Bignose

Just to kind of further was dave said here... once we got into symbolic only math, an explicit multiplication symbol was only used when it may have been ambiguous without it. I.e. if the equation was too long for a single line, there would often be an explicit multiplication symbol at the line break. But, crack open any university-level math text, or a math journal, and it is exceptionally rare to see any kind of explicit multiplication symbols. Using nothing is pretty much the standard today.

But, you see Solarist, you say things like: and you don't provide exactly what is meant by specific kinetic energy -- which is tough enough to define because of the frame dependence -- nor do you provide any kind of evidence for these statements. Can you cite any evidence that kinetic energy creates a magnetic field? And if it does, how hard do I have to hit a golf ball with my driver to create a magnetic field?

In this context there is no difference between [math]\cdot[/math] and [math]\times[/math], but in higher math there becomes a significant difference. These two equations have very different meanings: Given vectors [math]\mathbf{a}[/math] and [math]\mathbf{b}[/math], Let [math] \mathbf{c} = \mathbf{a} \times \mathbf{b} [/math] and let [math] d = \mathbf{a} \cdot \mathbf{b} [/math]. There is a very big difference between c and d, the biggest being that c is a vector, and d is a scalar. If there is a choice, Cap'n, I'd try to convert everything over to [math]\cdot[/math] or even no multiplication symbol at all (just implied), because the [math]\cdot[/math] symbol is much more consistent a wider range of mathematics.

I just don't see how. I'd look into some of the matrix multiplication routines that MATLAB uses. MATLAB is by and far the fastest matrix multiplication I've ever seen.

I don't really see how since with matrix multiplication in general [math] \mathbf{AC} \neq \mathbf{CA}[/math]. I looked at something like [math] ({\mathbf{CA}}^T)^T = (\mathbf{A}^T \mathbf{C}^T)^T = (\mathbf{A} \mathbf{C}^T)^T [/math] but you still have to compute the [math] (\mathbf{A} \mathbf{C}^T) [/math] which doesn't appear to save anything. For that matter, even if both A & B are symmetric, in general [math] \mathbf{AB} \neq \mathbf{BA}[/math]. [math] \mathbf{AB} = (\mathbf{BA})^T[/math]

No, what it means is that kinetic energy is completely, 100%, totally, and wholly determined by what reference frame you pick. And, since there is no such thing as a "preferred" reference frame, then any theories based on absolute values of the kinetic energy have to be able to take all reference frames into account. Specifically, you cannot specifically say "item x has y Joules of kinetic energy" -- you'd have to say "in reference frame z, item x has y Joules of KE" you have to be very specific about the reference frame.

No, you aren't understanding the concept of frames of reference. Start out with an empty space with only two balls in it, each with a mass of 1 kg. One ball is moving at 1 m/s the other at 2 m/s. It doesn't matter ball you choose as a frame of reference, but the one you pick as the reference has no kinetic energy and the other has 1 J of kinetic energy. So which one has more kinetic energy? You cannot say, any frame of reference is as valid as any other. No one frame is better than any other. Now, let's look at the Earth. It has kinetic energy from it's rotation about the Sun. But, there is also some from it's rotation from bring in the spiral arm of the Milky Way. The Milky Way is also moving through space, and space itself is expanding. The really important thing to note is that there is no fixed point in space where you can sit back and just tally these things up. So, what kinetic energy is important? The kinetic energy of the rotation about the sun? The rotation about the galactic center? How you even calculate these numbers is 100% completely dependent upon how you look at situation. Do you pick a frame where you are moving with the Milky Way? Then the movement of the galaxy itself means that there is no KE from that. How about a frame where the Sun is still... then there is only KE from the revolutions about the Sun. How about a frame where the Earth is still... then the earth has 0 KE. Finally, it is a fundamental principle of modern physics that there is no such thing as a preferred frame. That is to say, that there is no one "right" frame or any "wrong" frames. They are all equivalent. In this way, just basing a theory on a single kinetic energy is very troublesome, because there is no "right" KE or "wrong" KE -- it is all 100% dependent on your frame. And as near as we can tell, the laws of nature are frame independent -- the laws work no matter what frame of reference we use.

Since you've gotten your answer, I just wanted to make one final comment in case anyone else reads this thread. I think that smokey didn't get my first hint because he only considered mixing to be between two different substances. But mixing is a very general term, and in this case, it is the mixing of the warm and cool fluid. If no other considerations are important, then turbulent heat transfer is always more efficient because of the mixing inherent with turbulence. In this case, the turbulence brings the warm fluid in the center of the vessel to the cooling jacket much more efficiently than a laminar flow would. Because of the mixing, the fluid ends up more uniformly cooled than it otherwise would have.

smokey, I'm not going to just tell you an answer. I was hoping that my questions would help you explore the answer on your own. In that regard -- are you sure that mixing is unimportant? (I don't think that I can be any more unsubtle than this - *wink* *wink*)

