One last reminder about the upcoming WindPower conference. I've checked with AWEA and they will be opening up the WindPower Confererence exhibit hall to the public (free admission) between 4:00 & 6:00 pm on Tuesday May 17th. The venue is the Colorado Convention Center. More info on the converence is at http://www.awea.org/wp05.html Have a great summer! Tony Jimenez
Is there any way you could tell me what the first chapter in the book the final exam will cover? The final isn't cumulative and just covers "new" info right? I know you can't tell me everything that will be on the final because you may not know how far we will get in class, but do you have a rough idea? Exam 1 covered Ch. 1 and 2, and Exam 2 covered Ch. 3 and 4, so I expect the final will cover Ch. 5, 6, and whatever we do after that (which is likely to be *part* of Ch. 7, and *part* pf Ch. 10) I'll try to get more info about this up on the web page in a week or so. The final will try to focus "new" stuff, but as always, there's some "cumulative" nature because e.g. heat engines, fossil fuels, and so on from the early chapters are still an integral part of the later chapters. Hope this helps!
Hi Steve,
The "correct" answer in CAPA's problem one this week is incorrect.
We're given that an event has probability 10^-6/year, and we're asked to
find the number of years it would take for the event to be certain.
The naive approach is to divide the probability into one, resulting in
10^6 years (the CAPA correct answer). However, the probability of a
large release in a million years is actually:
1 - (1 - 10^-6)^(10^6)
Which is only about 63%. The number of years that would have to pass
for the event to be certain is, well, infinity.
Yes indeed, you are quite correct. It's an order of magnitude
estimate. It tells you the *expected* number of years between events
(or some such thing) I guess the point is, once the probability of a
large deadly explosion has reached 63%, it might as well be 1 for
practical purposes. The simplistic 1/Prob gives you the "typical
timescale" for the probability to start getting high.
By the logic I (and therefore CAPA) used, after flipping a coin twice
you'd have gotten heads with probability one. Obviously nonsense. The
logic *IS* good if you ask for any probability much less than one.
(E.g, if I tell you the probability is 1E-6 each year, then in 2 years,
it's darned close to 2E-6! This "linearity" continues (though it's only
approximate) until the probability starts approaching one.
But you knew that :-)
Steve P.
Prof. Pollock, What does MWe mean? I am sure I should know this but I can't find it's definition in the book and I cant find any conversions for it. Thanks. See you tomorrow. MW is Mega Watts. The little "e" means "electric". So if you're talking about the energy *output* of a power plant, you would say it's e.g. 250 MWe. (It outputs 250 MegaWatts of electric energy). The "e" distinguishes that it's the output we're talking about. (At 33% efficiency, that plant burns 750 MWt, where the "t" would be "thermal"). It's been used in the book and my notes, but now that you mention it I may have just slipped it in without ever clearly defining it. Hope that helps!
I was working on a physics problem that asked about the electric generating capacity of large dams in the US. I picked Hoover as an example. I found the nameplate rating on your site, and calculated the annual nameplate generation, 18.2 billion kWh. Actual generation last year was just 4.0 billion kWh. So, I'm finding a capacity factor of just 22%. This seems awfully low to me. My windfarm project has an estimated capacity of 42%. I looked at previous years, in case this was the result of the drought, but found that 22%, while low, is not out of the ballpark for Hoover. >The % is correct. Hoover Dam is used to provide peaking, some base load >and ancillary services (spinning reserve, voltage support, non-spinning >reserve, etc). So it runs at different capacities through out the day. >The real story goes back to the law of the Colorado River. > >Water storage, navigation, flood control all have a priority of power. > >If there is not a request by contracted water user for water we do not >release water. Since only about 9 million acre feet of water is >required by contracted users we only release 9 million acre feet of >water. As you calculated we could release almost 4 times that amount >with our generating capacity. > >When Hoover Dam was built in the early 1930 the Colorado River had no >big storage reservoirs up stream. We now have Glen Canyon Dam (Lake >Powell), and several others that store the Colorado River Prior to Glen >Canyon Dam (1964) Hoover would normally be able to meet the 9 million >down stream demand plus have about 4 million additional acre-feet of >water we needed to release through the units to meet the reservoirs >flood storage requirements. So, what's the story? I know Hoover is used for peaking, and Nevada has unusually large fluctuations in electricity use. >Most of our water and electricity goes to California. Las Vegas and >Southern Nevada get very little so their electrical demands do not affect us. I know you do preventative maintenance every year, taking out 25% (?) of your generation. >Each of our four penstocks has four units attached to it so when we do >penstock maintenance all four units come down. After 2007 we will only >do penstock maintenance every 10 years or so. that will mean the 25% >figure due to maintenance should drop dramatically. Still, Hoover is at the bottom of the list for BLM dams. >Not BLM we do not do burros-Hoover Dam was built by the Bureau of >Reclamation along with most of the large concrete dams in the West >(Shasta, Grand Coulee, Glen Canyon...). Anyway, I hope you have a few minutes to respond to this email. I've spent a lot of time looking at dam statistics tonight and find them fascinating. Perhaps you can recommend a good book on the subject. >Cadillac Desert, Hoover Dam by Stevens
