Nominal power - watt peak Question ??

karamyhosbany · June 2014

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Hello

I have calculated power needs for a solar project and i have a problem in finding the the nominal power , peak power.

Any one can help me ?

Cariboocoot · June 2014

Re: Nominal power - watt peak Question ??

Welcome to the forum.

Okay you see that first item? Three 1 ton A/C units pulling 1400 Watts each for 12 hours totaling 50 kW hours per day? That's where your project just ended. To provide that amount of electricity from solar using a grid-tie system would require over $60,000 in equipment. Item number two at 20kW hours per day is also going to add a like amount to the costs.

So let's look at the project from the beginning; what are you trying to achieve? Reduced electric bill? All off-grid? If the latter you'll need to be as rich as Bill Gates, and won't be after you buy all the equipment.

Now a bit about the power usage. Unless those numbers are actually measured with a meter they are probably wildly inaccurate. The process of using manufacturers' consumption numbers (Volts & Amps and/or Watts) times estimated hours of use rarely results in accurate Watt hours.

Maximum or peak Watts is the total of everything that would be on at the same time. This is necessary to size the inverter, and should be in Volt Amps (meaning it needs to include any power factor involved such as you get with induction motors).

Average or nominal Watts is the daily usage averaged out and is of no particular value. Average Watt hours is used to size battery bank and is the total Watt hours for the day divided by 24.

What is more is that allowances have to be made for large power using items. Motors kick on with a hefty start surge and any system must be able to handle that. This means sizing both inverter and batteries (if used) to take the sudden high current demand.

So again the question is: what are you trying to achieve?

karamyhosbany · June 2014

Re: Nominal power - watt peak Question ??

It's an Off-Grid system. I know if we want to convert to KWp we may use a factor expresses the sunshine duration in hours ( SSD )

KWp = Kw.h / SSD

Have you ever hear about it ?

Cariboocoot · June 2014

Re: Nominal power - watt peak Question ??

If this is an off-grid system we're back to "how many millions of dollars have you got?"

I know all about the hours of equivalent good sun and how it affects battery charging. But that does not negate the basic problem of trying to provide an estimated 97 kW hours per day from batteries. That is a lot of power usage even for a grid install.

Really, based on that number and standard available equipment you're looking at over 4,000 Amp hours of battery @ 48 Volts, five 80 Amp MPPT charge controllers, and at least 25,000 Watts of array.

Depending on how the power is used (how much at night which must be stored in batteries and how much during good sun hours) there is a small amount of flexibility.

It is evident that these usage numbers are estimated. Looks like "computer has 250 Watt power supply, there are four of them used 10 hours per day that's 10 kW hours". This may be quite inaccurate; just because the rating on something is 250 Watts does not mean it actually uses that much power, rather that is a maximum power rating. For much of the time something like a computer will use a lot less.

Likewise an A/C unit's (or other refrigeration) consumption will vary with ambient conditions. A water pump may draw more than expected. Considering the size of the proposed project you need more accurate consumption figures to begin with, as the sizing from estimates could be wildly off resulting in a too expensive system or one that can not provide the required power.

jonr · June 2014

Re: Nominal power - watt peak Question ??

When it comes to PE solar driving large AC, I'd look at using a water-to-water (usually geo-thermal) or air-to-water heat pump and only running it when the sun shines. Combined with a large water tank (thermal storage), it can eliminate the need for most of the batteries. Plus of course you would insulate and air seal extremely well.

I'm curious about what you are powering.

zoneblue · June 2014

Re: Nominal power - watt peak Question ??

Lets put it this way: what is your budget for this project? Are the funds approved and ready to go? If its more then $350,000, then you are good to go. Whats the application?

karamyhosbany · June 2014

Re: Nominal power - watt peak Question ??

Regardless of the mistakes and wrong calculations. Assume any number of KWh , How can i convert it to KWp ?

I know that the loads are inaccurate and disorganized.

Cariboocoot · June 2014

Re: Nominal power - watt peak Question ??

karamyhosbany wrote: »

Regardless of the mistakes and wrong calculations. Assume any number of KWh , How can i convert it to KWp ?

I know that the loads are inaccurate and disorganized.

You'd need a number for the amount of sunshine in your area. There is an international version of PV Watts - somewhere (I seem to be having trouble actually locating the URL. Probably due to the rain). There's another similar calculator (which Bill may be able to find) that is also international in nature. Either of these will give you some approximation of expected insolation for your area, from which you can determine a minimum amount of equivalent good sun hours.

