I have been given some coursework which has rather puzzled me, and I'm not sure how to tackle the probem: Any tips that would get me started would be ideal! I think I'm going to have to get this question clarified by the lecturer, as no-one else seems to understand this either - we haven't done much electronics on the course yet.
Its that simple? So 45Kva = 45000W... therfore the first one is running at 391.3A (assuming single phase)?
Ack... now I understand..... This thing is running at 3 phase 115VAC and I'll have to use some sort of Power Factor - I'm learning as I go along here, so I'll check this out. I need the Watt rating and kVA to work out the power factor, would I use the 200A and 28VDC to get the W for the first one?
I know that if would make sence that VA=V*A=W but according to http://www.technology.niagarac.on.ca/people/bgracey/voltamp.html 1W =1.4VA this UPS is the one I have, and as you can see it says 525 VA and only 300W. (and of course 300*1.4 != 525 ) However given that I can kill it in ~9 min I would think that I pull more than 300W. EDIT: yeah I bet there is something in there that is beyond me, so check it out some more.
True, but you'd expect the problem to give you the power factor if they want an accurate answer. which is your 1.4 (near enough). Or if it's 3-phase, http://www.electrical-contractor.net/ubb/Forum7/HTML/000272.html
OK, cheers I'll leave the Power Factor out for now and see how it goes - as I can't see any other way either, but the 2nd part of the equation seems to suggest it. Thanks for your help
Note that the specification doesn't include the capacity of the battery, the fact that you can drain the battery in 9 minutes doesn't mean that you're using more than 300W. 525VA is the apparent power rating of the inverter circuitry.
Be carefull... Watts and VA are two slightly different animals. If the load on the three phase AC supply is purely resistive then Watts will exactly equal VA. If its reactive somehow (as nearly all real life loads are) then VA will be larger than Watts. The ratio of Watts/VA is your power factor. Having said that... since they dont give you any information about what type of loads you are powering or give you a phase angle of the current, you can assume them to be "ideal" loads and use Watts and VA interchangably - for now. But be aware there is a difference as I'm sure you will come across before too long in your course.
I think the power factor is a LOAD parameter, not a supply parameter. Thats why for a pure resistive load, the power factor (pf) is 1. The 1.4 value suggested above is the typical power factor for a computer / monitor load. CORRECTION: I meant that (1 / 1.4, which is approx. 0.715) is the typical power factor for a computer / monitor load. As another user sais, the power factor is a number from 0 to 1. The extreme values, 0 and 1 are for ideal condition (purely inductive load or purely resistive load).
Just a note... The highest value the power factor can have is 1 or 100%. Also, in the link posted above, he mentions that VA is "just another unit for measuring power". VA is NOT another unit for measuring power. It is possible, in fact quite common, to have a load that draws 100A or more on a 347V/600V 3 phase power system, which works out to about 104kVA, and consumes near 0 Watts.
Sheesh, and you guys manage to not blow any electronics up?? I'm going to define power factor as two different things, you'll see why in a minute. 1) Power Factor is the cosine of the phase difference between the current and voltage with respect to the voltage. (IE: Current lags voltage, or leads voltage) 2) Power Factor is the ratio of average power [measured in W] over apparent power[measured in VA]. This is horrible simplified, but apparent power is how much power (watts) your load actually sees, and average power is how much power is actually available if the Pf=0. For a resistor, ohm's law tells us that I∟Θ°=V∟Θ°/Z where Θ is the phase angle of the voltage with respect to (wrt) a reference angle and Z is the impedance, which for a resistor, Z=R. Thus, the current and voltage are both in phase. Average power is equal to |V||I|cosΘ° [in W], Θ for our resistor is 0°, therefore cos(0°) = 1, and P=|V||I| which everyone knows. BTW - all my V & I values are RMS values. Now, for a capacitor, I=C∂V/∂t, those who know some calculus will know that if V is a sine wave, then ∂V/∂t is a cosine wave, and cos=sin-90°, thus the current leads the voltage by 90°. Now the Pf for a purely capacitive circuit is cos(90°) = 0 which is specified as '0 leading'. The average power is |V||I|cosΘ° [W], and the average power is 0 W. However, the apparent power is still |V||I|, which is not 0VA. This is where people get confused. Inductors, V=L∂I/∂t, thus current lags voltage by 90°. This time, cos(-90°) is again equal to 0. The Pf would be specified as '0 lagging'. Again the average power is 0W, but the apparent power is not 0VA Summary: Pf has range: 0 lagging - 1 - 0 leading VA is NOT equal to W Guessing the Pf is not good enough, as it can change your answer by a factor of 100%. There is no 'set' Pf, it is entirely dependant on the source/load conditions. I'm currently working out the equations used to describe how a synchronous AC generator can also be used for power factor correction at the source end, but that's a few posts away yet. Sorry for the circuits class, it bugs me when people say VA is the same as watts. Now, down to the actual problem. This is all completely irrevelant,is not needed. The question says that 1SHp = 750VA, thus you can do all of your calculations in VA units, you don't even need to touch watts. If the question asked for 45kW of power, then you'd be stuck.
Fandu, Thanks for the trouble of adding the math to support my statement... I guess I should have taken the time to do this.
Does this make sense then, as I used Watts: Integrated Drive Generator Assuming a single phase voltage and Power Factor to be unity (as the brief states “required power” and not actual power). Therefore, 45kVA = 45000W 200A * 28VDC = 5600W Total Power Required = [45000 / (0.85 * 0.85)] + [5600 / (0.85 * 0.85 * 0.90)] Total Power Required = 70895.8W = 95.072 SHP Total Power Dissipated during power generation = 70895.8 – (70895.8 * 0.852) Total Power Dissipated during power generation = 19673.6W Total Power Dissipated during power conversion = [(70895 * 0.852) – 45000] - 5600 Total Power Dissipated during power conversion = 622.2W Variable Frequency Generator Assuming a single phase voltage and Power Factor to be unity (as the brief states “required power” and not actual power). Therefore, 60kVA = 60000W 200A * 28VDC = 5600W Total Power Required = [60000 / 0.85] + [5600 / (0.85 * 0.90)] Total Power Required = 77908.5W = 104.477 SHP Total Power Dissipated during power generation = 77908.5 – (77908.5 * 0.85) Total Power Dissipated during power generation = 11686.275W Total Power Dissipated during power conversion = [(77908.5 * 0.85) – 60000] - 5600 Total Power Dissipated during power conversion = 622.2W Overall Efficiencies IDG Efficiency = 50600 / 70895.8 = 71.4% VFG Efficiency = 65600 / 77908.5 = 84.2%