Hi, I can't seem to get the first problem. I multiplied the dimensions of the lake to get the area and then used the conversion 1g/cm3 to get the mass. Then I used the formula mgh with that mass, the height of the water fall, and 9.8 m/s2. Then I multiplied this by .8 (the efficiency). Can you please tell me what I am doing wrong? Couple thoughts. You said you got "the area". Did you mean "the volume"? That would be in m^3, did you remember to convert that to cm^3 (in order to then use that conversion of 1 g/cm^3). What number did you get for the conversion of m^3 to cm^3, because that one is really easy to mess up... Then, did you convert from gram to kg (which is what you need for mgh). Those are the obvious things I can think of. Hope it helps!
Hi Professor. I just had a conversation with my roommate about how he had heard that leaving a computer on all day is actually cost saving compared to turning it on and off 5 or 6 times a day. I told him that there's no way that that's possible if one considers that 24 hours a day is a lot more than it would otherwise be on. He thinks that the actual act of turning on the computer takes enough energy to equal or supercede the 24 hours of continued use. I told him once again that I doubted it, and although I personally am quite convinced that turning off the computer does indeed save energy, I had to admit that I did not absolutely certain. Could you perhaps explain which is true? What got me confused is that I believe that computers are a bit different from light bulbs in that I think they do have a bit of a "charging up" period when you first turn them on. Your answer is correct. It is remotely POSSIBLE that "cycling" could use more energy, if the "turn on period" consumes considerably more power than normal running, and if you turn it off and on VERY frequently. One case where this may happen is with fluorescent bulbs. Before the device is fully on, the resistance is low, current drawn is very high, so power=v*i is high. If it's, say, 10 times higher current than normal operation, and "turn on" lasts for a few (3-6??) seconds, then you could imagine that you use as much power on turn-on as you would in, say, a minute of operation. So in that scenario, keeping it on continuously would be worse ONLY if you turned it on and off once or twice a minute. So that would say don't turn off the fluorescent bulbs if you're leaving the room for just 30 seconds or a minute. But for more than that, you save energy by turning it off. (I just found a use DOE site that corrects my numbers - they say turn-on for a modern bulb is more like one *cycle* of the 60 Hz power supply - so rather than 30 seconds, they claim 5 seconds!) But in general your argument has to hold if the current drawn is roughly similar. For your friends "6 times is worse than a whole day" claim, supposing that "turn on" lasts for, say 30 seconds (drive spinning up, screen warming up, machine warming up, that's as long as I can stretch my imagination that the operation is not 'normal'. Do you agree? ) Then, for his "6 times/day" claim to hold, you would have to draw as much current during that 30 second period as a normal "on" machine would in 4 hours (1/6 of a day) , a ratio of 480 to 1 in current. Since you're normally drawing perhaps 20-60 watts of power continuously (i.e. about 1/3-1 amp) your roommate is telling us the machine would draw 160-480 amps for 30 seconds. Alas, the wires in my house won't supply anything near that, the fuse blows at 12 Amps, so he's off by an order of magnitude or more. He just can't be right. I have been told many times to leave my computer (stereo, fluorescent lights, etc) on instead of turning them off for SHORT periods. I believe the underlying reason is not energy use (which most people really don't care about anyway), but lifetime of the device. Turn on for such devices does draw more current, and this can stress a variety of components. (Thermal, mostly) Flicking the fluorescent light on and off repeatedly will shorten its lifetime. Ditto for any electronics. So that's why I was told not to turn off my computer - not to save energy, but to decrease the likelihood of "blowing" some electronic part that's hard or impossible to fix. I found a link that says this might have been true 10 years ago, but isn't particularly relevant with modern computers. It's a random source, I don't know how reliable, but it struck me as quite plausible. http://reviews.cnet.com/5208-7586-0.html?forumID=68&threadID=19769&messageID=218068 The DOE web site (which is pretty comprehensive, much better discussion than the above) is at http://www.eere.energy.gov/consumerinfo/factsheets/ef3.html Their bottom line is that if you're going to leave it for more than 20 minutes, you turn off the monitor. For more than 2 hours, you turn off the PC itself. Take a look, they have pretty decent defense of their answers... Cheers, Steve P.
