Summary (Elective) TUe (3DEX0) Physics of energy: sources, transport and storage Full Revision Notes
38 vues 0 fois vendu
Cours
Physics of energy (3DEX0)
Établissement
Technische Universiteit Eindhoven (TUE)
This is a complete summary for the elective course "Physics of energy: sources, transport and storage" at TUe (Eindhoven University of Technology). The course code is 3DEX0. This summary contains all the theory needed and it also includes the corresponding reports for the 4 hands-in-activity.
(3DEX0) Physics of energy:
sources, transport and
storage
Full Notes!
1
,Contents INDEX Page number
-C1. Thermodynamics 3
-C2. Thermodynamic Processes and 2nd Law of Thermodynamics 9
-C3. Carnot Theorem, Carnot Cycle and Entropy 14
-C4. Focus Topic- Photovoltaics 18
-C5. Focus Topic- Batteries 34
-C6. Focus Topic- Nuclear Fusion 44
-C7. Focus Topic- Solar Fuels 52
-C8. Focus Topic- Thermal Storage 59
-C10. Stirling Engine Report 70
-C11. Heat Capacity of Gases Report 78
-C12. Characteristic Curves of a Solar Cell Report 83
-C13. The Characteristic Curve of a PEM Electrolyzer Report 89
2
, Thermodynamics
Thermodynamics
1st Law of Thermodynamics: Energy can be transformed from one form to
another but, it can be neither created nor destroyed.
Thermodynamic System: A system is separated from the surroundings by
means of well-defined boundaries. Exchanges in the form of work and heat
take place at the boundaries between the system and the surroundings.
Relationship between Work and Energy:
𝑊𝑜𝑟𝑘 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 W=JQ , where J is a constant J=4.186kJ/kcal
Internal Energy:
𝛥𝑈 = 𝑄 − 𝑊
Where 𝛥𝑈 is the change in internal energy in J, Q is the heat energy in J and W
is the work in J.
Q is positive when it enters the system, and negative when it leaves the
system. W is positive when work is done by the system and negative when
work is done on the system.
3
,Ideal Gas Equation:
𝑃𝑉 = 𝑛𝑅𝑇
Pressure in Pa, volume in m3, number of moles of gas in mol, R constant of the
gas (8.31 J/mol K) and temperature in K.
Number of Moles:
𝑀𝑎𝑠𝑠 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠
𝑛 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 =
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠
Avogadro ‘s Constant:
1 mole of gas contains 6.022 ⋅ 1023 molecules, 𝑁𝐴 = 6.022 ⋅ 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⋅ 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑎 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 = 𝑀 = 𝑁𝐴 𝑚
4
,Reversible and Irreversible Process:
Work Done By the System:
𝑉2
𝑊 = ∫ 𝑝 𝑑𝑉
𝑉1
Where W is the work done in J, V is the volume in m3 and p is the pressure in
Pa.
When the pressure is constant:
𝑊 = 𝑝𝛥𝑉
pV-Diagrams:
When a thermodynamic system changes from an initial state to a final state, it
passes through a series of intermediate states. This series of states is called a
path.
The area under a pV-curve is equal to the work done by the system during a
volume change.
As the area is different for each path, the work done depends on the path
taken.
5
,To calculate the work done for the squared/rectangular areas, we can simply
use the base times height method. Another option is using the formula above
for work done which involves the integral.
However, for the last area we can only to use the formula which involves the
integral. This is done like this:
𝑉2 𝑉2 𝑉2
𝑑𝑉 𝑑𝑉 𝑉2
𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑛𝑅𝑇 = 𝑛𝑅𝑇 ∫ = 𝑛𝑅𝑇 ln
𝑉 𝑉 𝑉1
𝑉1 𝑉1 𝑉1
This is done by substituting pressure with the ideal gas equation.
Relationship between Temperature and Internal Energy:
The internal energy depends only on the gas temperature. Hence, if a system
undergoes a process where its temperature doesn’t change, even though there
is a change in pressure and volume, the internal energy won’t change.
Meaning that 𝛥𝑈 = 𝑈2 − 𝑈1 = 0. All isolated processes will then satisfy this
statement. For cyclic processes this condition is also satisfied since you end up
where you started, as you will have the same temperature at the beginning
and at the end. Hence, for cyclic processes, considering all the steps, 𝛥𝑈 = 0.
6
,Specific Heat Capacity of a Substance:
This is for solids, liquids and gases.
𝑄 = 𝑛𝑐𝛥𝑇
Where Q is the heat energy in J, n is the number of moles in mol, 𝛥𝑇 is the
change in temperature and c is the specific heat capacity of a substance in
J/mol K.
Relation between Cp and Cv for Ideal Gases:
𝐶𝑝 = 𝐶𝑣 + 𝑅
Where 𝐶𝑝 is the specific heat capacity at a constant pressure, 𝐶𝑣 is the specific
heat capacity at a constant volume and R is the gas constant.
Equipartition of Energy Theorem:
1
𝐾𝑎𝑣𝑔 = (𝐷𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚) ⋅ 𝑘𝐵 𝑇
2
Where 𝐸𝑘−𝑎𝑣𝑔 is the average kinetic energy, T is the temperature in kelvin and
𝑘𝐵 is the Boltzmann constant which is 1.38 ⋅ 10−23 J/K.
