Solving Energy Consumption

RebReb Registered Users Posts: 5
Hello everyone I was tasked in my thesis to show the energy consumption/usage of every equipment of my system.
My System is 5kw grid interactive Solar Water Pumping System,,

 My Questions are:
1.How to solve the energy consumption of an inverter
2. How To solve the energy consumption of my variable frequency drive.

Thanks in advance for a response,, it would really help me in my thesis,, Ill have my defense on Monday,, Thanks

Comments

  • BB.BB. Super Moderators, Administrators Posts: 31,433 admin
    Welcome to the forum Reb.

    At this point, with so little time (the weekend), you probably are not going to get any measuring equipment from the school lab, let alone 3 phase (?) power equipment. So you are left with the specifications of the major components (solar panel output vs sun, the GT inverter, and the VFD).

    Is this pumping system actually connected to the utility grid? If so, you can use your utilty meter to do some of the measurements (if you can shut down the rest of the power in the building for an hour or two while testing). And find/borrow a DC Current Clamp RMS reading DMM (actually AC+DC clamp meter) to measure the DC input to the GT inverter. Something like this:

    https://www.amazon.com/gp/product/B019CY4FB4

    Depending on what specifically your degree is in--Are you familiar with the differences between Watts and VA ratings (Power Factor, Cosine/phase angle between voltage and current, different wave forms such as sine and square/modified square waves in AC power circuits and motors?)? I.e., how much detail/math do you want to go into here? What is your level of accuracy expected (in general, measuring using clamp/volt meters will get you to within 10% of the VA/kVA of the AC, and perhaps within 5% of the DC power (Watts). You will not be able to measure "Watts" (power) with a simple clamp meter--Only VA/kVA (need meter that can measure both voltage and current at the same time and "do the math" to give you the results). The utility meter should be able to give you Watts/kWatts/Watt*Hours/kWH (you will need to do some research on the type of meter you have and how to read it).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • RebReb Registered Users Posts: 5
    Good morning sir Bill,,
    Unfortunately I can't go to school due to lockdown,,,

    My inverter is single phase,,
    Then the only given specification is the Maximum capacity of the inverter.
    Which cannot be used in the formula for getting the energy consumption

    = Rated Power x no. Of hours being used

    The only way to get its consumption for now is through theory/estimated value/given specs, since I can't get the actual data.

    Sir, do you have any other way in solving its energy consumption using theory or using the specifications given?
    Thanks in advance
  • RebReb Registered Users Posts: 5
    Correct me if I'm wrong sir,,
    As what I understand I can't use the capacity given of the inverter specs, since it is the maximum capacity that an inverter can carry,,

    So without using actual data (since I can't get because it's lockdown here)
    How can I solve its energy consumption?
    The only data that I have here is the specification of the inverter,,
    So how can I get its energy consumption in a day?,
    Thanks sir in advance
  • BB.BB. Super Moderators, Administrators Posts: 31,433 admin
    Reb,

    I guess you are in the Philippines. Here is a neat program that can give you hour by hour (24 hours per day, 365 days per year) that you can load into a spread sheet and run your math on that.

    https://pvwatts.nrel.gov/pvwatts.php

    The data is based on real observations/measurements (estimates from nearest data sources). And uses "real day's solar irradiation" (they take 20+ years of data, and find a "representative day" and use that data). And the PVWatts program takes care of the solar intensity and solar angles (fixed arrays)--So you don't have to do that.

    There are lots of parameters that you can plug in (we typically use around 75% to 77% overall efficiency (or 25 to 23% for derating panels for heat by 81%, and ~95% GT inverter losses. 0.81*0.95=0.77), angles of array (south facing, tilt, if any trackers), dirt on array, temperature offsets, etc.

