itk-233-2_pvt behavior of fluid
TRANSCRIPT
DICKY DERMAWANwww.dickydermawan.net78 .net
d ickydermawan@gmai l .com
ITK-233Termodinamika Teknik
Kimia I
3 SKS
2 - PVT Behavior of Fluid, Equation of State
For the regions of the diagram where a single phase exist:
This means an equation of state exist relating P, V, T. An EOS may be solved for any one of the three quantities as a function of the other two, viz.: V = V(T,P)
Volume expansivity:
Isothermal compressibility:
Equation of State
0)T,V,P(f
dPPV
dTTV
dVTP
PTV
V1
TPV
V1
dPdTVdV
Example
For liquid acetone at 20oC & 1 bar:Β = 1.487 x 10-3 oC-1 κ = 62 x x 10-6 bar-1 V=
1.287 cm3 g-1
For aceton, find:
a. The value of b. The pressure generated by heating at constant V from 20oC & 1 bar to 30oCc. The change in volume for a change from 20oC & 1 bar to 0oC & 10 bar.
VTP
Problem 3.1
Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial derivatives.
The isothermal compressibility coefficient () of water at 50oC and 1 bar is 44.18 x 10-6 bar-1. To what pressure must water be compressed at 50oC to change its density by 1%? Assume that is independent of P.
Problem 3.2 & 3.3
Generally, volume expansivity β and isothermal compressibility κ depend on T and P.Prove that
The Tait equation for liquids is written for an isotherm as:
where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.
PT TP
PBAP
1VV 0
Problem 3.4
For liquid water the isothermal compressibility is given by:
where c & b are functions of temperature only. If 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bar at 60oC, how much work is required? At 60oC, b=2700 bar and c = 0.125 cm3 g-1
)bP(Vc
Problem 3.5
Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temperature of 32oF from 1 atm to 3000 atm. The isothermal compressibility of mercury at 32oF is:
κ/atm-1 = 3.9 x 10-6 – 0.1 x 10-9 P (atm)
Problem 3.6
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0oC to 20oC. Determine ΔVt, W, Q, and ΔUt
The properties for liquid carbon tetrachloride at 1 bar & ooC may be assumed independent of temperature:
β = 1.2 x 10-3 K-1
Cp = 0.84 kJ kg-1 K-1
ρ = 1590 kg m-3
Problem 3.7
A substance for which κ is a constant undergo a mechanically reversible process from initial state (P1,V1) to final state (P2,V2), where V is molar volume.
a. Starting with the definition of κ, show that the path of the process is described by
b. Determine an exact expression which gives the isothermal work done on 1 mol of this constant-κ substance
)Pexp()T(AV
Virial EOS & Ideal Gas
Virial expansion:
Alternative form:
Compressibility factor
For ideal gas: Z = 1, thus
...P'DP'CP'B1(RTPV 32
...VD
VC
VB
1(RTPV 32
RTPV
Z
RTPV
The Ideal Gas
The internal energy of a real gas is a function of pressure and temperature.
This pressure dependency is the result of forces between the molecules.
In an ideal gas, such forces does not exist. No energy would be required to alter the average intermolecular distance, and therefore no energy would be required to bring about volume & pressure changes in an ideal gas at constant temperature. In other word, the internal energy of an ideal gas is a function of temperature only. )T(UU
Implied Property Relations for an Ideal Gas
)T(CdTdU
TU
C VV
V
RT)T(UPVUH
Thus H is also a funcion of temperature only
)T(CdTdH
TH
C PP
P
R)T(C)T(CdTdH
C vPP
Isothermal ProcessClosed System, Ideal Gas, Mechanically Reversible
0U
0H
2
1
1
2
PP
lnRTVV
lnRTdVVRT
PdVW
WQ
PV = constant
Isobaric ProcessClosed System, Ideal Gas, Mechanically Reversible
dTCU V
)TT(R)VV(PdVPPdVW 1212
dTCH P
HQ
TV-1 = constant
Isochoric (Constant V) ProcessClosed System, Ideal Gas, Mechanically Reversible
dTCU V
0PdVW
dTCH P
UQ
TP-1 = constant
Adiabatic ProcessClosed System, Ideal Gas, Mechanically Reversible
dTCU V
dV
VRT
dTCdVP0dTCdTCdUdVPdW0dQ
vv
v
dTCH P
0Q
PCR
1
2
1
2
PP
TT
vCR
2
1
1
2
VV
TT
PRT
dPdTCv
v
P
CC
2
1
1
2
VV
PP
PdVW
ttanconPV
ttanconsPT
ttanconsVT1
1
V
P
CC
Polytropic ProcessClosed System, Ideal Gas, Mechanically Reversible
Polytropic process can be considered as general form of process.
