1. Lecture 3 Entropy and result of Euler’s homogenous theorem

2. The change of entropy over a Rankine cycle
Clausius Duhem entropy equation W
Turbine or
Qin A
Boiler
Condenser
-Qout B
An open system
Feed
water pump B A
Cycle process, a closed system
Feed
water tank
An open system
A=B
A is the initial state
B is the final state
T is the boundary layer temperature between the system and the surroundings
 = entropy generation [J/molK]

3. The entropy change S(B) – S(A) of a system
The entropy change of an isolated system is always a positive one or zero.
The entropy change of an open or a closed system may be a positive one, negative one or zero.
Combined entropy change of the system and the surroundings is always a positive one or zero => we always generate entropy in real processe.
Combined change of entropy is zero if nothing happens or the process is a reversible one (not possible in real life).

4. The change of entropy, s(T,p)
Total differential of entropy
where
and
=>
=>
If phase transition takes place it is included as follows

5. Example 1
Calculate the change of entropy for O2 from the initial sate A (p = 1bar, T = 300K)
to the final state B (p = 3bar, T = 450K). O2 can be treated as an ideal gas.
v = RT/p  = 1/T
=>
Average cp ( K) = J/(molK)
=>
Volumetric thermal expansion coefficient

6. The third law of thermodynamics
By Nernst, It is impossible to reach absolute zero of temperature with finite numbers of
operations => it can never be reached.
On the basis of the Nernst theorem can be “concluded” that the entropy of a pure,
perfectly crystalline substance (element or compound) is zero at zero kelvin.
The third law allows to calculate the absolute entropies for elements and compounds.
Understanding the third law properly would require sufficient knowledge of the statistical
thermodynamics.

7. Absolute entropy
On the basis of the third law the absolute entropy can be defined for elements and compounds.
For example, for N2, CO2 and Al, entropies at 0K and 1 bar (the standard state) are:
s°[N2(s), T = 0 K] = s°[CO2(s), T = 0 K] = s°[Al(cr), T = 0 K] = 0
This definition is used when absolute entropies of elements and compounds are calculated

8. Calculation of the absolute entropy for N2 at 100K and 1bar=
= J/(molK)

9. Absolute entropy of N2 in the JANAF table

10. Tabulated entropy values for H2O(g)
The absolute entropy so or So [J/(molk)] is used.
Pressure is 1 bar (standard state)
NOTE: These values have not been taken from
the Janaf table and therefore they may slighlty differ
from Janaf table values

11. Entropy of formation Sfo
For a compound the entropy of formation at 1bar and temperature T ( the standard state) is
Sf° (T) = s°(compound) - isi° (elements, the most stable state)
where s°is the absolute entropy at temperature T and i the stoichiometric coefficient
Entropy of formation is not known for all substances. For example, contrary to the enthalpy
of formation, the entropy of formation cannot be usually defined for solid fuels
(e.g peat or wood).

12. Example 2
What is the entropy of formation for CH4 at 500K at the standard state?
C + 2H2  CH4
Absolute entropies at 500K
so(C) = J/(molK)
so(H2) J/(molK)
so (CH4) = J/(molK)
Sfo(CH4) = so(CH4) - so(C) - 2 so(H2) =  = J/(molK)

13. Entropy change in chemical reactions at the standard state
For chemical reactions at the standard state, the entropy change is
S° =  isi°(products) -  i si°(reactants)
where si is the absolute entropy of the substance and i the stoichiometric coefficient

14. Example 3
Calculate the change of entropy for the following reaction at 500 K and 1 bar: CO + 1/2O2  CO2
So (500K) = so(CO2) – so(CO) – 1/2so(O2) = –  = J/(molK)
CO2 500K, 1bar, 1 mol/s
O2 500 K, 1 bar, 0.5 mol/s
water 90oC, 2 bar, 1.66 kg/s Q
CO 500 K, 1 bar, 1 mol/s
water 45oC, 2 bar, 1.66 kg/s
The entropy change (or entropy generation rate) over the ”boiler” is
S = nCO2sCO2 + mwsw2 – nCOsCO – nO2sO2 – mwsw1 = -1 1192 – 1.66703 = 736 W/K
=> The total entopy change is a positive one as it has to be.
Entropies for CO2 , CO and O2 have been taken from Janaf tables
Entropies for water have been taken from steam tables

15. The dependence of the specific heat on the pressure
=>
=>
v = molar volume [m3/mol]
Derivate of the product
dfg = f’g + g’f

16. Heat capacity's dependence on the pressure for an ideal gas, liquid and solid
For an ideal gas v(T,p) = RT/p
=>
=> cp(T,p) = cp(T)
Spefic heat capacities of liquids and solids only slightly depnd on the pressure.
With good accuracy:
cp,solid (T,p)  cp(T) and cp,liquid(T,p)  cp(T)

17. Heat capacity's dependence on the pressure for a real gas
For real gases
especially close to the critical point
and presure may have a remarkable effect on cp
For example, for water vapor close to the critical point (221bar and 647K)
Vapor: cp ( K, 1.0 bar) = kJ/kgK
Vapor: cp ( K, 75 bar) = kJ/kgK
Vapor cp ( K, 100 bar) = kJ/kgK
Vapor cp ( K, 190 bar) = kJ/kgK

18. Chemical system Chemical system T constant temperature [K] T, p
p constant total pressure [Pa]
ni the number of moles of each species (compound or element)
in the system [mol]
xi mole fractions of each species (compound or element)
in the system [-]
x1 + x2 +… xm = 1
T, p
n1, n2,... nm
(x1, x2....xm)
Enthalpy of the system
where
- hi is molar enthalpy
- Species in Finnish is “osalaji”

19. Euler’s theorem for homogenous functions in thermodynamics
Using Euler’s homogenous theorem it is possible to show that state functions can be expressed
as follows
where
where
where
where
Homogenous theorem considers if si , hi ... in a mixture
depends on other species.

20. Volume of mixture For solutions where
Molar volumes of components in solutions may differ from molar volumes of pure components
when they exist alone. For example, in ethanol-water solution molar volumes of ethanol and
water are not the same as molar volumes of pure ethanol and water. Therefore homogenous
theorem must be used in order to calculate total volume of a solution.
For ideal gases
where pi is the partial pressure of species in the gas mixture
Molar volumes of species in an ideal gas mixture does not obey Euler’s homogenous theorem.