Dean, We try not to just answer questions in the homework part of the forum. We do help the OPs by giving hints, or pointing in the right direction, or asking leading questions of the OP -- basically all-in-all a Socratic approach. But, we don't just give answers. On top of that, I don't think that you can say with certainty which one is better. Because I don't think that the OP has given enough information. I.e. is it more important to achieve the cooling? or is it more important to keep the flow nicely behaved? And, are we talking about the flow of the nylon, or the flow of the water inside the wall? What about the size/capabilities of the pump? For that matter, nylon being a polymer probably doesn't flow like a typical Newtonian fluid. Turbulence in non-Newtonian fluids isn't anywhere as well behaved as turbulence in Newtonian fluids -- not that turbulence in Newtonian fluids is well behaved in the first place. Furthermore, Dean, your example of a condenser isn't terribly appropriate, since the OP didn't mention needing to condense anything... Condensing something is very, very different from just cooling a fluid down. There are a lot of questions that need answers to fully define the problem. You have to think about some of the fundamental differences between laminar and turbulent flow. In particular, the major difference between the two is the amount of mixing. The question is whether you want that mixing or not. Or can you prevent the mixing/turbulence if you want to?

You can always "linearize" any non-linear system of equations and then use the linear methods of solution. This is the basis on most of the computational solvers out there. Like, the system of fluid mechanics equations -- the mass, linear momentum, and energy equations are highly non-linear. Especially if you start to introduce things like turbulence and temperature dependence on the physical properties of the fluid (both of which change the effective or actual viscosity of the fluid). Basically, to linearize a non-linear system, you "freeze" some set of the terms -- the non-linear parts -- and then solve the linear remaining part. Then, with the new solution, you update the non-linear parts with the new information, "freeze" them with the new information and re-solve. And repeat. Until some sort of convergence occurs. The real tricks are in how big of a step you take -- i.e. if the solution from the frozen set of equations differs a lit from the previous solution, how much of that difference do you accept? If you accept a large difference, then you may miss a lot of what the non-linearity in the problem was supposed to be doing. If you take too small of the difference, you are essentially resolving the same problem as the previous one which is wasted time. There are many incredibly non-linear problems that can be solved successfully this way. The entire field of computational solutions is based on it, and computational solutions is very, very rich. I am most familiar with the computational fluid dynamics side, but the solution of solid dynamics is equally rich, maybe even more so.

I haven't checked them out for a few years now, but the magazine U.S. News and World report does an annual ranking of colleges every year, and at least 5 or so years ago that included a ranking of chem e departments. The classically good ones: MIT, Minnesota, UC Berkeley, CalTech The ones that are certainly still good, but are living a little more on reputation and history than current achievements: Wisconsin, Illinois The ones that aren't traditionally well known, but are making a name for themselves with good younger researchers: Delaware, UC Davis The ones that have long been considered good, but never seem to break into the tip top elite group: Purdue, Ohio State, Florida, Iowa State, Arizona, Carnegie Mellon, Georgia Tech, Illinois Inst. of Technology, Michigan State, Northwestern, Pennsylvania, RPI, Texas, Washington. And then there are many very good programs. As much as anything, you want to visit the school and get a feel for the school, the people that work there, the other students and see if you'll fit in. Being happy and comfortable you'll get a lot more out a school than if you went to a school with a better name but you were miserable there. ** My apologies to the chemical engineering departments outside of the US, but being a life-long resident of the US, I am only most familiar with the US ones. I am sure that there are many, many fine non-US schools, but any rankings I would do would be based a lot on reputation/word of mouth, and nothing about the actual quality of the people that are there doing the work and how good the instruction the students get. edited to add: furthermore, if you are only just starting your engineering student career, I hope that you won't get too hung up on just chemical engineering. Your first year or so of classes, you wouldn't be taking any chem e-specific classes anyway. You'll just be taking the basic physics, math, etc. classes. Make sure that you explore all the different engineering flavors that are available -- go to a professional society meeting is as many of them as you can. You'll get a good idea of what people in those professions actually do by talking with the guest speakers they bring in. And you'll get a good idea of what kind of students are in each disciple. You might find something that just jives with you better than what you first thought would be best. In the long run, it is far, far better to be happy with what you are doing than to force yourself to complete a degree in a field you don't enjoy all that much.

Anubis boy, you are right. The 4 equations aren't linearly independent... I too get only trivial results from it -- angles equal to 0 and 180. I don't have a tremendous amount of time to think on this right now, but I'm going to keep mulling it over...

Here's how you do 2 equations with 2 unknowns: As an example, let's look at: x+y = 7.2 3x-2y = 10.6 Rearrange one of the equations to isolate a variable. I'm going to do the first equation: y = 7.2-x Now, plug this definition of y into the second equation: 3x -2(7.2-x) = 10.6 And solve for x Then, with the solution for x, you can compute y = 7.2-x You should do this and find that x=5 and y =2.2 Note that the choices are completely arbitrary. I could have used x = 7.2-y as the rearrangement. Or, I could have used the second equation. x = (1/3)*(2y + 10.6) or y = (1/2)*(-10.6+3x) Now, 4 equations with 4 unknowns is exactly the same. You just have to repeat it a lot more times. And be careful to write down every step and be accurate to make sure the results are correct.

what exactly can't you do?

