bcjm February 1, 2003 Share February 1, 2003 Can you calculate these unit from one to another? I am interested getting an air pump. The spec on the pump says 1/8 HP, 2 AMP. The manufacture's site says it only consumes 90W. I am just very surprised that a motor uses 2A only takes 90W. Link to comment Share on other sites More sharing options...
Aquariareview February 1, 2003 Share February 1, 2003 A watt is an electrical unit of power. This term is commonly used to rate appliances using relatively small amounts of electricity. The standard formula is: Wattage = Amps x Voltage 2 amps X 120 volts = 240 watts hope that helps Link to comment Share on other sites More sharing options...
Coral Hind February 1, 2003 Share February 1, 2003 When the outermost shell or valence shell contains less then eight valence electrons the attraction of pos charge protons and neg charge electrons.... Never mind! :D One HP = 746 watts or One Watt = .00134 HP HP is used for motors while watts is used to rate generators, regulators, and household appliances. Link to comment Share on other sites More sharing options...
Aquariareview February 1, 2003 Share February 1, 2003 Does anyone know how many photons are generated by the $5 airpump that runs the undergravel filter in my 2 gal betta tank. I think it needs new dylitheum crystals Link to comment Share on other sites More sharing options...
bcjm February 1, 2003 Author Share February 1, 2003 1/8HP divided by 746 = 93.25 Watt. So it is correct then that this 1/8 HP pump uses 90Watt like the factory said. Why does it say 2 AMP on the pump? Maybe it is the starting AMP or the AMP under the max. pressure. Link to comment Share on other sites More sharing options...
Coral Hind February 1, 2003 Share February 1, 2003 The 2A rating is likely the max amp draw for the motor and is for over current protection purposes. Link to comment Share on other sites More sharing options...
Guest Scott324 February 2, 2003 Share February 2, 2003 The air pump can draw 90 watts at 2 amps if the voltage is 45VAC. This is unlikely but the formular (Ohm's Law) is P = I x E where P is power(watts), I is current (amps) and E is voltage (volts). I would tend to lean on the towards the max or load amperage, but anything is possible. It should tell you what the operating voltage is. Disclaimer: The Power formula (p=ie) is only used as a reference, many other things are involved to figuring power or any part of the equation. Impedance(resistance) is a big one, but when using ballasts (transformers), there is the power factor which greatly concerns us. HTH, electricity 101, LOL Link to comment Share on other sites More sharing options...
Coral Hind February 2, 2003 Share February 2, 2003 Ohm's law has a "P" in it? E=voltage I=current(amps) R=resistance(ohms) W=watts The German scientist stated, "1 volt is the pressure required to force 1 ampere of current through a resistance of 1 ohm." If you know any two of the values you can find the other. Here is a calc for ohms law. http://www.anderson-bolds.com/calculator.htm The power factor=true power/apparent power applies to any AC circuit not just ballasts. HTH electricity 202 :D Link to comment Share on other sites More sharing options...
pez February 2, 2003 Share February 2, 2003 Ohms law is V=IR. There are many ways to write it, but W=VA (the West Virginia law) is common for figuring out total "power". W = watts V = voltage I = current A = current (amps) R = resistence Horse Power can not be related directly to electrical power without a lot of assumptions (most of which are generally valid for what we are doing with our tanks). So David's normalization factor for HP to P is probably accurate for us. He definately seems to know more than me about it. -T Link to comment Share on other sites More sharing options...
Guest Scott324 February 2, 2003 Share February 2, 2003 Fred, I guess that's why you are still an electrician and I am not anymore :D :D P stands for power which is a derived from Ohm's law, P=IE or at least that is the equation I have always been taught. I guess you could use W=IE too. We didn't do much with power factor when I was in, but I learned a lot from my timer burn up last summer. The advance people and a couple electrical engineers told me it was only applicable to transformers and ballasts. They said it was the correction of the Sin wave because it was disturbed during the magnetic field transfer that occurs in most ballasts. I guess this was just a lamens definition, but am unsure. In a normal AC circuit, how would true power and apparent power be different? Link to comment Share on other sites More sharing options...
Coral Hind February 2, 2003 Share February 2, 2003 This is a simple example as I don't want to do math all day. Scott if you still have CDC vol 2 elec. fundamentals I think it is in there. A motor is hooked up to an amp, volt and watt meter. We see 5A, 120V, 550W on the meters Being from West by God Va I'll use Pez's W=VA. 120Vx5A=600W Power Factor=true(measured)/apparent(equation) This motors PF is 550Wx100/600W=91.6% The cause is the out of phase condition of the sine caused by capacitance or inductance (reactive power) in the circuit. There will always be a slight PF but it only plays a big role on coiled wire such as motors and transformers. The reason I said the 2A of the air pump was for overcurrent protection is because if you use the above equation the pump would have a 37.5% PF. Very bad! 62.5% of the energy is wasted. Link to comment Share on other sites More sharing options...
Guest Scott324 February 2, 2003 Share February 2, 2003 Fred, you should know I had a nice book burning March 13 2001. The day after I got out of the military. The CDC's are the last thing I would keep on my bookshelf. :D :D :D :D :D Link to comment Share on other sites More sharing options...
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