The other factor involved is the end-to-end efficiency of the system. This varies widely depending on the particular components used and (most important) when the loads occur.

Around here we would use this formula for your power needs:
97 kW hours / 4 hours equivalent good sun = 24.25 kW hours 'harvested' per hour / 0.52 over-all efficiency = 47 kW array needed.

That's $50,000+ in panels alone.

jonr · June 2014

Re: Nominal power - watt peak Question ??

You might want to look into using DC directly without an external inverter for increased efficiency. Computers and inverter based AC units are just going to change it to DC anyway. Maybe the searchlights (electronic ballast?) can accept DC too.

BB. · June 2014

Re: Nominal power - watt peak Question ??

Two that I find easy to use:

http://rredc.nrel.gov/solar/calculators/pvwatts/version1/ (limited international sites)
http://solarelectricityhandbook.com/solar-irradiance.html (lots of international sites)

I will copy and paste a post for Off Grid sizing with minimal information that was really a Grid Tied sizing question (oh-well, now the post can be useful for somebody). Note I am answering the question for a 10kWatt peak load (and no other information) on how I would start sizing the system.

You have a 97.3 kWH per day average load calculated--That is huge for an off grid system. The system I have penciled out is around 20 kWH per day--You can re-do the math, or simply multiply by ~5x to get a ~100 kWH per day answer for your loads:

Conservation and optimizing your energy usage (both peak and average) is key to having any sort of "affordable" off grid power system.

Can you design a system that will run 10-15 kWatt peak average load... Sure. It will just cost as much as a house in many parts of the country.

And, there is a big difference between peak power, and average kWatt*Hours per day of loads. In general, if you design for your kWH per day loads, you will usually be able to manage your peak loads too (i.e., a system large enough to provide XX kWH per day will start/run most any specific combination of loads you "normally" run).

If you have unusual needs (high peak loads, and low average loads)--Then that may be an issue. But until you start with the basics (how many kWH per day do you want to run)--A single peak load number is usually not enough to really get started on designing your system other in the most basic of rules of thumbs...

For example, a 10 kWatt system will need a minimum of 1,000 AH @ 48 volt battery bank.

A 48 volt battery bank will need 5% to 13% (roughly) to be properly charged:

1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 3,831 watt array minimum
1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 7,662 watt array nominal
1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.13 rate of charge = 9,961 watt array "cost effective" maximum

Assuming you are full time off grid and use the "nominal" array. And live around Amman Jordan, using Solar Electricity Handbook link for fixed array 58 degrees from vertical:

Amman
Average Solar Insolation figures

Measured in kWh/m²/day onto a solar panel set at a 58° angle:
(For best year-round performance)

Jan
Feb
Mar
Apr
May
Jun

3.71
4.30
5.29
5.99
6.53
6.73

Jul
Aug
Sep
Oct
Nov
Dec

6.72
6.61
6.35
5.44
4.27
3.54

Say you get a minimum of 5 hours of sun per day on "sunny days" were you need the cooling:

7,662 Watt Array * 0.52 off grid system eff * 5 hours min sun = 19,921 WH = 19.9 kWH per day

A 1,000 AH @ 48 battery bank used for 2 days of storage and 50% maximum discharge will supply:

1,000 AH * 48 volts * 0.85 inverter eff * 1/2 days * 0.50 max discharge = 8,160 WH = 8.2 kWH of average daily storage

Making a "balanced" system design requires you to both size for your "peak loads" (minimum size battery bank and AC inverter) and your "average loads" (Watts*Hours per day--Again size of battery bank and solar array to replace energy used).

Does this help?

-Bill

karamyhosbany · June 2014

Re: Nominal power - watt peak Question ??

Cariboocoot wrote: »

You'd need a number for the amount of sunshine in your area. There is an international version of PV Watts - somewhere (I seem to be having trouble actually locating the URL. Probably due to the rain). There's another similar calculator (which Bill may be able to find) that is also international in nature. Either of these will give you some approximation of expected insolation for your area, from which you can determine a minimum amount of equivalent good sun hours.

The other factor involved is the end-to-end efficiency of the system. This varies widely depending on the particular components used and (most important) when the loads occur.

Around here we would use this formula for your power needs:
97 kW hours / 4 hours equivalent good sun = 24.25 kW hours 'harvested' per hour / 0.52 over-all efficiency = 47 kW array needed.

That's $50,000+ in panels alone.

Thanks alot. The overall efficiency is a main factor which was absent from my mind.

karamyhosbany · June 2014

Re: Nominal power - watt peak Question ??