    More or less, GT Inverters are "constant power" devices... So, whatever comes from the array (Volts*Amps--For most part, Vmp is ~80-90% of Vmp-std, and Current is based on solar intensity and angle to sun (cosine of sun to panel). And feeds the maximum available array current out to the AC mains:

    Solar power in * 0.95 eff = GT Power out to grid

    GT inverters do not (in general) interact in any way with the local loads. Think of the GT inverter as "charging a giant AC Battery bank", and as the sun goes down, the local AC loads slowly discharge that "giant AC battery bank". Very much like your car electrical system works (engine to alternator to battery bank to DC loads). Because the Utility "giant AC battery bank" capacity is huge, there is no need for a "charge controller" with the GT inverter (utility can absorb energy or supply energy as needed at 230 VAC @ 50 Hz). Lots of details and simplifications, but these should be "good enough" to get you started.

    Assume VFD is 95% efficient... The pump motor efficiency will depend on the VFD waveform (sine wave most efficient, square wave is much less efficient/induction motors runs hotter). And if you can identify the water pump (centrifugal or positive displacement, find a flow vs rpm vs power vs pressure chart for that pump, back pressure, etc.).

    Just to save you a heart attack, there are lots of losses in solar power systems... I would not be surprised to see:

    • 0.81 (hot panels) * 0.95 typical GT inverter eff * 0.95 typical VFD eff * 0.8 induction motor eff (if sine wave, perhaps down to 0.60 for SqWave VFD) = 0.58 best case to 0.44 SqWave eff (panels to motor shaft).
    You also have to take into account the pump type... Centrifugal pumps, use most energy with maximum water flow and minimum back pressure. And use the less energy the more the flow is restricted (adding a gate valve on the output pipe is frequently used to reduce flow/increase back pressure to reduce peak motor power).

    For positive displacement pumps, they use the least amount of power with low back pressure, and as back pressure increases, the pump requires more torque to pump (basically) same flow rate against higher pressure.

    VFDs can be programmed with a pressure sensor (vary RPM based on pump output pressure--I.e., slow pump down when not much water flow, speed up when lots of water flow needed).

    You will probably need to make lots of assumptions. Just document those assumptions, and printout/keep links to any data you use for your modeling.

    The above is based on using a Grid Tied (utility interactive) AC inverter (solar panels -> GT inverter -> AC mains/utility power).

    There are VFDs that take DC input power directly from the solar array and convert it to power for the motor/pump unit. In this case, the VFD will "soft start" the pump in the morning, and slowly increase RPM until around local solar noon (maximum power from array), and then as the sun heads west, the VFD will monitor the available energy from the array (Vmp*Imp=Pmp) and slowly decrease pump RPM (to reduce power usage). is disss

    Good luck Reb on your classes and degree!
    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • mcgivormcgivor Solar Expert Posts: 3,724 ✭✭✭✭✭✭
    Probably the most accurate means would be to use a power data logging devices on the AC input, inverter AC output and the solar DC input. Short of doing that it would be mostly assumptions and guess work due to the devices not having linear performance curves, most inverters have similar profiles however, being less efficient at the lower end, this would be the same for a frequency drive as it too converts DC to variable frequency AC, even if it has an AC input it is rectified first. 

    Below is a graph showing the efficiency of some inverters, note how below ~30% the efficiency drops, so for a 5 Kw inverter anything below 1500 W would impact the efficiency to varying degrees depending on manufacturer.

       
    1500W, 6× Schutten 250W Poly panels , Schneider MPPT 60 150 CC, Schneider SW 2524 inverter, 400Ah LFP 24V nominal battery with Battery Bodyguard BMS 
    Second system 1890W  3 × 300W No name brand poly, 3×330 Sunsolar Poly panels, Morningstar TS 60 PWM controller, no name 2000W inverter 400Ah LFP 24V nominal battery with Daly BMS, used for water pumping and day time air conditioning.  
    5Kw Yanmar clone single cylinder air cooled diesel generator for rare emergency charging and welding.
  • RebReb Registered Users Posts: 5
    Thanks sir, Bill, for that information,, 
     Sir do you have any,, 
    Summarize formula for that?
    For example

    Inverter Energy Consumption  = Eff. * V * A

    If you have sir,, can I asked? Thank you and Bless you always,, it would be a big help in my study, 
  • RebReb Registered Users Posts: 5
    @ Sir mcgivor thanks,, unfortunately I can't performing data logging, since I can't go to the system, 
  • BB.BB. Super Moderators, Administrators Posts: 31,433 admin
    Reb, I do not have the formulas for what you are asking as I have no information/details on the equipment you are using, the load factors, etc...

