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G B Reid MIMechE, SIMarEST

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G B Reid MIMechE, SIMarEST last won the day on December 19 2019

G B Reid MIMechE, SIMarEST had the most liked content!

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    Benfleet, England
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    [NOTE: No longer SIMarEST, but connot remove!] Mechanical engineering "generalist", experience in Design (industrial, mechanical, aerospace grade components), Rail, Underground, defence, process flow optimisation, Project Engineering and Mechanical Systems 3D CAD user (various) since 1992. Electrical design installation and verification.

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  1. Its worth just trying out a standard thermal expansion coefficient calculation. (You have not stated the material that you are measuring) Take the material thermal expansion coefficient of the material you are working with and calculate its diameter for 30 degrees C...then calculate the same for 60 degrees C...I think you'll find a good correlation. Then just use the 60 degree figure for your measurements. Hope this helsp
  2. With any motor, the power is from P = IV The power (in Watts) used is proportional to the voltage (in Volts) AND the Current (in Amps).....If you up your voltage, the ampage should drop significalty...as you've experimentally discovered, if the voltage is too low, the ability to supply a sufficient current (ampage) becomes the limiting factor. New (DC) motors are rated for a no load speed at a specific VOLTAGE...the current increases subject to the load to match the power requirements...and CRITICALLY the wire requirements are rated on CURRENT. If you run your motor at too low a voltage, your wires will overheat - both externally and internally - the varnish will evaporate from the winding wires inside the motor and the motor will either burnout through a resulting internal shortcircuit or catch fire! Hope this helps I shan't - at the moment - expand further as your experimentally discovering lots on your own which is by far the best way to learn....but I'll be happy to look again for additional questions. Good luck!
  3. Okay...enough is enough...my turn! Assuming the chain links form a cyclic and therefore have a natural load-spread between both legs and no stress concentration: Stress = F/A Stress = 75 000 000 N/m^2, F = 50 000 N A = 50 000 / 75 000 000 = 0.000 667 m^2 for both legs A = (pi r^2)/2 per leg so r = root (0.000 333 / Pi) = 0.010 301 m d = 2r = 0.020 601 m = 20.601 mm - always round to the safety side....so: Therefore use 21mm dia stock minimum!
  4. I'd agree with DrD....that said, ball-bearings are inappropriate in this application....I'd use taper roller as it has both an axial and radial component of force, otherwise - due to the mass of air, mass of the assembly and the resultant action - you will have premature bearing failure!
  5. Marin; NEVER feel ashamed for attempting something and making an error....being ashamed should be left to those who do not try - and I include myself in that (ie I have not tried and that IS shameful!!!). I've been an Engineer now for 30 years, and the one thing I can say without fear of contradiction is you NEVER learn anything from getting things right...that just generates hubris...it reassures that the calculations are done correctly, but that can occur without the deepened understanding needed to be successful in the field! Celebrate mistakes, especially if you learn from them! And to tackle my "Shame": Using the H7/u6 details on the quoted text (reference docs not available) Generally on a "nominal" 120mm dia shaft, allow for a "runout" of 0.1, so all dimensions should be a minimum of the runout on both sides (120 - (0.1*2)) = 119.8 - ie the maximum diameter of the shaft MUST be below this value. Therefore assuming a "base" of 119.6mm to allow for the +.166 of the shaft... Therefore shaft at u6 = 119.744 - 119.766 (on a "base" of 119.6mm - +.144+.166) Hole on H7 = 119.6 - 119.635 (on a "base" of 119.6 - +0.000/+.0.035) It's definitely an interference fit...now, where did I leave the nitrogen...? I may be wrong; I look forward to Saurabh's answers! Warmest wishes! Bruce
  6. There are a couple of different types of "Eco Cooler" design....one utilising water - sometimes known as an evaporative cooler works by taking advantage of the energy of evaporation of water....basically as water (or any liquid for that matter ) evaporates, it absorbs a certain amount of energy... If this utilises natural airflow through an opened window, it will have the effect of cooling the air enterring, but will increaee the humidity within. The other form (inverted cones) takes advantages of the ideal gas laws...it effectively forces a minor pressure drop in the air which reduces the air temperature, but this may not be sufficient to actually notice. Both of these do work to some degree, however if utilising a powered driver there will be energy consumed there. The best form of heating/cooling is that of a "heat pump" which moves heat from one area to another and only "cost" is the power to do so...they can have an energy multipier effect - sometimes of as much as 8x - where for a power input of - say - 1000 Watts - a heating/cooling effect of 8000Watts can be achieved.
  7. Just perusing the answers to this quiz...I find them...well....."interesting" answers to say the least and...er.."worryingly" consistent. If the shaft starts at 120mm you need to think on the maximum size of the shaft available....are you going to spray-weld the shaft to increase its size...before you even begin? If you look at the question it states the SHAFT diameter is 120mm....
  8. Balaji, you have rounded down - ie to the "unsafe" side. In fasteners of any kind, you would always round up...always "enhance" safety, not detract from it.
  9. I may no t be understanding you r question properly, but it looks like you may be misunderstanding what the "static pressure" actually is.... When the flow increases through an enclose space - be it pipe or venturi - the presure reduces proportionally to the velocity of the flow. Basically, the faster the flow rate, the lower the pressure is. This flow rate presure (was historically....there are now sensors that can....)/(is) difficult to measure directly, so there's a "trick" that can be used to infer - with a very high degree of accuracy - the velocity of the fluid - be it gas or liquid - in the pipe....that is the measurement of the static pressure. Effectively, the static pressure is the pressure to which the internal flow presure drops. This is from memory, so I may get it upside down, but it still works anyway as it is a ratio of proportionality....it's: rho1/v1 = rho2/v2 The static presure measurement is the measurement of the rho(1 or 2). It's basically the mass flow calculation at two different speeds for the same mass...rho = m/v This then becomes rho1/v1 = m and rho2/v2 = m and, as m is the same, rho1/v1 = rho2/v2 Therefore, if the flow velocity increases, the pressure drops - this is measured by a static tube measuring hte drop in the static pressure as a result of the change in velocity - so therefore any change in velocity DOES affect the static pressure. Hope this helps