Isobaric process : δ = 0Isothermal process : δ = 1Adiabatic process : δ = γIsochoric process : δ = ∞
ttanconPV
ttanconsPT
ttanconsVT1
1
Example 3.2
Air is compressed from an initial condition of 1 bar & 25oC to a final state of 5 bar & 25oC by three different mechanically reversible processes in a closed system:
a. Heating at constant volume followed by cooling at constant pressure
b. Isothermal compressionc. Adiabatic compression followed by cooling at constant
volume.Assume air to be an ideal gas with the constant heat
capacities: CV = 5/2 R, CP = 7/2 R.
Sketch the process in a PV diagram & calculate the work required, heat transferred, and the changes in internal energy & entalphy of the air for each processes
Example 3.3
An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system:
a. From an initial state of 70oC & 1 bar, it is compressed adiabatically to 150oC
b. It is then cooled from 150oC to 70oC at constant pressure
c. Finally, it is expanded isothermally to its original state
Sketch the process in a PV diagram & calculate W, Q, U, and H for each of the three processes and for the entire cycle.
Take CV = 3/2 R and CP = 5/2R
Problem 3.8
One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 8 bar & T1 = 600 K to P2 = 1 bar by each of the following path:
(a) Constant volume(b) Contant temperature(c) AdiabaticallyAssuming mechanical reversibility, calculate
W, Q, U, and H for each of the three processes.
Sketch each path in a single PV diagram
Problem 3.9
An ideal gas initially at 600 K & 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 1-2 pressure decreases isothermally to 3 bar; in step 2-3 pressure decreases at constant volume to 2 bar; in step 3-4 volume decreases at constant pressure; and in step 4-1 the gas returns adiabatically to its initial step.
a. Sketch the cycle on a PV diagramb. Determine (where unknown) both T & P for states
1, 2, 3, and 4c. Calculate W, Q, U, and H for each step of the
cycle
Data: CV = 5/2 R, CP = 7/2 R
Problem 3.10
An ideal gas, CV = 3/2 R & CP = 5/2 R is changed from P1 = 1 bar & Vt
1 = 12 m3 to P2 = 12 bar & Vt2 = 1 m3 by the
following mechanically reversible processes:a. Isothermal compressionb. Adiabatic compression followed by cooling at constant
pressurec. Adiabatic compression followed by cooling at constant
volumed. Heating at constant volume followed by cooling at
constant pressuree. Cooling at constant pressure followed by heating at
constant volume
Calculate W, Q, Ut, and Ht for each of these processes, and sketch the paths of all processes on a single PV diagram
Example 3.4
A 400 gram mass of nitrogen at 27oC is held in a vertical cylinder by a frictionless piston. The weight of the piston makes the pressure of the nitrogen 0.35 bar higher than that of the surroundings atmosphere, which is at 1 bar & 27oC. Thus the nitrogen is initially at a pressure of 1.35 bar, and is in mechanical & thermal equilibrium with its surroundings. Consider the following sequence of process:
a. The apparatus is immersed in an ice/water bath and is allowed to come to equilibrium
b. A variable force is slowly applied to the piston so that the nitrogen is compressed reversibly at the constant temperature of 0oC until the gas volume reaches one-half the value at the end of step a. At this point the piston is held in place by latches.