BB. wrote: »

Two that I find easy to use:

1- If you have unusual needs (high peak loads, and low average loads)--Then that may be an issue. But until you start with the basics (how many kWH per day do you want to run)--A single peak load number is usually not enough to really get started on designing your system other in the most basic of rules of thumbs...

2 - For example, a 10 kWatt system will need a minimum of 1,000 AH @ 48 volt battery bank.

A 48 volt battery bank will need 5% to 13% (roughly) to be properly charged:
1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 3,831 watt array minimum

1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 7,662 watt array nominal

1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.13 rate of charge = 9,961 watt array "cost effective" maximum

Assuming you are full time off grid and use the "nominal" array. And live around Amman Jordan, using Solar Electricity Handbook link for fixed array 58 degrees from vertical:

[h=3]Amman
Average Solar Insolation figures[/h] Measured in kWh/m²/day onto a solar panel set at a 58° angle:
(For best year-round performance)

Jan
Feb
Mar
Apr
May
Jun

3.71

4.30

5.29

5.99

6.53

6.73

Jul
Aug
Sep
Oct
Nov
Dec

6.72

6.61

6.35

5.44

4.27

3.54

Say you get a minimum of 5 hours of sun per day on "sunny days" were you need the cooling:

7,662 kWH per day * 0.52 off grid system eff * 5 hours min sun = 19,921 WH = 19.9 kWH per day

A 1,000 AH @ 48 battery bank used for 2 days of storage and 50% maximum discharge will supply:
1,000 AH * 48 volts * 0.85 inverter eff * 1/2 days * 0.50 max discharge = 8,160 WH = 8.2 kWH of average daily storage

Making a "balanced" system design requires you to both size for your "peak loads" (minimum size battery bank and AC inverter) and your "average loads" (Watts*Hours per day--Again size of battery bank and solar array to replace energy used).

Does this help?

-Bill

Great Bill.

1- 10 kWatt system will need a minimum of 1,000 AH @ 48 volt battery bank.

How did you find it 1000 AH ? !!

2 - 1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 3,831 watt array minimum

Look to the units :
( AH ) * ( Volt ) * (No unit for eff ) * ( rate of change unit ?? ) = ( watt )

3 - 7,662 kWH per day * 0.52 off grid system eff * 5 hours min sun = 19,921 WH = 19.9 kWH per day

( KW.H / day ) * ( No unit ) * ( H ) = KW.H / day ????????

BB. · June 2014

Re: Nominal power - watt peak Question ??

Some explinations and corrections

:

karamyhosbany wrote: »

1- 10 kWatt system will need a minimum of 1,000 AH @ 48 volt battery bank.

How did you find it 1000 AH ? !!

This one of our "rules of thumbs"--First suggested by Solar Guppy--A designer of off grid power systems:

100 AH @ 48 volt per 1kWatt of inverter output
200 AH @ 24 volt (lower voltage battery bank)
400 AH @ 12 volts (even lower voltage--Note P=V*I -- So power is the same in all cases)

And, from what I have seen working the math backwards to other rules--This works out to, for flooded cell deep cycle batteries, a maximum short term continuous of C/5 discharge rate for a battery bank (C/8 is really our recommended long term maximum discharge rate). And a C/2.5 maximum surge discharge rate (based on a "good" AC inverter can output ~2x rated power during surge--such as motor starting). All of these work backwards (within round off error) to the 100 AH @ 48 volt for 1,000 Watt AC inverter sizing.

2 - 1,000 AH * 59 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 3,831 watt array minimum

Look to the units :
( AH ) * ( Volt ) * (No unit for eff ) * ( rate of change unit ?? ) = ( watt )

Yea--This is where units get into a bit of issue... We use the xx% rate of discharge based on the battery's AH capacity... I guess I could justify it as:

C/20 hour rate = 5%/Hour rate--That converts AH*V/H to AV = Watts.

3 - 7,662 kWH per day * 0.52 off grid system eff * 5 hours min sun = 19,921 WH = 19.9 kWH per day

( KW.H / day ) * ( No unit ) * ( H ) = KW.H / day ????????

Should have been Watt array, not kWH per day (starting with XYZ Watt array and finding kWH per day available power):

7,662 W array * 0.52 off grid system eff * 5 hours min sun = 19,921 WH = 19.9 kWH per day

I was slicing an equation WH or kWH per day, to Array Wattage--And I missed the units. Go up and fix now.

Thank you,
-Bill

karamyhosbany · June 2014

Re: Nominal power - watt peak Question ??

Thank you Bill.

Thanks all

Nominal power - watt peak Question ??

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