Solar Panel Energy Calculators

[Crossy]

Something we've been missing is a link to online solar "calculators" (really estimators), so here are a couple to get us started.

 

A nice simple one:- https://www.whatnextnow.com/home/solar/solar-panel-calculator

 

And something more complex:- https://re.jrc.ec.europa.eu/pvg_tools/en/tools.html#SA

 

Please feel free to add your own favourite.

 

[MJCM]

To calculate the Solar Panel tilt Angle of your Solar Panels use one of the above calculators @Crossy linked to or use this rough guide

 

Use the latitude of your location and multiply by 0,87

 

For example BKK 13.7563° N, 100.5018° E.

 

13.7563 * 0.87 = 11,96° = 12°

 

If the latitude value is between 25 and 50, it is multiplied by 0.87 and added 3.1 degrees to the result. The ideal angle can be obtained at approximately 45 degrees if the latitude value is 50 degrees and above.

 

https://lightingequipmentsales.com/how-to-calculate-solar-panel-tilt-angle.html

 

 

[Crossy]

The actual "ideal" tilt varies throughout the year, for example on April 26th and August 16th 2022 the ideal tilt in Bangkok was dead flat (sun directly overhead at the zenith).

 

In reality being this close to the equator the tilt angle actually makes very little practical difference (try playing with the calculators and adjusting the tilt angle). Other factors like the weather make far greater variations.

 

If your roof isn't ideally tilted just spend what you would have spent on fancy tilted or adjustable mounts on an extra panel or two.

 

"Tilt a bit and point roughly south" works remarkably well ???? 

 

 

[MJCM]

4 minutes ago, Crossy said:

"Tilt a bit and point roughly south" works remarkably well

????????

 

Edit: What the rough guide actually doesn't take into consideration (and not mention) is the direction, which Crossy wisely says "roughly South"

[Encid]

6 hours ago, Crossy said:

And something more complex:- https://re.jrc.ec.europa.eu/pvg_tools/en/tools.html#SA

I must be doing something wrong... 

 

 

I just plugged in the location of our farm in Isaan, added 10 x 450W panels (4.5kWp) to a flat roof, and also 2 x 9kWh batteries (18 kWh), and then punched in 3 kWh as our daily consumption (based on our current usage).

 

The results above show a lot of excess energy not captured.

 

Am I missing something?

[Crossy]

11 hours ago, Encid said:

I just plugged in the location of our farm in Isaan, added 10 x 450W panels (4.5kWp) to a flat roof, and also 2 x 9kWh batteries (18 kWh), and then punched in 3 kWh as our daily consumption (based on our current usage).

 

The results above show a lot of excess energy not captured.

 

Am I missing something?

 

Nope, you are waaaaay over producing if you really are using only 3kWh.

 

Your panels will be generating around 15kWh per day. Your batteries will run you for 6 days in the dark.

[motdaeng]

13 hours ago, Encid said:

... 3 kWh as our daily consumption (based on our current usage).

are you sure about 3 kwh per day? that's 90 kwh per month!

a few years back using under 90 kwh per month was free ...,

don't know about today.

 

your bill must be  around 400 thb per month ...  ????

[MJCM]

31 minutes ago, motdaeng said:

a few years back using under 90 kwh per month was free ...,

don't know about today.

it's 50kWh a month but only if you use a 5/15kW meter, and then only after the first 3 months you use less then (total) 150 kWh.

 

 

Rate2015.pdf

[Encid]

6 hours ago, Crossy said:

Nope, you are waaaaay over producing if you really are using only 3kWh.

Yes I see now where I made an error... our current average monthly consumption is about 1800kWh, or 1,800,000 Wh, divided by 30 days = 60,000 Wh/day.

 

The results look vastly different with the correct daily consumption figures plugged in.

 

[Sophon]

I wonder if the calculator from the European Commission is biased towards colder climates:

 

If we compare the numbers from January and August, I would be producing more than 4 kWh per day more in January than I would in August. That is reflected in the fact that the calculator expects me to fully charge my battery 29 days in January against only 22 days in August. However, the calculator for some reason also expects me to run out of battery every single day in January but only on 21 days in August. That makes no sense to me.

