how can i control the torque of a dc generator to setting point while varying shaft RPM?
107875
Limit the load current.\nEDIT: This is often done by adjusting the stator field current if it's a generator with an electromagnet (not permanent magnet) field.
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.
117405
Generators (AC or DC) can be manufactured to produce output from either the rotor or the stator.  If from the stator, the output is AC and one or more diodes are needed if DC is wanted.  If from the rotor, the current within the rotor is AC and can be (1) picked off with slip rings, providing AC, and converted to DC with diodes if needed, or (2) picked off with commutator and brushes if DC is needed.\nSo one answer to the question would be, ...except that any DC generator must use commutator/brushes or diodes.\nI guess you have listed multiple-choice options.  If so the only correct answer is #2.
1061376
When the shunt motor reaches full rpm, its speed will remain fairly constant. The reason the speed remains constant is due to the load characteristics of the armature and shunt coil. You should remember that the speed of a series motor could not be controlled since it was totally dependent on the size of the load in comparison to the size of the motor. If the load was very large for the motor size, the speed of the armature would be very slow. If the load was light compared to the motor, the armature shaft speed would be much faster, and if no load was present on the shaft, the motor could run away.\n\nThe shunt motor's speed can be controlled. The ability of the motor to maintain a set rpm at high speed when the load changes is due to the characteristic of the shunt field and armature. Since the armature begins to produce back EMF as soon as it starts to rotate, it will use the back EMF to maintain its rpm at high speed. If the load increases slightly and causes the armature shaft to slow down, less back EMF will be produced. This will allow the difference between the back EMF and applied voltage to become larger, which will cause more current to flow. The extra current provides the motor with the extra torque required to regain its rpm when this load is increased slightly.\n\nThe shunt motor's speed can be varied in two different ways. These include varying the amount of current supplied to the shunt field and controlling the amount of current supplied to the armature. Controlling the current to the shunt field allows the rpm to be changed 10-20% when the motor is at full rpm.\n\nThis type of speed control regulation is accomplished by slightly increasing or decreasing the voltage applied to the field. The armature continues to have full voltage applied to it while the current to the shunt field is regulated by a rheostat that is connected in series with the shunt field. When the shunt field's current is decreased, the motor's rpm will increase slightly. When the shunt field's current is reduced, the armature must rotate faster to produce the same amount of back EMF to keep the load turning. If the shunt field current is increased slightly, the armature can rotate at a slower rpm and maintain the amount of back EMF to produce the armature current to drive the load. The field current can be adjusted with a field rheostat or an SCR current control.\n\nThe shunt motor's rpm can also be controlled by regulating the voltage that is applied to the motor armature. This means that if the motor is operated on less voltage than is shown on its data plate rating, it will run at less than full rpm. You must remember that the shunt motor's efficiency will drop off drastically when it is operated below its rated voltage. The motor will tend to overheat when it is operated below full voltage, so motor ventilation must be provided. You should also be aware that the motor's torque is reduced when it is operated below the full voltage level.\n\nSince the armature draws more current than the shunt field, the control resistors were much larger than those used for the field rheostat. During the 1950s and 1960s SCRs were used for this type of current control. The SCR was able to control the armature current since it was capable of controlling several hundred amperes. \nDc motors have a maximum standard speed range by field control is 3:1, and this occurs only at low base speeds. Special dc motors have greater speed ranges, but if the dc motor's speed range is much greater than 3:1, some other control method is used for at least part of the range
617956
http://1accordion.net/tuning.html
1187149
By "improve", I assume you want total power factor seen on the power "inputs" to your business or plant to more closely equal 1 (unity).  \n\nBy playing with the field excitation of your synchronous motor, you can make the unloaded synchronous motor either act as a capacitive load or inductive load.  Typically, other loads in a plant (transformers, induction motors, etc.) are slightly inductive, and the desired power factor correction needs to act as a capacitive load.  In this case, you would want to overexcite the synchronous motor field (on the rotor) so that the rotor "leads" stator current (actually, unity power factor on a synchronous motor is realized when the rotor field lags the stator field current by 90 degrees, thus "leading" here means that the rotor lags the field current by less than 90 degrees).  Play with your motor field excitation until the out-of-phase current in the rest of your plant is canceled by the leading current being produced by the synchronous motor.  For more information, look up keyword "synchronous condenser" on the web.
