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THERMODYNAMICS 1
The First Law and Other Basic Concepts (part 2)
Department of Chemical Engineering, Semarang State University
Dhoni Hartanto S.T., M.T., M.Sc.
Equilibrium
Have you ever cooked?
Equilibrium (cont.)
Equilibrium is a word denoting a static condition, the absence of change
In thermodynamics, it means not only the absence of change but also the
absence of any tendency toward on macroscopic scale
Other definition, equilibrium is static condition in which no changes occur
in the macroscopic properties of a system with time.
But, in microscopic properties, the condition is not static
Equilibrium condition  all forces in exact balance
kondensor
heat
Equilibrium (cont.)
A
B
Vapor
phase
Liquid
phase
t = 0 minute
(in a certain “A” composition mixtures,
P, and T)
Liquid
phase
t = C minutes
(in a certain “B” composition mixtures,
P, and T)
Condition :
Macroscopic no changes, static
Microscopic changes, not static
Phase Rule
When two intensive thermodynamic properties are set at definite values,
the state of a pure homogeneous fluid is fixed.
In contrast, when two phases are in equilibrium, the state of the system is
fixed when only a single property is specified
101.325
kPa
373.15 K
A mixture of steam + liquid water in
equilibrium
Changing temperature
will also change the
pressure if vapor-liquid
are in equilibrium
Phase Rule (cont.)
Let’s check it out in HYSYS
Go to fluid
packages
Choose H2O as selected component
Choose material stream
Choose Peng-Robinson as property package
Go to simulation environment
Phase Rule (cont.)
Let’s check it out in HYSYS
Set pressure to 101.325 kPa
Set temperature to 100 oC
Take basis molar flow 1 kgmole/h
Go to composition, then fill H2O mole fraction with 1, then OK
Phase Rule (cont.)
Let’s check it out in HYSYS
With set the temperature and pressure, HYSYS automatically change the vapor phase
to be 1, it means the water in vapor phase
Phase Rule (cont.)
Cook water until boiling point in mountain
When the water boil
Pressure less than 1
atm (101.32 kPa)
Temperature also
less than 100 oC
What happen when you boil water in high area such as mountain which the pressure is
less than 1 atm (101.32 kPa)
Phase Rule (cont.)
Cook water until boiling point in deep blue sea
Tugas HYSYS :
Cek campuran
etanol (1)+ air (2)
T (K)
konsentrasi mol
fraction 0-1 dengan
incremen 0.1 (0,0.1,
x1
0,2 dst.)
Vapor fraction = 1
Pada P = 1 atm
Bagaimana
When the water boil perubahan
temperatur,
Pressure more than Buat grafiknya
1 atm (101.32 kPa) (mole fraction
etanol vs
Temperature also
temperature)
increase to more
than 100 oC
What happen when you boil water in high pressure area with pressure more than 1
atm (101.32 kPa)
Phase Rule (cont.)
Degree of freedom of the system
For any system at equilibrium, the number of independent variables that must be
arbitrarily to establish its intensive state is given by the phase-rule
The phase-rule is intensive property
F  2   N
N : number of chemical species
F : degree of freedom
phi : phase
Example :
Various phase can coexist , it must be in equilibrium
Three-phase system at equilibrium is a saturated aqueous solution at its boiling
point with excess salt crystals present.
 = 3 (three phase) are crystalline salt, the saturated aqueous solution
N = 2 (two chemical species) are water and salt
So, degree of freedom
F  2 3 21
Phase Rule (cont.)
Degree of freedom of the system
The intensive state of a system at equilibrium is established when its
temperature, pressure, and the composition of all phase are fixed
The phase rule gives the number of variables from this set which
must be arbitrarily specified to fix all remaining phase-rule variables
The minimum degree of freeedom for any system is zero
When F = 0
The system is invariant
Equation becomes   2  N
Value of phi is the maximum number of phase which can coexist at equilibrium
for a system containing N chemical species
Phase Rule (cont.)
Degree of freedom of the system
For example :
The triple point of water where liquid, vapor, and the common from ice exist
together in equilibrium at 273.16 K (0.01oC) and 0.0061 bar
Any change from these condition causes at least one phase to dissapear
  2 N
  2 1
liquid
 3 phase
Water at
0.01oC
0.0061 bar
ice
vapor
Phase Rule (cont.)
How many degrees of freedom has each of the following system :
a) Liquid water in equilibrium with its vapor
b) Liquid water in equilibrium with a mixture of water vapor and nitrogen
c) A liquid solution of alcohol in water in equilibrium with its vapor
Answer :
a) F  2    N  2  2  1  1
In fact, temperature or pressure but not both may be specified for a system
of water in equilibrium with its vapor
b) F  2    N  2  2  2  2
The addition of an inert gas to a system of water in equilibrium with its
vapor changes the characteristic of the system. Temperature and pressure
maybe independtly varied
c) F  2    N  2  2  2  2
The phase-rule variables are temperature, pressure, and the phase
composition
Fixing the mole fraction of water in liquid phase automatically fixes the
mole fraction of the alcohol
The Reversible Process
A process is reversible when its direction can be reversed at any point
by an infinitesimal change in external conditions.
The Reversible Process (cont.)
When heated, CaCO3 decompossed forms CaO and CO2
When weight is increased, CO2 pressure is increased and CO2 combines with
CaO to form CaCO3 allowing the weight to fall slowly
The Reversible Process (cont.)
Summary :
A reversible process has the following condition :
1. Is frictionless
2. Is never more than differentially removed from equilibrium
3. Traverses a succession of equilibrium states
4. Is driven by forces whose imbalance is differential in magnitude
5. Can be reversed at any point by a differential change in external conditions
6. When reversed, retraces its forward path, and restores the initial state of
system and surroundings
The Reversible Process (cont.)
Mechanically reversible
V2t
W    t P dV t
V1
Example :
A horizontal piston/cylinder arrangement is placed in a constant-temperature
bath. The piston slides in the cylinder with negligible friction, and an external
force holds it in place against an initial gas pressure of 14 bar. The initial gas
volume is 0.003 m3 . The external force on the piston is reduced gradually and
the gas expands isothermally as its volume doubles. If the volume of the gas is
related to its pressure so that the product PVt is constant, what is the work
done by the gas in moving the external force?
How much work would be done if the external force were suddenly reduced to
half its initial value instead of being gradually reduced?
The Reversible Process (cont.)
Solution :
The process is mechanically reversible
If PVt = k , then P=k/Vt
V2t
V2t
W    t P dV  k 
t
V1t
V1
V1t  0.03 m3
dV t
V2t
 k ln t
t
V
V1
;V2t  0.06 m3
k  PV t  P1V1  (14 . 105 )(0.03)  42000 J
t
W  42000 ln 2  29112 J
Final pressure :
k 42000
P2  t 
 700000 Pa  7 bar
V2
0.06
The Reversible Process (cont.)
Solution :
In the second case, a half of the initial force has been removed
The gas under goes a sudden expansion against a constant force equivalent to
pressure of 7 bar. Thus V t
is the same as before and the net work
accomplished equals the equivalent external pressure times the volume change.
W  (7 .105 )(0.06  0.03)  21000 J
This second case is irreversible, and compared with reversible one the
efficiency is
efficiency 
21000
 0.721 or 72.1%
29112
Constant V and Constant P Process
Energy balance for a homogeneous closed system of n moles :
d (nU )  dQ  dW
Work in mechanically reversible :
dW   Pd (nV )
Combine this two equation, yield :
d (nU )  dQ  Pd (nV )
General first-law equation for
mechanically reversible and closed
system
Constant V (volume) Process
In constant total volume process, the work is 0, thus the equation will be :
dQ  d (nU )
Q  n U
Thus for a mechanically reversible, constant-volume, closed-system process,
the heat transferred is equal to the internal-energy change of the system.
Constant P (Pressure) Process
Arrange the equation below to solve dQ :
d (nU )  dQ  Pd (nV )
yield,
dQ  d (nU )  Pd (nV )
For constant pressure
dQ  d (nU )  Pd (nV )  d[n(U  PV )]
Where U + PV is the definition of enthalpy
H  U  PV
The equation become
dQ  d (nH )
Q  n H
Thus for a mechanically reversible, constant-pressure, closed-system process,
the heat transferred is equal to the enthalpy change of the system.
Enthalpy
Unit of enthalpy (H) : energy per mole or unit mass
Enthalpy is state function due to U, V, and P are state function
Enthalpi is intensive property
dH  dU  d ( PV )
H  U  ( PV )
These equation apply to a unit mass or a mole of substance
Heat Capacity
Heat has relation with its effect on the object
This is the origin of the idea that a body has capacity
The smaller the temperature change in a body caused by the transfer of a given
quantity of heat, the greater its capacity
C
dQ
dT
In fact, there are 2 kind of heat capacities are in common use for homogeneous
fluids. Both are state function
There are :
1. Heat capacity at constant volume (Cv)
2. Heat capacity at constant pressure (Cp)
Heat Capacity at constant volume (Cv)
 Q 
CV  


