Chapter 6 - Thermal Energy

Question 1

How was the Earth created?

Answer 1

From differentiating accretion of planetary dust.

Question 2

How are elements with different densities dispersed in the Earth?

Answer 2

High density elements predominant in earth center, minerals with lower density predominant in outer crust.

Question 3

Which six shells can we separate the Earth into? What are their phases?

Answer 3

Inner core (solid), outer core (liquid), lower mantle (liquid), transition zone (viscous), upper mantle (viscous) and crust (solid).

Question 4

How come the inner core is solid, even though the temperature is higher than on the surface of the sun?

Answer 4

Because of the high pressure.

Question 5

What is superrotation?

Answer 5

The fact that the inner core of the Earth rotates quicker (about 0.3-0.5 degrees / year) faster than the mantle. This adds up to one additional turn in about 1000 years.

Question 6

Where is the earth magnetic field generated?

Answer 6

In the liquid outer core.

Question 7

How do we divide the Earth crust?

Answer 7

Into oceanic crust (5-10 km) and continental crust (30-60 km).

Question 8

Where does the crust “swim”?

Answer 8

On top of the asthenosphere, which is the upper part of the upper mantle.

Question 9

How much oxygen does the crust contain?

Answer 9

About 50 at%.

Question 10

How far have we been able to drill down into the crust?

Answer 10

12 km, in Russia.

Question 11

What is the average energy flux through the Earth surface from below?

Answer 11

60 mW/m^2 (compare to 1000 W / m^2 incoming solar irradiation).

Question 12

What are the sources of the earth heat flux?

Answer 12

1) 50% from continuous cooling of inner earth due to heat conduction.
2) 50% from decay of radioactive elements.

Question 13

What is the average temperature of the inner earth?

Answer 13

2000K. This was the kinetic and potential energy transformed into heat during the accretion of the earth.

Question 14

What is the thermal conductance of the crust?

Answer 14

lambda ≈ 2 W / Km

Question 15

What is the thermal gradient through the crust (geothermal gradient)?`

Answer 15

dQ/dt = 60 mW / m^2 (heat flux), lambda = 2 W / K m (heat conductance)

=> ∆T / ∆x = dQ/dt * 1/lambda = 0.03 K /m = 3K / 100 m.

Question 16

Where can we find strong deviations from the average heat flux of the earth?

Answer 16

At volcanic anomalies, where liquid magma reaches very close to the earth surface. Here dQ/dt can reach above 300 mW / m^2.

Question 17

Which radioactive isotopes are the most important for the heat flux of the Earth? What kind of radiation are we talking about?

Answer 17

Alpha decay: 235U, 238U, 232Th

Beta decay: 40K.

Question 18

What is the total heat generated by radioactive decays in the Earth?

Answer 18

The heat flux is about 3 µW / m^3, which amounts to about 30 mW / m^2. Total all over the world is then about 16TW.

Question 19

What is the total stored thermal energy in the Earth?

Answer 19

99% of the Earth is hotter than 1000C. THis means that the total stored thermal energy is ≈ 3*10^15 TWh. This could supply the human consumption for 10^10 years (more than the Earth will exist).

Question 20

What are the three main forms of geothermal power?

Answer 20

• Geothermal anomalies (high enthalpy sites)
• Hot-dry-rock process, aquifers
• Heat pumps

Question 21

What are high enthalpy sites?

Answer 21

These are sites that exist close to volcanoes or special “hot spots”. They consist of water-vapor mixtures (so-called “two-phase” mixture) with temperatures between 100-300 C. Pressures ranges between 10 and 100 bar, which makes the boiling point of water 300 C at the highest.

Even extinct volcanoes there can exist hot magma pipes for millions of years.

These sites can be used directly (for heating) or in combination with turbines to produce electricity.

Question 22

Name one problem with high enthalpy sites.

Answer 22

The water is contaminated with sulfur, which gives unwanted corrosion and smell. For this we have to use closed heat exchangers.

Question 23

What are low enthalpy sites?

Answer 23

These are sites characterized by temperatures below 100C. Requires use of aquifers, or can use the hot-dry-rock method which employs a region of fractured rock and water injection.

Question 24

What is the Law of Darcy?

Answer 24

It describes the transport (perlocation) of water in porous media.

v = sigma_w * H/L

sigma_w is the hydraulic conductance, H is the height of water column above the flow region and L is the length of flow region.

Question 25

How can one calculate the energy content of a cavern used for heat extraction with the HDR-method?

Answer 25

Q = ∆T * m * C_v,

where m is the mass of the rocks, ∆T is the temperature difference between hot reservoir and temperature after cooling, and C_v is the heat capacity of the rock.

Question 26

If we make a HDR heat plant, and extract all the heat, how long before it is reheated?

Answer 26

It will take along time due to the low thermal conductance.

For the example with the granite cavern of 100 x 100 x 100 m, with a surface area of 6*10^4 m^2 and a temperature gradient of 1 K / m it would take 100 years.

Question 27

What are heat pumps? Draw a schematic.

Answer 27

They are reversely operated heat engines. See schematic on page 5.

Question 28

What is the efficiency of a heat pump?