I can't locate the energy storage capacity for concrete in my book. Is this also the R-value or does it involve that number because the book does list the R-values for two different types of concrete. I'm really not sure where to find the correct value. Thank you. The number is in the "passive solar" section (top of p. 103) You're looking for heat capacity, not thermal conductivity (which is what R values would tell you) Hope this helps!
My parents have had a PV system installed at their not sure where to find the correct value. Thank you.A student's house, here's a link to some pictures of it if you're interested. http://homepage.mac.com/fivestar/PhotoAlbum13.html I do not know how much it cost, except that it was on a similar magnitude to a car. I do know that since the system was installed in July of last year a little less than 1000 kwh have been generated, so with currently 60% of the planned capacity installed the solar panels are covering about a quarter of the total electricity use for the house. It is important to remember that these numbers do not include any of May or June, and about half of July. These after all are the sunniest times of the year, so it is reasonable to assume that the average generation so far is lower than it will be in the long run. Also, the most generated in one day has been 9 kwh. Very cool, thanks. The last pix say "1750 Watts", which at $5/Watt would be $9000. Plus the inverter (and any costs for the wiring and installation) So your guess of "similar magnitude to a car" sounds just right. The numbers also make sense: 5 "peak" hours at 1750 Watts = 9000 W hours, or 9 kw hr, that's your "most generated in one day". (Here in Colorado at $.10/kw hr that's worth $1! ) The total of 1000 kwh so far is worth $100 (and of course you're right, the numbers so far are mostly for the winter - the highest return should be coming in the summer). I bet it'll easily return $300/year. It's a long payback (maybe 30-40 years) unless your dad got any sort of tax rebates, or support from the power company. (Do you know about either of those?) But in any case they look beautiful, and it must be so cool for him to watch the meter run backwards on sunny days :-)
On Tues., Mar. 1, the Herd, the student arm of the CU Alumni Association, will e-mail all students to ask for nominations for the 2005 Teaching Recognition Awards. Students will be able to vote at: http://www.cualum.org/theherd/TRA2005.htm Students must cast their vote before March 9. Please encourage your students to participate in this program. The winners for the drawing and the Teaching Recognition Award will be notified March 18, 2005. Since 1962, students have voted for outstanding teachers in the only CU-Boulder faculty awards program chosen and administered by students. The Herd, the student arm of the CU Alumni Association, conducts the program based on votes from of current students. If you have questions about this award, please contact the Alumni Association at 303-492-8484.
How does CU sell their ecess power back to the power plant? The power company is able to monitor the flow of electrical energy through the grid. At your home, there's a little meter (the one with the spinning wheel) that indicates the electric usage in your house. If you put solar panels on your roof, you'd use less energy from the outside world, and pay less money to the power company. If your panels were good enough, you might produce *more* than you used, in which case electrical energy would flow out of your house and into the grid, which means your neighbors could use it. The meter on your house would then run *backwards*. In some cases (I'm not sure if it's true for *individual* homes in Colorado, but I believe it is true in e.g. California) the power company will then credit you, effectively paying YOU for the electricity you produced and output. They might pay you a different rate per kW hr than you pay them... I'm assuming it works the same way for CU, but I don't know any of the technical or logistical details! If you're curious, I bet there must be a public relations person over at the power plant who could fill you in! (And of course, let me know if you find out...) Hope this helps a little - Is there anyone else in the class reading this who knows more? Let me know, and I'll post it here!
Are the pre-test questions supposed to be on the reading assignment due for monday? I wouldn't think so based on the content of the reading assignment. The reason I ask is because these questions seem to be on topics from future reading assignments that haven't been assigned. Is that the intent? The pre-test questions are getting ahead, on purpose. I really want to know what people's "background" knowledge is *before* we cover it (even in readings) So, you should not feel like you need to know this stuff, I'm trying to get a sense for how much people know before we even get started, to target the level and speed of future weeks classes! Hope that makes sense, and thanks for participating, it's very helpful to me. Hopefully makes the class work a little better too!
