how the single phase induction motor speed controll with the sgnal coming frome pressure sencer frequency cont
85096
The induction motor speed can be varied by varying the supply voltage as well as supply frequency. This can be done using variable speed drives. These drives accept 4 to 20 mA signal from process and vary speed from 0 to 100 % depending on the signal.This speed control is stepless.\nVariable speed drives work using IGBTs and Pulse width modulation techniques in firing of the thyristors to obtain variable frequency and voltage output.
309387
Since no-one else has answered the question yet, I'll take a stab at it -\n\nI think the term you're looking for is Hysteresis
759590
Shaft Speed of a motor= 120xF/P -Slip\n\n Slip for NEMA B Motor = 3 to 5% of Base Speed which is 1800 RPM at Full Load \n \n F = Frequency applied to the motor\nP = Number of motor poles \nExample: \n Shaft Speed = 120x60 Hz/4 -Slip\nWe can conveniently adjust the speed of a motor by changing the frequency applied to the motor.
207200
I found a couple of sites listed below maybe one of these will help you out.\n\nhttp://www.uoguelph.ca/~antoon/circ/hv/classic/classic.html\n\nhttp://spazioinwind.libero.it/gabinetto_di_fisica/elmag/elmaginduction.htm#modrocchetto\n\nhttp://physics.kenyon.edu/EarlyApparatus/Electricity/Induction_Coil/Induction_Coil.html
583473
Pretty simple. If you are going to employ a variable frequency drive or inverter to vary the motor speed you must then select an inverter duty motor. The reason is because motors are designed to cool themselves while running. Conventional motors are designed to dissipate heat at full speed (60 Hz). If a conventional motor ran at lower speeds then eventually it would get too hot causing insulation breakdown and then.....bad things happen. \n\nInverter duty motors have more insulation and can therefore handle the heat at slower motor speeds.
87578
P = EI (Power equals voltage E times current I)\n\nMany electrical properties can be explained as analogous to water flow.  For example, a transformer "transforms" high voltage and low current into low voltage and high current, keeping the power constant (except for minor losses) at full load.\n\nThis would be equivalent to a high pressure water pump supplying low water flow to the primary winding and taking out high water flow at low pressure from the secondary winding to drive a low pressure water turbine.\n\nYou could represent the transformer core as a hollow square of wood or even thick cardboard.  On one side wrap say ten turns of thin clear plastic tube around the core to represent the primary (first) winding.  On the other side wrap say five turns of larger diameter plastic tube around the core to represent the secondary (second) winding.  Fill the primary tubing with water and red food coloring to represent "hot" high voltage and the secondary tubing with water and blue food coloring to represent "cooler" low voltage.\n\nThe explanation of course is that the high voltage primary winding will have more windings of thin tube (similar to more thinner copper wire windings) because the current will be less than in the secondary winding.\n\nThere will be less windings for the secondary coil using larger tubing because the current will be higher at low pressure.\n\nSupply the secondary tube with a shut off valve (or clip) to represent the fact that there will be no current unless the circuit is open.  In the formula, if current is zero power is also zero (on both sides of the transformer).\n\nYou can represent the pump and turbine with pictures (or drawings) of same but it would not be impossible to rig an actual pump and turbine, just not necessary to "see" what is happening.\n\nThe turbine can use up more and more power until the maximum power of the primary winding is reached but not more.  The magnet of the core will be "saturated at full power."\n\nThe rest is explaining magnetic flux in the core and how it is produced by the primary winding and cuts through the wires of the secondary to produce voltage and current if there is a load.  Each winding of a coil adds to the effect.  A transformer with equal primary and secondary windings would have the same voltage and current under load.  I believe that the phase would shift 180 degrees because of lead/lag.   Good luck.
620886
(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
584010
E = 1/2 M x v^2 x F x H \n \nWhere v is the velocity of the load hitting the surface and H is the time period in which the frequency continues.
