# How to estimate Apparent Power

#### DickCappels

Joined Aug 21, 2008
9,614
If you have many similar loads you can probably do just fine by measuring the RMS voltage and RMS current of one unit and then multiply the apparent power factor by the number of units.

You would use either a RMS voltmeter or RMS ammeter (which could be a voltmeter measuring the voltage across an low resistance shunt resistor).

You can also buy watt meters that show real and apparent power or real power and power factor.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,695
They are EXACTLTY the same thing, physically and numerically. It is why we use RMS values when doing computations with sinusoidal waveforms.
I'm really sorry, I'm very ignorant regarding this subject. I never liked this subject and I struggle to understand it.
I don't understand how that value of 14.07 can be the kVA value I'm looking for! If I need to have an Apparent Power of 14.07kVA, I'm completely screwed. But I don't think it is!

Then you can take the straightforward kVA rating of the power supply as one thing you will be powering. Do you really expect to have every device you own turned on at once? If you are going to run the mining rig 24x7 then you can consider that the baseline and work out the peak demand you expect from the other devices.

In other words, there is no reason to apply a derating to the power supply, just take the kVA rating and use it as a given baseline requirement, then add on the other devices to decide what your peak load will be, and what you can tolerate. Either you are going to buy enough power for the peak load, or you are going to have to manage your use to ensure you don’t exceed your capacity.

If you are buying for peak, then you will just need to add up all the load (including accounting for power factor). Your cooker, for example, will be purely resistive, with a 1.0PF, so that’s just pure kW. Other things will vary. But your very best bet is to get a power meter that can measure current and PF and empirically test everything you will be plugging in.
I don't expect to have many devices working at full load for much long. The fact is that I currently have 3.45kVA hired and I experience this problem quite often. I can give you some examples.
- Let's say I have the washing machine working. If I turn on the oven, puff, the Electric board shuts down.
- If I have the electric water boiler on, the oven and I use the microwave heater, electric board shuts down.

So, I really need to have this into account if I want to have a powerful machine as an ASIC, working at home, 24/7 and if I need to use the oven for an hour or two. Or if I want to have this ASICm machine and I need to use the electric water boiler and the microwave heater

If you have many similar loads you can probably do just fine by measuring the RMS voltage and RMS current of one unit and then multiply the apparent power factor by the number of units.

You would use either a RMS voltmeter or RMS ammeter (which could be a voltmeter measuring the voltage across an low resistance shunt resistor).

You can also buy watt meters that show real and apparent power or real power and power factor.
It's not that the case. I don't have many similar loads. I just want to prevent that if I have this ASIC machine working at home, I can still use other appliances simultaneously without the Electric board shuts down itself!

Edited;

For instance, knowing that 3.45kVA is kinda the lower limit I can afford to have without the ASIC device, if I want to purchase this ASIC and be safe that I can use it and also use other appliances simultaneously without the Electric board shuts down, would 6.9kVA be enough?

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#### Ya’akov

Joined Jan 27, 2019
7,017
It sounds like you are already underpowered. So take the mining rig requirements as a baseline of additional needs, and add somewhat more than you currently have on top of it.

Also, get a UPS for the mining rig that can give you at least 5 minutes to restore the power so you don't have it crash if the power goes out.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,695
It sounds like you are already underpowered. So take the mining rig requirements as a baseline of additional needs, and add somewhat more than you currently have on top of it.

Also, get a UPS for the mining rig that can give you at least 5 minutes to restore the power so you don't have it crash if the power goes out.
Yes, I have that impression too. But it was what I found when I came to this apartment.
I will probably upgrade this to 6.9kVA and will probably be enough to handle the ASIC, the oven, lights, 2 laptops and maybe one more appliance. This should be the most common situation of most appliances on, simultaneously.

My Electricity provider shows us, in a link of their site, how to estimate the Apparent Power needed for a given needed consumption power. They just add up the individual power ratings and multiply by the amount of each device that might be on, simultaneously. And then, chose the closest nominal Apparent Power value available, for hire.

So, let's say 1x ASIC, 1x oven, 1x lights, 1x refrigerator, 2x laptops and 1x some other appliance.
This will add up something like 3100 W (ASIC) + 2000 W (??, not sure the power of my oven) + 200 W (lights/illumination, it's less but nevermind), + 200 W (refrigerator) + 2 x 250 W (2 laptops) + 2400 W (electric water boiler or 1200 for the microwave heater) = 8100 W. That would mean I need 10kVA to be safe. But I'm not willing to have this Apparent Power, so, I'll go for the nominal level below.

@DaviBrons I'm not sure I understood everything you mentioned but I think the values you wrote are for a single device, right? And based on their nominal current and mains voltage!

So, for the ASIC device, the nominal current is 20A and at my mains voltage of 239V/240V, that would be 4800W.
My question here is how that relates to the nominal power stated by the ASIC manufacturer of 3068W??

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