How does CU sell their ecess power back to the power plant? The power company is able to monitor the flow of electrical energy through the grid. At your home, there's a little meter (the one with the spinning wheel) that indicates the electric usage in your house. If you put solar panels on your roof, you'd use less energy from the outside world, and pay less money to the power company. If your panels were good enough, you might produce *more* than you used, in which case electrical energy would flow out of your house and into the grid, which means your neighbors could use it. The meter on your house would then run *backwards*. In some cases (I'm not sure if it's true for *individual* homes in Colorado, but I believe it is true in e.g. California) the power company will then credit you, effectively paying YOU for the electricity you produced and output. They might pay you a different rate per kW hr than you pay them... I'm assuming it works the same way for CU, but I don't know any of the technical or logistical details! If you're curious, I bet there must be a public relations person over at the power plant who could fill you in! (And of course, let me know if you find out...) Hope this helps a little - Is there anyone else in the class reading this who knows more? Let me know, and I'll post it here!
I just don't see how that is a physically correct statement and I don't recall us discussing anything like this in class or reading anything about it in the text. Did I miss something or should it be intuitive, or are we suppose to research it ourselves? It's basically the last couple of sentences of the first paragraph of section 3.6. I was hoping you could put this together by thinking about that section - it's not "intuitive", really, but it's kind of the crux of that section. We should have covered this in class by now (when we did that "handout" on carnot efficiency last week, it was in there) but we really didn't get to it - I plan on doing this tomorrow (too late for the homework, but the homework should help make the conversation in class go better, I hope!) Anyway, maybe knowing which section it is might help you make a good stab. Let me know...
Page 81 example 3.1 where do they get 12.3 watts? COP is defined to be Qh/W. They're talking about "for every watt of power used", which means they're setting W=1. So that makes Qh = 13.3 W. By conservation of energy, Qc + W = Qh, so Qc + 1 = 13.3, leaving 12.3 for Qc That's where the 12.3 came from! And when the gas is going into the turbine,and combusting, how is it turning the turbine, is it the pressure of the combustion and if so where is the O2 comming from, and intake valve or something? I guess what exactly is happeningin that gas turbine for cogeneration on page 84. I'm not 100% sure, so you might want to do some web hunting (like "how things work"?) but my understanding is that Natural gas and O2 are mixed together right in the input to the turbine. (I.e. you pump in air with the natural gas) But I can't tell you the details of intake valves and so on! In jets, the air is quite simply getting sucked in the front of the engine, burned with the fuel, and the resulting gases ejected out the back (so you ALSO get thrust that way!)
We are having some trouble with capa #2. if EEF=heat removed/energy consumed and my EEF is 8 and i want to find energy into kitchcn per energy put in, it would seem that heat removed and per unit energy put in would be equal to the EEF. am i thinknig about this wrong? So there is "heat removed" (from the cold box) which I'll call Qc, "heat added" (to the room) which I'll call Qh, and "energy consumed" (from the electric plug) which I'll call W. (Do you understand why I use those particular symbols for each quantity?) What we know is heat removed / energy consumed = 8, or Qc/ W = 8 The problem says, if you put in ONE unit of work (so W=1), how much energy goes into the kitchen, i.e. what is Qh. (Note that it is not asking for Qc, which is indeed just 8 in this case) How are W, Qc, and Qh related (by conservation of energy)? Hope this helps! By the way, we'll talk about it in class tomorrow, and this set is now not due unti Friday...
This is a quick question about #4 from this week's capa assignment. It's been a while since I've had chemistry, and I don't remember how to attempt a problem like that. Do I need to convert to mols then to tons and figure it out that way? I know it seems like a relatively easy problem but any info you could give would help a lot. ThanksP. 65 of the text talks a little about this, but yeah, you still need to remember a bit of chemistry. But not much - the main point is this: CH4 + 2 O2 yields CO2 + 2 H2O That equation tells you what happens in a single reaction. Every single molecule of CH4 (that's methane) will yield ONE molecule of CO2 (and TWO molecules of water) If you know the weight of one molecule of CH4, and the weight of one molecule of CO2, then the RATIO will be the same no matter whether it's one molecule, or one mole, or one ton... (Do you see that? If one CH4 becomes one CO2, then 10 CH4's become 10 CO2's...) That means we just need to know the relative weights of those two things. The weight of molecules can be found from the periodic table. Every atom has an atomic weight, and then you just add the weight of all the atoms that make up your molecule. Some of them I just tend to remember - Hydrogen has weight 1 (in a certain scale. For this problem it doesn't even matter "one what"!) Helium has weight 4. Carbon is 12, Oxygen is 16, Nitrogen is 14. Others can be found in any periodic table! Let me do a different example from the above equation. Every TWO oxygen molecules will yield TWO H2O molecules (that's the 2's on the left...) So it's still "one for one", if you burn one oxygen molecule, you'll get one water molecule out. (Can you see that?) Now, the weight of an oxygen MOLECULE (which is O2) is not the same as an oxygen ATOM (which is O) because the molecule has 2 oxygens (that the 2 *after* the O in O2) Let's figure out the weight of an oxygen molecule. It contains two O's, so it weighs 2*16 = 32. How about a water molecule? It has 2 H's and one O, which weighs 2*1 + 16 = 18. (This making sense?) Thus, for every 32 "weight units" of Oxygen molecules input, you'll get 18 "weight units" of water coming out. So if you oxidize 32 pounds of Oxygen gas, you'll get 18 pounds of water. If you oxidize 32 tons of Oxygen, you'll get 18 tons of water. Do you see how all you need is the ratio of atomic weights? CAPA is asking: if you burn one TON of methane, how many tons of CO2 will you get? Hope this gets you going, ask if it still doesn't make sense.