497883
Hey lets make it simple...\n\nwhenever u use only proportional controller and tune it to good gain then also u will have an offset error i.e ur Process value will never reach the setpoint, so if u introduce I (Integral) controller it will reduce this offset error or even makes it zero, but u gota tune it porperly, If u increase more time valve of I controller they system will become slow so be care ful if want more details then refer to  http://en.wikipedia.org/wiki/PID_controller
973544
It appears that you have asked about the synchronous motors that are used to run flourmills. \n\nThese motors can be initiated at a particular RPM only. For that, a small induction motor runs the synchronous motor and brings it to a particular RPM. As soon as it happens, the main synchronous motor is switched on.\n\nYou might have seen a lever switch for this setup… the operator operates the lever to the upper position for starting induction motor… then on assessing that the motor has reached the minimum required RPM he brings down the lever to the downward position for switching on the main synchronous motor.
730275
It's called 'impulse' and any good, lower division, physics book will have an explanation of what it is and how it works.\n\n\nDoug
175584
http://www.willamette.edu/~fthompso/MgmtCon/Control_Systems.html\n\n\nhttp://www.isd.mel.nist.gov/documents/falco/ITSecurityProcess.pdf#search='applications%20of%20digital%20control%20systems%20on%20different%20industries'\n\nhttp://www.cs.uidaho.edu/~krings/publications/SACT-2002-D.pdf#search='digital%20control%20systems%20in%20pipelines'\n\nTry them...
126809
first the back plate give you the rated of the motor,the coil risistance calculated by 4/1.1 is the resistance of coil in work that diffrent from that calculated in V/I methodes becuse in 4/1.1 the effect of armature reaction and armature back voltage therefore the coil resistance calculated in V/I is the right means 4.6 ohm,\nto calculate the turke (T=Po/w) where Po:output power wich approx. equil input power (100% effecency)=V/I,(( w=2*pie*N N:speed in rps
772586
You have to place a pressure regulator on the infeed line to each cylinder and then adjust the regulators until each cylinder operates at the same speed and time
231710
its electrical properties
721433
Edit: Perhaps I should revamp my answer to provide a clearer (actual) answer, that explains exactly what plays a role and, in effect, why I am right. I also added a couple sources you should read (in case you don't believe me ;-).\n\nAs asked at the first source (which has very good information about electric motors), "How can I make the motor spin faster or slower?". The answer is that it has to do with the power of the magnetic field. Because electicity is magnetic, you can either effect the speed by using a more powerful magnet or a more powerfully magnetic electrical current.\n\nTo quote Wikipedia about what a volt is, "The volt is the SI derived unit of electric potential difference". To truly understand what the "electric potential difference" means, you must first better understand the volt. The volt is a measurement of electromotive force (emf) and is just defined in terms of amperes and watts.\n\nAn amp is the constant current which produces a force equal to 2x10^-7N per meter, when travelling across two parellel wires one meter apart. It is one coulomb of charge per second (Q/t). It is a base unit upon which other units of electicity are defined (such as the volt).\nA watt is the unit of power equal to one joule of energy per second (P=W/t).\n\nHmm. Power? Power of a magnetic field? So how can you get more power in your circuit? If your power source is not rated by power output, then you can figure out the power using convertion:\nA=W/V=C/s\n\nIf you want a higher wattage, you can increase the voltage (electrical potential) or the ampage (current) or both. Or you can increase the charge - the amount of electricity (coulombs).\n\nSo the fact remains that voltage is *not* directly related to the speed of a motor. It can be directly converted in a way that makes it useful for that purpose, but any number of other things can be done to increase the power - the real goal you are looking for.\n\nThe answer is a definite "No." and a half-assed "Yes.".\n\n\n---Original Answer---\nNo. You might want to read more about the electric motor ( http://en.wikipedia.org/wiki/Electric_motor ) and magnetic properties of electricity ( http://en.wikipedia.org/wiki/Electromagnetism ). Also see the definitions of volt ( http://en.wikipedia.org/wiki/Volt ), watt ( http://en.wikipedia.org/wiki/Watt ), and amp ( http://en.wikipedia.org/wiki/Ampere ).
193990
temperature affects the frequency..\n\nPV = MRT\nP: pressure\nV: volume\nM: mass\nR: rate of mole\nT: temperature