    The best first guess for inverter and VFD efficiencies I would use is something like (AC Inverters and VFDs are very similar devices in general--The control and feedback systems behave differently).

    You can try and curve fit an equation to what McGivor gave you above... Or try something like:
    • Actual Array Power * 0.975 Efficiency - (Inverter Max power rating * 0.025 tare losses) = Power Out to utility
    • If Actual Array is equal too or less than 2.5% of inverter max power rating, then set GT inverter output to Zero Watts
    My guess is that roughly the inverter needs (very roughly) 2.5% of rated power to operate (electronics, switching losses of transistors), and the other 2.5% of the losses are load related (resistance of wiring, transformer/inductor/switching losses).


    That gives you an efficiency profile something like McGivor has provided. Notice that there is a slight peak in overall efficiency at ~60%... Probably because electrical heating (as a result of current flow) can be modeled as Power=Current^2*R (the heating effect of current goes up with the square of the current). Probably not a second order effect (i.e., not a "trivial" effect on the efficiency curve). But you will have to decide how closely you need to model the effects for the accuracy you need--And frankly what you exact degree is, and how closely you are expected to know about the internal workings of AC inveters, VFDs, induction motors, and water pumps. And how much you "know" about the installation you are modeling.

    AC Power Math is, in detail, not a trivial subject either. And the effects of sine vs square wave power into induction motors, phase angles between voltage and current (also affected by frequency, loading, motor design, etc.) are not trivial either.

    https://www.arrow.com/en/research-and-events/articles/real-vs-reactive-power

    One engineer I worked with decades ago got his doctorate in Switching Power Supplies for computers, then started a company making switching computer power supplies.

    Getting into all the details and getting them "right" down to 1% or better accuracy is not something you are going to do in a weekend or with just in a couple of pages... To be honest, with what we do here helping people with off grid and GT systems--If their predicted output/harvest is within 10% of what the measure--That is "dead on accuracy" with standard DMM, current clamp/shut meters, etc. You are not going to get better measurements than that unless using specialized lab gear costing $10,000 of USD. And even then, down at 1% or better accuracy, your lab equipment measurements can disagree between brands/models.

    The GT inverter (and other power equipment displays are typically in the 2% to 6% accuracy range. And (as I recall), in the USA, utility meters are guaranteed to something like 2% accuracy. And that is with "Sine Wave" current/voltages... Throw in non-sine wave, and you need meters that do RMS (root mean square) calculations (i.e., the area under an arbitrary curve) which needs "fast(er)" sampling to accurately measure the "true" curves, and math to model the area under the curve. And again, limits, faster sampling is more expensive and more accurate, but also more costly and higher battery drain on portable instruments.

    https://en.wikipedia.org/wiki/Root_mean_square

    You have to accept some approximations and focus on what your degree "needs"

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Dave AngeliniDave Angelini Solar Expert Posts: 5,805 ✭✭✭✭✭
    Why can't she go to the site? Even when we were in "stay home mode" essential services were free to roam about safely. I kept this in my
    truck. If this is important she needs to meter it. The math could be off, as it sometimes is.




    .
    "we go where power lines don't" Sierra Mountains near Mariposa/Yosemite CA
     http://members.sti.net/offgridsolar/
    E-mail [email protected]

  • BB.BB. Super Moderators, Administrators Posts: 31,433 admin
    Reb is in the Philippines... Don't know what rules will apply there.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Dave AngeliniDave Angelini Solar Expert Posts: 5,805 ✭✭✭✭✭
    Hopefully, common sense.
    "we go where power lines don't" Sierra Mountains near Mariposa/Yosemite CA
     http://members.sti.net/offgridsolar/
    E-mail [email protected]

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