Hi Steve, Elevator Question: If I take the elevator up to turn off the lights I will, presumably, take it down again for I still havent eaten lunch. Therefore, I would use 2 minutes of elevator time, not one. So I think the correct answer is 198.4, not 149.2. I think the main point here is that 20 hp = 15,000 Watts, which is 150 times 100 Watts. So fundamentally, every second in the elevator is roughly equivalent to 150 seconds (=2.5 minutes) of the office light bulbs... From this point out, it's just counting and comparing... I timed it, the Gamow elevator takes a bit under 30 seconds to go up to the 10th floor, so I was kind of already doubling that, just to do an order of magnitude game. I really wanted to know "is it a few seconds worth of lights, a few minutes, a few hours, or a few days..." There are lots of other issues - does the elevator have to get "called down" to you? Is anyone else riding it? Does it use as much energy on the way down as on the way up? Etc. I just thought it was kind of interesting, and (like many things in our energy based society) impressive how much energy we use without thinking a split second about it. I ride the elevator to the 4th floor probably half a dozen times a day without batting an eye. (I usually take the stairs down though :-) I got into a discussion with colleagues about this question of *eating* those calories to walk up. If you DID have to consume extra calories -> the "mgh" calories, by eating an extra candy bar or something, with your body's 10-20% efficiency and all the fossil fuels wasted in producing processed food, you might not actually win over using the elevator, which converts electricity with high efficiency, and the electricity is made at our campus power plant which is itself about as efficient as any fossil fuel burning facility anywhere. (mid 30%'s) BUT, we metabolize at a rate of about 100 W whether we move, climb, sit, sleep, or whatever. Walking up slowly, you might use mgh = 60 kg * 10 m/s^2 * 30 m = 18,000 J. If you do that in 5 minutes = 300 sec, that's a power of 60 Watts. So yes, you'd be burning fuel at a slightly higher rate than normal. But it would be good for you, and you might just spend 5 minutes less at the gym, keeping your daily calorie intake fixed overall :-) I guess I'm not convinced you'd necessarily have to consistently eat more if you walked up the tower a couple times a day, because your body would probably compensate in other ways to keep your daily input/output the same. (It's possible your body would re-equilibrate to a lower weight after a while, that's one reason that people exercise :-) But this does add some complication to the story that I can't resolve fully. People could make a similar argument about, e.g. my biking to work. If you figure all the calories I consume for that ride, and just ADD that to my diet "otherwise", the amount of fossil fuel consumed in producing those food calories may indeed be roughly comparable (order of magnitude) to the fossil fuels a nice efficient car like my Civic would burn. (Depends on whether it's an apple from a tree in my yard, which involves zero fossil fuels, or a "power bar" that involves grains, processing, shipping, packaging, etc.) But I think the whole argument is fallacious, because if I didn't bike, I'd probably eat the *same* amount and just burn the extra calories either in an artificial sporting activity, or perhaps by having a less efficient metabolism because I'd be out of shape! In fact, I suspect I eat *fewer* calories when I'm exercising a lot. (That's more biophysics than freshman physics, and necessarily more complicated!) The Carnot equation is for an ideal heat machine. Could we generalize the equation for all machines by substituting <= in the equation? For *any* real heat engine, operating between two baths of temperature Tc and Th (with any mechanism, any working fluids, etc) the efficiency (work out/energy input) will always be less than or equal to 1-Tc/Th. (So I guess this means Qc/Qh >= Tc/Th. You could always dump MORE heat into the cold bath (at the expense of LESS work coming out) but you can't go the other way, you can never run the engine with LESS heat being expelled than the Carnot amount . The proof of this is subtle (but accessible, take a look at any freshman physics book. It involves thinking first about ideal Carnot heat engines, then running them "backwards" to create an ideal Carnot refrigerator. If you ever managed to build a real heat engine that was more efficient than the Carnot one, you could COMBINE it with an ideal Carnot refrigerator and end up violating the 2nd law of thermodynamics).