Monoatomic gases have 3 degrees of freedom corresponding to the three
orthogonal directions x, y, z along which each atom can translate. Hence, for
3
monoatomic gases, 𝐾𝑎𝑣𝑔 = 𝑘𝐵 𝑇.
2
Diatomic gases have 3 degrees of freedom corresponding to the three
orthogonal directions along which the molecule can translate, plus 2 rotational
5
degrees of freedom. Hence, for diatomic gases, 𝐾𝑎𝑣𝑔 = 𝑘𝐵 𝑇.
2
Cv and Cp of Monoatomic and Diatomic Gases:
After some rearrangement of the equations above, we can obtain values of Cv
and Cp for monoatomic and diatomic gases.
For Monoatomic Gases:
3
𝐶𝑣 = 𝑅 = 12.5 𝐽/𝑚𝑜𝑙 𝐾
2
𝐶𝑝 = 𝐶𝑣 + 𝑅 = 20,8 𝐽/𝑚𝑜𝑙 𝐾
, Thermodynamic Processes and 2nd Law of Thermodynamics
Thermodynamic Processes:
Adiabatic Process:
An adiabatic process is defined as one with no heat transfer into or out of a
system. We can prevent heat flow either by surrounding the system with
thermally insulating material or by carrying out the process so quickly that
there is not enough time for appreciable heat flow.
Therefore, for an adiabatic process 𝑄 = 0, this means that,
𝛥𝑈 = 𝑈2 − 𝑈1 = −𝑊
When a system expands adiabatically, W is positive as the work is done by the
system on the surroundings. So the change in internal energy is negative
meaning that the internal energy decreases. However, when a system is
compressed adiabatically, W is negative as the work is done on the system by
the surroundings. So the change in internal energy is positive, meaning that
the internal energy increases. Usually in many systems an increase of internal
energy is accompanied by an increase in temperature and vice versa.
Isochoric Process:
An isochoric process is a process that occurs at a constant volume. When the
volume of a thermodynamic system is constant, it does no work on its
surroundings.
Therefore, for an isochoric process, 𝑊 = 0, this means that,
𝛥𝑈 = 𝑈2 − 𝑈1 = 𝑄
In an isochoric process, all the energy added as heat remains in the system as
an increase in internal energy. Heating a gas in a closed constant-volume
container is an example of an isochoric process.
For an isochoric process, 𝑄 is calculated using the molar heat capacity formula,
𝑄 = 𝑛𝐶𝑣 𝛥𝑇
9
, Isobaric Process:
An isobaric process is a process that occurs at a constant pressure. We know
𝑉
that 𝑊 = ∫𝑉 2 𝑝 𝑑𝑉 , and that at a constant pressure 𝑊 = 𝑝𝛥𝑉. Therefore, for
1
an isobaric process, 𝛥𝑈 = 𝑈2 − 𝑈1 = 𝑄 − 𝑊 = 𝑄 − 𝑝(𝑉2 − 𝑉1 ).
For an isobaric process, 𝑄 is calculated using the specific heat capacity formula,
𝑄 = 𝑛𝐶𝑝 𝛥𝑇
Boiling water at a constant temperature is an example of an isobaric process.
Isothermal Process:
An isothermal process is a process that occurs at a constant temperature. For a
process to be isothermal, any heat flow in or out of the system must occur
slowly enough that thermal equilibrium is maintained. For processes in which
the system is an ideal gas, which is the ones we usually treat, if the
temperature is constant then the internal energy is also constant.
Therefore, for an isothermal process, 𝛥𝑈 = 0, this means that,
𝑊=𝑄
Any energy entering a system as heat must leave as work done by the system.
The path followed in an adiabatic process (a to 1) is called an adiabat. A vertical
line (constant volume) is an isochor, a horizontal line (constant pressure) is an
isobar, and a curve of constant temperature (shown as light blue lines) is an
isotherm.
10
Les avantages d'acheter des résumés chez Stuvia:
Qualité garantie par les avis des clients
Les clients de Stuvia ont évalués plus de 700 000 résumés. C'est comme ça que vous savez que vous achetez les meilleurs documents.
L’achat facile et rapide
Vous pouvez payer rapidement avec iDeal, carte de crédit ou Stuvia-crédit pour les résumés. Il n'y a pas d'adhésion nécessaire.
Focus sur l’essentiel
Vos camarades écrivent eux-mêmes les notes d’étude, c’est pourquoi les documents sont toujours fiables et à jour. Cela garantit que vous arrivez rapidement au coeur du matériel.
Foire aux questions
Qu'est-ce que j'obtiens en achetant ce document ?
Vous obtenez un PDF, disponible immédiatement après votre achat. Le document acheté est accessible à tout moment, n'importe où et indéfiniment via votre profil.
Garantie de remboursement : comment ça marche ?
Notre garantie de satisfaction garantit que vous trouverez toujours un document d'étude qui vous convient. Vous remplissez un formulaire et notre équipe du service client s'occupe du reste.
Auprès de qui est-ce que j'achète ce résumé ?
Stuvia est une place de marché. Alors, vous n'achetez donc pas ce document chez nous, mais auprès du vendeur alextrevina500. Stuvia facilite les paiements au vendeur.
Est-ce que j'aurai un abonnement?
Non, vous n'achetez ce résumé que pour 10,00 €. Vous n'êtes lié à rien après votre achat.