T

V
dU  CV dT
T2
U   CV dT
T1
For mechanically reversible at constant volume process
T2
Q  n U  n  CV dT
T1
Heat Capacity at constant pressure (Cp)
 H 
CP  


T

P
dH  C P dT
T2
H   C P dT
T1
For mechanically reversible at constant pressure process :
T2
Q  n H  n  CP dT
T1
Open System Energy Balance
W
Q
m
system
First Law:
E(system) + E(surrounding) = 0
Per unit mass containing energy:
1 2
U  u  zg
2
Total energy carried out:
1 2


m U  u  zg 
2


Open System Energy Balance
Energy in the system can change due to accumulation or loss :
d mU 
dt
Thus:
d mU 
1 2
1 2



  


  m j U j  u j  z j g    mi U i  ui  zi g   Q  W
dt
2
2

 i


j
(input)
(output)
Open System Energy Balance
Work: caused by fluid pushing in and out or piston (Wf)
and shaft work (Ws)
W  W  W
f
s
W f   PjV j m j   PiVi m i
j
i
input
output
d mU 
1
1




  m j U j  u 2j  z j g    m i U i  ui2  zi g   Q   PjV j m j   PiVi m i  W s
dt
2
2

 i


j
j
i
Remember:
H  U  PV
Open System Energy Balance
d mU 
1
1




  m i  H i  ui2  zi g    m j  H j  u 2j  z j g   Q  W s
dt
2
2

 j


i
input
output
In general:
d mU 
0
•Steady state:
dt
•one inlet and outlet stream: m i  m j  m
1 2


m  H  u  gz   Q  W s
2


1
H  u 2  gz  Q  Ws
2
Rate of energy
Rate of energy per
unit mass or mole
Thank you