The saturation domes for R-134a and isopentane shown in Figures 13.4
are quite asymmetric, with steep (R-134a) or even re curving (isopentane) saturation curves. Water, on the other hand, has a relatively symmetric saturation dome (Figure 12.9)
Explain why this feature makes R-134a and isopentane more suitable fluids for low-temperature heat extraction devices and engines than water. Why is water so widely used for higher-temperature systems such as steam turbines? What kind of shape for the saturation dome would be better for a phase-change fluid in a higher-temperature system?
Consider a parabolic trough where the height of the reflector is identical to the height of the center of the absorber, as depicted in the figure
The concentrator width is 8 m and the absorbing tube of radius 0.5 m is centered along the focal line at a height of 2 m above the bottom of the trough. Incident sunlight hits the concentrator from a direction parallel to the line of symmetry, with an intensity of 1000 W/m2.
(a) What is the effective concentration of the concentrator?
(b) What is the acceptance angle within which all light hits the absorber?
(c) Assume that the tube carries a fluid that is heated and removes some of the incoming energy. The tube then radiates as a black body at a temperature of 150 ◦C. Compute the net rate at which energy is collected and transferred to the fluid, for each meter of length of the trough.
The thickness of Earth’s ozone layer at any location is measured in Dobson Units (DU), where 1 DU corresponds to a gas layer of thickness 10 μm if the gas pressure were raised to 1 atm. Show that 1 DU corresponds to 2.69 × 1016 molecules/cm2. Data on the absorption cross section for ozone are given in Figure 23.13
Normally the thickness of the ozone layer is ≈ 300 DU, but due to ozone destruction by chlorofluorocarbons it had dropped as low as ≈ 90 DU over Antarctica during the 1990 s. How low would the ozone level have to drop to allow 10−6 of the incident UV flux at λ ≈ 250 nm or 220 nm to reach sea level? (You will need to use the result of Problem 23.5.)
A flat panel solar collector is to be tilted at an angle θ to the horizontal to maximize the amount of sunlight it can collect over the whole year. Once tilted, it is oriented toward the south (in the Northern Hemisphere). One might think that θ should be chosen to equal the latitude λ of the installation so that the Sun would be directly above the panel at noon on the equinoxes. Instead, if total insolation were the only concern, θ would be chosen to be less than λ. Explain why. Describe (but do not attempt to compute) how you would compute the optimal angle for a given location to maximize annual collection. Explain why, despite these considerations, in many practical situations θ may be chosen to be equal to or greater than λ.
Consider a linear 2D compound parabolic concentrator built from parabolas tilted at 10◦ to the vertical, with a trough of width 3 m, and an absorber width of 0.5 m. Compute the concentration C of the concentrator. If the incoming radiation has intensity I0 = 1000 W/m2 and the (blackbody) absorber is kept at a temperature of 100 ◦C by circulation of a thermal fluid, compute the rate of energy transfer to the fluid. Compare this rate of energy transfer to that for a linear parabolic concentrator with the same absorber area and acceptance angle; assume that the parabolic concentrator has height equal to the focal length (as in Example 24.4), and that the absorber has an area covering the lower half of a cylinder centered on the focal line, and is kept at the same temperature of 100 ◦C.
The absorption coefficient of silicon has a strong dependence on photon energy, as shown in Figure 25.16
For simplicity, consider an idealized material, material S, similar to crystalline silicon, with an absorption coefficient of κ = 7×103 m−1 for light at any wavelength. What fraction of (normally incident) light will be absorbed by a wafer of material S that is 200 μm thick? If we could fabricate a layer of material S just 1 μm thick, what fraction of incident radiation would it absorb? The absorption coefficient of amorphous silicon is significantly higher than that of crystalline silicon. If we had another material, material A, with an absorption coefficient κ = 1.5×106 m−1 (again independent of wavelength), what fraction would be absorbed by a 1 μm layer?
Derive the Fermi–Dirac (25.8)
Start by considering a single electron state of energy E that is either occupied (n = 1) or not occupied (n = 0) by an electron. Now, consider coupling this two-state system to a thermal reservoir at temperature T so that not only energy but also particles can move between the smaller system and the reservoir. The entropy of the reservoir S (U, N) now depends on the total energy and total number of particles in the reservoir. Define EF = −T(∂S/∂N)|U . Use an argument similar to that used to derive the Boltzmann in §8 to derive the Fermi–Dirac (25.8)
Assume that in a given scenario with no significant increase in nuclear power usage and gradually increasing reliance on renewables over the next century, atmospheric CO2 levels reach a maximum of 700 ppmv and then stabilize. Now assume that this scenario is varied by building a thousand 1 GW nuclear power plants at a steady rate over the next century. Estimate the decrease in radiative forcing and average surface temperatures assuming that these nuclear plants replace coal power plants. How would you compare the risks and environmental hazards associated with the nuclear power plants against the risks posed by the marginal warming offset by the nuclear plants?