Are the pre-test questions supposed to be on the reading assignment due for monday? I wouldn't think so based on the content of the reading assignment. The reason I ask is because these questions seem to be on topics from future reading assignments that haven't been assigned. Is that the intent? The pre-test questions are getting ahead, on purpose. I really want to know what people's "background" knowledge is *before* we cover it (even in readings) So, you should not feel like you need to know this stuff, I'm trying to get a sense for how much people know before we even get started, to target the level and speed of future weeks classes! Hope that makes sense, and thanks for participating, it's very helpful to me. Hopefully makes the class work a little better too!
Hi, When reading over the short answer questions for this week, I was a little confused. For problem one it states, "In question three, discuss your answer a bit." What question three are you referring to, the one from the book, or the one from the CAPA homework. I assume you're referring to the book, but the book does not mention both petroleum and oil shale as the problem statement does. No, I just meant question 3 from the CAPA set - the one about shale!
FYI: An email I received (might be useful for projects!) The spring term has begun, and as papers and projects come up your students may want an alternative resource to Norlin. The Environmental Library inside the Environmental Center (UMC 355) has a 2000 piece collection of books, videos, speeches etc. that may be useful. This term we are implementing a web based catalog with Library of Congress call numbers, searchable by title, author, and keywords. Thank you, E-Center Library, Matt Heck, Student Library Chair -- Environmental Center University of Colorado Student Union University Memorial Center room 355 207 UCB Boulder, CO 80309-0207 (303)492-8308 www.colorado.edu/ecenter
Prof. Pollock, On the first two problems of the third problem set, how do we take into account the 70% efficiency? Suppose I said you need 100 Btu to heat a certain pot of water to a desired temperature. But then I tell you that your heating system/device loses some energy to the room - it's inefficient. Let's say, e.g. that it's only 25% efficient. That means that 25% of the energy you input [ pay for ] goes into heating the water. All the rest just turns into useless room heat. So how much energy would you NEED to provide to heat up that water now? You'd need 4 times as much, 400 Btu! (Because you're only effectively getting 25% of it!) Does this help you? Let me know if it's still confusing!
I was trying to find a partner for the project and I was thinking of doing something about how much energy goes into making hybrid cars, cleaner fuel sources for cars, such as hydrogen cells or biodiesel or something along those lines. I will appreciate any suggestions on the idea and any email address of anyone interested. I'm a little unclear on your topic - is it two different things (1) energy used in making cars, 2) clean fuel sources for cars) or did you have in mind somehow connecting them, e.g. looking at how much energy is used when generating biodiesel, hydrogen fuel, or hydrogen cells. Item 1 alone is probably a little too limited for a full project - you'll dig up some numbers, try to assess their reliability - that'd be a nice piece or start, but I'm sure you can dig much deeper. Looking into "energy cost" for manufacturing a broader variety of "green" items (solar panels, biodiesel, etc) would be pretty cool, and could become part of a project to see how much fossil fuel is needed to make the transition to a renewable society... Just some thoughts, it's of course up to you, just want to give you my impressions about what the scope might be.
Hi, I've come up with an idea for a topic for the project however I am not sure how to get " in depth" enough and also am looking for partners. The topic is fairly broad, alternative fuels for combustion engines. The focus would be on biodiesel and/or ethanol, and perhaps measure practicability as well as environmental impacts of both production and use. So if you have any thoughts or suggestions they would be welcome, and as I mentioned above I am also looking for other people to work with. I think this is a good starting place - alternative fuels is something that'll likely play an increasing role in the future. You could look at costs, distribution, pollution,... I think "energy costs" is also a really interesting part of the story - how much fossil fuels are used in the production of ethanol? (I have read really different stories about this, it'd be nice to dig in and try to sort it out - figure out what approximations/assumptions different people make who claim that it's a net "energy loser" or a net "energy win".) There's a lot of "hype" around this - people who are pro biodiesel make lots of strong claims, and it'll be cool to try to figure out what the detailed story looks like.
