My generator on my ship if parallel votage drop but the rpm and frequency normal/steady how to solve.?
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.
71357
Volts-drop test.\n\nPut the armature on a V-block powered by a regulated voltage source.  Check the voltage across 2 adjacent com segments and repeat as you rotate thru 360 degrees.  They should all be the same.
526693
Yes. The FIELD windings (not the armature windings) of a generator quite often can over heat and "short" together, thus reducing the capable out put under load, no mater how fast the armature is rotated.
285339
i would check out the boots on the plug wires and the ign. box for leaks.you could have it running and squirt the meantioned items and see when it quits.
602130
This happened so long ago to suggest there should have been a standard is funny.  The frequency has more to do with the speed of the turbine and the number of poles on the generator.  Early US generation was actually 25Hz.  The larger the frequency the smaller the transformers and motors are, but also the higher the losses (inductive reactance is proportional to frequency).  So there isn't necessarily a perfect frequency to use.
739136
Go get a new 450+W PSU. This way, there are no worries. A calculator is linked below.
720670
It's not diufferent frequencies, it's just different times of 'beaming".  you can hear the magnetron kicking in. If you put it at 70%, you can hear it turning off and on.  WHne you do it at 100% it stays on all the time or at least more often.  The fan stays on I believe, but the waves are going off.
188032
possibly if you add more details- what are you working with?\n\nsend me a wiring diagram if you have one and I can help you- or tell me what wires (how many, colors, etc)- what kind of boat you are hooking it up in- all that jazz.\n\ngo to the site below and send an email to "Christian" and it will get forwarded to me. (at the forum)
551707
i have a presario 2500 and my fan runs hard on sertain oocasions. yours will be in overdrive due to the work the cpu is foing feeding the graphics card to run the game. What i do with mine is make sure its well ventilated underneath, either with a laptop stand, or a book so the famns underneath can intake as much air as possible and that the exhaust ducts are not blocked with dust or obstructed either. I have a stand with two cooling fans and mine works fine. As for the restarting, it could be a motherboard/power problem, or possibly you mighthave a virus.
738632
What do you mean by the outlet temperature not being uniform. Do you mean not the same over time? Or are there multiple outlets and the temperature from each outlet is not the same?\n\nIf the inlet and outlet channels or headers are not designed carefully to allow for good fluid dynamics then there can certainly be portions of the exchanger that is not getting the same flow as other portions.\n\nWithout more information, I can not see how an energy balance will help solve the problem.
903919
Generator sizing becomes more difficult when electric motors are involved, due to their large starting demand.  Typically, there are two motor styles: CODE G requires 3 times running watts to start, and CODE L requires 5 to 6 times running watts to start.\n\nIt is not practical to think that all loads will start at the same time.  Nor is it practical to size a generator using all starting watts for all selected motors, as some motor loads will already be running (or OFF), when others are ready to start.  As you can see, it can be difficult to exactly size a generator for a group of specific loads, as they are constantly changing in their "ON" or "OFF" cycle.  Always try to limit your load selections to the bare essential items.\n\nTO FIND:\n1 PHASE KW = V * A * PF / 1000\n3 PHASE KW = V * A * 1.73 * 0.8 / 1000\n1 PHASE KVA = V * A * 1000\n3 PHASE KVA = V * A * 1.73 / 1000\nMOTOR - REQD 1 PHASE KW = ( MOTOR HP) * (EFF)\nMOTOR - REQD 3 PHASE KW = (HP) * (0.746) * (EFF)\nMOTOR - REQD 1 PHASE KVA = (MOTOR HP) * (EFF) * (0.8)\nMOTOR - REQD 3 PHASE KVA = (HP) * (0.746) * (EFF) * (0.8)\n1 PHASE PF = 3 PHASE PF = KW / KVA\n1 PHASE AMPS = (KW * 1000) / (VOLTS * PF) =  KVA * 1000 / VOLTS\n3 PHASE AMPS = (KW * 1000) / (1.73 * VOLTS * PF) = (KVA * 1000) / (1.73 * VOLTS)\n \nREQUIRED PRIME MOVER (ENGINE) HORSE POWER\n KW / (GENERATOR EFF. X .746)\n \nFREQUENCY (HERTZ)\n NUMBER OF POLES X R.P.M / 120\n \n ~ ~\nEXAMPLES IN GENERATOR SIZING FOR LARGE ELECTRIC MOTORS \n\nAll AC electric motors require large amounts of electric power to start up, due to it's starting winding. After approximately 3/4 - 1-1/2 seconds, the starting winding drops out, and the running winding continues, at a much lower power demand.  A generator must be sized to handle the load of the starting winding.\n\nSizing a generator for three phase motor starting LRA : \nThe formula is: Motor HP x KVA per HP x (1000 / (Volts x 1.73)) = LRA. \nGiven the size of your machines, estimate KVA/HP at approximately 6.\nTherefore, for 11 kW ... LRA = 134.6 = 135 amps. Generator must produce this current at 380 V, so (1.73 * V * A * 0.8 / 1000) = KW = 71.\nSimilarly, for 15 kW .... LRA = 184 A and KW = 97.\n\nBOTH SINGLE PHASE AND THREE PHASE MOTOR SIZING EXAMPLES ARE BASED ON INFORMATION IN NATIONAL ELECTRICAL CODE HANDBOOK, ARTICLE 430.7 AND IS INTENDED FOR MAXIMUM VOLTAGE DIP OF 25% UPON INITIAL MOTOR START-UP.  IF 35% VOLTAGE DIP IS ALLOWABLE, REDUCE LRA, THEREFORE KW SIZE, BY 25% (LRA or KW X .75). IT IS NOT RECOMMENDED TO CHOOSE ANY GENERATOR KW SIZING THAT YIELDS MORE THAN 35% VOLTAGE DIP.\n\nStarting more than one motor, at one time: make a list of all motors, totaling all running watts.  Calculate the starting demand when the largest motor is started while all others are running.  Add the largest motor starting watts, to the running watts of all smaller motors already running.
541382
vertical oscillator/driver shot. shorted capacitor, or maybe transistor/ic? \nnowadays, you might as well trash it if you can get a halfway decent deal on a flat panel. or give it to your cousin who's MIT material to see if he can fix it.
694954
Variables: 150 rev/min  2.2h = 140 min \n(a) Difference in velocity (150 to 0) over difference in time (0 minutes to 140)\ndelta v / delta t   (150-0)/(0-140) = -1.0714 rev/min/min\nAcceleration is negative since it is slowing down.\n(b) during the slowdown? one less than the answer to (c)?\n(c) (v not times delta t) / acceleration absolute value  (150*140)/|-1.0741|  =  19553.07 revs\n(d) and (e) email if you really want, but this stuff should give you a start. Hope this isn't due for a couple days tho.
957415
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
12356
Sounds like static electricty, which tends to develop in windy dry conditions.  Check it out.
617956
http://1accordion.net/tuning.html
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.