 

The only reason I can think of is that the calculator expects me to use more energy in the winter for heating (it has the battery running out every night in November, December and January), or that it somehow expects that the battery is used for an extended time because of longer nights. That is not really the case in Thailand, where we as a general rule probably use more energy during the night in August (Air condition), and where the difference in day/night hours doesn't wary much from summer to winter.

 

Maybe the calculator has some constants in the calculations that are more relevant for Europe than Thailand. It is after all an calculator supplied by the European Commission.

 

Can anyone come up with a plausible explanation?

[Crossy]

1 hour ago, RichardGravener01 said:

Globally a formula E = A x r x H x PR is followed to estimate the electricity generated in output of a photovoltaic system. Example : the solar panel yield of a PV module of 250 Wp with an area of 1.6 m² is 15.6% .

 

Some definition of what the variables are would be helpful ???? 

[MJCM]

29 minutes ago, Crossy said:

 

Some definition of what the variables are would be helpful ???? 

 

Quote

 

E is Energy (kWh), A is total Area of the panel (m²), r is solar panel yield (%), H is annual average solar radiation on tilted panels and PR = Performance ratio, constant for losses (range between 0.5 and 0.9, default value = 0.75).  r is the yield of the solar panel given by the ratio : electrical power (in kWp) of one solar panel divided by the area of one panel

 

Example : the solar panel yield of a PV module of 250 Wp with an area of 1.6 m² is 15.6% . It is worth mentioning that this nominal ratio is given for standard test conditions (STC) : radiation=1000 W/m², cell temperature=25 °C, Wind speed=1 m/s, AM=1.5 The unit of the nominal power of the photovoltaic panel in these conditions is called “Watt-peak” (Wp or kWp=1000 Wp or MWp=1000000 Wp).

 

Now you have to find the global annual irradiation incident on your PV panels with your specific inclination (slope, tilt) and orientation (azimuth) to calculate H.

 

PR: estimates the quality of a photovoltaic installation as it gives the performance of the installation independently of the orientation, inclination of the panel. It includes all losses which depend on the size of the system, technology used and the site.

The DC Current generated undergoes a series of losses before it can finally become AC Current and used by us.

 

Example of losses that gives the PR value
  * shadow losses
  * Temperature losses
  * DC cables losses
  * AC cables losses
  * Inverter losses
  * Losses due to dust

 

https://www.saurenergy.com/solar-energy-blog/here-is-how-you-can-calculate-the-annual-solar-energy-output-of-a-photovoltaic-system

[Crossy]

All very interesting ^^^.

 

But I'm sure most of us would be happy to plug in location and pull out azimuth and tilt angle with a guesstimate of how big a system would be adequate for our usage ????

 

[MJCM]

15 minutes ago, Crossy said:

All very interesting ^^^.

 

But I'm sure most of us would be happy to plug in location and pull out azimuth and tilt angle with a guesstimate of how big a system would be adequate for our usage ????

 

My guess it's only for the likes like @Muhendis @Crossy and @Encid 55555555555 ????

[Muhendis]

2 hours ago, Crossy said:

All very interesting ^^^.

 

But I'm sure most of us would be happy to plug in location and pull out azimuth and tilt angle with a guesstimate of how big a system would be adequate for our usage ????

 

If you ever get a bit bored you could try getting your head round this

https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface

[Encid]

On 8/28/2022 at 11:11 AM, Muhendis said:

If you ever get a bit bored you could try getting your head round this

https://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-on-a-tilted-surface

Now we're getting serious... 

[Muhendis]

36 minutes ago, Encid said:

Now we're getting serious... 

Cutting a long story short

Power = sine of angle of incidence x 1000 

So if angle of incidence = 90° then sine =1 and power =1kw

easy peasy.

 

[Encid]

5 minutes ago, Muhendis said:

Cutting a long story short

Power = sine of angle of incidence x 1000 

So if angle of incidence = 90° then sine =1 and power =1kw

easy peasy.

Awww... that takes all the fun out of it... 

 

[Muhendis]

8 minutes ago, Encid said:

Awww... that takes all the fun out of it... 

 

There's still plenty of fun to be had finding what is the angle of incidence.

To do that hold a stick so that one end is touching the panel and point the other end in the direction of the sun and adjust 'til there is no shadow.

Use an angle finder/protractor to measure the smallest angle 'twixt stick and panel. and away you go.