Another topic idea I had...in case this person already found a partner is to take the current National Energy Policy for 2005 and reasarch cost and efficiency comparison between that of what the NEP development group suggests agains what it would cost to invest that same dollar amount into sustainable energy solutions. Sounds good! It's a little less "personal" and it may be rather easy to get lost in a sea of predictions about what sort of cost and payoff "sustainable energy solutions" will involve... But if it's something you're interested in, I think you could certainly make a great project out of it. (As usual - if you read this and are interested, email me and I'll get you in touch with the person who sent it)
I don't yet have a partner, but I do have a possible topic. In relation to making one's home more cost-efficient, environmentally friendly, or 'off the grid', I'd like to focus on different housing types that may be relevant - such as geodesic dome homes, or underground homes, for example. The energy-saving potential of some of these structures could be very beneficial. Please let me know what you think. I think digging into the details of home energy efficiency could be a really good basis for a project. Lots of directions you could go - including issues of modifying existing homes as well as building new ones. You could look at costs, but also other layers of impact. There are lots of organizations that could help provide info - in fact, Boulder's "greenpoints" system for new construction could be interesting to look into and incorporate. There used to be a local environmental center that had annual "green houses tours" - you might see if they still exist (I think it was the "BECC" or something)
I have an idea for a project, but don't know how to go about finding a partner. I would like to take that pie graph that you showed in class, and update the percentage of US energy that goes towards what aspects of society and where it comes from. I would like to see what portion of renewables is being used in the different sectors. any ideas? I think it's a good rough idea, though I might flesh it out a little. Just looking up the data probably won't be interesting (or challenging) enough, you might want to use this data as part of a larger picture/question. What are you after here - trying to figure out where renewables could best be utilized? Which sectors are most "ripe" for change? How much of a difference it would make (vs. how much it would cost?) Those all lead you down interesting paths that will require data, but also go beyond it to doing some "analysis", which is what we're after here. I think you're on a good track! As far as finding a partner - if anyone finds this interesting, let me know and I'll put the two of you in contact! (Or, just speak up before class)
I've tried everything to get my hands on a textbook somehow, but I've had no success. Most of the problems involve conversions with help that can be found on the inside cover of the textbook. I've searched to find these conversions on the internet, but nothing seems to be helping me find the right answers. What can I do? I talked with the bookstore today, they claim they're fed-ex'ing new texts, so they should be in tomorrow or Friday. Which doesn't help you much on this set! Yes it is tougher to do these problems without the textbook, but do your best - learning to dig up useful information for yourself is actually an enormously useful skill which I'm hoping to encourage in Phys 3070! Basic unit conversions can be found in many textbooks. In fact, I claim all you need for this CAPA are 2 conversions (Joules/Btu and Joules/(ft*lb). (For problems 4 and 5 and 7) That kind of metric to English conversion is all over the web - hunt a bit more, you will find them!! There's one more number you need for problem 8, (amount of coal consumed/yr in the US) but that is in my on-line lecture notes (page 3/15 of the first batch, see our "resources" link) That's the only number from the book you need for either problem 8 or 9! Please note that there are two copies of the text on reserve in the math/physics library (right across the hall from our classroom). Because of your email, though, I have decided to extend the due date on this CAPA by 24 hours, so that I have a chance to tell everyone in lecture Wed about the books on reserve. So you'll have a chance tomorrow before or after class to go dig up the numbers if you still need them (and maybe photocopy the front flyleaf! ) Hope this helps, really sorry for the bookstore hassle. Let me know if you have more questions!
Professor Pollock, I am having trouble with problem 5, my approach so far is using E=mc^2 w/ the mass given in the problem to get an energy of 9*10^10 J then converting this to BTU's w/ the conversion in the front of the text, is this the correct method?? I'm afraid E=mc^2 gives you "rest mass energy" - which is not what we're after here. (Rest energy is always astronomically huge, but also not something that's easy to "get at". It's generally relevant in questions about nuclear energy) This question asks for "kinetic energy", which is much more practical and everyday - it's given by a simple formula (see your textbook, or my online notes). Since everything is given in metric, this formula will tell you the kinetic energy in JOULES, not Btu, so then you just have to convert. Hope this helps, let me know if not!
