºC; P=25 MPa  h) Vapor superaquecido - Fluido
 i) Líquido comprimido - Fluido
 i) P=50 MPa; T=25 ºC

Ex. 3.2) Encontre o volume específico
dos estados “b”, “d” e “h” do exercício
anterior.
b) P=1,0 MPa; T=107,5 ºC  vvl=0,001050
(utilizar referência T=107,5ºC)
d) P=1,0 MPa; T=179,91 ºC; x=0,45  v=0,08812
[v=(1-x)vl+x(vv)]
h) T=500 ºC; P=25 MPa  v=0,011123 m3/kg
(Tabela A-1.3 Vapor Superaquecido)

Ex. 3.4) Amônia a P=150 kPa, T=0 ºC se
encontra na região de vapor
superaquecido e tem um volume
específico e entalpia de 0,8697 m3/kg e
1469,8 kJ/kg, respectivamente.
Determine sua energia interna
específica neste estado.
u =1339,345 kJ/kg
(u = h - P v)