Mechanical Engineering

# G B Reid MIMechE, SIMarEST

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1. ## Write one line on centrifugal pump ?

They work just like the 1980s cartoon series "Battle of the Planets"....G-Force!

3. ## Determine the amount of cfm flow required to provide as in put tp main pipe (branched out into perforated pipes)

I shan't give you the answer, as that is doing the job for you, but I will pont you in the right direction...(besides, there is sufficient ambiguity in your description to make this rather awkward without lots of clarification) If you ignore compressability, model this as a "Constant Volume", you know the diameters of the pipes, and you also know how much total area the outlets are...the back-calculation is very simple from there! Good luck
4. ## What is the torque output force of this machine?

This made me Smile! As I said in my post, I have discussed this and had the detail in the "Seapower" sketches expanded on in the past, so I understand where the diagram is coming from and the idea itself...it is a little "unclear" from the above but the concept IS the "Seapower" diagram, the umbrella, a potential drive facilitator for the bubbles.
5. ## Force applied to fulcrum

I've just perused this again and had a thought....you're making a punch...have you calculated whether the force required to punch through the metal can be achieved? It's a CSA calculation (circumference x thickness) and the shear strength of the material in Pa Take the CSA (in m^2) and multiply by the shear strength in Pa to get the required force in N. Good Luck!
6. ## How to calculate the force required?

Hi Govardhanan! You will likely have problems with this design due to the foodstuffs not flowing sufficiently. If the foodstuff is too viscous, it will not exit through the holes at the bottom...in fact in some instances, the container will fail before it does. If you look at dispensers available commercially, you will find that the majority have a cone at the bottom to facilitate exiting of the material. Whilst you acknowledge that you are a student of mechanics, I am sure you will be familiar with the phrase "every action has an equal and opposite reaction." (Isaac Newton) . Basically, in this situation, all of the force from the piston is reflected back into the piston with nothing facilitating flow through the exit holes. If this exit was altered to a cone of - say - 60 degrees (radially), 86.6% of the force would push down and 50% of the force would be deflected to "feeding" the material through towards the target exit. [Yes that does exceed 100%, but that's trigonometric mechanical advantage] Note: I have not sketched this out and checked as have no paper at the moment, so it's a "finger in the air" calc. but at very least the principle is right if the figures are wrong... Good luck! Bruce
7. ## Vibration switch

Failing the accelerometer - which would be very successful if on a drum - you could connect a microphone to a transistor to switch on the light. It is a very simple circuit and is reproduced on a number of differing electronics websites. Basically the "switch" would be the microphone. They generally have a resistor in the circuit...if this was a variable resistor, it could be adjustable. daisy chain this to a timer circuit and you're laughing....or drumming! Good luck!
8. ## What is the torque output force of this machine?

I think the best option is to ask "what sort of buoyancy force could this generate....?" The buoyancy gives the force the speed of which would be limited hydrodynamically - basically by the speed of the bubbles rising. Asking "Torque" is not really helpful as this will depend entirely on the size of the cog used to drive the unit. Basically Torque is Force x distance. You need to first know the force and the systems can be modified for the torque sought....the speed you have already calculated. You have done most of the work on this....the buoyancy is identical to "The sum of" "the volume of the bubbles x the mass of water x g"; you have already calculated these at various depths: Basically, calculate the volumes, add them up and multiply by 1000kgm^-3 and then 9.81ms^-2 (Metric). This would calculate down and cancel to give "kg m s^-2" - which is Newtons - the fundamental unit of force in SI units. I apologise, we have communed on this in the past, I just haven't had the time to look into this in detail, and time I would need. I confess to being a "Metric only" guy, which basically means I'd have to spend the time recalculating everything to make head nor tail of it and verify your original calcs (which I would not convert, but redo to check). [Have to confess - I took your name, "tooldtocare" as being "tool'd to care"....I guess I must be a cup half full guy!🤣] Hope this helps Bruce
9. ## Question on Degree of freedom of wet region

This answer is from a kinematics perspective - if you are querying something other, please be clear in your query. In a system, if there is 1 DoF; there is 1 DoF. The "wetness" of the surface affects the coefficient of friction (mu) and is a non-dimensional multiplier...it will therefore have no effect on the number of degrees of freedom. I hope this helps.
10. ## Find the rate of change

<sigh>...I'll give you a clue...you're looking for "dI/dt"... (note: that's not an "L")
11. ## I need help

The most likely reason is "fatigue" (high quality close-up images of the broken surface could confirm this). If you bend a bit of metal back and fourth, it will snap...that's an extreme example of fatigue. Fatigue happens when an item is loaded and unloaded multiple times and can cause items to fail when there is very little actual use due to the cyclical nature...it is effectively crack propagation within the lattice structure of the metal's crystalline structure. [its not even "real" cracks, it is incomplete crystals shunting individual atoms across internally within the lattice...they multiply until they become significant.] It is a little complex in reality, but a general rule of thumb is if the item subject to fatigue is designed to Load/0.3, it should significantly increase the lifespan....and use the yield point for the calcs, not the UTS. Good luck Bruce
12. ## Which tensile strength should I use?

Hi Pablo Alas, I have no first hand experience with the Jominy, so I will not comment on this. (I only comment where I have knowledge or first hand experience...) That said, on the other, yes, use the figures for the 100mm, if incorrect, the error will be on the "safe" side. Bruce
13. ## Which tensile strength should I use?

Hi AliceIn....apologies for the delay, just read your message! ALWAYS err on the side of caution and use the lowest value unless absolutely sure! Warmest wishes Bruce

16. ## Temperature compensation formula

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
17. ## How to Measure Torque Required for motor to spin a rotating shaft?

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!
18. ## Mechanical Numerical of the day

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!
19. ## Shaft Design & Analysis

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!
20. ## Mechanical Numerical of the day : May 18 2021

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
21. ## Eco-cooler

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.
22. ## Mechanical Numerical of the day : May 18 2021

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....
23. ## Mechanical Numerical of the day

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.
24. ## Differential Pressure measurement

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
25. ## Alternative for thermal spray welding in scarcity of oxygen

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!
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