exergy and fundamental thermodynamics - r1.pdf · direct conversion of chemical energy into...
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Exergy and Fundamental Thermodynamics
Electrochemical Cells
Direct conversion of chemical energy into electricity
In principle, the electrical energy out equals the exergy available – it’s reversible
Not all reactions can be implemented as an electrochemical cell
Taking current from the cell causes irreversible effects to appear (exergy loss)
Some systems are inherently problematic – especially gas electrodes and specifically oxygen electrodes
Temperature and Entropy
Idea of temperature is fundamental to our experience
What about a formal definition?
Associated with the flow of heat
Easy to put temperatures in order
How about a scale?
How do we get a fundamental scale?
Heat Engine
Takes in heat at higher temperature
Rejects heat at lower temperature
Produces work
Assume we have one that operates reversibly
Reversible Heat Engine
T1
T2
q1w
q2
q1 = q2 + wHow about w/q1 ?
Basic Heat Power Cycle
Heat in
Heat out
Power out
Power inPump Motor
Pressure high
Pressure low
Reversible Heat Engines
T1
T2
1
q1w
q2
2
q3
q4
Reversible Heat Engines
All reversible heat engines operating between the same pair of reservoir temperatures must have the same efficiency.
What is that efficiency?
Reversible Heat Engines
T1
T2
1
q1w1
q2
T32q3
q2
w2
3
q1w3
q3
Engine Efficiencies
Must be a function of the temperatures
Say: q2 /q1 = F(T2 , T1 ) Engine 1
So: q3 /q2 = F(T3 , T2 ) Engine 2
and: q3 /q1 = F(T3 , T1 ) Engine 3
But (q3 /q2 ) x (q2 /q1 ) = (q3 /q1 )
So: F(T3 , T2 ) x F(T2 , T1 ) = F(T3 , T1 )
F(T2 , T1 ) = f(T2 ) / f(T1 )
Thermodynamic Temperature Scale
A scale such that for a reversible heat engine:
heat in/heat out = Tin /Tout
Implies the existence of a thermodynamic zero temperature
To define degree size - water freezing point 273.15K
William Thomson
(1824 – 1907)
Entropy
The entropy of heat q at temperature T is defined as: q/T
For a reversible heat engine, qin /Tin = qout /Tout
The entropy of the upper reservoir decreases, and that of the lower increases
No change in overall entropy
We know that for real processes, there is a tendency to turn work into heat – irreversible processes increase entropy
Third Law
The entropy of pure crystalline substances at absolute zero is zero
Absolute zero is unattainable in a finite number of operations
Cooling
How do we achieve cooling in the absence of a reservoir cooler than the temperature we want?
Refrigerator – evaporate a fluid by reducing pressure
R22 refrigerant: -20C no problem
Nitrogen: boiling point (1 atm) 77K, can get lower temperatures by reducing the pressure
Helium: boiling point (1 atm) 4.22K, can get to about 0.8K by reducing the pressure
Super-Low Temperatures
Below 0.8K or so, evaporative cooling is not practical
Magnetic cooling is used
Paramagnetic material can be placed in strong magnetic field, cooled and the magnetic field then removed
The magnetic moments of the material were lined up with the magnetic field – they then become random
This is in some ways analogous to evaporation – it requires energy
Cooling
Temperature
Entropy
1
2
Exergy Analysis – the details
Construct flow diagram of process
Separate all parts for which you want to know the exergy loss
Power Plant Exergy Flows and Destruction
27Fuel 92
Stack 2
Steam 43
7Shaft Power 32
2Steam 3
Other Losses 1
Cooling Water 1
Turbine
Condenser
Combustion
2065
HeatTransfer
Exergy Analysis – one element
Mysteryprocess
S1 , H1
S2 , H2
S3 , H3
S4 , H4
S5 , H5
Enthalpy Estimation
Water – liquid or steam
Use steam tables (available and accurate)
Gases – almost always in mixtures
Enthalpies are additive for ideal gases
Enthalpies are independent of pressure for ideal gases
These are not bad assumptions for most real gases if the pressure is 1 atm or below, temperature 20C and up.
If you want really accurate numbers, sophisticated methods (software) will be necessary
Otherwise, one can use relatively simple calculation methods, details of which are available on the internet
Example – Combustion of Methane
Assume stoichiometric combustion
Assume air is 80% N2 and 20% O2
Combustion reaction is:
CH4 + 2O2 →
CO2 + 2H2 O
So for each mole of methane burned, 8 moles of nitrogen and 2 moles of oxygen (mixed as air) are needed
The product gas will contain 8 moles of nitrogen, 1 of carbon dioxide and 2 of water vapour
Example – Combustion of Methane
To get reactant enthalpies
Look up enthalpies of components at inlet temperature
Add in correct proportions
To get product enthalpies
We need to know the temperature
If we don’t know it, it’s the one at which the total product enthalpy equals the input material enthalpy
(Make sure the enthalpies include heats of formation)
This is an iterative process, as the enthalpy equations are non-linear
Entropy of Mixing
For ideal gases, it has nothing to do with mixing
It’s all about partial pressures of components
Entropy of Mixing
1 atm N2
1 atm O2
1 atm O2
1 atm N2
Example – Combustion of Methane Entropy
Once we have the product temperature, we can get the entropy
Entropies calculated from the usual equations are for 1 atm pressure
If the pressure of a component is not 1 atm, then the entropy will have to be adjusted by including another term in the molar entropy
-R*ln(P) where P is in atmospheres and R is the gas constant [8.314 J/(mol.K)]
Entropy Calculation
In this case, the product gas contains 1 part in 11 of CO2 , so the calculated molar entropy for CO2 must be adjusted by –R*ln(1/11) [19.9 J/(mol.K) increase]
Likewise, the product gas contains 8 parts of nitrogen and 2 parts of water vapour, so their calculated molar entropies must be adjusted by –R*ln(8/11) and –R*ln(2/11) respectively
The same must be done for the oxygen and nitrogen in the combustion air
Entropy Change Calculation
We can now calculate the entropy change caused by combustion
Subtract the total input gas entropy from the after reaction entropy
The result, multiplied by the reference temperature, gives the exergy destruction in that process
Summary
Calculate entropies of flows exiting and sum them
Calculate entropies of flows entering and sum them
Subtract to find the increase in entropy
Multiply by the reference temperature
The parts of the system should then be assembled to make sure that all the exergy numbers are consistent
At every node: Out = In – Destruction
Final Comments
Exergy analysis provides a method to show where the Second Law losses occur in systems
It provides a method to assess the value of different types of energy on a consistent basis
The choice of a reference temperature matters – if you choose absolute zero then there is never any loss! It should be consistent with the purpose of the analysis you are doing