Do you know anything about this? Here is a link in case you have never heard of TDP: http://www.sovereignty.org.uk/features/footnmouth/zwaste2.html Seems to be the answer to peak oil production. Of course, it will only make global warming worse. It looks like a variant of "biomass", which is real, and will surely become one of the many ingredients in trying to deal with peak oil production. Calling it "the answer" is probably overly optimistic, though. Most of the "fuel" described in that article is biological or petrochemical waste, so this plant is not really generating energy, its recycling. That's obviously very helpful - the more you can recycle, the less you need to dig from the ground. At 85% efficiency (which they claim, though I'd really want to look at the numbers, I'm skeptical its really that high) you could certainly have some impact, IF it can be scaled up, IF the cost isn't beyond a (few?) hundred dollars/barrel, IF it doesn't produce toxic wastes, etc... I just ran across an interesting (if perhaps rather glum) page that addresses these kinds of issues pretty well. It's "http://www.lifeaftertheoilcrash.net/" Take it with the same skepticism you take the "we've solved the upcoming oil crisis" articles - (although they've done a decent job of documenting their claims.) By the way, interestingly enough, biofuel does produce CO2, but the products it uses as fuel are themselves biological, and hence consumed the CO2 from the atmosphere when they were formed. In the long run, it's overall "CO2 budget" neutral, so it should NOT necessarily contribute to global warming concerns!
Steve, I registered my new clicker last night, and I found out that I can borrow a friends so I don't have to spend the billion dollars for the new huge one that I just bought. How do I go about registering and getting rid of the first one? Thanks Thanks for letting me know, glad you found a better deal! Just be careful that the one you can borrow works ok (it needs to have two bulbs on the front, not one) Just go ahead and register the second one online in the usual way. Registering two shouldn't hurt anything. If you do get rid of the expensive one, however, it would help in about one or two weeks time if you sent me an email with your name, student ID #, and the ID # of the clicker that you ended up deciding to use every day. I'll fix it all up then. Cheers, Steve P.
The CU Environmental Center is holding several large events this Spring semester. Green Careers Conference Tuesday, January 18, 2005, 5-7 pm Humanities 250 Learn more about environmental career opportunities in consulting, industry, green business, sustainable design, and environmental protection. Two panels of environmental employers and career experts will share employment opportunities, career advice, and inside tips in a two-hour satellite broadcast for college students. Spring Volunteer Interest Meeting Wednesday, January 19 6:00pm Hellems 252 Come to the interest meeting for some pizza and discussion of how to become involved with the environmental center and other environmental groups. It will be a great venue to find out what's going on and how you can get in on the action. William McDonough Thursday, February 3 7:00pm UMC Glenn Miller Ballroom William McDonough is a renowned architect as well as the author of "Cradle to Cradle," where he blends ecological, social and economical philosophies into design practices for a more sustainable world. His presentation is the featured keynote of the Colorado Sustainability Summit and will address a new conceptual framework for institutional sustainability. Admission to Mr. McDonough's presentation is included for Summit registrants, is free for all CU students (tickets available at the door), and is also open to the public (tickets are $10 each. Available in advance at the Boulder Theater Box Office, www.bouldertheater.com or 303-786-7030; or at the door of the event. the week of 02/03/05 - 02/04/05 Colorado Sustainability Summit: Forging Solutions at Colorado's Colleges and Universities UMC The Colorado Sustainability Summit will provide a forum for representatives from each of Colorado's campuses to network and learn about a variety of environmental issues facing institutions, explore ways of improving campus practices and policies, and ultimately work toward environmental and economic sustainability. The Summit will help make Colorado's colleges and universities national leaders in campus sustainability. Featured presentations include renowned architect William McDonough on sustainable development and a plenary panel composed of leading campus administrators addressing Higher Education and the Economics of Sustainability Environmental Center University of Colorado Student Union University Memorial Center room 355 207 UCB Boulder, CO 80309-0207 (303)492-8308 www.colorado.edu/ecenter
Is there any way you could tell me what the first chapter in the book the final exam will cover? The final isn't cumulative and just covers "new" info right? I know you can't tell me everything that will be on the final because you may not know how far we will get in class, but do you have a rough idea? Exam 1 covered Ch. 1 and 2, and Exam 2 covered Ch. 3 and 4, so I expect the final will cover Ch. 5, 6, and whatever we do after that (which is likely to be *part* of Ch. 7, and *part* pf Ch. 10) I'll try to get more info about this up on the web page in a week or so. The final will try to focus "new" stuff, but as always, there's some "cumulative" nature because e.g. heat engines, fossil fuels, and so on from the early chapters are still an integral part of the later chapters. Hope this helps!
Physics 3070 home page.