Write your answers in the space underneath each question. Put your name on every page. You are welcome to use extra paper. Extra paper will be provided (put your name on every page). If writing in pencil, make sure it is dark enough to be scanned.
You may use a single-sided note card of size 8.5” x 11” (letter paper). Use this note card to write down any information you want to access during the exam.
You may use a calculator. Alternatively, full credit will be given if you do arithmetic by hand and make a reasonable effort to get a final answer within \(\pm 20\%\). For example, \(\pi\) can be rounded to \(3\), and \(g\) can be rounded to \(10\text{ m/s}^2\).
One goal of this course is that you become comfortable choosing a reasonable value for a quantity. Therefore, the questions won't tell you every quantity. If you get stuck because you cannot choose a reasonable value, please ask. The instructor will tell you the quantity, and make a small reduction in the points available for that question.
The course instructor will also be available to answer clarifying questions during the exam. You can ask the instructor any question you wish. They may or may not answer.
State any assumptions you make to solve the problem. Show the mathematics that you use to solve the problem. Show units when working with the numerical values of physical quantities. Because time is limited, you are not required/expected to write very many words explaining your reasoning. However, using words to explain your reasoning can allow the grader to distinguish small mistakes from big ones.
If a question asks for a quantitative answer, do not expect partial credit for a conceptual answer.
Very big or very small numbers must be expressed in scientific notation (for example, \(1.2\times 10^6\)). You will lose points if you use decimal notation to express numbers that are greater than \(10^6\) or less than \(10^{-3}\). You will also lose points if you use E-notation (for example, do not write 1.2E6).
The average kinetic energy of one gas atom is \(KE = \frac{3}{2}k_\text{B}T\) (which is \(U/N\)). So, for a atom at room temperature (293 K), the \(KE = \frac{3}{2}[1.38\times10^{-23}\text{ J/K}][293\text{ K}] = 6.06\times10^{-21}\text{ J}\).
This is similar to the problem we did in class Wind energy.
We start by finding the mass of air passing the windmill in time \(\Delta t\). \begin{align} \text{mass of air} = \rho v \Delta t A \end{align} then we find the kinetic energy of that air. \begin{align} \text{KE} = \frac12 \text{[mass of air] } v^2 \end{align} which is a crude way to estimate the energy we could get per unit time (\(\text{KE}/\Delta t\)) from a single windmill. To make the estimate better, multiply by the fraction 16/27 to find Betz's limit. This is the fundamental upper limit for wind energy. To be even more realistic, go to the manufacturer's website and find the detailed specs of the windmill (no windmill can exceed Betz's limit, but some designs come within 80%).
Assuming Betz's limit, the rate at which energy is produced by one windmill is \begin{align} \text{rate of energy production } = \frac{16}{27}\cdot \frac{1}{2}\rho_\text{air}Av_{i}^3 \end{align} Since the area is the circle swept out by the windmill blades, we find \begin{align} \text{rate of energy production} &= \frac{16}{27}\cdot\frac12[1.2\text{ kg/m}^3] \pi \left[40\text{ m}\right]^2[7\text{ m/s}]^3 \\ &\approx 0.6\times10^{6} \text{ J/s} \end{align} Using this value we can find the total energy of the farm by multiplying by the total number of turbines, 200. This means the upper limit for energy produced by Bigelow Canyon (at this wind speed) is \(1.2\times10^8\text{ J/s}\).
Imagine a future, 20 years from now, in which Oregon builds a 1-gigawatt nuclear power plant beside the Columbia River because Facebook wants more electricity for its data centers! The nuclear furnace produces 3 GJ/s, which is sent to a heat engine that produces 1 GJ/s of electrical energy.
Now we have to calculate the temperature change per second of the water in the Columbia River. \begin{align} (\text{The flow rate of the Columbia}) \times (\text{density of water}) = \text{mass of water flowing per second} \end{align} \begin{align} (\text{mass of water flowing per second}) \times (\text{specific heat capacity of water})\times \text{temperature raise}= \text{energy per second used to heat the water} \end{align} Note: It is important to notice how the final answer is looking for change in temperature and thus we must divide the energy per second going into the river by the Joules required to raise the temperature of water by 1 Kelvin Second \begin{align} (\text{Joules per second going into the River}) \div (\text{Joules needed to raise the water 1\(^\circ K\) per second}) = \text{Change in temperature in \(^\circ K\)} \end{align} Plugging in values we get: \begin{align} (7500\text{ m}^3/\text{s}) \times (10^6 \text{ g}/\text{m}^3) = 7.5\times10^9 \text{g/s flowing} \end{align} \begin{align} (7.5\times10^9 \text{ g/s}) \times (4.2 \text{ J/(gK)})\Delta T = 10^{9}\text{ J/s} \end{align} \begin{align} \frac{2\times 10^8\text{ J/s}}{10^{9}\text{ J/Ks}} = 0.2\text{ K} \end{align} So we're not having a huge effect on the river, but it should definitely be observable, and we could imagine it affecting life in the river.
Notes: The flow rate of the Columbia River is about \(7500 \text{ m}^3\text{/s}\). The specific heat capacity of water is \(4.2\text{ J/(g K}\)).