973437
You should use pulse width modulation (PWM).  PWM is essentially an on-off sequence in time, like a square wave alternating between 0 and 1.\n\nClearly full voltage will give you maximum speed.  This happens when the voltage is always on.  Now, suppose you only wanted to go half the maximum speed.  You could send a a square wave with 50% duty cycle (on half the time and off half the time) at high enough frequency that the motor will spin at a constant 50% maximum speed.  The motor is like a lowpass filter and therefore, a square wave at high enough frequency will produce a steady response without oscillation.\n\nIf you want the motor to go slower, say, at 25% maximum velocity then send a pulse that is on 25% of the time and off 75% of the time.\n\nIn reality, if you want 25% speed you may have to send something like a 27% duty cycle because of the time delay between commanding 'on' and the system actually turning 'on'.  In this application, though, I don't think you will have this porblem.
466769
Many times, generators are governed to maintain voltage and frequency. If this is your case, it is not surprising that RPM (and thus frequency) are constant. If your generator does not have a constant voltage regulator on it, and most don't for good reasons, then you would expect to see the voltage drop as load (current) in the busses goes up. By having a voltage regulator that droops when loaded, it allows two generators to safely be hooked together (paralelled). You could test this by starting and stopping loads on your system to see if the voltage rises and lowers during the test.
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
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
531796
You would need multiple alternators plus an inverter to accomplish this.  A common Delco GM alternator puts out 120A at 13.5-14V DC at an engine RMP above 3000.  You would have to put the DC through an RV inverter (for use camping where you hook up a battery bank and get AC out).  Assuming inverter efficiency of 85%, you would need 7475 (or more if e is less than 85%) DC watts.  You would need 5 alternators to power your inverter.  The problem is the inverter is VERY expensive, as in thousands of dollars.  \n\nThe problem with using a car alternator as an AC source is it is THREE PHASE.  I have had people disable the regulator and get high AC voltage out, but it is three phase.  You would have a difficult time clocking multiple auto alternators so the phases all matched, and then you would only have 3 phase voltage (probably less than 220V but 120V) and it wouldn't be regulated.\n\nFor 6500 watts you need a belt driven gen head.  See northern tool link to belt driven 25KW head.  Also look at their inverters for taking DC and making AC (yeah, I know--americans use 120 but I'm sure you could find a supplier to go to 240)\n\nBut dollar for dollar, you would spend more to fool around with DC alternators than just getting a gen head which was rated at the wattage and voltage with self-regulation.  ALSO you must REGULATE the RPM.  SO just hooking it to a car engine isn't going to work without a governor, which car engines don't have unless you preserve the cruise control option and hook up the speed sensor (which may give weird results at high engine RPM such as 3600 for 60HZ or 3000 RPM for 50Hz).  NOTE: my links are for 120VAC at 60Hz.  Europe will have different specs.
1077557
82 kVp with 25mAs  (3 pulse or better generator) and GOOD positioning
446286
This type of ac motor uses the shading coil (the shaded pole or poles) to change the phase shift of the magnetic flux or field.The rotor is usually a laminated cylinder.Whe the induced currents in the rotor are out-of-phase a torque is produced on the rotor.The electric meter on a house uses this type , and the rotor is the aluminum disc that rotates.It can be shown that when ther is a phase shift caused by the closed copper turn(s) that a torque is produced.
96754
Im not really clear what you mean by settings but the best physical part swap for acceleration in ring and pinion swap to lower the final drive ratio which means the engine revs faster hope i could help
1023698
You do not.  Get an AC to DC converter then run the motor.
153367
most certainly you can get one from one of these suppliers:\nhttp://www.mmacactuators.com/\nhttp://www.motioncontrol.com/\nhttp://www.rotork.com/products/linear/index.cfm\n\ni found these by just searching around on the web.  the catalogues might specify the technical details you want to know, but in almost all cases like this, the thing you want to do is CALL the place up and ask to speak to a sales engineer.  they are a person who works in sales but has engineering training so they won't go "durrr" when you start talking about Newtons and so on.  their job is to help you in exactly the way you need.\n\nanyway good luck!
89951
Just as with any other generator, electrical power is proportional to shaft torque and speed.  A wind turbine generator is designed to operate in a certain speed and torque range.  The design of the turbine and pitching of the blade angle keep it operating in this range.  It's actually wind speed that forces the control of torque/speed up to a certain maximum level, so yes, electrical output is directly coupled to wind speed and energy capture of the wind.  \n\nThe generators in wind turbines do generaly require some amount of excitation current to regulate power output.  This is really what allows wind turbines to operate at different power outputs and not a single fixed rating.