Just for the record, my mathematical translation of statement 1 is wrong... It should be P(A')P(B') = 0.07 multiplication, not addition (must have been thinking of something else)

Well, there aren't any restrictions on how to solve it. So I graphed it. There is an answer. I don't know if there are any pure analytical ways to get to it -- but numerical solutions are just as valid as any other method.

I guess you could get really fancy and work based off the Navier-Stokes equations. But, i_a's got it right in that for the vast majority of applications, Bernoulli's equation is going to be more than sufficient. Especially if you find the Bernoulli equation with friction terms in it which will help you account for the influence of any bends.

4 equations 4 unknowns: right most small triangle: 5x + 3x + BDC = 180 left most small triangle: 4x + ABD + ADB = 180 large triangle: 4x + ABD + 5x + 3x = 180 line: ADB + BDC = 180 Now it's just equation solving... ------------------------- I have a small side question based on the original question of the OP. Would knowing that a triangle has 180 degrees be something known "without trig" I mean, trig obviously knows that 180 degrees is special -- sin(180 degrees) = 0, cos(180 degrees) = -1, etc. But, is that 180 degrees something known without trig? To a certain extent, didn't trig help define exactly what a degree, and what a radian is? I don't know the history of math well enough.... these are just ponderances of mine.

Ummmmm, no. 3x + 4x + (5x + something) = 180. This is along the right idea -- using the sum of the internal angles must be equal to 180 degrees -- but 5x isn't the entire angle of the top of the large triangle. Using the idea that the angles of a triangle must add to 180 you can write a bunch of equations. Also, the angles along a straight line (not so subtle hint, that'd be angle ABD and angle BDC) also add to 180. Write out all the equations for the triangle and the straight line and you should have more than enough equations to determine any unknowns.

Since you have solved it, here's the equations I had: Let A = Mary goes shopping. Let A' ("A prime") denote she doesn't. Let B and B' denote Jane does or does not go shopping, respectively. Statement 1 mathematically is: P(A') + P(B') = 0.07 Statement 2 is: P(A) + P(A') = 1 Statement 3 is: P(B) + P(B') = 1 and Statement 4, Bayes' (please note the correct spelling, the guy who came up with it, his last name was Bayes, NOT Baye, so the possessive is Bayes') theorem is P(B|A)P(A) = P(A|B)P(B) Since you have been given the conditional probabilities, P(A|B) and P(B|A) in the problem statement, there are 4 equations with only 4 unknowns, which should be solvable.

When I write everything out, I get 4 equations and 4 unknowns. Here are the statements in words (I'll leave the math to you) 1) The probability neither goes is 0.07. 2) The probability that Mary either goes shopping or doesn't is 1. 3) The probability that Jane either goes shopping or doesn't is also 1. 4) Bayes' theorem that relate conditional probabilities http://en.wikipedia.org/wiki/Bayes'_theorem 4 statements, 4 unknowns (I'll leave you to figure out the 4 unknowns, unless you get really stuck and ask for help later). Should be solvable. Get to know Bayes' Theorem. It is an exceptionally powerful tool.

So, here's my particular thought.... why not think a little outside the box. You have several requirements here: (A) goes through (1,20) (2,4) (5,3) (6,2) and (10,1) (B) 3 points of inflection © At least 1 local minimum and local maximum (D) At least one Critical point (eg max/min/intercept etc) is not at a given point (E) The curve is continuous and differentiable throughout (F) The equation is not a single polynomial, but must be a piecewise-defined function 1) Fit a polynomial to go through those 6 points. This takes care of (A). Limit the polynomial to the range x=(0,11) 2) Then piecewise append functions in the negative quadrant of x, and x>11 to fulfill the other requirements. That is, if you've written out the entire problem statement, there isn't a given limit on how much of x you can use. So, use the rest of [math] (-\infty,0)[/math] and [math](11,+\infty)[/math] Put the necessary points of inflection at x = 100,200,500. Or something like that. The only real difficulty here is to pick good functions at the piecewise breaks to fulfill the continuity and differentiability requirements.

I agree with this. It is possible (not easy, but possible) to take carbon dioxide and create polymers, but I don't think that it is right to say that polymers are "trapped" carbon dioxide molecules. The polymer has completely different properties because it is a completely different molecule. It is a completely different substance. Graviphoton, you're going to have to provide a lot more evidence than just saying "it stands to reason"

I guess you don't want: [math]f(t) = \begin{cases} e t^2 \text{ if } t <= 1,\\ e^t \text{ if } t >1 \end{cases}[/math] Like Aeternus said, you probably need to provide more info to get better answers