Answer 28

It is Qw/W. Since we extract more heat than we use to compress the gas, the efficiency is always above 1. (if not, it would not make sense to do this).

Question 29

What is the working principle of a heat pump? Draw a schematic.

Answer 29

1) Evaporation of working medium. Heat of evaporation Qc is taken from the environment.
2) Compression by mechanical work, W, close to adiabatic.
3) Liquiefaction of working medium by giving the heat of condensation Qw to environment.
4) Decrease of pressure in a throttle.

See schematic on page 6.

Question 30

What are typical efficiencies of heat pumps, and what does it depend on?

Answer 30

eta = 2-6.

Depends on the temperature spread Tw-Tc. This leads to a divergence between avilable heating power and the necessary heating power (works better when we don’t need it).

Question 31

In a heat pump, what can be the cold reservoir Tc?

Answer 31

The surrounding air, ground water or soil (typical power extraction 10-40 W/m^2).

Question 32

What are typical working medium used in heat pumps?

Answer 32

Propane, NH3, CO2. CFC gases used before, but illegal now.

Question 33

How are photons absorbed? Draw a schematic.

Answer 33

Photons excite electrons through Coulombic forces. The excited electron than undergoes an electron-phonon coupling where a phonon mode is excited. See page 7 for schematic.

Question 34

What are the principal processes occuring for incoming photons?

Answer 34

Reflection: coherent process at a defined interface described by Fresnel equations.

Scattering: incoherent process by photon interaction with atoms or molecules (Rayleigh-scattering, Mie-scattering).

Absorption: incoherent process via transformation of photons into excited electrons and/or phonons.

Transmission: part of incoming flux is neither reflected, scattered or absorbed.

Question 35

Which energy conservation equation holds for incoming photons on a material?

Answer 35

R + S + A + T = 1.

Question 36

How is the Lamber-Beer law defined? What does it tell us?

Answer 36

PHI(x) = (1-R)PHI_0 exp[-alhpa x]

with alpha = alpha[E_photon] being the absorption coefficient. PHI(X) is the photon flux at depth x in an absorber, PHI_0 is the incoming photon flux.

It tells us how much of the photon flux still remains after having travelled a distance x in the absorber.

Question 37

Which of the photon interaction processes are important for solar thermal energy?

Answer 37

All of them.

Absorption: ideal absorbers have A = 1 for all wavelength., selective absorbers have A = 1 for wavelengths under a critical wavelength, and A = 0 above.

Reflection: mirrors have R = 1, A= 0 and are used for concentration of solar radiation. R = 0 for perfect antireflex coating, but here it is decoupled from A.

Scattering: at the surface or in materials and environment.

Question 38

What is a problem regarding the emissivity of a black body with a finite temperature, when speaking about solar thermal energy?

Answer 38

The fact that it always emits radiation, even without incoming radiation. The hotter it gets, the more it radiates.

Question 39

How does the amount of scattered light change when we change the angle of incidence?

Answer 39

The amount of Rayleigh and Mie scattering that occurs when we change from normal incidence to 5% increases a lot. This is diffuse radiation, and cannot be concentrated.

Question 40

What things does the average irradiation depend upon?

Answer 40

Time of year and latitude.

Question 41

What type of trackers can we install?

Answer 41

1-axis trackers, that tracks the sun over the course of the day.
2-axis trackers, that also tracks the sun depending on time of year.

Question 42

What are the main applications of solar thermal energy?

Answer 42

i) Architecture (winter garden, low energy house)
ii) Non-concentrating solar collectors.
iii) Concetrating solar recievers

Question 43

How are non-concentrating solar collectors best realized?

Answer 43

By combining glass windows (selective transmission) with selective absorbers (prevents emission losses since emission is low, which it is because absorption is very low in the IR-range).

Question 44

Why are selective absorbers good?

Answer 44

Because they absorb in the frequency range of the incoming photons, but they do not absorb in the frequency range of IR. This is important since, through Kirchhoff’s law of radiation, the emissivity is large when the absorption is large. The typical range the absorber would emit would be in the IR-range, and since the absorption is low here, the emissivity is also low.

Question 45

Give examples of some frequently used selective absorbers. What are their typical absorption data?

Answer 45

Black chromium, black nickel and TiNOx.

Typical values:
for lambda 3µm: A = 5-10%, T = 0, R = 90-95%.

Question 46

What are concentrating solar recievers?

Answer 46

They use concentrated sunlight, concentrated by linear or parabolic mirrors, to heat a working media (steam, Stirling engines, phase change salts) to high temperatures of 400-1000 C.

Question 47

What is the maximum concentration for point focussing?

Answer 47

This is given by the opening angle of sun as seen from Earth.

alpha = arcsin R_sun / R_SE = 16’ = 0.27 deg = 0.0046 rad

Maximum concentration is given by

C_max = 1/alpha^2 = 46211.

For one-axis linear concentration, this is given as:

C_max, lin = sqrt(C_max) = 215

Question 48

What does the maximum concentration of point focussing determine?

Answer 48

It determines the maximum reachable absorber temperature, assuming ideal thermal isolation and e = 1.

C * SC = sigma T^4

T = (C*SC/sigma)^1/4 => T = 5780 K (same as surface of sun)