  10. Now this is entirely theoretical!! It may be possible to use "arc-spatter" to advantage. (My welder's knackered at the moment so I cannot try it to see...) If a (semi-sacrificial) earthed-point is located immediately proud of a MIG-tip, and the power and gas-flow ramped up to "silly levels", the spatter from the feed wire "should" spray on to the target metal. It will likely take some fine tuning, but - despite the increased use of gas - it may be a workable solution. Basically, the "earthed-point" will spark the arc...if the power is high enough, the metal should near-vapourise, forming a fine spray, being easily blown - by the shielding gas - onto the workpiece where it will solidify. If it works, it's an oxygen free solution. Please do feed-back if this works - good luck! My thoughts and best wishes are with all in India at the moment....and indeed all who have been either directly or indirectly affected by this terrible disease wherever they are in the world!
  11. Hi there "Alwayslearning"! I've had a similar error in the past using ANSYS (or may have been NASTRAN...?)...when I gave up after banging my head against the wall for about two days, I got on to the Technical support guys... ...it turned out in my case that it was a migrogap between two edges (where there should not have been a gap). This resulted in the model be discontinuous from the analysis perspective. The model was redrawn and the problem solved. Hope this helps.
  12. Hi there...it is just a simple moment....on the fulcrum itself, it is the total force applied in the down direction. I'd calculate the moment on one side then the other, then total the vertical forces and the "reaction" is the sum. I would do both as - occasionally - the moment forces can exceed. typed in a rush sop hopefully makes sense - good luck!
  13. This made my day Dudley! The kid in me: - Fantastic!!!! The Engineer in me: - It may be able to run, but would you want it to...? The way I (have always) see(n) it is it has far too high a centre of gravity, far too narrow a track and - from the animation - a hyper extending knee joint, so if it did run, the would likely break its front legs, or fall over which - even stationary - is not a "nice" concept, but at speed... I never could "square" the design being an "all terrain attack transport".......Now the Walkers on the other hand... (AT-ST)...seem a far more "considered" design! Thank you for sharing this, you really have put a broad grin on my chops! Bruce
  14. I believe this is known as a "hydraulic bouyancy lifting engine" (the Hindenburg was an Air-buoyancy lifting engine) The easiest way to calculate the usable force is to try it out with a tube set vertically in water and catch the water that comes out over a set period of time (say 10 seconds)...the usable energy (Work) can then be calculated back from the volume of water that has been ejected. Work done is Force times distance (W = Fs). The volume of the water lifted would be linked to the mass of the water x G...this would be calculated as energy gained (PE) = mgh. With regards to the amount of force each balloon applies, it is a buoyancy, so it is the volume of the balloon that displaces water - the volume of water displaced (less the mass of the balloon and whatever is tethered to it) is the upward force. Compared to air, the water is very viscous, so the terminal velocity of the rising balloon will be limited suite quickly by the resistance of the water it is pushing through (your diagram states 3ft per second...?) The energy for the lowest balloon will be near as dammit equal to the pressure at the point of origin - very slightly higher - though it may have to be significantly higher depending on the elasticity of the balloon...the "danger" would be the balloons bursting as they near the surface due to the expansion of the air as the pressure drops. For reference "Electrical answers " is not quite accurate. Energy is measured in Joules. Watts is a measure of energy applied per second, so 1 Watt would be one Joule per second. 800 Watts is 800 Joules per second. in my first paragraph, the "mgh" is measured in joules. In the air-con unit, it benefits from a "free" energy boost as a result of an evaporative (or condensing) phase resulting in a significantly greater usable energy output than that input. Work done is also measured in Joules. The difference between the usable work done and the energy used is the efficiency. (Note for equations "s" is "Spacial displacement" - ie distance. Time is always "t". I know many new to the field get "s" and seconds confused, so its worth mentioning.) Unfortunately, this is nowhere near enough information for you to go on - I understand - but hopefully it is a start in the right direction. Unfortunately, there is only so much time that can be spent on "interest items" when work beckons! If I can spare some more time and expand on this, I will do, but my visits of late tend to be sporadic and short-lived due to other commitments, so there are no promises made! Good luck...
  15. This is a system that has been used for many years to provide an upthrust in dredging. This displaced water at the top provides a negative pressure at the bottom which is automatically pressure adjusted, and very - often dangerously - powerful! The pressure needed to create the bubbles at depth is slightly greater than the depth of the water, so there is still a net loss...but that does not mean that the system cannot be utilised do a specific end...but it should not be seen as a route to perpetual motion...e.g....if you can get compressed air bottled at a sufficiently low cost, you could use this system to displace water to power a hydro turbine. There are methods that do produce a net gain in energy utlilsiable...heat pumps utilise a phase transition to give an energy boost and a "net gain" against input energy. e.g. an aircon unit is an example of a heat pump...to quote one I've just looked ups...it takes an ~800W input and can give a heating or cooling output of about 3.3-3.4kW, a significant net gain! {Note: Not typos...British-English spellings!}
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