How to calculate power requirement of a Rotary AirLock/Rotary Feeder for conveying dust?
473367
I have no experience in rotary airlock design but having several years experience with 'baghouse' filter systems in cement manufacturing plants, I can describe them. Email me if you get no answers.
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
777054
The product of flow and pressure put out by the power pack is approximately the power available to do work at the remote actuators.\n\nIn a real system, piping, fittings, filters, etc. cause some pressure drop, so the actual power available is somewhat less.
560205
Please refer this link \nhttp://www.sepialine.com/docs/html/Power%20Logic%20Connector%20Manual.htm
511307
well you need the power factor as well, From O/P KW to I/P KW by I/P KW = O/P KW / Efficiency; I/P KVA = I/P KW / power factor; I/P Current = I/P KVA / (V3 x 480); V3 is a 3 phase motor is used. so simple.
1084947
about 6000, it takes 1500 horse power just to turn the blower, they come off the line pulling 5 G's.  The exhaust alone gives the rear wheels something like 200 lbs. of downforce.  0-100 mph in less than .8 of a second.
279585
http://www.mkssupply.com/manuals.asp\nThis should do it for you.
1033944
Efficiency=Poutput/Psupplied\nEfficiency=3760/4000=0.94\nEfficiency(%)=94%\nThe answer is B
290101
Join the EVGray yahoo egroup.  There you will find people who can not only answer this question but teach you how to make it a Rotoverter that runs on a fraction of the wattage and can eventually be made to run "Overunity" with an identical motor for a generator putting out more power than it requires to run.  These Rotoverters are 3Phase motors wired for 3phase but run on normal house 110vac at lower horsepower rating but generating more power than they consume.
1032301
Fan, then chicken wire, then hay, then chicken wire.\nMake the hay and wire the same size square as the fan.\nThe hay should be about 6" thick.\nPlastic pipe with 3/64 holes every 6" across the top of the hay.\nConnect a hose to 1 end and cap the other.\nTurn on water little by little until you get a trickle from the holes.\nTurn on the fan.\nSwamp cooler!
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
627569
It's a simple matter of adding up the power required by everything, EVERYTHING, you want connect to your supply.\n\nMachines, tools, heating, cooling/air conditioning, ventilation fans, lighting, computers, printers, copiers, telephones, fax machines, answering machines, refirgerators, freezers, water coolers, security devices, etc, etc, etc.\n\nChoose a generator with a capacity at least a bit greater than your total load, but preferably quite a bit more to allow for future load increase.
564626
Which depends on the stres acted on the load and material used
499722
Conveyor belt systems may be used to fill hoppers which feed the burners in boilers.  The most common common way to get coal actually into the combustion chamber is with an auger feed system.  Like a meat grinder, the conveyor will feed the hopper, and the augers will deliver it into the burner.  Other systems combine the conveyor, hopper, and auger with a grinding system to make coal dust which is "blown" into the chamber with a fan.  Check out the website.
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
531203
prooly need ifrared thingy\ntry a industrial magnet
99832
i believe it is around 9.5%.\n\nBecause Efficiency = O/p Power / Input power * 100.\n\nOutput is the given one \n= 4 * 104 J/min * 3 * 60 min = 74880 J \nInput is  = 75 * 1076 * 9.8 J ( mgh - mg * 0 = Potential energey) \nthefore, eff. is 9.5%\n\nplease reply at autoprash90@yahoo.co.in