A 5.0 horse power would be equivalent to an electric mower rated at what power? (it will be measured in watts)
442636
While you can convert HP to Watts, electric motors are also rated in HP.  A 5HP motor is pretty big, and won't use a normal wall outlet (15A, 120V).
756742
kilo is the prefix meaning 1,000, thus one kilowatt would equal one thousand watts.... if this is for a class you will probably run into some more numerical prefixes so ill give u a few more...\n\nassuming the standard unit is 1:\ncenti-100th\nmilli-1000th\nmicro-1,000,000th\nnano1,000,000,000th
897028
1/500th or less. You can check this out by looking at an auto supply catalog, finding out how much current is drawn by the fan under loaded conditions, determine the current draw (presumably at 12 volts), and find the wattage.\n\nThis can then be converted into horsepower by multiplying watts X 0.00134.
950109
In engineering practice we normally use kilowatt units for calculations. But horsepower is used for commercial terms and to categorize a motor. The conversion factor is 0.746 KW = 1 HP and 1KW  = 1.3405 HP
171175
P=IV\n =12.5*120\n =1500W\n =1.5KW
717868
5 A is wished at the least. It depends strongly of your electrical needs.
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.
348003
Have a look at http://science.howstuffworks.com/power.htm
204829
2.6096\n1 horsepower= 0.7456 kilowatt
63042
Electric power has two components:\n\nActive power, which produces work.\nReactive power, which is needed to generate magnetic fields required for operation of inductive electrical equipment, but performs no useful work. \nActive power is measured in KW (1000 Watts) \nReactive power is measured in KVAR (1000 Volt-Amperes Reactive) \nTotal power is measured in KVA (1000 Volt-Amperes)\n\nPower Factor is the measure of the relationship between Active and Reactive power at the point of consumption.\n\ncheck out powerfactorhelp.com -- pretty clear and simplistic
1186419
That would depend on the voltage. Current = Watts / Volts\nA couple of examples:\n110V - 90.9 Amps\n220V - 45.45 Amps\n240V - 41.67 Amps\n440V - 22.7 Amps\n10,000V - 1 Amp
280553
You cannot compare those two together. One watt-hour is equivalent to one watt of power used for one hour. Watt alone refers to the power produced which equals the voltage multiplied by current therefore:\n\n1 megawatt hour = 1000 Kwh\n1 megawatt  = 1000 Kw
340440
power is the amount of work done in unit time \nit is measured in Joules per second , dude
662957
Many people are often confused about power and energy.\nLet's start from the beginning, with some basic equations:\n\nEnergy = Force * distance\n\nFor a mass m moving at a speed v, the energy E is:\n\nE = 0.5 * m * v * v\n\nEnergy can be measured in Joules (J) or British Thermal units (BTU). 1 BTU is equal to roughly 1055 Joules. If a 2100 pound car is travelling at 60miles/hour, the energy is about 342600 Joules or 325 BTUs. This is the same amount of energy required to accelerate the car from 0 to 60 miles/hour on level pavement. This is independent of acceleration, it doesn't matter if it takes 6 seconds or 60 seconds, it's the same amount of energy.Energy is always conserved, meaning it can't created or destroyed, it can\nonly change form.\n\nPower = rate of energy\n\nPower can be measured in Watts (W) or Horsepower (HP). 1 HP is the same as 746 Watts.\n\nPower, unlike energy, is not always conserved. If we have a system where energy is flowing at a constant rate, then power is conserved, meaning the power in is equal to the power out. For standard cars, power from the engine is delivered to the wheels through a transmission. There are some losses due to the transmission, air conditioner, alternator, and other systems, but this is not considered a significant loss. If the air conditioner is turned off, your headlights are off, and you have a manual\ntransmission, the losses will be not more than a few percent. Automatic transmissions only increase losses.\n\nThe gear ratio is a transmission does not change power output. Remember,power is conserved in this case. There might be some slight change in power loss for different gear ratios, but for the most part, it's insignificant.\n\nSince we know power is the rate of energy, we can divide the energy calculated above by the time required to produce the energy. If it takes your car 6 seconds to go from 0 to 60 miles/hour, then the power is about 57000 Watts or 76.6 HP. Keep in mind, this is the average power, it doesn't\ntell us the power at any time during the acceleration. We would need detailed speed measurements from start to finish to calculate that.\n\nGiven these numbers, I would guess the engine is capable of producing 150 HP or more. How did I come up with that number? Internal combustion engines, like gasoline, Diesel, and even turbines, do not have a flat power curve. The power output is zero at zero RPM and increases more or less\nlinearly until some peak. In stock gasoline engines, the peak power is at 5000-6000 RPM or so. As a rule of thumb, the peak is generally 1000 RPM below red line. So if you could run your engine at 1000 RPM below red line,\nthen you could achieve near peak performance. If watch your tachometer, I would guess that the RPM averages much less than that, maybe even half.\nThat's why you are getting only half of the power the engine is capable of producing.\n\nRPM, power and torque are all related. Torque is simply power divided by RPM. The power out of a transmission is the same as the power in (minus small losses), but the RPM is different depending on the gear ratio. If the output shaft spins at half the speed of the input, then the output torque is double that of the input. RPM doesn't really matter in the end as far as power is concerned, remember, power is conserved. Changing gear ratios does not change the power, it only affects torque and RPM. It's easier to start with calculate power and work out RPM and torque if you need it. You can do it the other way, but it's easy to make mistakes.\n\nThe reason most cars have a transmission at all is because gasoline engines cannot produce torque over a wide range of RPM, particularly at zero RPM.Diesel electric locomotives and subways do not have transmissions with variable gear ratios because they use electric traction motors instead of\ntransmissions. Electric motors can produce torque over a wide range, even at zero RPM. This is known as the stall torque
865403
Alright Alright...I'll calculate this for you! Whatever! Ok...What 4x4 are we talking about here?\n\nAssumption:\n4x4 = JK Jeep Wrangler.  Here are the specs from Wikipedia:\nA 3.8 L OHV V6 producing 205 hp (153 kW) and 240 ft·lbf (325 N·m) will be the base engine, replacing the venerable AMC Straight-6 engine. \n\n205hp=153kW!!!!!  I can tell you straight up that to produce this amount of power from solar panels would be ridiculous!  If you go up to Home Depot and ask them for a 153kW System for your house, they will tell you to get lost! The average solar panel at PEAK Efficieny can produce roughly between 60 - 110 Watts! \n\nThus lets take the best case scenario.  At peak efficiency:\n153,000 Watts/110 = 1,390 Solar Panels!!\n\nSo...to power up your 4x4(Of course that 205hp to the crank and not to the wheels!) you would need around 1,390 solar panels.  Not that bad though!
147090
I believe the rating is 345 HP
573659
"Pounds of Thrust" is a measure of Force.\n"Horse Power" is a measure of power.\nThey are two different units, you cannot convert them directly.\n\nPower is calculated as energy per unit time (Joules / second).\n\nA force applied over a distance will do work (ie. expend energy).  If you knew the distance over which the constant force was applied, you could calculate the work done.  By dividing this work by the time it took to preform it, you could calculate the power output.\n\nA force will cause a mas to accelerate.\nForce = mass * acceleration\nIf a constant force is applied to a mass, it will cause a constant acceleration.\nIf a mass is undergoing a constant acceleration with some known initial velocity (even zero), one could calculate the time it took for it to travel a given distance.\nd = 1/2 at^2 + v*t\nwhere d is the distance traveled, a is the acceleration, t is the time, and v is the initial velocity in that direction.