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dijinj

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About dijinj

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  1. thanks. I think I should at-least do one round of calculation to satisfy my quest
  2. what is different in solution case, is the Coulomb forces are the obstacle. how to calculate energy balance in such cases? In water solution Na+ will be surrounded by polar H2O molecule in one way and OH- will be surrounded by polar H2O molecule by other way; all these are in equilibrium. To separate Na+ we should break this equilibrium and Coulomb forces between +ve and -ve ions will be the obstacle. can you give good references with worked out example to calculate energy balance of such system
  3. John Cuthber can you please be more specific on subject, what I learned is ionization needs energy, so De-ionization releases energy. is this correct in Na+ s case?
  4. http://www.scienceforums.net/topic/37193-no-you-cant-make-sodium/ in the above discussion it says that Na could be deposited on electrode and immediately reacts with water to form NaOH. from my knowledge, dissolution of NaOH and reacting Na with H2O; both are exothermic. if my memory serves correctly the forming elements from its ions is also exothermic ( or produce electric potential). so how does this whole set of reactions are not favorable to produce H2 and OH in concern with energy balance. in first look it seems like a free energy device to produce hydrogen. please reply with supporting evidence like links, sci paper or references
  5. Prime concern is to recycle NaOH, so if nacent Na reacts with H2O and form NaOH is no problem. But it should not burn the apparatus and burn H2(side product from that reaction) . Making H2 is most important
  6. is there any way to separate "Na+" from NaOH and H2O solution, like electrolysis, electromagnetic or electrostatic etc. I think Coulomb forces between Na+ and OH- will be the bottleneck to separate them
  7. will electrolysis donate an electron to Na+ and can Na be deposited on cathode. I know this require some energy but how to calculate it, can you suggest good references for calculating energy balance?
  8. if both reactions are exothermic. then is this a free energy(not Gibbs free energy) method to split water. first reaction gives OH and second reaction gives H2 and NaOH is recycled and we will get heat energy as bonus. Can somebody give enthalpy of formation (del H f)to show that this reactions are truly exothermic or is there any catch like you cant separate Na+ and OH- from H2O (in the first reaction)
  9. NaOH(s) + H2O(l) => Na+ + OH- + H2O + HEAT 2 Na(s) + 2 H2O(l) => 2 NaOH(aq) + H2(g) Is the second reaction exothermic? How to calculate energy balance of both reactions?
  10. these are not just made up numbers these are from calculations i made earlier I am quoting both calculations for your reference these results are put in following calculations
  11. We know the below relation between work and momentum p=(W/s)t There are 3 stages of momentum one is initial momentum (p1) which is zero. Second is just after initial path independent force (p2) and third is after path dependent force (p3). p1 - p2 - p3 = 0 if momentum is conserved Assume path independent force is applied in both stages then from the line integral and regular work example done above; we know that work is 4 and -4 for each case respectively. And s is 2 and assume time is equal to 1 then inserting values into (W/s)*t. we get 0 - 4/2 - (-4/2)=0-2+2=0 So momentum is conserved if both forces are path independent. Now take the case with one path independent and one path dependent. From the path integral example done above we know that work of path independent force is 4 and work of path dependent force through arc is -6.28. Inserting these values in equation, 0 - 4/2 - (-6.28/2) = 0 – 2 + 3.14 = 1.14 So momentum is not conserved. So there could be thrust.
  12. if you agree with how laws of physics are connected and you can make use of them for good; then you must explain the following since s( displacement) does not change. and work is non conserved then momentum must be non-conserved. correct me if I am wrong with a good illustration if possible. don't just stick to conservation of momentum I know that law. The fundamental law to be breached if this concept has to work is: "center of of gravity of a system of objects will not change its initial velocity or position(if stationary) if external force is not applied" a part of conservation of momentum
  13. The victory, 200 Miles s Daytona race, march 1970. There was no faster or better route to that goal than to put one of America's most illustrious racers on the motorcycle and have him ride it for 200 miles at top speed to win America's most prestigious motorcycle race. That's exactly what Honda did. In March of 1970, Dick Mann won the Daytona 200. http://www.daytona70.com/English/history.htm After a break of almost 12 years from racing, Honda rejoined the World Championships in the late 1970s and by 1983 they had changed their philosophy from using 4-stroke machinery to build the V3 500 2-stroke, known as the NS500, on which Freddie Spencer took the 500cc World title – his first championship win and the first for Honda since their return to Grand Prix(MotoGP) http://www.motogp.com/en/MotoGP+Basics/history 2006–2007The eighth generation RR was introduced in 2006 and offered incremental advancements over the earlier model with more power, better handling and less weight. Changes for 2006 included: New intake and exhaust porting (higher flow, reduced chamber volume) Higher compression ratio (from 11.9:1 to 12.2:1) Higher redline (from 11,650 rpm to 12,200 rpm)
  14. I will illustrate an example of how physical laws are connected W=F.s Power is = W/t that is power= F.s/t you know, s/t=v so power is = F.v ( force X Velocity) in 1970s Honda used this principle to squeeze more power out of 'limited volume' racing engines. ie they increased rpm of engine there by increasing velocity of piston thus getting more power out of their engines( I agree there are some overheads). they slaughtered competition and came into lime light
  15. Thanks. however. Line integral of Non conservative Vector field is the loop hole I am looking for. I think inclusion of friction in CFD software may be useful since CFD is primarily a vector calculus since in CFD Curl etc has to be calculated.
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