Example 3.4 (cont’)
c. The apparatus is removed from ice/water bath and comes to thermal equilibrium with the surrounding atmosphere at 27oC
d. The latches are removed, and the apparatus is allowed to return to complete equilibrium with its surroundings
Sketch the entire cycle on a PV diagram, and calculate Q, W, U & H for the nitrogen for each step of the cycle. Nitrogen may be considered an ideal gas for which CV = 5/2 R and CP = 7/2 R
Cubic Equation of State:van der Waals Equation (1873)
2Va
bVRT
P
C
2C
2
PTR
6427
a
C
C
PRT
81
b
C
CC P
RT83
V
83
TRVP
ZC
CCC
Theorem of Corresponding State; Accentric Factor
All fluids, when compared at the same reduced temperature & reduced pressure, have approximately the same compressibility factor, and all deviate from ideal gas behavior to about the sam degree
Cr T
TT
Cr P
PP
Accentric Factor(Pitzer)
7.0Tsatr rPlog0.1
Critical Constants & Accentric Factors:
Olefin & Miscellaneous Organics
Tc/K Pc/bar Vc/10-6m3.mol-1 Zc
Critical Constants & Accentric Factors:
Miscellaneous Organic Compounds
Tc/K Pc/bar Vc/10-6m3.mol-1 Zc
Critical Constants & Accentric Factors:
Miscellaneous Inorganic Compounds
Tc/K Pc/bar Vc/10-6m3.mol-1 Zc
Cubic Equation of State:Redlich/Kwong Equation (1949)
)bV(V)T(a
bVRT
P
C
2C
2
rC
2C
2
r PTR
T42748.0PTR
)T()T(a 21
C
C
C
C
PRT
08664.0PRT
b
31
TRVP
ZC
CCC
Cubic Equation of State:Soave/Redlich/Kwong Equation (1972)
)bV(V)T(a
bVRT
P
C
2C
2
rC
2C
2
r PTR
),T(42748.0PTR
),T()T(a
C
C
C
C
PRT
08664.0PRT
b
31
TRVP
ZC
CCC
22/1r
2r )T1()176.0574.1480.0(1),T(
Cubic Equation of State:Peng - Robinson Equation (1976)
)bV(V)T(a
bVRT
P
C
2C
2
rC
2C
2
r PTR
),T(45724.0PTR
),T()T(a
C
C
C
C
PRT
07779.0PRT
b
30740.0TRVP
ZC
CCC
22/1r
2r )T1()26992.054226.137464.0(1),T(
Example 3.8
Given that the vapor pressure of n-butane at 350 K & 9.4573 bar, find the molar volumes of saturated vapor and saturated liquid n-butane at these conditions as given by:
a. Van der Waalsb. Redlich/Kwongc. Soave/Redlich/Kwongd. Peng-Robinson
Application of The Virial Equations
At low to moderate pressure, it is common to use truncated virial equation:
At pressure above the range of applicability of the above eqn, the appropriate form is:
Benedict/Webb/Rubin equation:
and its modifications are inspired by volume expansion virial eqn and are used in the petroleum & natural gas industries for light hydrocarbons.
RTBP
1RTPV
Z
2VC
VB
1RTPV
Z
2223632
2000
Vexp
V1
TVc
Va
VabRT
VT/CARTB
VRT
P
Example 3.7
Reported values for the virial coefficients of isopropanol vapor at 200oC are:B = -388 cm3 mol-1 C = -26000 cm6 mol-2
Calculate V and Z for isopropanol at 200oC & 10 bar by:
a. The ideal gas equationb. Two term truncated pressure expansion virial
equationc. Three term truncated volume expansion virial
equation
Pitzer Correlation for the Second Virial Coefficient
r
r
C
C
TP
RTBP
1RTBP
1Z
10
C
C BBRTBP
6.1r
0
T422.0
083.0B
2.4r
1
T172.0
139.0B
Real Gas, EOS
Calculate Z and V for steam at 250oC and 1,800 kPa by the following:
a. Ideal gas equationb. Truncated virial equation with the following
experimental values of virial coefficient:
B = -152.5 cm3 mol-1 C = -5,800 cm6
mol-2
c. Pitzer correlation for virial equationd. Pitzer-type correlation of Lee – Kesslere. Steam tablef. van der Waals equation g. Redlich/Kwong equationh. Soave/Redlich/Kwong equationi. Peng-Robinson Equation
Problem 3.45
A 30 m3 tank contains 14 m3 of liquid n-butane in equilibrium with its vapor at 25oC. Estimate the total mass of n-butane in the tank. The vapor pressure if n-butane at the given temperature is 2.43 bar.
Ideal Gas
A cylinder pressure vessel having an inside diameter of 50 cm and height of 1.25 cm. Amount of the ammonia gas is confined in that a pressure vessel. Measured pressure and temperature of gas are 30 bars and 200oC. What are the specific volume and amount of the ammonia gas in pressure vessel (in SI unit), assuming ammonia is an ideal gas?
Real Gas, EOS
Calculate Z and for steam at 250oC and 1,800 kPa by the following:
The ideal gas equationThe truncated virial equation with the following
experimental values of virial coefficient:B = -152.5 cm3 mol-1 C = -5,800
cm6mol-2The van der Waals equation The Redlich/Kwong equationThe steam tableData: critical point of steam, TC = 647.1 K; PC =
220.55 bar; = 55.9 cm3mol-1
1st law after diagram
1 kmole of ideal gas (Cv = 3R/2) undergoes a three-step mechanically reversible cycle in a closed system.
Gas in initial state at 500 kPa and 27 oC is heated at constant pressure
Followed by adiabatic expansion until the its pressure become 120 kPa
Finally step, gas is isothermally compressed to initial condition
From that process description:Illustrate the cycle process at P-V diagram in T parameter!
b. Calculate H, U, Q and W for each step of process and total!